MATHEMATICS HIP-HOP
CBC O-Level Mathematics Question Bank
20 Original Chain-Based Sets • 120 Extended Items • Student Tasks • Worked Answer Guides
Welcome to the Mathematics Hip-Hop CBC Mathematics Question Bank. These sets follow the supplied sample's structure: Section A contains Items 1–2, while Section B has Part I (Items 3–4) and Part II (Items 5–6). Each item is deliberately written as a connected mathematical chain, with later tasks building on earlier results.
MATHEMATICS HIP-HOP
CBC MATHEMATICS — 20 ORIGINAL CHAIN-BASED SETS
Lengthy Student Items • Explicit Tasks • Worked Answer Guides
These sets follow the supplied sample's examination structure: Section A contains Items 1–2; Section B Part I contains Items 3–4; Part II contains Items 5–6. Each item is deliberately chained: later tasks reuse results from earlier tasks. The questions are original rather than copies.
Answer guides are placed after each item so the document can also serve as a teacher reference.
SET 1: HEALTH CENTRE DEVELOPMENT
ITEM 1
The local authority in Health Centre Development has secured land for a community project. The project team has 20 acres available. The plan reserves 2/5 of the total land for the main buildings.
After the building area is removed, 1/4 of the land that remains is to be landscaped or planted. The rest is divided among staff facilities, a public waiting area and parking in the ratio 3:2:4.
The project engineer estimates that one vehicle needs 0.02 acres of parking space. However, 10% of the full parking capacity must remain free for emergency access and movement.
The landscaping must be cleared before construction. Two workers can clear half an acre in one day when working at the same rate. Each worker is paid Shs 18,000 for one day's work.
- Determine the area allocated to the main buildings.
- Determine the area remaining after the buildings and hence find the area reserved for landscaping or planting.
- Use the given ratio to determine the areas for staff facilities, the public waiting area and parking.
- Determine the theoretical maximum number of vehicles that the parking area can accommodate.
- After keeping 10% of the capacity free, determine the maximum number of ordinary vehicles that should be admitted at one time.
- Determine the number of workers required to clear the landscaping area in one day.
- Calculate the total amount required to pay those workers for the one-day job.
(a) Buildings = 20 × 2/5 = 8 acres.
(b) Remaining = 12 acres. Landscaping = 3 acres. Land still available after landscaping = 9 acres.
(c) Total ratio parts = 9. Staff = 3 acres; waiting area = 2 acres; parking = 4 acres.
(d) Parking capacity = 4 ÷ 0.02 = 200 vehicles.
(e) Emergency reserve = 10% × 200 = 20.0 vehicle spaces. Ordinary capacity = 90% × 200 = 180 whole vehicles.
(f) At 0.25 acre per worker-day, workers required = 3 ÷ 0.25 = 12.00, so 12 workers are required.
(g) Labour cost = 12 × Shs 18,000 = Shs 216,000.
ITEM 2
A small enterprise in Health Centre Development uses two energy systems. Let S be the daily maintenance cost of the solar system and B the daily maintenance cost of the backup system.
Records show that S plus twice B is Shs 28,000. They also show that three times S exceeds twice B by Shs 4,000.
The enterprise produces two products, X and Y. Let x and y be the numbers produced per day. It must produce at least 20 units of X, X must be at least twice Y, total production cannot exceed 120, and the production budget is at most Shs 33,000.
Each X uses 3 units of a special input and each Y uses 4 units. At most 450 input units are available. Production costs are Shs 200 per X and Shs 300 per Y, while profits are Shs 500 and Shs 600 respectively.
- Form two simultaneous equations from the maintenance records and determine S and B.
- Define x and y and write all the production restrictions as inequalities, including non-negativity.
- Write the daily profit function P in terms of x and y.
- Determine the important feasible corner points and calculate the profit at each point.
- Identify the production plan that gives the greatest feasible profit.
- For that plan, calculate the total production cost and the total amount of special input used.
- Calculate the corresponding revenue and verify the profit.
(a) S + 2B = 28000; 3S − 2B = 4000. Adding gives 4S = 32000, so S = 8000. Then B = 10000. Thus solar maintenance is Shs 8,000 and backup maintenance is Shs 10,000.
(b) x ≥ 20; x ≥ 2y; x + y ≤ 120; 200x + 300y ≤ 33000; 3x + 4y ≤ 450; x ≥ 0, y ≥ 0.
(c) P = 500x + 600y.
(d) The key feasible corner points are (20,0), (20,10), (120,0) and (80,40). Their profits are Shs 10,000, Shs 16,000, Shs 60,000 and Shs 64,000 respectively.
(e) Therefore the maximum occurs at x = 80, y = 40, giving a maximum daily profit of Shs 64,000.
(f) Production cost = 200(80) + 300(40) = Shs 28,000. Input used = 3(80) + 4(40) = 400 units.
(g) Revenue = production cost + profit = Shs 28,000 + Shs 64,000 = Shs 92,000. Subtracting the cost from the revenue again gives Shs 64,000 profit.
ITEM 3
A management team in Health Centre Development records the number of customers served by 50 staff members during a trial period. Management will continue the arrangement if the average number served exceeds 45.
Staff members who serve at most that threshold are also reviewed for additional support or possible reassignment.
The recorded values are listed below:
20, 22, 24, 26, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 49, 50, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 59, 59, 60, 61, 62, 63, 64, 65, 66, 67, 69.
- Arrange the data into a frequency distribution using class intervals of 10, beginning with the smallest observation. Include cumulative frequency and class midpoint.
- Determine the total number of observations and the total number of customers served, then calculate the mean.
- Use the management rule to state whether the average target is exceeded.
- Determine the number and percentage of staff who served at most 45 customers.
- Determine the number and percentage of staff who served at least 60 customers.
- Use the ordered data to determine the median and identify the median class.
- Explain how the frequency distribution, mean and percentages can help management make a decision.
(a) Frequency table: 20–29: f=6, cf=6, midpoint=24.5; 30–39: f=10, cf=16, midpoint=34.5; 40–49: f=12, cf=28, midpoint=44.5; 50–59: f=13, cf=41, midpoint=54.5; 60–69: f=9, cf=50, midpoint=64.5
(b) There are 50 staff members. Total customers served = 2318. Mean = 2318 ÷ 50 = 46.36.
(c) Since 46.36 > 45, the stated average target is exceeded.
(d) 23 staff served at most 45. Percentage = 23/50 × 100 = 46.0%.
(e) 9 staff served at least 60. Percentage = 9/50 × 100 = 18.0%.
(f) The 25th and 26th ordered observations are 47 and 48; median = 47.5. The median lies in the 40–49 class.
(g) The table shows concentration across ranges, the mean gives an overall average, and the percentages identify groups needing attention or further support.
ITEM 4
A class committee in Health Centre Development wants to choose one student representative from three candidates A, B and C. There are 120 students in the class.
16 students support A only, 28 support B only and 14 support C only. In addition, 10 support both A and B only, 12 support B and C only, 8 support A and C only, and 6 support all three.
Students who support none of the candidates are recorded separately. The teacher will select the candidate with the largest total support.
- Determine the number of students who support none of the three candidates.
- Determine the total number of supporters of A, B and C respectively.
- Identify the candidate with the greatest support and find the difference between the highest and second-highest totals.
- Determine the percentage of the whole class supporting the selected candidate.
- Determine the number and percentage of students who do not support the selected candidate.
- State whether the selected candidate has support from more than half the class and whether fewer than 50% oppose the candidate.
- Explain why the overlapping groups must be counted when making the final decision.
(a) Known regions total = 94; therefore none = 120 − 94 = 26 students.
(b) A = 40; B = 56; C = 40. Each total includes the relevant overlapping regions.
(c) Candidate B has the greatest support with 56 students. The second-highest total is 40, so the difference is 16.
(d) Support percentage = 56/120 × 100 = 46.67%.
(e) Non-supporters = 64. Percentage = 53.33%.
(f) Half the class is 60.0. Since 56 is greater than half, the selected candidate has majority support. The opposition percentage is 53.33%, which is below 50%.
(g) A student in an overlap supports more than one candidate. Ignoring these intersections would undercount candidates' totals and could change the result.
ITEM 5
A recreation facility in Health Centre Development is creating a paved common region where two circular lawns overlap. The radii of the circles are 2.00 m and 1.50 m.
The distance between the two intersection points is 1.60 m. The common region alone will be paved.
Each rectangular paver measures 4.0 cm by 8.0 cm. The supplier recommends buying an additional 5% above the calculated number because some pavers will be cut or damaged. Take π = 3.14.
- Determine the perpendicular distance from the centre of each circle to the common chord.
- Hence determine the distance between the centres of the two circles.
- Determine the central angle subtended by the common chord at each centre.
- Find the area of the common region by calculating the two circular segments.
- Convert the area of one paver into square metres and determine the minimum number of pavers required before the allowance.
- Allowing 5% extra, determine the number of pavers to purchase.
- Determine the total cost and explain why the final number must be rounded upward.
(a) Half-chord = 0.800 m. Distance for radius 2.00 m = √(2.00²−0.800²) = 1.833 m. Distance for radius 1.50 m = 1.269 m.
(b) Centre distance = 1.833 + 1.269 = 3.102 m.
(c) Central angles are approximately 47.16° and 64.46°.
(d) Using sector area minus triangle area for each segment, the common area is approximately 0.429 m².
(e) One paver area = 0.040 × 0.080 = 0.00320 m². Minimum before allowance = 0.429 ÷ 0.00320 ≈ 134 pavers.
(f) With 5% extra: 134 × (1 + 0.05) = 140.70, so 141 pavers should be purchased.
(g) Cost = 141 × Shs 200 = Shs 28,200. The result must be rounded upward because a fractional paver cannot be purchased and rounding downward would leave part of the region uncovered.
ITEM 6
A small manufacturing group in Health Centre Development produces open-topped ice-cream cones. One original cone has capacity 32 cm³ and height 4.00 cm.
The original cone costs Shs 50 to make and is sold for Shs 200. A new order requires cones with capacity 8 times the original capacity. The new cones are similar to the original cones.
The selling price is directly proportional to capacity, while production cost is directly proportional to the curved surface area of the cone. Use π = 3.14.
- Determine the radius and diameter of the original cone.
- Determine the slant height and curved surface area of the original cone.
- Determine the linear scale factor for the new cone, given that its capacity is 8 times the original.
- Determine the new height and diameter.
- Determine the curved surface area of one new cone and hence the total material area for an order of 250 cones.
- Determine the production cost of one new cone and the selling price of one new cone.
- Determine the profit per cone and the total profit on an order of 250 cones.
- Explain why capacity changes by a different factor from diameter.
(a) Using V = ⅓πr²h: r = √(3V/(πh)) = 2.765 cm. Diameter = 5.529 cm.
(b) Slant height l = √(r²+h²) = 4.862 cm. Curved surface area = πrl ≈ 42.211 cm².
(c) Volume factor = 8, so linear scale factor k = ∛8 = 2.000.
(d) New height = 4.00 × 2.000 = 8.000 cm. New diameter = 11.059 cm.
(e) New curved surface area = 168.844 cm² per cone. For 250 cones, total material area ≈ 4.221 m².
(f) Surface area scales as k², so new production cost = Shs 50 × 2.000² ≈ Shs 200. Selling price scales directly with capacity, so new selling price = Shs 200 × 8 = Shs 1,600.
(g) Profit per cone ≈ Shs 1,400. Total profit for 250 cones ≈ Shs 350,000.
(h) Capacity is a volume, so it changes with the cube of the linear scale factor, whereas diameter changes only linearly. Surface area changes with the square of the scale factor.
SET 2: VOCATIONAL SCHOOL EXPANSION
ITEM 1
The local authority in Vocational School Expansion has secured land for a community project. The project team has 18 acres available. The plan reserves 5/9 of the total land for the main buildings.
After the building area is removed, 1/3 of the land that remains is to be landscaped or planted. The rest is divided among staff facilities, a public waiting area and parking in the ratio 2:3:4.
The project engineer estimates that one vehicle needs 0.02 acres of parking space. However, 10% of the full parking capacity must remain free for emergency access and movement.
The landscaping must be cleared before construction. Two workers can clear half an acre in one day when working at the same rate. Each worker is paid Shs 20,000 for one day's work.
- Determine the area allocated to the main buildings.
- Determine the area remaining after the buildings and hence find the area reserved for landscaping or planting.
- Use the given ratio to determine the areas for staff facilities, the public waiting area and parking.
- Determine the theoretical maximum number of vehicles that the parking area can accommodate.
- After keeping 10% of the capacity free, determine the maximum number of ordinary vehicles that should be admitted at one time.
- Determine the number of workers required to clear the landscaping area in one day.
- Calculate the total amount required to pay those workers for the one-day job.
(a) Buildings = 18 × 5/9 = 10 acres.
(b) Remaining = 8 acres. Landscaping = 2.667 acres. Land still available after landscaping = 5.333 acres.
(c) Total ratio parts = 9. Staff = 1.185 acres; waiting area = 1.778 acres; parking = 2.37 acres.
(d) Parking capacity = 2.37 ÷ 0.02 = 120 vehicles.
(e) Emergency reserve = 10% × 120 = 12.0 vehicle spaces. Ordinary capacity = 90% × 120 = 108 whole vehicles.
(f) At 0.25 acre per worker-day, workers required = 2.667 ÷ 0.25 = 10.67, so 11 workers are required.
(g) Labour cost = 11 × Shs 20,000 = Shs 220,000.
ITEM 2
A small enterprise in Vocational School Expansion uses two energy systems. Let S be the daily maintenance cost of the solar system and B the daily maintenance cost of the backup system.
Records show that S plus twice B is Shs 29,500. They also show that three times S exceeds twice B by Shs 4,500.
The enterprise produces two products, X and Y. Let x and y be the numbers produced per day. It must produce at least 20 units of X, X must be at least twice Y, total production cannot exceed 126, and the production budget is at most Shs 34,400.
Each X uses 3 units of a special input and each Y uses 4 units. At most 470 input units are available. Production costs are Shs 200 per X and Shs 300 per Y, while profits are Shs 500 and Shs 600 respectively.
- Form two simultaneous equations from the maintenance records and determine S and B.
- Define x and y and write all the production restrictions as inequalities, including non-negativity.
- Write the daily profit function P in terms of x and y.
- Determine the important feasible corner points and calculate the profit at each point.
- Identify the production plan that gives the greatest feasible profit.
- For that plan, calculate the total production cost and the total amount of special input used.
- Calculate the corresponding revenue and verify the profit.
(a) S + 2B = 29500; 3S − 2B = 4500. Adding gives 4S = 34000, so S = 8500. Then B = 10500. Thus solar maintenance is Shs 8,500 and backup maintenance is Shs 10,500.
(b) x ≥ 20; x ≥ 2y; x + y ≤ 126; 200x + 300y ≤ 34400; 3x + 4y ≤ 470; x ≥ 0, y ≥ 0.
(c) P = 500x + 600y.
(d) The key feasible corner points are (20,0), (20,10), (126,0) and (84,42). Their profits are Shs 10,000, Shs 16,000, Shs 63,000 and Shs 67,200 respectively.
(e) Therefore the maximum occurs at x = 84, y = 42, giving a maximum daily profit of Shs 67,200.
(f) Production cost = 200(84) + 300(42) = Shs 29,400. Input used = 3(84) + 4(42) = 420 units.
(g) Revenue = production cost + profit = Shs 29,400 + Shs 67,200 = Shs 96,600. Subtracting the cost from the revenue again gives Shs 67,200 profit.
ITEM 3
A management team in Vocational School Expansion records the number of customers served by 50 staff members during a trial period. Management will continue the arrangement if the average number served exceeds 46.
Staff members who serve at most that threshold are also reviewed for additional support or possible reassignment.
The recorded values are listed below:
21, 23, 25, 27, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 50, 51, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 60, 60, 61, 62, 63, 64, 65, 66, 67, 68, 70.
- Arrange the data into a frequency distribution using class intervals of 10, beginning with the smallest observation. Include cumulative frequency and class midpoint.
- Determine the total number of observations and the total number of customers served, then calculate the mean.
- Use the management rule to state whether the average target is exceeded.
- Determine the number and percentage of staff who served at most 46 customers.
- Determine the number and percentage of staff who served at least 61 customers.
- Use the ordered data to determine the median and identify the median class.
- Explain how the frequency distribution, mean and percentages can help management make a decision.
(a) Frequency table: 21–30: f=6, cf=6, midpoint=25.5; 31–40: f=10, cf=16, midpoint=35.5; 41–50: f=12, cf=28, midpoint=45.5; 51–60: f=13, cf=41, midpoint=55.5; 61–70: f=9, cf=50, midpoint=65.5
(b) There are 50 staff members. Total customers served = 2368. Mean = 2368 ÷ 50 = 47.36.
(c) Since 47.36 > 46, the stated average target is exceeded.
(d) 23 staff served at most 46. Percentage = 23/50 × 100 = 46.0%.
(e) 9 staff served at least 61. Percentage = 9/50 × 100 = 18.0%.
(f) The 25th and 26th ordered observations are 48 and 49; median = 48.5. The median lies in the 41–50 class.
(g) The table shows concentration across ranges, the mean gives an overall average, and the percentages identify groups needing attention or further support.
ITEM 4
A class committee in Vocational School Expansion wants to choose one student representative from three candidates A, B and C. There are 122 students in the class.
17 students support A only, 29 support B only and 15 support C only. In addition, 11 support both A and B only, 13 support B and C only, 9 support A and C only, and 7 support all three.
Students who support none of the candidates are recorded separately. The teacher will select the candidate with the largest total support.
- Determine the number of students who support none of the three candidates.
- Determine the total number of supporters of A, B and C respectively.
- Identify the candidate with the greatest support and find the difference between the highest and second-highest totals.
- Determine the percentage of the whole class supporting the selected candidate.
- Determine the number and percentage of students who do not support the selected candidate.
- State whether the selected candidate has support from more than half the class and whether fewer than 50% oppose the candidate.
- Explain why the overlapping groups must be counted when making the final decision.
(a) Known regions total = 101; therefore none = 122 − 101 = 21 students.
(b) A = 44; B = 60; C = 44. Each total includes the relevant overlapping regions.
(c) Candidate B has the greatest support with 60 students. The second-highest total is 44, so the difference is 16.
(d) Support percentage = 60/122 × 100 = 49.18%.
(e) Non-supporters = 62. Percentage = 50.82%.
(f) Half the class is 61.0. Since 60 is greater than half, the selected candidate has majority support. The opposition percentage is 50.82%, which is below 50%.
(g) A student in an overlap supports more than one candidate. Ignoring these intersections would undercount candidates' totals and could change the result.
ITEM 5
A recreation facility in Vocational School Expansion is creating a paved common region where two circular lawns overlap. The radii of the circles are 2.05 m and 1.50 m.
The distance between the two intersection points is 1.65 m. The common region alone will be paved.
Each rectangular paver measures 5.0 cm by 10.0 cm. The supplier recommends buying an additional 6% above the calculated number because some pavers will be cut or damaged. Take π = 3.14.
- Determine the perpendicular distance from the centre of each circle to the common chord.
- Hence determine the distance between the centres of the two circles.
- Determine the central angle subtended by the common chord at each centre.
- Find the area of the common region by calculating the two circular segments.
- Convert the area of one paver into square metres and determine the minimum number of pavers required before the allowance.
- Allowing 6% extra, determine the number of pavers to purchase.
- Determine the total cost and explain why the final number must be rounded upward.
(a) Half-chord = 0.825 m. Distance for radius 2.05 m = √(2.05²−0.825²) = 1.877 m. Distance for radius 1.50 m = 1.253 m.
(b) Centre distance = 1.877 + 1.253 = 3.129 m.
(c) Central angles are approximately 47.46° and 66.73°.
(d) Using sector area minus triangle area for each segment, the common area is approximately 0.468 m².
(e) One paver area = 0.050 × 0.100 = 0.00500 m². Minimum before allowance = 0.468 ÷ 0.00500 ≈ 94 pavers.
(f) With 6% extra: 94 × (1 + 0.06) = 99.64, so 100 pavers should be purchased.
(g) Cost = 100 × Shs 225 = Shs 22,500. The result must be rounded upward because a fractional paver cannot be purchased and rounding downward would leave part of the region uncovered.
ITEM 6
A small manufacturing group in Vocational School Expansion produces open-topped ice-cream cones. One original cone has capacity 34 cm³ and height 4.25 cm.
The original cone costs Shs 55 to make and is sold for Shs 220. A new order requires cones with capacity 27 times the original capacity. The new cones are similar to the original cones.
The selling price is directly proportional to capacity, while production cost is directly proportional to the curved surface area of the cone. Use π = 3.14.
- Determine the radius and diameter of the original cone.
- Determine the slant height and curved surface area of the original cone.
- Determine the linear scale factor for the new cone, given that its capacity is 27 times the original.
- Determine the new height and diameter.
- Determine the curved surface area of one new cone and hence the total material area for an order of 250 cones.
- Determine the production cost of one new cone and the selling price of one new cone.
- Determine the profit per cone and the total profit on an order of 250 cones.
- Explain why capacity changes by a different factor from diameter.
(a) Using V = ⅓πr²h: r = √(3V/(πh)) = 2.765 cm. Diameter = 5.529 cm.
(b) Slant height l = √(r²+h²) = 5.070 cm. Curved surface area = πrl ≈ 44.014 cm².
(c) Volume factor = 27, so linear scale factor k = ∛27 = 3.000.
(d) New height = 4.25 × 3.000 = 12.750 cm. New diameter = 16.588 cm.
(e) New curved surface area = 396.122 cm² per cone. For 250 cones, total material area ≈ 9.903 m².
(f) Surface area scales as k², so new production cost = Shs 55 × 3.000² ≈ Shs 495. Selling price scales directly with capacity, so new selling price = Shs 220 × 27 = Shs 5,940.
(g) Profit per cone ≈ Shs 5,445. Total profit for 250 cones ≈ Shs 1,361,250.
(h) Capacity is a volume, so it changes with the cube of the linear scale factor, whereas diameter changes only linearly. Surface area changes with the square of the scale factor.
SET 3: MARKET UPGRADE PROJECT
ITEM 1
The local authority in Market Upgrade Project has secured land for a community project. The project team has 27 acres available. The plan reserves 4/9 of the total land for the main buildings.
After the building area is removed, 1/5 of the land that remains is to be landscaped or planted. The rest is divided among staff facilities, a public waiting area and parking in the ratio 4:3:2.
The project engineer estimates that one vehicle needs 0.033 acres of parking space. However, 10% of the full parking capacity must remain free for emergency access and movement.
The landscaping must be cleared before construction. Two workers can clear half an acre in one day when working at the same rate. Each worker is paid Shs 15,000 for one day's work.
- Determine the area allocated to the main buildings.
- Determine the area remaining after the buildings and hence find the area reserved for landscaping or planting.
- Use the given ratio to determine the areas for staff facilities, the public waiting area and parking.
- Determine the theoretical maximum number of vehicles that the parking area can accommodate.
- After keeping 10% of the capacity free, determine the maximum number of ordinary vehicles that should be admitted at one time.
- Determine the number of workers required to clear the landscaping area in one day.
- Calculate the total amount required to pay those workers for the one-day job.
(a) Buildings = 27 × 4/9 = 12 acres.
(b) Remaining = 15 acres. Landscaping = 3 acres. Land still available after landscaping = 12 acres.
(c) Total ratio parts = 9. Staff = 5.333 acres; waiting area = 4 acres; parking = 2.667 acres.
(d) Parking capacity = 2.667 ÷ 0.033 = 80 vehicles.
(e) Emergency reserve = 10% × 80 = 8.0 vehicle spaces. Ordinary capacity = 90% × 80 = 72 whole vehicles.
(f) At 0.25 acre per worker-day, workers required = 3 ÷ 0.25 = 12.00, so 12 workers are required.
(g) Labour cost = 12 × Shs 15,000 = Shs 180,000.
ITEM 2
A small enterprise in Market Upgrade Project uses two energy systems. Let S be the daily maintenance cost of the solar system and B the daily maintenance cost of the backup system.
Records show that S plus twice B is Shs 31,000. They also show that three times S exceeds twice B by Shs 5,000.
The enterprise produces two products, X and Y. Let x and y be the numbers produced per day. It must produce at least 20 units of X, X must be at least twice Y, total production cannot exceed 132, and the production budget is at most Shs 35,800.
Each X uses 3 units of a special input and each Y uses 4 units. At most 490 input units are available. Production costs are Shs 200 per X and Shs 300 per Y, while profits are Shs 500 and Shs 600 respectively.
- Form two simultaneous equations from the maintenance records and determine S and B.
- Define x and y and write all the production restrictions as inequalities, including non-negativity.
- Write the daily profit function P in terms of x and y.
- Determine the important feasible corner points and calculate the profit at each point.
- Identify the production plan that gives the greatest feasible profit.
- For that plan, calculate the total production cost and the total amount of special input used.
- Calculate the corresponding revenue and verify the profit.
(a) S + 2B = 31000; 3S − 2B = 5000. Adding gives 4S = 36000, so S = 9000. Then B = 11000. Thus solar maintenance is Shs 9,000 and backup maintenance is Shs 11,000.
(b) x ≥ 20; x ≥ 2y; x + y ≤ 132; 200x + 300y ≤ 35800; 3x + 4y ≤ 490; x ≥ 0, y ≥ 0.
(c) P = 500x + 600y.
(d) The key feasible corner points are (20,0), (20,10), (132,0) and (88,44). Their profits are Shs 10,000, Shs 16,000, Shs 66,000 and Shs 70,400 respectively.
(e) Therefore the maximum occurs at x = 88, y = 44, giving a maximum daily profit of Shs 70,400.
(f) Production cost = 200(88) + 300(44) = Shs 30,800. Input used = 3(88) + 4(44) = 440 units.
(g) Revenue = production cost + profit = Shs 30,800 + Shs 70,400 = Shs 101,200. Subtracting the cost from the revenue again gives Shs 70,400 profit.
ITEM 3
A management team in Market Upgrade Project records the number of customers served by 50 staff members during a trial period. Management will continue the arrangement if the average number served exceeds 47.
Staff members who serve at most that threshold are also reviewed for additional support or possible reassignment.
The recorded values are listed below:
22, 24, 26, 28, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 51, 52, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 61, 61, 62, 63, 64, 65, 66, 67, 68, 69, 71.
- Arrange the data into a frequency distribution using class intervals of 10, beginning with the smallest observation. Include cumulative frequency and class midpoint.
- Determine the total number of observations and the total number of customers served, then calculate the mean.
- Use the management rule to state whether the average target is exceeded.
- Determine the number and percentage of staff who served at most 47 customers.
- Determine the number and percentage of staff who served at least 62 customers.
- Use the ordered data to determine the median and identify the median class.
- Explain how the frequency distribution, mean and percentages can help management make a decision.
(a) Frequency table: 22–31: f=6, cf=6, midpoint=26.5; 32–41: f=10, cf=16, midpoint=36.5; 42–51: f=12, cf=28, midpoint=46.5; 52–61: f=13, cf=41, midpoint=56.5; 62–71: f=9, cf=50, midpoint=66.5
(b) There are 50 staff members. Total customers served = 2418. Mean = 2418 ÷ 50 = 48.36.
(c) Since 48.36 > 47, the stated average target is exceeded.
(d) 23 staff served at most 47. Percentage = 23/50 × 100 = 46.0%.
(e) 9 staff served at least 62. Percentage = 9/50 × 100 = 18.0%.
(f) The 25th and 26th ordered observations are 49 and 50; median = 49.5. The median lies in the 42–51 class.
(g) The table shows concentration across ranges, the mean gives an overall average, and the percentages identify groups needing attention or further support.
ITEM 4
A class committee in Market Upgrade Project wants to choose one student representative from three candidates A, B and C. There are 124 students in the class.
18 students support A only, 30 support B only and 16 support C only. In addition, 12 support both A and B only, 14 support B and C only, 10 support A and C only, and 8 support all three.
Students who support none of the candidates are recorded separately. The teacher will select the candidate with the largest total support.
- Determine the number of students who support none of the three candidates.
- Determine the total number of supporters of A, B and C respectively.
- Identify the candidate with the greatest support and find the difference between the highest and second-highest totals.
- Determine the percentage of the whole class supporting the selected candidate.
- Determine the number and percentage of students who do not support the selected candidate.
- State whether the selected candidate has support from more than half the class and whether fewer than 50% oppose the candidate.
- Explain why the overlapping groups must be counted when making the final decision.
(a) Known regions total = 108; therefore none = 124 − 108 = 16 students.
(b) A = 48; B = 64; C = 48. Each total includes the relevant overlapping regions.
(c) Candidate B has the greatest support with 64 students. The second-highest total is 48, so the difference is 16.
(d) Support percentage = 64/124 × 100 = 51.61%.
(e) Non-supporters = 60. Percentage = 48.39%.
(f) Half the class is 62.0. Since 64 is greater than half, the selected candidate has majority support. The opposition percentage is 48.39%, which is below 50%.
(g) A student in an overlap supports more than one candidate. Ignoring these intersections would undercount candidates' totals and could change the result.
ITEM 5
A recreation facility in Market Upgrade Project is creating a paved common region where two circular lawns overlap. The radii of the circles are 2.10 m and 1.50 m.
The distance between the two intersection points is 1.70 m. The common region alone will be paved.
Each rectangular paver measures 6.0 cm by 8.0 cm. The supplier recommends buying an additional 7% above the calculated number because some pavers will be cut or damaged. Take π = 3.14.
- Determine the perpendicular distance from the centre of each circle to the common chord.
- Hence determine the distance between the centres of the two circles.
- Determine the central angle subtended by the common chord at each centre.
- Find the area of the common region by calculating the two circular segments.
- Convert the area of one paver into square metres and determine the minimum number of pavers required before the allowance.
- Allowing 7% extra, determine the number of pavers to purchase.
- Determine the total cost and explain why the final number must be rounded upward.
(a) Half-chord = 0.850 m. Distance for radius 2.10 m = √(2.10²−0.850²) = 1.920 m. Distance for radius 1.50 m = 1.236 m.
(b) Centre distance = 1.920 + 1.236 = 3.156 m.
(c) Central angles are approximately 47.75° and 69.04°.
(d) Using sector area minus triangle area for each segment, the common area is approximately 0.509 m².
(e) One paver area = 0.060 × 0.080 = 0.00480 m². Minimum before allowance = 0.509 ÷ 0.00480 ≈ 107 pavers.
(f) With 7% extra: 107 × (1 + 0.07) = 114.49, so 115 pavers should be purchased.
(g) Cost = 115 × Shs 250 = Shs 28,750. The result must be rounded upward because a fractional paver cannot be purchased and rounding downward would leave part of the region uncovered.
ITEM 6
A small manufacturing group in Market Upgrade Project produces open-topped ice-cream cones. One original cone has capacity 36 cm³ and height 4.50 cm.
The original cone costs Shs 60 to make and is sold for Shs 240. A new order requires cones with capacity 8 times the original capacity. The new cones are similar to the original cones.
The selling price is directly proportional to capacity, while production cost is directly proportional to the curved surface area of the cone. Use π = 3.14.
- Determine the radius and diameter of the original cone.
- Determine the slant height and curved surface area of the original cone.
- Determine the linear scale factor for the new cone, given that its capacity is 8 times the original.
- Determine the new height and diameter.
- Determine the curved surface area of one new cone and hence the total material area for an order of 250 cones.
- Determine the production cost of one new cone and the selling price of one new cone.
- Determine the profit per cone and the total profit on an order of 250 cones.
- Explain why capacity changes by a different factor from diameter.
(a) Using V = ⅓πr²h: r = √(3V/(πh)) = 2.765 cm. Diameter = 5.529 cm.
(b) Slant height l = √(r²+h²) = 5.281 cm. Curved surface area = πrl ≈ 45.848 cm².
(c) Volume factor = 8, so linear scale factor k = ∛8 = 2.000.
(d) New height = 4.50 × 2.000 = 9.000 cm. New diameter = 11.059 cm.
(e) New curved surface area = 183.392 cm² per cone. For 250 cones, total material area ≈ 4.585 m².
(f) Surface area scales as k², so new production cost = Shs 60 × 2.000² ≈ Shs 240. Selling price scales directly with capacity, so new selling price = Shs 240 × 8 = Shs 1,920.
(g) Profit per cone ≈ Shs 1,680. Total profit for 250 cones ≈ Shs 420,000.
(h) Capacity is a volume, so it changes with the cube of the linear scale factor, whereas diameter changes only linearly. Surface area changes with the square of the scale factor.
SET 4: SPORTS ACADEMY
ITEM 1
The local authority in Sports Academy has secured land for a community project. The project team has 22 acres available. The plan reserves 1/2 of the total land for the main buildings.
After the building area is removed, 1/4 of the land that remains is to be landscaped or planted. The rest is divided among staff facilities, a public waiting area and parking in the ratio 3:1:2.
The project engineer estimates that one vehicle needs 0.018 acres of parking space. However, 10% of the full parking capacity must remain free for emergency access and movement.
The landscaping must be cleared before construction. Two workers can clear half an acre in one day when working at the same rate. Each worker is paid Shs 22,000 for one day's work.
- Determine the area allocated to the main buildings.
- Determine the area remaining after the buildings and hence find the area reserved for landscaping or planting.
- Use the given ratio to determine the areas for staff facilities, the public waiting area and parking.
- Determine the theoretical maximum number of vehicles that the parking area can accommodate.
- After keeping 10% of the capacity free, determine the maximum number of ordinary vehicles that should be admitted at one time.
- Determine the number of workers required to clear the landscaping area in one day.
- Calculate the total amount required to pay those workers for the one-day job.
(a) Buildings = 22 × 1/2 = 11 acres.
(b) Remaining = 11 acres. Landscaping = 2.75 acres. Land still available after landscaping = 8.25 acres.
(c) Total ratio parts = 6. Staff = 4.125 acres; waiting area = 1.375 acres; parking = 2.75 acres.
(d) Parking capacity = 2.75 ÷ 0.018 = 150 vehicles.
(e) Emergency reserve = 10% × 150 = 15.0 vehicle spaces. Ordinary capacity = 90% × 150 = 135 whole vehicles.
(f) At 0.25 acre per worker-day, workers required = 2.75 ÷ 0.25 = 11.00, so 11 workers are required.
(g) Labour cost = 11 × Shs 22,000 = Shs 242,000.
ITEM 2
A small enterprise in Sports Academy uses two energy systems. Let S be the daily maintenance cost of the solar system and B the daily maintenance cost of the backup system.
Records show that S plus twice B is Shs 32,500. They also show that three times S exceeds twice B by Shs 5,500.
The enterprise produces two products, X and Y. Let x and y be the numbers produced per day. It must produce at least 20 units of X, X must be at least twice Y, total production cannot exceed 138, and the production budget is at most Shs 37,200.
Each X uses 3 units of a special input and each Y uses 4 units. At most 510 input units are available. Production costs are Shs 200 per X and Shs 300 per Y, while profits are Shs 500 and Shs 600 respectively.
- Form two simultaneous equations from the maintenance records and determine S and B.
- Define x and y and write all the production restrictions as inequalities, including non-negativity.
- Write the daily profit function P in terms of x and y.
- Determine the important feasible corner points and calculate the profit at each point.
- Identify the production plan that gives the greatest feasible profit.
- For that plan, calculate the total production cost and the total amount of special input used.
- Calculate the corresponding revenue and verify the profit.
(a) S + 2B = 32500; 3S − 2B = 5500. Adding gives 4S = 38000, so S = 9500. Then B = 11500. Thus solar maintenance is Shs 9,500 and backup maintenance is Shs 11,500.
(b) x ≥ 20; x ≥ 2y; x + y ≤ 138; 200x + 300y ≤ 37200; 3x + 4y ≤ 510; x ≥ 0, y ≥ 0.
(c) P = 500x + 600y.
(d) The key feasible corner points are (20,0), (20,10), (138,0) and (92,46). Their profits are Shs 10,000, Shs 16,000, Shs 69,000 and Shs 73,600 respectively.
(e) Therefore the maximum occurs at x = 92, y = 46, giving a maximum daily profit of Shs 73,600.
(f) Production cost = 200(92) + 300(46) = Shs 32,200. Input used = 3(92) + 4(46) = 460 units.
(g) Revenue = production cost + profit = Shs 32,200 + Shs 73,600 = Shs 105,800. Subtracting the cost from the revenue again gives Shs 73,600 profit.
ITEM 3
A management team in Sports Academy records the number of customers served by 50 staff members during a trial period. Management will continue the arrangement if the average number served exceeds 48.
Staff members who serve at most that threshold are also reviewed for additional support or possible reassignment.
The recorded values are listed below:
23, 25, 27, 29, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 52, 53, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 62, 62, 63, 64, 65, 66, 67, 68, 69, 70, 72.
- Arrange the data into a frequency distribution using class intervals of 10, beginning with the smallest observation. Include cumulative frequency and class midpoint.
- Determine the total number of observations and the total number of customers served, then calculate the mean.
- Use the management rule to state whether the average target is exceeded.
- Determine the number and percentage of staff who served at most 48 customers.
- Determine the number and percentage of staff who served at least 63 customers.
- Use the ordered data to determine the median and identify the median class.
- Explain how the frequency distribution, mean and percentages can help management make a decision.
(a) Frequency table: 23–32: f=6, cf=6, midpoint=27.5; 33–42: f=10, cf=16, midpoint=37.5; 43–52: f=12, cf=28, midpoint=47.5; 53–62: f=13, cf=41, midpoint=57.5; 63–72: f=9, cf=50, midpoint=67.5
(b) There are 50 staff members. Total customers served = 2468. Mean = 2468 ÷ 50 = 49.36.
(c) Since 49.36 > 48, the stated average target is exceeded.
(d) 23 staff served at most 48. Percentage = 23/50 × 100 = 46.0%.
(e) 9 staff served at least 63. Percentage = 9/50 × 100 = 18.0%.
(f) The 25th and 26th ordered observations are 50 and 51; median = 50.5. The median lies in the 43–52 class.
(g) The table shows concentration across ranges, the mean gives an overall average, and the percentages identify groups needing attention or further support.
ITEM 4
A class committee in Sports Academy wants to choose one student representative from three candidates A, B and C. There are 126 students in the class.
19 students support A only, 31 support B only and 17 support C only. In addition, 10 support both A and B only, 15 support B and C only, 11 support A and C only, and 6 support all three.
Students who support none of the candidates are recorded separately. The teacher will select the candidate with the largest total support.
- Determine the number of students who support none of the three candidates.
- Determine the total number of supporters of A, B and C respectively.
- Identify the candidate with the greatest support and find the difference between the highest and second-highest totals.
- Determine the percentage of the whole class supporting the selected candidate.
- Determine the number and percentage of students who do not support the selected candidate.
- State whether the selected candidate has support from more than half the class and whether fewer than 50% oppose the candidate.
- Explain why the overlapping groups must be counted when making the final decision.
(a) Known regions total = 109; therefore none = 126 − 109 = 17 students.
(b) A = 46; B = 62; C = 49. Each total includes the relevant overlapping regions.
(c) Candidate B has the greatest support with 62 students. The second-highest total is 49, so the difference is 13.
(d) Support percentage = 62/126 × 100 = 49.21%.
(e) Non-supporters = 64. Percentage = 50.79%.
(f) Half the class is 63.0. Since 62 is greater than half, the selected candidate has majority support. The opposition percentage is 50.79%, which is below 50%.
(g) A student in an overlap supports more than one candidate. Ignoring these intersections would undercount candidates' totals and could change the result.
ITEM 5
A recreation facility in Sports Academy is creating a paved common region where two circular lawns overlap. The radii of the circles are 2.15 m and 1.50 m.
The distance between the two intersection points is 1.75 m. The common region alone will be paved.
Each rectangular paver measures 4.0 cm by 10.0 cm. The supplier recommends buying an additional 8% above the calculated number because some pavers will be cut or damaged. Take π = 3.14.
- Determine the perpendicular distance from the centre of each circle to the common chord.
- Hence determine the distance between the centres of the two circles.
- Determine the central angle subtended by the common chord at each centre.
- Find the area of the common region by calculating the two circular segments.
- Convert the area of one paver into square metres and determine the minimum number of pavers required before the allowance.
- Allowing 8% extra, determine the number of pavers to purchase.
- Determine the total cost and explain why the final number must be rounded upward.
(a) Half-chord = 0.875 m. Distance for radius 2.15 m = √(2.15²−0.875²) = 1.964 m. Distance for radius 1.50 m = 1.218 m.
(b) Centre distance = 1.964 + 1.218 = 3.182 m.
(c) Central angles are approximately 48.03° and 71.37°.
(d) Using sector area minus triangle area for each segment, the common area is approximately 0.553 m².
(e) One paver area = 0.040 × 0.100 = 0.00400 m². Minimum before allowance = 0.553 ÷ 0.00400 ≈ 139 pavers.
(f) With 8% extra: 139 × (1 + 0.08) = 150.12, so 151 pavers should be purchased.
(g) Cost = 151 × Shs 275 = Shs 41,525. The result must be rounded upward because a fractional paver cannot be purchased and rounding downward would leave part of the region uncovered.
ITEM 6
A small manufacturing group in Sports Academy produces open-topped ice-cream cones. One original cone has capacity 38 cm³ and height 4.75 cm.
The original cone costs Shs 65 to make and is sold for Shs 260. A new order requires cones with capacity 4 times the original capacity. The new cones are similar to the original cones.
The selling price is directly proportional to capacity, while production cost is directly proportional to the curved surface area of the cone. Use π = 3.14.
- Determine the radius and diameter of the original cone.
- Determine the slant height and curved surface area of the original cone.
- Determine the linear scale factor for the new cone, given that its capacity is 4 times the original.
- Determine the new height and diameter.
- Determine the curved surface area of one new cone and hence the total material area for an order of 250 cones.
- Determine the production cost of one new cone and the selling price of one new cone.
- Determine the profit per cone and the total profit on an order of 250 cones.
- Explain why capacity changes by a different factor from diameter.
(a) Using V = ⅓πr²h: r = √(3V/(πh)) = 2.765 cm. Diameter = 5.529 cm.
(b) Slant height l = √(r²+h²) = 5.496 cm. Curved surface area = πrl ≈ 47.711 cm².
(c) Volume factor = 4, so linear scale factor k = ∛4 = 1.587.
(d) New height = 4.75 × 1.587 = 7.540 cm. New diameter = 8.777 cm.
(e) New curved surface area = 120.223 cm² per cone. For 250 cones, total material area ≈ 3.006 m².
(f) Surface area scales as k², so new production cost = Shs 65 × 1.587² ≈ Shs 163.79. Selling price scales directly with capacity, so new selling price = Shs 260 × 4 = Shs 1,040.
(g) Profit per cone ≈ Shs 876.21. Total profit for 250 cones ≈ Shs 219,052.57.
(h) Capacity is a volume, so it changes with the cube of the linear scale factor, whereas diameter changes only linearly. Surface area changes with the square of the scale factor.
SET 5: COMMUNITY LIBRARY
ITEM 1
The local authority in Community Library has secured land for a community project. The project team has 16 acres available. The plan reserves 3/8 of the total land for the main buildings.
After the building area is removed, 1/4 of the land that remains is to be landscaped or planted. The rest is divided among staff facilities, a public waiting area and parking in the ratio 2:3:5.
The project engineer estimates that one vehicle needs 0.025 acres of parking space. However, 10% of the full parking capacity must remain free for emergency access and movement.
The landscaping must be cleared before construction. Two workers can clear half an acre in one day when working at the same rate. Each worker is paid Shs 16,000 for one day's work.
- Determine the area allocated to the main buildings.
- Determine the area remaining after the buildings and hence find the area reserved for landscaping or planting.
- Use the given ratio to determine the areas for staff facilities, the public waiting area and parking.
- Determine the theoretical maximum number of vehicles that the parking area can accommodate.
- After keeping 10% of the capacity free, determine the maximum number of ordinary vehicles that should be admitted at one time.
- Determine the number of workers required to clear the landscaping area in one day.
- Calculate the total amount required to pay those workers for the one-day job.
(a) Buildings = 16 × 3/8 = 6 acres.
(b) Remaining = 10 acres. Landscaping = 2.5 acres. Land still available after landscaping = 7.5 acres.
(c) Total ratio parts = 10. Staff = 1.5 acres; waiting area = 2.25 acres; parking = 3.75 acres.
(d) Parking capacity = 3.75 ÷ 0.025 = 150 vehicles.
(e) Emergency reserve = 10% × 150 = 15.0 vehicle spaces. Ordinary capacity = 90% × 150 = 135 whole vehicles.
(f) At 0.25 acre per worker-day, workers required = 2.5 ÷ 0.25 = 10.00, so 10 workers are required.
(g) Labour cost = 10 × Shs 16,000 = Shs 160,000.
ITEM 2
A small enterprise in Community Library uses two energy systems. Let S be the daily maintenance cost of the solar system and B the daily maintenance cost of the backup system.
Records show that S plus twice B is Shs 34,000. They also show that three times S exceeds twice B by Shs 6,000.
The enterprise produces two products, X and Y. Let x and y be the numbers produced per day. It must produce at least 20 units of X, X must be at least twice Y, total production cannot exceed 144, and the production budget is at most Shs 38,600.
Each X uses 3 units of a special input and each Y uses 4 units. At most 530 input units are available. Production costs are Shs 200 per X and Shs 300 per Y, while profits are Shs 500 and Shs 600 respectively.
- Form two simultaneous equations from the maintenance records and determine S and B.
- Define x and y and write all the production restrictions as inequalities, including non-negativity.
- Write the daily profit function P in terms of x and y.
- Determine the important feasible corner points and calculate the profit at each point.
- Identify the production plan that gives the greatest feasible profit.
- For that plan, calculate the total production cost and the total amount of special input used.
- Calculate the corresponding revenue and verify the profit.
(a) S + 2B = 34000; 3S − 2B = 6000. Adding gives 4S = 40000, so S = 10000. Then B = 12000. Thus solar maintenance is Shs 10,000 and backup maintenance is Shs 12,000.
(b) x ≥ 20; x ≥ 2y; x + y ≤ 144; 200x + 300y ≤ 38600; 3x + 4y ≤ 530; x ≥ 0, y ≥ 0.
(c) P = 500x + 600y.
(d) The key feasible corner points are (20,0), (20,10), (144,0) and (96,48). Their profits are Shs 10,000, Shs 16,000, Shs 72,000 and Shs 76,800 respectively.
(e) Therefore the maximum occurs at x = 96, y = 48, giving a maximum daily profit of Shs 76,800.
(f) Production cost = 200(96) + 300(48) = Shs 33,600. Input used = 3(96) + 4(48) = 480 units.
(g) Revenue = production cost + profit = Shs 33,600 + Shs 76,800 = Shs 110,400. Subtracting the cost from the revenue again gives Shs 76,800 profit.
ITEM 3
A management team in Community Library records the number of customers served by 50 staff members during a trial period. Management will continue the arrangement if the average number served exceeds 49.
Staff members who serve at most that threshold are also reviewed for additional support or possible reassignment.
The recorded values are listed below:
24, 26, 28, 30, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 53, 54, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 63, 63, 64, 65, 66, 67, 68, 69, 70, 71, 73.
- Arrange the data into a frequency distribution using class intervals of 10, beginning with the smallest observation. Include cumulative frequency and class midpoint.
- Determine the total number of observations and the total number of customers served, then calculate the mean.
- Use the management rule to state whether the average target is exceeded.
- Determine the number and percentage of staff who served at most 49 customers.
- Determine the number and percentage of staff who served at least 64 customers.
- Use the ordered data to determine the median and identify the median class.
- Explain how the frequency distribution, mean and percentages can help management make a decision.
(a) Frequency table: 24–33: f=6, cf=6, midpoint=28.5; 34–43: f=10, cf=16, midpoint=38.5; 44–53: f=12, cf=28, midpoint=48.5; 54–63: f=13, cf=41, midpoint=58.5; 64–73: f=9, cf=50, midpoint=68.5
(b) There are 50 staff members. Total customers served = 2518. Mean = 2518 ÷ 50 = 50.36.
(c) Since 50.36 > 49, the stated average target is exceeded.
(d) 23 staff served at most 49. Percentage = 23/50 × 100 = 46.0%.
(e) 9 staff served at least 64. Percentage = 9/50 × 100 = 18.0%.
(f) The 25th and 26th ordered observations are 51 and 52; median = 51.5. The median lies in the 44–53 class.
(g) The table shows concentration across ranges, the mean gives an overall average, and the percentages identify groups needing attention or further support.
ITEM 4
A class committee in Community Library wants to choose one student representative from three candidates A, B and C. There are 128 students in the class.
20 students support A only, 28 support B only and 18 support C only. In addition, 11 support both A and B only, 12 support B and C only, 12 support A and C only, and 7 support all three.
Students who support none of the candidates are recorded separately. The teacher will select the candidate with the largest total support.
- Determine the number of students who support none of the three candidates.
- Determine the total number of supporters of A, B and C respectively.
- Identify the candidate with the greatest support and find the difference between the highest and second-highest totals.
- Determine the percentage of the whole class supporting the selected candidate.
- Determine the number and percentage of students who do not support the selected candidate.
- State whether the selected candidate has support from more than half the class and whether fewer than 50% oppose the candidate.
- Explain why the overlapping groups must be counted when making the final decision.
(a) Known regions total = 108; therefore none = 128 − 108 = 20 students.
(b) A = 50; B = 58; C = 49. Each total includes the relevant overlapping regions.
(c) Candidate B has the greatest support with 58 students. The second-highest total is 50, so the difference is 8.
(d) Support percentage = 58/128 × 100 = 45.31%.
(e) Non-supporters = 70. Percentage = 54.69%.
(f) Half the class is 64.0. Since 58 is greater than half, the selected candidate has majority support. The opposition percentage is 54.69%, which is below 50%.
(g) A student in an overlap supports more than one candidate. Ignoring these intersections would undercount candidates' totals and could change the result.
ITEM 5
A recreation facility in Community Library is creating a paved common region where two circular lawns overlap. The radii of the circles are 2.20 m and 1.50 m.
The distance between the two intersection points is 1.80 m. The common region alone will be paved.
Each rectangular paver measures 5.0 cm by 8.0 cm. The supplier recommends buying an additional 5% above the calculated number because some pavers will be cut or damaged. Take π = 3.14.
- Determine the perpendicular distance from the centre of each circle to the common chord.
- Hence determine the distance between the centres of the two circles.
- Determine the central angle subtended by the common chord at each centre.
- Find the area of the common region by calculating the two circular segments.
- Convert the area of one paver into square metres and determine the minimum number of pavers required before the allowance.
- Allowing 5% extra, determine the number of pavers to purchase.
- Determine the total cost and explain why the final number must be rounded upward.
(a) Half-chord = 0.900 m. Distance for radius 2.20 m = √(2.20²−0.900²) = 2.007 m. Distance for radius 1.50 m = 1.200 m.
(b) Centre distance = 2.007 + 1.200 = 3.207 m.
(c) Central angles are approximately 48.30° and 73.74°.
(d) Using sector area minus triangle area for each segment, the common area is approximately 0.599 m².
(e) One paver area = 0.050 × 0.080 = 0.00400 m². Minimum before allowance = 0.599 ÷ 0.00400 ≈ 150 pavers.
(f) With 5% extra: 150 × (1 + 0.05) = 157.50, so 158 pavers should be purchased.
(g) Cost = 158 × Shs 300 = Shs 47,400. The result must be rounded upward because a fractional paver cannot be purchased and rounding downward would leave part of the region uncovered.
ITEM 6
A small manufacturing group in Community Library produces open-topped ice-cream cones. One original cone has capacity 40 cm³ and height 4.00 cm.
The original cone costs Shs 70 to make and is sold for Shs 280. A new order requires cones with capacity 27 times the original capacity. The new cones are similar to the original cones.
The selling price is directly proportional to capacity, while production cost is directly proportional to the curved surface area of the cone. Use π = 3.14.
- Determine the radius and diameter of the original cone.
- Determine the slant height and curved surface area of the original cone.
- Determine the linear scale factor for the new cone, given that its capacity is 27 times the original.
- Determine the new height and diameter.
- Determine the curved surface area of one new cone and hence the total material area for an order of 250 cones.
- Determine the production cost of one new cone and the selling price of one new cone.
- Determine the profit per cone and the total profit on an order of 250 cones.
- Explain why capacity changes by a different factor from diameter.
(a) Using V = ⅓πr²h: r = √(3V/(πh)) = 3.091 cm. Diameter = 6.182 cm.
(b) Slant height l = √(r²+h²) = 5.055 cm. Curved surface area = πrl ≈ 49.063 cm².
(c) Volume factor = 27, so linear scale factor k = ∛27 = 3.000.
(d) New height = 4.00 × 3.000 = 12.000 cm. New diameter = 18.546 cm.
(e) New curved surface area = 441.569 cm² per cone. For 250 cones, total material area ≈ 11.039 m².
(f) Surface area scales as k², so new production cost = Shs 70 × 3.000² ≈ Shs 630. Selling price scales directly with capacity, so new selling price = Shs 280 × 27 = Shs 7,560.
(g) Profit per cone ≈ Shs 6,930. Total profit for 250 cones ≈ Shs 1,732,500.
(h) Capacity is a volume, so it changes with the cube of the linear scale factor, whereas diameter changes only linearly. Surface area changes with the square of the scale factor.
SET 6: IRRIGATION FARM
ITEM 1
The local authority in Irrigation Farm has secured land for a community project. The project team has 30 acres available. The plan reserves 9/20 of the total land for the main buildings.
After the building area is removed, 1/5 of the land that remains is to be landscaped or planted. The rest is divided among staff facilities, a public waiting area and parking in the ratio 3:2:5.
The project engineer estimates that one vehicle needs 0.03 acres of parking space. However, 10% of the full parking capacity must remain free for emergency access and movement.
The landscaping must be cleared before construction. Two workers can clear half an acre in one day when working at the same rate. Each worker is paid Shs 19,000 for one day's work.
- Determine the area allocated to the main buildings.
- Determine the area remaining after the buildings and hence find the area reserved for landscaping or planting.
- Use the given ratio to determine the areas for staff facilities, the public waiting area and parking.
- Determine the theoretical maximum number of vehicles that the parking area can accommodate.
- After keeping 10% of the capacity free, determine the maximum number of ordinary vehicles that should be admitted at one time.
- Determine the number of workers required to clear the landscaping area in one day.
- Calculate the total amount required to pay those workers for the one-day job.
(a) Buildings = 30 × 9/20 = 13.5 acres.
(b) Remaining = 16.5 acres. Landscaping = 3.3 acres. Land still available after landscaping = 13.2 acres.
(c) Total ratio parts = 10. Staff = 3.96 acres; waiting area = 2.64 acres; parking = 6.6 acres.
(d) Parking capacity = 6.6 ÷ 0.03 = 220 vehicles.
(e) Emergency reserve = 10% × 220 = 22.0 vehicle spaces. Ordinary capacity = 90% × 220 = 198 whole vehicles.
(f) At 0.25 acre per worker-day, workers required = 3.3 ÷ 0.25 = 13.20, so 14 workers are required.
(g) Labour cost = 14 × Shs 19,000 = Shs 266,000.
ITEM 2
A small enterprise in Irrigation Farm uses two energy systems. Let S be the daily maintenance cost of the solar system and B the daily maintenance cost of the backup system.
Records show that S plus twice B is Shs 35,500. They also show that three times S exceeds twice B by Shs 6,500.
The enterprise produces two products, X and Y. Let x and y be the numbers produced per day. It must produce at least 20 units of X, X must be at least twice Y, total production cannot exceed 150, and the production budget is at most Shs 40,000.
Each X uses 3 units of a special input and each Y uses 4 units. At most 550 input units are available. Production costs are Shs 200 per X and Shs 300 per Y, while profits are Shs 500 and Shs 600 respectively.
- Form two simultaneous equations from the maintenance records and determine S and B.
- Define x and y and write all the production restrictions as inequalities, including non-negativity.
- Write the daily profit function P in terms of x and y.
- Determine the important feasible corner points and calculate the profit at each point.
- Identify the production plan that gives the greatest feasible profit.
- For that plan, calculate the total production cost and the total amount of special input used.
- Calculate the corresponding revenue and verify the profit.
(a) S + 2B = 35500; 3S − 2B = 6500. Adding gives 4S = 42000, so S = 10500. Then B = 12500. Thus solar maintenance is Shs 10,500 and backup maintenance is Shs 12,500.
(b) x ≥ 20; x ≥ 2y; x + y ≤ 150; 200x + 300y ≤ 40000; 3x + 4y ≤ 550; x ≥ 0, y ≥ 0.
(c) P = 500x + 600y.
(d) The key feasible corner points are (20,0), (20,10), (150,0) and (100,50). Their profits are Shs 10,000, Shs 16,000, Shs 75,000 and Shs 80,000 respectively.
(e) Therefore the maximum occurs at x = 100, y = 50, giving a maximum daily profit of Shs 80,000.
(f) Production cost = 200(100) + 300(50) = Shs 35,000. Input used = 3(100) + 4(50) = 500 units.
(g) Revenue = production cost + profit = Shs 35,000 + Shs 80,000 = Shs 115,000. Subtracting the cost from the revenue again gives Shs 80,000 profit.
ITEM 3
A management team in Irrigation Farm records the number of customers served by 50 staff members during a trial period. Management will continue the arrangement if the average number served exceeds 50.
Staff members who serve at most that threshold are also reviewed for additional support or possible reassignment.
The recorded values are listed below:
25, 27, 29, 31, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 54, 55, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 64, 64, 65, 66, 67, 68, 69, 70, 71, 72, 74.
- Arrange the data into a frequency distribution using class intervals of 10, beginning with the smallest observation. Include cumulative frequency and class midpoint.
- Determine the total number of observations and the total number of customers served, then calculate the mean.
- Use the management rule to state whether the average target is exceeded.
- Determine the number and percentage of staff who served at most 50 customers.
- Determine the number and percentage of staff who served at least 65 customers.
- Use the ordered data to determine the median and identify the median class.
- Explain how the frequency distribution, mean and percentages can help management make a decision.
(a) Frequency table: 25–34: f=6, cf=6, midpoint=29.5; 35–44: f=10, cf=16, midpoint=39.5; 45–54: f=12, cf=28, midpoint=49.5; 55–64: f=13, cf=41, midpoint=59.5; 65–74: f=9, cf=50, midpoint=69.5
(b) There are 50 staff members. Total customers served = 2568. Mean = 2568 ÷ 50 = 51.36.
(c) Since 51.36 > 50, the stated average target is exceeded.
(d) 23 staff served at most 50. Percentage = 23/50 × 100 = 46.0%.
(e) 9 staff served at least 65. Percentage = 9/50 × 100 = 18.0%.
(f) The 25th and 26th ordered observations are 52 and 53; median = 52.5. The median lies in the 45–54 class.
(g) The table shows concentration across ranges, the mean gives an overall average, and the percentages identify groups needing attention or further support.
ITEM 4
A class committee in Irrigation Farm wants to choose one student representative from three candidates A, B and C. There are 130 students in the class.
16 students support A only, 29 support B only and 19 support C only. In addition, 12 support both A and B only, 13 support B and C only, 8 support A and C only, and 8 support all three.
Students who support none of the candidates are recorded separately. The teacher will select the candidate with the largest total support.
- Determine the number of students who support none of the three candidates.
- Determine the total number of supporters of A, B and C respectively.
- Identify the candidate with the greatest support and find the difference between the highest and second-highest totals.
- Determine the percentage of the whole class supporting the selected candidate.
- Determine the number and percentage of students who do not support the selected candidate.
- State whether the selected candidate has support from more than half the class and whether fewer than 50% oppose the candidate.
- Explain why the overlapping groups must be counted when making the final decision.
(a) Known regions total = 105; therefore none = 130 − 105 = 25 students.
(b) A = 44; B = 62; C = 48. Each total includes the relevant overlapping regions.
(c) Candidate B has the greatest support with 62 students. The second-highest total is 48, so the difference is 14.
(d) Support percentage = 62/130 × 100 = 47.69%.
(e) Non-supporters = 68. Percentage = 52.31%.
(f) Half the class is 65.0. Since 62 is greater than half, the selected candidate has majority support. The opposition percentage is 52.31%, which is below 50%.
(g) A student in an overlap supports more than one candidate. Ignoring these intersections would undercount candidates' totals and could change the result.
ITEM 5
A recreation facility in Irrigation Farm is creating a paved common region where two circular lawns overlap. The radii of the circles are 2.00 m and 1.55 m.
The distance between the two intersection points is 1.85 m. The common region alone will be paved.
Each rectangular paver measures 6.0 cm by 10.0 cm. The supplier recommends buying an additional 6% above the calculated number because some pavers will be cut or damaged. Take π = 3.14.
- Determine the perpendicular distance from the centre of each circle to the common chord.
- Hence determine the distance between the centres of the two circles.
- Determine the central angle subtended by the common chord at each centre.
- Find the area of the common region by calculating the two circular segments.
- Convert the area of one paver into square metres and determine the minimum number of pavers required before the allowance.
- Allowing 6% extra, determine the number of pavers to purchase.
- Determine the total cost and explain why the final number must be rounded upward.
(a) Half-chord = 0.925 m. Distance for radius 2.00 m = √(2.00²−0.925²) = 1.773 m. Distance for radius 1.55 m = 1.244 m.
(b) Centre distance = 1.773 + 1.244 = 3.017 m.
(c) Central angles are approximately 55.10° and 73.28°.
(d) Using sector area minus triangle area for each segment, the common area is approximately 0.667 m².
(e) One paver area = 0.060 × 0.100 = 0.00600 m². Minimum before allowance = 0.667 ÷ 0.00600 ≈ 112 pavers.
(f) With 6% extra: 112 × (1 + 0.06) = 118.72, so 119 pavers should be purchased.
(g) Cost = 119 × Shs 200 = Shs 23,800. The result must be rounded upward because a fractional paver cannot be purchased and rounding downward would leave part of the region uncovered.
ITEM 6
A small manufacturing group in Irrigation Farm produces open-topped ice-cream cones. One original cone has capacity 32 cm³ and height 4.25 cm.
The original cone costs Shs 50 to make and is sold for Shs 200. A new order requires cones with capacity 8 times the original capacity. The new cones are similar to the original cones.
The selling price is directly proportional to capacity, while production cost is directly proportional to the curved surface area of the cone. Use π = 3.14.
- Determine the radius and diameter of the original cone.
- Determine the slant height and curved surface area of the original cone.
- Determine the linear scale factor for the new cone, given that its capacity is 8 times the original.
- Determine the new height and diameter.
- Determine the curved surface area of one new cone and hence the total material area for an order of 250 cones.
- Determine the production cost of one new cone and the selling price of one new cone.
- Determine the profit per cone and the total profit on an order of 250 cones.
- Explain why capacity changes by a different factor from diameter.
(a) Using V = ⅓πr²h: r = √(3V/(πh)) = 2.682 cm. Diameter = 5.364 cm.
(b) Slant height l = √(r²+h²) = 5.026 cm. Curved surface area = πrl ≈ 42.324 cm².
(c) Volume factor = 8, so linear scale factor k = ∛8 = 2.000.
(d) New height = 4.25 × 2.000 = 8.500 cm. New diameter = 10.728 cm.
(e) New curved surface area = 169.297 cm² per cone. For 250 cones, total material area ≈ 4.232 m².
(f) Surface area scales as k², so new production cost = Shs 50 × 2.000² ≈ Shs 200. Selling price scales directly with capacity, so new selling price = Shs 200 × 8 = Shs 1,600.
(g) Profit per cone ≈ Shs 1,400. Total profit for 250 cones ≈ Shs 350,000.
(h) Capacity is a volume, so it changes with the cube of the linear scale factor, whereas diameter changes only linearly. Surface area changes with the square of the scale factor.
SET 7: BUS TERMINAL
ITEM 1
The local authority in Bus Terminal has secured land for a community project. The project team has 25 acres available. The plan reserves 9/25 of the total land for the main buildings.
After the building area is removed, 1/4 of the land that remains is to be landscaped or planted. The rest is divided among staff facilities, a public waiting area and parking in the ratio 4:3:3.
The project engineer estimates that one vehicle needs 0.02 acres of parking space. However, 10% of the full parking capacity must remain free for emergency access and movement.
The landscaping must be cleared before construction. Two workers can clear half an acre in one day when working at the same rate. Each worker is paid Shs 21,000 for one day's work.
- Determine the area allocated to the main buildings.
- Determine the area remaining after the buildings and hence find the area reserved for landscaping or planting.
- Use the given ratio to determine the areas for staff facilities, the public waiting area and parking.
- Determine the theoretical maximum number of vehicles that the parking area can accommodate.
- After keeping 10% of the capacity free, determine the maximum number of ordinary vehicles that should be admitted at one time.
- Determine the number of workers required to clear the landscaping area in one day.
- Calculate the total amount required to pay those workers for the one-day job.
(a) Buildings = 25 × 9/25 = 9 acres.
(b) Remaining = 16 acres. Landscaping = 4 acres. Land still available after landscaping = 12 acres.
(c) Total ratio parts = 10. Staff = 4.8 acres; waiting area = 3.6 acres; parking = 3.6 acres.
(d) Parking capacity = 3.6 ÷ 0.02 = 180 vehicles.
(e) Emergency reserve = 10% × 180 = 18.0 vehicle spaces. Ordinary capacity = 90% × 180 = 162 whole vehicles.
(f) At 0.25 acre per worker-day, workers required = 4 ÷ 0.25 = 16.00, so 16 workers are required.
(g) Labour cost = 16 × Shs 21,000 = Shs 336,000.
ITEM 2
A small enterprise in Bus Terminal uses two energy systems. Let S be the daily maintenance cost of the solar system and B the daily maintenance cost of the backup system.
Records show that S plus twice B is Shs 37,000. They also show that three times S exceeds twice B by Shs 7,000.
The enterprise produces two products, X and Y. Let x and y be the numbers produced per day. It must produce at least 20 units of X, X must be at least twice Y, total production cannot exceed 156, and the production budget is at most Shs 41,400.
Each X uses 3 units of a special input and each Y uses 4 units. At most 570 input units are available. Production costs are Shs 200 per X and Shs 300 per Y, while profits are Shs 500 and Shs 600 respectively.
- Form two simultaneous equations from the maintenance records and determine S and B.
- Define x and y and write all the production restrictions as inequalities, including non-negativity.
- Write the daily profit function P in terms of x and y.
- Determine the important feasible corner points and calculate the profit at each point.
- Identify the production plan that gives the greatest feasible profit.
- For that plan, calculate the total production cost and the total amount of special input used.
- Calculate the corresponding revenue and verify the profit.
(a) S + 2B = 37000; 3S − 2B = 7000. Adding gives 4S = 44000, so S = 11000. Then B = 13000. Thus solar maintenance is Shs 11,000 and backup maintenance is Shs 13,000.
(b) x ≥ 20; x ≥ 2y; x + y ≤ 156; 200x + 300y ≤ 41400; 3x + 4y ≤ 570; x ≥ 0, y ≥ 0.
(c) P = 500x + 600y.
(d) The key feasible corner points are (20,0), (20,10), (156,0) and (104,52). Their profits are Shs 10,000, Shs 16,000, Shs 78,000 and Shs 83,200 respectively.
(e) Therefore the maximum occurs at x = 104, y = 52, giving a maximum daily profit of Shs 83,200.
(f) Production cost = 200(104) + 300(52) = Shs 36,400. Input used = 3(104) + 4(52) = 520 units.
(g) Revenue = production cost + profit = Shs 36,400 + Shs 83,200 = Shs 119,600. Subtracting the cost from the revenue again gives Shs 83,200 profit.
ITEM 3
A management team in Bus Terminal records the number of customers served by 50 staff members during a trial period. Management will continue the arrangement if the average number served exceeds 51.
Staff members who serve at most that threshold are also reviewed for additional support or possible reassignment.
The recorded values are listed below:
26, 28, 30, 32, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 55, 56, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 65, 65, 66, 67, 68, 69, 70, 71, 72, 73, 75.
- Arrange the data into a frequency distribution using class intervals of 10, beginning with the smallest observation. Include cumulative frequency and class midpoint.
- Determine the total number of observations and the total number of customers served, then calculate the mean.
- Use the management rule to state whether the average target is exceeded.
- Determine the number and percentage of staff who served at most 51 customers.
- Determine the number and percentage of staff who served at least 66 customers.
- Use the ordered data to determine the median and identify the median class.
- Explain how the frequency distribution, mean and percentages can help management make a decision.
(a) Frequency table: 26–35: f=6, cf=6, midpoint=30.5; 36–45: f=10, cf=16, midpoint=40.5; 46–55: f=12, cf=28, midpoint=50.5; 56–65: f=13, cf=41, midpoint=60.5; 66–75: f=9, cf=50, midpoint=70.5
(b) There are 50 staff members. Total customers served = 2618. Mean = 2618 ÷ 50 = 52.36.
(c) Since 52.36 > 51, the stated average target is exceeded.
(d) 23 staff served at most 51. Percentage = 23/50 × 100 = 46.0%.
(e) 9 staff served at least 66. Percentage = 9/50 × 100 = 18.0%.
(f) The 25th and 26th ordered observations are 53 and 54; median = 53.5. The median lies in the 46–55 class.
(g) The table shows concentration across ranges, the mean gives an overall average, and the percentages identify groups needing attention or further support.
ITEM 4
A class committee in Bus Terminal wants to choose one student representative from three candidates A, B and C. There are 132 students in the class.
17 students support A only, 30 support B only and 14 support C only. In addition, 10 support both A and B only, 14 support B and C only, 9 support A and C only, and 6 support all three.
Students who support none of the candidates are recorded separately. The teacher will select the candidate with the largest total support.
- Determine the number of students who support none of the three candidates.
- Determine the total number of supporters of A, B and C respectively.
- Identify the candidate with the greatest support and find the difference between the highest and second-highest totals.
- Determine the percentage of the whole class supporting the selected candidate.
- Determine the number and percentage of students who do not support the selected candidate.
- State whether the selected candidate has support from more than half the class and whether fewer than 50% oppose the candidate.
- Explain why the overlapping groups must be counted when making the final decision.
(a) Known regions total = 100; therefore none = 132 − 100 = 32 students.
(b) A = 42; B = 60; C = 43. Each total includes the relevant overlapping regions.
(c) Candidate B has the greatest support with 60 students. The second-highest total is 43, so the difference is 17.
(d) Support percentage = 60/132 × 100 = 45.45%.
(e) Non-supporters = 72. Percentage = 54.55%.
(f) Half the class is 66.0. Since 60 is greater than half, the selected candidate has majority support. The opposition percentage is 54.55%, which is below 50%.
(g) A student in an overlap supports more than one candidate. Ignoring these intersections would undercount candidates' totals and could change the result.
ITEM 5
A recreation facility in Bus Terminal is creating a paved common region where two circular lawns overlap. The radii of the circles are 2.05 m and 1.55 m.
The distance between the two intersection points is 1.90 m. The common region alone will be paved.
Each rectangular paver measures 4.0 cm by 8.0 cm. The supplier recommends buying an additional 7% above the calculated number because some pavers will be cut or damaged. Take π = 3.14.
- Determine the perpendicular distance from the centre of each circle to the common chord.
- Hence determine the distance between the centres of the two circles.
- Determine the central angle subtended by the common chord at each centre.
- Find the area of the common region by calculating the two circular segments.
- Convert the area of one paver into square metres and determine the minimum number of pavers required before the allowance.
- Allowing 7% extra, determine the number of pavers to purchase.
- Determine the total cost and explain why the final number must be rounded upward.
(a) Half-chord = 0.950 m. Distance for radius 2.05 m = √(2.05²−0.950²) = 1.817 m. Distance for radius 1.55 m = 1.225 m.
(b) Centre distance = 1.817 + 1.225 = 3.041 m.
(c) Central angles are approximately 55.22° and 75.60°.
(d) Using sector area minus triangle area for each segment, the common area is approximately 0.719 m².
(e) One paver area = 0.040 × 0.080 = 0.00320 m². Minimum before allowance = 0.719 ÷ 0.00320 ≈ 225 pavers.
(f) With 7% extra: 225 × (1 + 0.07) = 240.75, so 241 pavers should be purchased.
(g) Cost = 241 × Shs 225 = Shs 54,225. The result must be rounded upward because a fractional paver cannot be purchased and rounding downward would leave part of the region uncovered.
ITEM 6
A small manufacturing group in Bus Terminal produces open-topped ice-cream cones. One original cone has capacity 34 cm³ and height 4.50 cm.
The original cone costs Shs 55 to make and is sold for Shs 220. A new order requires cones with capacity 4 times the original capacity. The new cones are similar to the original cones.
The selling price is directly proportional to capacity, while production cost is directly proportional to the curved surface area of the cone. Use π = 3.14.
- Determine the radius and diameter of the original cone.
- Determine the slant height and curved surface area of the original cone.
- Determine the linear scale factor for the new cone, given that its capacity is 4 times the original.
- Determine the new height and diameter.
- Determine the curved surface area of one new cone and hence the total material area for an order of 250 cones.
- Determine the production cost of one new cone and the selling price of one new cone.
- Determine the profit per cone and the total profit on an order of 250 cones.
- Explain why capacity changes by a different factor from diameter.
(a) Using V = ⅓πr²h: r = √(3V/(πh)) = 2.687 cm. Diameter = 5.374 cm.
(b) Slant height l = √(r²+h²) = 5.241 cm. Curved surface area = πrl ≈ 44.216 cm².
(c) Volume factor = 4, so linear scale factor k = ∛4 = 1.587.
(d) New height = 4.50 × 1.587 = 7.143 cm. New diameter = 8.530 cm.
(e) New curved surface area = 111.417 cm² per cone. For 250 cones, total material area ≈ 2.785 m².
(f) Surface area scales as k², so new production cost = Shs 55 × 1.587² ≈ Shs 138.59. Selling price scales directly with capacity, so new selling price = Shs 220 × 4 = Shs 880.
(g) Profit per cone ≈ Shs 741.41. Total profit for 250 cones ≈ Shs 185,352.17.
(h) Capacity is a volume, so it changes with the cube of the linear scale factor, whereas diameter changes only linearly. Surface area changes with the square of the scale factor.
SET 8: YOUTH CENTRE
ITEM 1
The local authority in Youth Centre has secured land for a community project. The project team has 24 acres available. The plan reserves 5/12 of the total land for the main buildings.
After the building area is removed, 1/3 of the land that remains is to be landscaped or planted. The rest is divided among staff facilities, a public waiting area and parking in the ratio 3:2:4.
The project engineer estimates that one vehicle needs 0.021 acres of parking space. However, 10% of the full parking capacity must remain free for emergency access and movement.
The landscaping must be cleared before construction. Two workers can clear half an acre in one day when working at the same rate. Each worker is paid Shs 18,000 for one day's work.
- Determine the area allocated to the main buildings.
- Determine the area remaining after the buildings and hence find the area reserved for landscaping or planting.
- Use the given ratio to determine the areas for staff facilities, the public waiting area and parking.
- Determine the theoretical maximum number of vehicles that the parking area can accommodate.
- After keeping 10% of the capacity free, determine the maximum number of ordinary vehicles that should be admitted at one time.
- Determine the number of workers required to clear the landscaping area in one day.
- Calculate the total amount required to pay those workers for the one-day job.
(a) Buildings = 24 × 5/12 = 10 acres.
(b) Remaining = 14 acres. Landscaping = 4.667 acres. Land still available after landscaping = 9.333 acres.
(c) Total ratio parts = 9. Staff = 3.111 acres; waiting area = 2.074 acres; parking = 4.148 acres.
(d) Parking capacity = 4.148 ÷ 0.021 = 200 vehicles.
(e) Emergency reserve = 10% × 200 = 20.0 vehicle spaces. Ordinary capacity = 90% × 200 = 180 whole vehicles.
(f) At 0.25 acre per worker-day, workers required = 4.667 ÷ 0.25 = 18.67, so 19 workers are required.
(g) Labour cost = 19 × Shs 18,000 = Shs 342,000.
ITEM 2
A small enterprise in Youth Centre uses two energy systems. Let S be the daily maintenance cost of the solar system and B the daily maintenance cost of the backup system.
Records show that S plus twice B is Shs 38,500. They also show that three times S exceeds twice B by Shs 7,500.
The enterprise produces two products, X and Y. Let x and y be the numbers produced per day. It must produce at least 20 units of X, X must be at least twice Y, total production cannot exceed 162, and the production budget is at most Shs 42,800.
Each X uses 3 units of a special input and each Y uses 4 units. At most 590 input units are available. Production costs are Shs 200 per X and Shs 300 per Y, while profits are Shs 500 and Shs 600 respectively.
- Form two simultaneous equations from the maintenance records and determine S and B.
- Define x and y and write all the production restrictions as inequalities, including non-negativity.
- Write the daily profit function P in terms of x and y.
- Determine the important feasible corner points and calculate the profit at each point.
- Identify the production plan that gives the greatest feasible profit.
- For that plan, calculate the total production cost and the total amount of special input used.
- Calculate the corresponding revenue and verify the profit.
(a) S + 2B = 38500; 3S − 2B = 7500. Adding gives 4S = 46000, so S = 11500. Then B = 13500. Thus solar maintenance is Shs 11,500 and backup maintenance is Shs 13,500.
(b) x ≥ 20; x ≥ 2y; x + y ≤ 162; 200x + 300y ≤ 42800; 3x + 4y ≤ 590; x ≥ 0, y ≥ 0.
(c) P = 500x + 600y.
(d) The key feasible corner points are (20,0), (20,10), (162,0) and (108,54). Their profits are Shs 10,000, Shs 16,000, Shs 81,000 and Shs 86,400 respectively.
(e) Therefore the maximum occurs at x = 108, y = 54, giving a maximum daily profit of Shs 86,400.
(f) Production cost = 200(108) + 300(54) = Shs 37,800. Input used = 3(108) + 4(54) = 540 units.
(g) Revenue = production cost + profit = Shs 37,800 + Shs 86,400 = Shs 124,200. Subtracting the cost from the revenue again gives Shs 86,400 profit.
ITEM 3
A management team in Youth Centre records the number of customers served by 50 staff members during a trial period. Management will continue the arrangement if the average number served exceeds 52.
Staff members who serve at most that threshold are also reviewed for additional support or possible reassignment.
The recorded values are listed below:
27, 29, 31, 33, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 56, 57, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 66, 66, 67, 68, 69, 70, 71, 72, 73, 74, 76.
- Arrange the data into a frequency distribution using class intervals of 10, beginning with the smallest observation. Include cumulative frequency and class midpoint.
- Determine the total number of observations and the total number of customers served, then calculate the mean.
- Use the management rule to state whether the average target is exceeded.
- Determine the number and percentage of staff who served at most 52 customers.
- Determine the number and percentage of staff who served at least 67 customers.
- Use the ordered data to determine the median and identify the median class.
- Explain how the frequency distribution, mean and percentages can help management make a decision.
(a) Frequency table: 27–36: f=6, cf=6, midpoint=31.5; 37–46: f=10, cf=16, midpoint=41.5; 47–56: f=12, cf=28, midpoint=51.5; 57–66: f=13, cf=41, midpoint=61.5; 67–76: f=9, cf=50, midpoint=71.5
(b) There are 50 staff members. Total customers served = 2668. Mean = 2668 ÷ 50 = 53.36.
(c) Since 53.36 > 52, the stated average target is exceeded.
(d) 23 staff served at most 52. Percentage = 23/50 × 100 = 46.0%.
(e) 9 staff served at least 67. Percentage = 9/50 × 100 = 18.0%.
(f) The 25th and 26th ordered observations are 54 and 55; median = 54.5. The median lies in the 47–56 class.
(g) The table shows concentration across ranges, the mean gives an overall average, and the percentages identify groups needing attention or further support.
ITEM 4
A class committee in Youth Centre wants to choose one student representative from three candidates A, B and C. There are 134 students in the class.
18 students support A only, 31 support B only and 15 support C only. In addition, 11 support both A and B only, 15 support B and C only, 10 support A and C only, and 7 support all three.
Students who support none of the candidates are recorded separately. The teacher will select the candidate with the largest total support.
- Determine the number of students who support none of the three candidates.
- Determine the total number of supporters of A, B and C respectively.
- Identify the candidate with the greatest support and find the difference between the highest and second-highest totals.
- Determine the percentage of the whole class supporting the selected candidate.
- Determine the number and percentage of students who do not support the selected candidate.
- State whether the selected candidate has support from more than half the class and whether fewer than 50% oppose the candidate.
- Explain why the overlapping groups must be counted when making the final decision.
(a) Known regions total = 107; therefore none = 134 − 107 = 27 students.
(b) A = 46; B = 64; C = 47. Each total includes the relevant overlapping regions.
(c) Candidate B has the greatest support with 64 students. The second-highest total is 47, so the difference is 17.
(d) Support percentage = 64/134 × 100 = 47.76%.
(e) Non-supporters = 70. Percentage = 52.24%.
(f) Half the class is 67.0. Since 64 is greater than half, the selected candidate has majority support. The opposition percentage is 52.24%, which is below 50%.
(g) A student in an overlap supports more than one candidate. Ignoring these intersections would undercount candidates' totals and could change the result.
ITEM 5
A recreation facility in Youth Centre is creating a paved common region where two circular lawns overlap. The radii of the circles are 2.10 m and 1.55 m.
The distance between the two intersection points is 1.60 m. The common region alone will be paved.
Each rectangular paver measures 5.0 cm by 10.0 cm. The supplier recommends buying an additional 8% above the calculated number because some pavers will be cut or damaged. Take π = 3.14.
- Determine the perpendicular distance from the centre of each circle to the common chord.
- Hence determine the distance between the centres of the two circles.
- Determine the central angle subtended by the common chord at each centre.
- Find the area of the common region by calculating the two circular segments.
- Convert the area of one paver into square metres and determine the minimum number of pavers required before the allowance.
- Allowing 8% extra, determine the number of pavers to purchase.
- Determine the total cost and explain why the final number must be rounded upward.
(a) Half-chord = 0.800 m. Distance for radius 2.10 m = √(2.10²−0.800²) = 1.942 m. Distance for radius 1.55 m = 1.328 m.
(b) Centre distance = 1.942 + 1.328 = 3.269 m.
(c) Central angles are approximately 44.79° and 62.15°.
(d) Using sector area minus triangle area for each segment, the common area is approximately 0.410 m².
(e) One paver area = 0.050 × 0.100 = 0.00500 m². Minimum before allowance = 0.410 ÷ 0.00500 ≈ 82 pavers.
(f) With 8% extra: 82 × (1 + 0.08) = 88.56, so 89 pavers should be purchased.
(g) Cost = 89 × Shs 250 = Shs 22,250. The result must be rounded upward because a fractional paver cannot be purchased and rounding downward would leave part of the region uncovered.
ITEM 6
A small manufacturing group in Youth Centre produces open-topped ice-cream cones. One original cone has capacity 36 cm³ and height 4.75 cm.
The original cone costs Shs 60 to make and is sold for Shs 240. A new order requires cones with capacity 8 times the original capacity. The new cones are similar to the original cones.
The selling price is directly proportional to capacity, while production cost is directly proportional to the curved surface area of the cone. Use π = 3.14.
- Determine the radius and diameter of the original cone.
- Determine the slant height and curved surface area of the original cone.
- Determine the linear scale factor for the new cone, given that its capacity is 8 times the original.
- Determine the new height and diameter.
- Determine the curved surface area of one new cone and hence the total material area for an order of 250 cones.
- Determine the production cost of one new cone and the selling price of one new cone.
- Determine the profit per cone and the total profit on an order of 250 cones.
- Explain why capacity changes by a different factor from diameter.
(a) Using V = ⅓πr²h: r = √(3V/(πh)) = 2.691 cm. Diameter = 5.382 cm.
(b) Slant height l = √(r²+h²) = 5.459 cm. Curved surface area = πrl ≈ 46.128 cm².
(c) Volume factor = 8, so linear scale factor k = ∛8 = 2.000.
(d) New height = 4.75 × 2.000 = 9.500 cm. New diameter = 10.764 cm.
(e) New curved surface area = 184.512 cm² per cone. For 250 cones, total material area ≈ 4.613 m².
(f) Surface area scales as k², so new production cost = Shs 60 × 2.000² ≈ Shs 240. Selling price scales directly with capacity, so new selling price = Shs 240 × 8 = Shs 1,920.
(g) Profit per cone ≈ Shs 1,680. Total profit for 250 cones ≈ Shs 420,000.
(h) Capacity is a volume, so it changes with the cube of the linear scale factor, whereas diameter changes only linearly. Surface area changes with the square of the scale factor.
SET 9: SCHOOL CAMPUS
ITEM 1
The local authority in School Campus has secured land for a community project. The project team has 21 acres available. The plan reserves 2/7 of the total land for the main buildings.
After the building area is removed, 1/5 of the land that remains is to be landscaped or planted. The rest is divided among staff facilities, a public waiting area and parking in the ratio 2:4:3.
The project engineer estimates that one vehicle needs 0.02 acres of parking space. However, 10% of the full parking capacity must remain free for emergency access and movement.
The landscaping must be cleared before construction. Two workers can clear half an acre in one day when working at the same rate. Each worker is paid Shs 20,000 for one day's work.
- Determine the area allocated to the main buildings.
- Determine the area remaining after the buildings and hence find the area reserved for landscaping or planting.
- Use the given ratio to determine the areas for staff facilities, the public waiting area and parking.
- Determine the theoretical maximum number of vehicles that the parking area can accommodate.
- After keeping 10% of the capacity free, determine the maximum number of ordinary vehicles that should be admitted at one time.
- Determine the number of workers required to clear the landscaping area in one day.
- Calculate the total amount required to pay those workers for the one-day job.
(a) Buildings = 21 × 2/7 = 6 acres.
(b) Remaining = 15 acres. Landscaping = 3 acres. Land still available after landscaping = 12 acres.
(c) Total ratio parts = 9. Staff = 2.667 acres; waiting area = 5.333 acres; parking = 4 acres.
(d) Parking capacity = 4 ÷ 0.02 = 200 vehicles.
(e) Emergency reserve = 10% × 200 = 20.0 vehicle spaces. Ordinary capacity = 90% × 200 = 180 whole vehicles.
(f) At 0.25 acre per worker-day, workers required = 3 ÷ 0.25 = 12.00, so 12 workers are required.
(g) Labour cost = 12 × Shs 20,000 = Shs 240,000.
ITEM 2
A small enterprise in School Campus uses two energy systems. Let S be the daily maintenance cost of the solar system and B the daily maintenance cost of the backup system.
Records show that S plus twice B is Shs 40,000. They also show that three times S exceeds twice B by Shs 8,000.
The enterprise produces two products, X and Y. Let x and y be the numbers produced per day. It must produce at least 20 units of X, X must be at least twice Y, total production cannot exceed 168, and the production budget is at most Shs 44,200.
Each X uses 3 units of a special input and each Y uses 4 units. At most 610 input units are available. Production costs are Shs 200 per X and Shs 300 per Y, while profits are Shs 500 and Shs 600 respectively.
- Form two simultaneous equations from the maintenance records and determine S and B.
- Define x and y and write all the production restrictions as inequalities, including non-negativity.
- Write the daily profit function P in terms of x and y.
- Determine the important feasible corner points and calculate the profit at each point.
- Identify the production plan that gives the greatest feasible profit.
- For that plan, calculate the total production cost and the total amount of special input used.
- Calculate the corresponding revenue and verify the profit.
(a) S + 2B = 40000; 3S − 2B = 8000. Adding gives 4S = 48000, so S = 12000. Then B = 14000. Thus solar maintenance is Shs 12,000 and backup maintenance is Shs 14,000.
(b) x ≥ 20; x ≥ 2y; x + y ≤ 168; 200x + 300y ≤ 44200; 3x + 4y ≤ 610; x ≥ 0, y ≥ 0.
(c) P = 500x + 600y.
(d) The key feasible corner points are (20,0), (20,10), (168,0) and (112,56). Their profits are Shs 10,000, Shs 16,000, Shs 84,000 and Shs 89,600 respectively.
(e) Therefore the maximum occurs at x = 112, y = 56, giving a maximum daily profit of Shs 89,600.
(f) Production cost = 200(112) + 300(56) = Shs 39,200. Input used = 3(112) + 4(56) = 560 units.
(g) Revenue = production cost + profit = Shs 39,200 + Shs 89,600 = Shs 128,800. Subtracting the cost from the revenue again gives Shs 89,600 profit.
ITEM 3
A management team in School Campus records the number of customers served by 50 staff members during a trial period. Management will continue the arrangement if the average number served exceeds 53.
Staff members who serve at most that threshold are also reviewed for additional support or possible reassignment.
The recorded values are listed below:
28, 30, 32, 34, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 57, 58, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 67, 67, 68, 69, 70, 71, 72, 73, 74, 75, 77.
- Arrange the data into a frequency distribution using class intervals of 10, beginning with the smallest observation. Include cumulative frequency and class midpoint.
- Determine the total number of observations and the total number of customers served, then calculate the mean.
- Use the management rule to state whether the average target is exceeded.
- Determine the number and percentage of staff who served at most 53 customers.
- Determine the number and percentage of staff who served at least 68 customers.
- Use the ordered data to determine the median and identify the median class.
- Explain how the frequency distribution, mean and percentages can help management make a decision.
(a) Frequency table: 28–37: f=6, cf=6, midpoint=32.5; 38–47: f=10, cf=16, midpoint=42.5; 48–57: f=12, cf=28, midpoint=52.5; 58–67: f=13, cf=41, midpoint=62.5; 68–77: f=9, cf=50, midpoint=72.5
(b) There are 50 staff members. Total customers served = 2718. Mean = 2718 ÷ 50 = 54.36.
(c) Since 54.36 > 53, the stated average target is exceeded.
(d) 23 staff served at most 53. Percentage = 23/50 × 100 = 46.0%.
(e) 9 staff served at least 68. Percentage = 9/50 × 100 = 18.0%.
(f) The 25th and 26th ordered observations are 55 and 56; median = 55.5. The median lies in the 48–57 class.
(g) The table shows concentration across ranges, the mean gives an overall average, and the percentages identify groups needing attention or further support.
ITEM 4
A class committee in School Campus wants to choose one student representative from three candidates A, B and C. There are 136 students in the class.
19 students support A only, 28 support B only and 16 support C only. In addition, 12 support both A and B only, 12 support B and C only, 11 support A and C only, and 8 support all three.
Students who support none of the candidates are recorded separately. The teacher will select the candidate with the largest total support.
- Determine the number of students who support none of the three candidates.
- Determine the total number of supporters of A, B and C respectively.
- Identify the candidate with the greatest support and find the difference between the highest and second-highest totals.
- Determine the percentage of the whole class supporting the selected candidate.
- Determine the number and percentage of students who do not support the selected candidate.
- State whether the selected candidate has support from more than half the class and whether fewer than 50% oppose the candidate.
- Explain why the overlapping groups must be counted when making the final decision.
(a) Known regions total = 106; therefore none = 136 − 106 = 30 students.
(b) A = 50; B = 60; C = 47. Each total includes the relevant overlapping regions.
(c) Candidate B has the greatest support with 60 students. The second-highest total is 50, so the difference is 10.
(d) Support percentage = 60/136 × 100 = 44.12%.
(e) Non-supporters = 76. Percentage = 55.88%.
(f) Half the class is 68.0. Since 60 is greater than half, the selected candidate has majority support. The opposition percentage is 55.88%, which is below 50%.
(g) A student in an overlap supports more than one candidate. Ignoring these intersections would undercount candidates' totals and could change the result.
ITEM 5
A recreation facility in School Campus is creating a paved common region where two circular lawns overlap. The radii of the circles are 2.15 m and 1.55 m.
The distance between the two intersection points is 1.65 m. The common region alone will be paved.
Each rectangular paver measures 6.0 cm by 8.0 cm. The supplier recommends buying an additional 5% above the calculated number because some pavers will be cut or damaged. Take π = 3.14.
- Determine the perpendicular distance from the centre of each circle to the common chord.
- Hence determine the distance between the centres of the two circles.
- Determine the central angle subtended by the common chord at each centre.
- Find the area of the common region by calculating the two circular segments.
- Convert the area of one paver into square metres and determine the minimum number of pavers required before the allowance.
- Allowing 5% extra, determine the number of pavers to purchase.
- Determine the total cost and explain why the final number must be rounded upward.
(a) Half-chord = 0.825 m. Distance for radius 2.15 m = √(2.15²−0.825²) = 1.985 m. Distance for radius 1.55 m = 1.312 m.
(b) Centre distance = 1.985 + 1.312 = 3.298 m.
(c) Central angles are approximately 45.13° and 64.32°.
(d) Using sector area minus triangle area for each segment, the common area is approximately 0.447 m².
(e) One paver area = 0.060 × 0.080 = 0.00480 m². Minimum before allowance = 0.447 ÷ 0.00480 ≈ 94 pavers.
(f) With 5% extra: 94 × (1 + 0.05) = 98.70, so 99 pavers should be purchased.
(g) Cost = 99 × Shs 275 = Shs 27,225. The result must be rounded upward because a fractional paver cannot be purchased and rounding downward would leave part of the region uncovered.
ITEM 6
A small manufacturing group in School Campus produces open-topped ice-cream cones. One original cone has capacity 38 cm³ and height 4.00 cm.
The original cone costs Shs 65 to make and is sold for Shs 260. A new order requires cones with capacity 27 times the original capacity. The new cones are similar to the original cones.
The selling price is directly proportional to capacity, while production cost is directly proportional to the curved surface area of the cone. Use π = 3.14.
- Determine the radius and diameter of the original cone.
- Determine the slant height and curved surface area of the original cone.
- Determine the linear scale factor for the new cone, given that its capacity is 27 times the original.
- Determine the new height and diameter.
- Determine the curved surface area of one new cone and hence the total material area for an order of 250 cones.
- Determine the production cost of one new cone and the selling price of one new cone.
- Determine the profit per cone and the total profit on an order of 250 cones.
- Explain why capacity changes by a different factor from diameter.
(a) Using V = ⅓πr²h: r = √(3V/(πh)) = 3.013 cm. Diameter = 6.025 cm.
(b) Slant height l = √(r²+h²) = 5.008 cm. Curved surface area = πrl ≈ 47.372 cm².
(c) Volume factor = 27, so linear scale factor k = ∛27 = 3.000.
(d) New height = 4.00 × 3.000 = 12.000 cm. New diameter = 18.076 cm.
(e) New curved surface area = 426.346 cm² per cone. For 250 cones, total material area ≈ 10.659 m².
(f) Surface area scales as k², so new production cost = Shs 65 × 3.000² ≈ Shs 585. Selling price scales directly with capacity, so new selling price = Shs 260 × 27 = Shs 7,020.
(g) Profit per cone ≈ Shs 6,435. Total profit for 250 cones ≈ Shs 1,608,750.
(h) Capacity is a volume, so it changes with the cube of the linear scale factor, whereas diameter changes only linearly. Surface area changes with the square of the scale factor.
SET 10: CLINIC EXPANSION
ITEM 1
The local authority in Clinic Expansion has secured land for a community project. The project team has 28 acres available. The plan reserves 2/5 of the total land for the main buildings.
After the building area is removed, 1/6 of the land that remains is to be landscaped or planted. The rest is divided among staff facilities, a public waiting area and parking in the ratio 5:2:3.
The project engineer estimates that one vehicle needs 0.02 acres of parking space. However, 10% of the full parking capacity must remain free for emergency access and movement.
The landscaping must be cleared before construction. Two workers can clear half an acre in one day when working at the same rate. Each worker is paid Shs 17,500 for one day's work.
- Determine the area allocated to the main buildings.
- Determine the area remaining after the buildings and hence find the area reserved for landscaping or planting.
- Use the given ratio to determine the areas for staff facilities, the public waiting area and parking.
- Determine the theoretical maximum number of vehicles that the parking area can accommodate.
- After keeping 10% of the capacity free, determine the maximum number of ordinary vehicles that should be admitted at one time.
- Determine the number of workers required to clear the landscaping area in one day.
- Calculate the total amount required to pay those workers for the one-day job.
(a) Buildings = 28 × 2/5 = 11.2 acres.
(b) Remaining = 16.8 acres. Landscaping = 2.8 acres. Land still available after landscaping = 14 acres.
(c) Total ratio parts = 10. Staff = 7 acres; waiting area = 2.8 acres; parking = 4.2 acres.
(d) Parking capacity = 4.2 ÷ 0.02 = 210 vehicles.
(e) Emergency reserve = 10% × 210 = 21.0 vehicle spaces. Ordinary capacity = 90% × 210 = 189 whole vehicles.
(f) At 0.25 acre per worker-day, workers required = 2.8 ÷ 0.25 = 11.20, so 12 workers are required.
(g) Labour cost = 12 × Shs 17,500 = Shs 210,000.
ITEM 2
A small enterprise in Clinic Expansion uses two energy systems. Let S be the daily maintenance cost of the solar system and B the daily maintenance cost of the backup system.
Records show that S plus twice B is Shs 41,500. They also show that three times S exceeds twice B by Shs 8,500.
The enterprise produces two products, X and Y. Let x and y be the numbers produced per day. It must produce at least 20 units of X, X must be at least twice Y, total production cannot exceed 174, and the production budget is at most Shs 45,600.
Each X uses 3 units of a special input and each Y uses 4 units. At most 630 input units are available. Production costs are Shs 200 per X and Shs 300 per Y, while profits are Shs 500 and Shs 600 respectively.
- Form two simultaneous equations from the maintenance records and determine S and B.
- Define x and y and write all the production restrictions as inequalities, including non-negativity.
- Write the daily profit function P in terms of x and y.
- Determine the important feasible corner points and calculate the profit at each point.
- Identify the production plan that gives the greatest feasible profit.
- For that plan, calculate the total production cost and the total amount of special input used.
- Calculate the corresponding revenue and verify the profit.
(a) S + 2B = 41500; 3S − 2B = 8500. Adding gives 4S = 50000, so S = 12500. Then B = 14500. Thus solar maintenance is Shs 12,500 and backup maintenance is Shs 14,500.
(b) x ≥ 20; x ≥ 2y; x + y ≤ 174; 200x + 300y ≤ 45600; 3x + 4y ≤ 630; x ≥ 0, y ≥ 0.
(c) P = 500x + 600y.
(d) The key feasible corner points are (20,0), (20,10), (174,0) and (116,58). Their profits are Shs 10,000, Shs 16,000, Shs 87,000 and Shs 92,800 respectively.
(e) Therefore the maximum occurs at x = 116, y = 58, giving a maximum daily profit of Shs 92,800.
(f) Production cost = 200(116) + 300(58) = Shs 40,600. Input used = 3(116) + 4(58) = 580 units.
(g) Revenue = production cost + profit = Shs 40,600 + Shs 92,800 = Shs 133,400. Subtracting the cost from the revenue again gives Shs 92,800 profit.
ITEM 3
A management team in Clinic Expansion records the number of customers served by 50 staff members during a trial period. Management will continue the arrangement if the average number served exceeds 54.
Staff members who serve at most that threshold are also reviewed for additional support or possible reassignment.
The recorded values are listed below:
29, 31, 33, 35, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 58, 59, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 68, 68, 69, 70, 71, 72, 73, 74, 75, 76, 78.
- Arrange the data into a frequency distribution using class intervals of 10, beginning with the smallest observation. Include cumulative frequency and class midpoint.
- Determine the total number of observations and the total number of customers served, then calculate the mean.
- Use the management rule to state whether the average target is exceeded.
- Determine the number and percentage of staff who served at most 54 customers.
- Determine the number and percentage of staff who served at least 69 customers.
- Use the ordered data to determine the median and identify the median class.
- Explain how the frequency distribution, mean and percentages can help management make a decision.
(a) Frequency table: 29–38: f=6, cf=6, midpoint=33.5; 39–48: f=10, cf=16, midpoint=43.5; 49–58: f=12, cf=28, midpoint=53.5; 59–68: f=13, cf=41, midpoint=63.5; 69–78: f=9, cf=50, midpoint=73.5
(b) There are 50 staff members. Total customers served = 2768. Mean = 2768 ÷ 50 = 55.36.
(c) Since 55.36 > 54, the stated average target is exceeded.
(d) 23 staff served at most 54. Percentage = 23/50 × 100 = 46.0%.
(e) 9 staff served at least 69. Percentage = 9/50 × 100 = 18.0%.
(f) The 25th and 26th ordered observations are 56 and 57; median = 56.5. The median lies in the 49–58 class.
(g) The table shows concentration across ranges, the mean gives an overall average, and the percentages identify groups needing attention or further support.
ITEM 4
A class committee in Clinic Expansion wants to choose one student representative from three candidates A, B and C. There are 138 students in the class.
20 students support A only, 29 support B only and 17 support C only. In addition, 10 support both A and B only, 13 support B and C only, 12 support A and C only, and 6 support all three.
Students who support none of the candidates are recorded separately. The teacher will select the candidate with the largest total support.
- Determine the number of students who support none of the three candidates.
- Determine the total number of supporters of A, B and C respectively.
- Identify the candidate with the greatest support and find the difference between the highest and second-highest totals.
- Determine the percentage of the whole class supporting the selected candidate.
- Determine the number and percentage of students who do not support the selected candidate.
- State whether the selected candidate has support from more than half the class and whether fewer than 50% oppose the candidate.
- Explain why the overlapping groups must be counted when making the final decision.
(a) Known regions total = 107; therefore none = 138 − 107 = 31 students.
(b) A = 48; B = 58; C = 48. Each total includes the relevant overlapping regions.
(c) Candidate B has the greatest support with 58 students. The second-highest total is 48, so the difference is 10.
(d) Support percentage = 58/138 × 100 = 42.03%.
(e) Non-supporters = 80. Percentage = 57.97%.
(f) Half the class is 69.0. Since 58 is greater than half, the selected candidate has majority support. The opposition percentage is 57.97%, which is below 50%.
(g) A student in an overlap supports more than one candidate. Ignoring these intersections would undercount candidates' totals and could change the result.
ITEM 5
A recreation facility in Clinic Expansion is creating a paved common region where two circular lawns overlap. The radii of the circles are 2.20 m and 1.55 m.
The distance between the two intersection points is 1.70 m. The common region alone will be paved.
Each rectangular paver measures 4.0 cm by 10.0 cm. The supplier recommends buying an additional 6% above the calculated number because some pavers will be cut or damaged. Take π = 3.14.
- Determine the perpendicular distance from the centre of each circle to the common chord.
- Hence determine the distance between the centres of the two circles.
- Determine the central angle subtended by the common chord at each centre.
- Find the area of the common region by calculating the two circular segments.
- Convert the area of one paver into square metres and determine the minimum number of pavers required before the allowance.
- Allowing 6% extra, determine the number of pavers to purchase.
- Determine the total cost and explain why the final number must be rounded upward.
(a) Half-chord = 0.850 m. Distance for radius 2.20 m = √(2.20²−0.850²) = 2.029 m. Distance for radius 1.55 m = 1.296 m.
(b) Centre distance = 2.029 + 1.296 = 3.325 m.
(c) Central angles are approximately 45.46° and 66.51°.
(d) Using sector area minus triangle area for each segment, the common area is approximately 0.486 m².
(e) One paver area = 0.040 × 0.100 = 0.00400 m². Minimum before allowance = 0.486 ÷ 0.00400 ≈ 122 pavers.
(f) With 6% extra: 122 × (1 + 0.06) = 129.32, so 130 pavers should be purchased.
(g) Cost = 130 × Shs 300 = Shs 39,000. The result must be rounded upward because a fractional paver cannot be purchased and rounding downward would leave part of the region uncovered.
ITEM 6
A small manufacturing group in Clinic Expansion produces open-topped ice-cream cones. One original cone has capacity 40 cm³ and height 4.25 cm.
The original cone costs Shs 70 to make and is sold for Shs 280. A new order requires cones with capacity 8 times the original capacity. The new cones are similar to the original cones.
The selling price is directly proportional to capacity, while production cost is directly proportional to the curved surface area of the cone. Use π = 3.14.
- Determine the radius and diameter of the original cone.
- Determine the slant height and curved surface area of the original cone.
- Determine the linear scale factor for the new cone, given that its capacity is 8 times the original.
- Determine the new height and diameter.
- Determine the curved surface area of one new cone and hence the total material area for an order of 250 cones.
- Determine the production cost of one new cone and the selling price of one new cone.
- Determine the profit per cone and the total profit on an order of 250 cones.
- Explain why capacity changes by a different factor from diameter.
(a) Using V = ⅓πr²h: r = √(3V/(πh)) = 2.999 cm. Diameter = 5.997 cm.
(b) Slant height l = √(r²+h²) = 5.201 cm. Curved surface area = πrl ≈ 48.976 cm².
(c) Volume factor = 8, so linear scale factor k = ∛8 = 2.000.
(d) New height = 4.25 × 2.000 = 8.500 cm. New diameter = 11.995 cm.
(e) New curved surface area = 195.903 cm² per cone. For 250 cones, total material area ≈ 4.898 m².
(f) Surface area scales as k², so new production cost = Shs 70 × 2.000² ≈ Shs 280. Selling price scales directly with capacity, so new selling price = Shs 280 × 8 = Shs 2,240.
(g) Profit per cone ≈ Shs 1,960. Total profit for 250 cones ≈ Shs 490,000.
(h) Capacity is a volume, so it changes with the cube of the linear scale factor, whereas diameter changes only linearly. Surface area changes with the square of the scale factor.
SET 11: AGRICULTURAL TRAINING FARM
ITEM 1
The local authority in Agricultural Training Farm has secured land for a community project. The project team has 32 acres available. The plan reserves 3/8 of the total land for the main buildings.
After the building area is removed, 1/4 of the land that remains is to be landscaped or planted. The rest is divided among staff facilities, a public waiting area and parking in the ratio 3:4:5.
The project engineer estimates that one vehicle needs 0.025 acres of parking space. However, 10% of the full parking capacity must remain free for emergency access and movement.
The landscaping must be cleared before construction. Two workers can clear half an acre in one day when working at the same rate. Each worker is paid Shs 18,500 for one day's work.
- Determine the area allocated to the main buildings.
- Determine the area remaining after the buildings and hence find the area reserved for landscaping or planting.
- Use the given ratio to determine the areas for staff facilities, the public waiting area and parking.
- Determine the theoretical maximum number of vehicles that the parking area can accommodate.
- After keeping 10% of the capacity free, determine the maximum number of ordinary vehicles that should be admitted at one time.
- Determine the number of workers required to clear the landscaping area in one day.
- Calculate the total amount required to pay those workers for the one-day job.
(a) Buildings = 32 × 3/8 = 12 acres.
(b) Remaining = 20 acres. Landscaping = 5 acres. Land still available after landscaping = 15 acres.
(c) Total ratio parts = 12. Staff = 3.75 acres; waiting area = 5 acres; parking = 6.25 acres.
(d) Parking capacity = 6.25 ÷ 0.025 = 250 vehicles.
(e) Emergency reserve = 10% × 250 = 25.0 vehicle spaces. Ordinary capacity = 90% × 250 = 225 whole vehicles.
(f) At 0.25 acre per worker-day, workers required = 5 ÷ 0.25 = 20.00, so 20 workers are required.
(g) Labour cost = 20 × Shs 18,500 = Shs 370,000.
ITEM 2
A small enterprise in Agricultural Training Farm uses two energy systems. Let S be the daily maintenance cost of the solar system and B the daily maintenance cost of the backup system.
Records show that S plus twice B is Shs 43,000. They also show that three times S exceeds twice B by Shs 9,000.
The enterprise produces two products, X and Y. Let x and y be the numbers produced per day. It must produce at least 20 units of X, X must be at least twice Y, total production cannot exceed 180, and the production budget is at most Shs 47,000.
Each X uses 3 units of a special input and each Y uses 4 units. At most 650 input units are available. Production costs are Shs 200 per X and Shs 300 per Y, while profits are Shs 500 and Shs 600 respectively.
- Form two simultaneous equations from the maintenance records and determine S and B.
- Define x and y and write all the production restrictions as inequalities, including non-negativity.
- Write the daily profit function P in terms of x and y.
- Determine the important feasible corner points and calculate the profit at each point.
- Identify the production plan that gives the greatest feasible profit.
- For that plan, calculate the total production cost and the total amount of special input used.
- Calculate the corresponding revenue and verify the profit.
(a) S + 2B = 43000; 3S − 2B = 9000. Adding gives 4S = 52000, so S = 13000. Then B = 15000. Thus solar maintenance is Shs 13,000 and backup maintenance is Shs 15,000.
(b) x ≥ 20; x ≥ 2y; x + y ≤ 180; 200x + 300y ≤ 47000; 3x + 4y ≤ 650; x ≥ 0, y ≥ 0.
(c) P = 500x + 600y.
(d) The key feasible corner points are (20,0), (20,10), (180,0) and (120,60). Their profits are Shs 10,000, Shs 16,000, Shs 90,000 and Shs 96,000 respectively.
(e) Therefore the maximum occurs at x = 120, y = 60, giving a maximum daily profit of Shs 96,000.
(f) Production cost = 200(120) + 300(60) = Shs 42,000. Input used = 3(120) + 4(60) = 600 units.
(g) Revenue = production cost + profit = Shs 42,000 + Shs 96,000 = Shs 138,000. Subtracting the cost from the revenue again gives Shs 96,000 profit.
ITEM 3
A management team in Agricultural Training Farm records the number of customers served by 50 staff members during a trial period. Management will continue the arrangement if the average number served exceeds 55.
Staff members who serve at most that threshold are also reviewed for additional support or possible reassignment.
The recorded values are listed below:
30, 32, 34, 36, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 59, 60, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 69, 69, 70, 71, 72, 73, 74, 75, 76, 77, 79.
- Arrange the data into a frequency distribution using class intervals of 10, beginning with the smallest observation. Include cumulative frequency and class midpoint.
- Determine the total number of observations and the total number of customers served, then calculate the mean.
- Use the management rule to state whether the average target is exceeded.
- Determine the number and percentage of staff who served at most 55 customers.
- Determine the number and percentage of staff who served at least 70 customers.
- Use the ordered data to determine the median and identify the median class.
- Explain how the frequency distribution, mean and percentages can help management make a decision.
(a) Frequency table: 30–39: f=6, cf=6, midpoint=34.5; 40–49: f=10, cf=16, midpoint=44.5; 50–59: f=12, cf=28, midpoint=54.5; 60–69: f=13, cf=41, midpoint=64.5; 70–79: f=9, cf=50, midpoint=74.5
(b) There are 50 staff members. Total customers served = 2818. Mean = 2818 ÷ 50 = 56.36.
(c) Since 56.36 > 55, the stated average target is exceeded.
(d) 23 staff served at most 55. Percentage = 23/50 × 100 = 46.0%.
(e) 9 staff served at least 70. Percentage = 9/50 × 100 = 18.0%.
(f) The 25th and 26th ordered observations are 57 and 58; median = 57.5. The median lies in the 50–59 class.
(g) The table shows concentration across ranges, the mean gives an overall average, and the percentages identify groups needing attention or further support.
ITEM 4
A class committee in Agricultural Training Farm wants to choose one student representative from three candidates A, B and C. There are 140 students in the class.
16 students support A only, 30 support B only and 18 support C only. In addition, 11 support both A and B only, 14 support B and C only, 8 support A and C only, and 7 support all three.
Students who support none of the candidates are recorded separately. The teacher will select the candidate with the largest total support.
- Determine the number of students who support none of the three candidates.
- Determine the total number of supporters of A, B and C respectively.
- Identify the candidate with the greatest support and find the difference between the highest and second-highest totals.
- Determine the percentage of the whole class supporting the selected candidate.
- Determine the number and percentage of students who do not support the selected candidate.
- State whether the selected candidate has support from more than half the class and whether fewer than 50% oppose the candidate.
- Explain why the overlapping groups must be counted when making the final decision.
(a) Known regions total = 104; therefore none = 140 − 104 = 36 students.
(b) A = 42; B = 62; C = 47. Each total includes the relevant overlapping regions.
(c) Candidate B has the greatest support with 62 students. The second-highest total is 47, so the difference is 15.
(d) Support percentage = 62/140 × 100 = 44.29%.
(e) Non-supporters = 78. Percentage = 55.71%.
(f) Half the class is 70.0. Since 62 is greater than half, the selected candidate has majority support. The opposition percentage is 55.71%, which is below 50%.
(g) A student in an overlap supports more than one candidate. Ignoring these intersections would undercount candidates' totals and could change the result.
ITEM 5
A recreation facility in Agricultural Training Farm is creating a paved common region where two circular lawns overlap. The radii of the circles are 2.00 m and 1.60 m.
The distance between the two intersection points is 1.75 m. The common region alone will be paved.
Each rectangular paver measures 5.0 cm by 8.0 cm. The supplier recommends buying an additional 7% above the calculated number because some pavers will be cut or damaged. Take π = 3.14.
- Determine the perpendicular distance from the centre of each circle to the common chord.
- Hence determine the distance between the centres of the two circles.
- Determine the central angle subtended by the common chord at each centre.
- Find the area of the common region by calculating the two circular segments.
- Convert the area of one paver into square metres and determine the minimum number of pavers required before the allowance.
- Allowing 7% extra, determine the number of pavers to purchase.
- Determine the total cost and explain why the final number must be rounded upward.
(a) Half-chord = 0.875 m. Distance for radius 2.00 m = √(2.00²−0.875²) = 1.798 m. Distance for radius 1.60 m = 1.340 m.
(b) Centre distance = 1.798 + 1.340 = 3.138 m.
(c) Central angles are approximately 51.89° and 66.31°.
(d) Using sector area minus triangle area for each segment, the common area is approximately 0.545 m².
(e) One paver area = 0.050 × 0.080 = 0.00400 m². Minimum before allowance = 0.545 ÷ 0.00400 ≈ 137 pavers.
(f) With 7% extra: 137 × (1 + 0.07) = 146.59, so 147 pavers should be purchased.
(g) Cost = 147 × Shs 200 = Shs 29,400. The result must be rounded upward because a fractional paver cannot be purchased and rounding downward would leave part of the region uncovered.
ITEM 6
A small manufacturing group in Agricultural Training Farm produces open-topped ice-cream cones. One original cone has capacity 32 cm³ and height 4.50 cm.
The original cone costs Shs 50 to make and is sold for Shs 200. A new order requires cones with capacity 4 times the original capacity. The new cones are similar to the original cones.
The selling price is directly proportional to capacity, while production cost is directly proportional to the curved surface area of the cone. Use π = 3.14.
- Determine the radius and diameter of the original cone.
- Determine the slant height and curved surface area of the original cone.
- Determine the linear scale factor for the new cone, given that its capacity is 4 times the original.
- Determine the new height and diameter.
- Determine the curved surface area of one new cone and hence the total material area for an order of 250 cones.
- Determine the production cost of one new cone and the selling price of one new cone.
- Determine the profit per cone and the total profit on an order of 250 cones.
- Explain why capacity changes by a different factor from diameter.
(a) Using V = ⅓πr²h: r = √(3V/(πh)) = 2.607 cm. Diameter = 5.213 cm.
(b) Slant height l = √(r²+h²) = 5.200 cm. Curved surface area = πrl ≈ 42.563 cm².
(c) Volume factor = 4, so linear scale factor k = ∛4 = 1.587.
(d) New height = 4.50 × 1.587 = 7.143 cm. New diameter = 8.275 cm.
(e) New curved surface area = 107.252 cm² per cone. For 250 cones, total material area ≈ 2.681 m².
(f) Surface area scales as k², so new production cost = Shs 50 × 1.587² ≈ Shs 125.99. Selling price scales directly with capacity, so new selling price = Shs 200 × 4 = Shs 800.
(g) Profit per cone ≈ Shs 674.01. Total profit for 250 cones ≈ Shs 168,501.97.
(h) Capacity is a volume, so it changes with the cube of the linear scale factor, whereas diameter changes only linearly. Surface area changes with the square of the scale factor.
SET 12: TOURISM CENTRE
ITEM 1
The local authority in Tourism Centre has secured land for a community project. The project team has 24 acres available. The plan reserves 5/12 of the total land for the main buildings.
After the building area is removed, 1/4 of the land that remains is to be landscaped or planted. The rest is divided among staff facilities, a public waiting area and parking in the ratio 2:3:4.
The project engineer estimates that one vehicle needs 0.023 acres of parking space. However, 10% of the full parking capacity must remain free for emergency access and movement.
The landscaping must be cleared before construction. Two workers can clear half an acre in one day when working at the same rate. Each worker is paid Shs 17,000 for one day's work.
- Determine the area allocated to the main buildings.
- Determine the area remaining after the buildings and hence find the area reserved for landscaping or planting.
- Use the given ratio to determine the areas for staff facilities, the public waiting area and parking.
- Determine the theoretical maximum number of vehicles that the parking area can accommodate.
- After keeping 10% of the capacity free, determine the maximum number of ordinary vehicles that should be admitted at one time.
- Determine the number of workers required to clear the landscaping area in one day.
- Calculate the total amount required to pay those workers for the one-day job.
(a) Buildings = 24 × 5/12 = 10 acres.
(b) Remaining = 14 acres. Landscaping = 3.5 acres. Land still available after landscaping = 10.5 acres.
(c) Total ratio parts = 9. Staff = 2.333 acres; waiting area = 3.5 acres; parking = 4.667 acres.
(d) Parking capacity = 4.667 ÷ 0.023 = 200 vehicles.
(e) Emergency reserve = 10% × 200 = 20.0 vehicle spaces. Ordinary capacity = 90% × 200 = 180 whole vehicles.
(f) At 0.25 acre per worker-day, workers required = 3.5 ÷ 0.25 = 14.00, so 14 workers are required.
(g) Labour cost = 14 × Shs 17,000 = Shs 238,000.
ITEM 2
A small enterprise in Tourism Centre uses two energy systems. Let S be the daily maintenance cost of the solar system and B the daily maintenance cost of the backup system.
Records show that S plus twice B is Shs 44,500. They also show that three times S exceeds twice B by Shs 9,500.
The enterprise produces two products, X and Y. Let x and y be the numbers produced per day. It must produce at least 20 units of X, X must be at least twice Y, total production cannot exceed 186, and the production budget is at most Shs 48,400.
Each X uses 3 units of a special input and each Y uses 4 units. At most 670 input units are available. Production costs are Shs 200 per X and Shs 300 per Y, while profits are Shs 500 and Shs 600 respectively.
- Form two simultaneous equations from the maintenance records and determine S and B.
- Define x and y and write all the production restrictions as inequalities, including non-negativity.
- Write the daily profit function P in terms of x and y.
- Determine the important feasible corner points and calculate the profit at each point.
- Identify the production plan that gives the greatest feasible profit.
- For that plan, calculate the total production cost and the total amount of special input used.
- Calculate the corresponding revenue and verify the profit.
(a) S + 2B = 44500; 3S − 2B = 9500. Adding gives 4S = 54000, so S = 13500. Then B = 15500. Thus solar maintenance is Shs 13,500 and backup maintenance is Shs 15,500.
(b) x ≥ 20; x ≥ 2y; x + y ≤ 186; 200x + 300y ≤ 48400; 3x + 4y ≤ 670; x ≥ 0, y ≥ 0.
(c) P = 500x + 600y.
(d) The key feasible corner points are (20,0), (20,10), (186,0) and (124,62). Their profits are Shs 10,000, Shs 16,000, Shs 93,000 and Shs 99,200 respectively.
(e) Therefore the maximum occurs at x = 124, y = 62, giving a maximum daily profit of Shs 99,200.
(f) Production cost = 200(124) + 300(62) = Shs 43,400. Input used = 3(124) + 4(62) = 620 units.
(g) Revenue = production cost + profit = Shs 43,400 + Shs 99,200 = Shs 142,600. Subtracting the cost from the revenue again gives Shs 99,200 profit.
ITEM 3
A management team in Tourism Centre records the number of customers served by 50 staff members during a trial period. Management will continue the arrangement if the average number served exceeds 56.
Staff members who serve at most that threshold are also reviewed for additional support or possible reassignment.
The recorded values are listed below:
31, 33, 35, 37, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 60, 61, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 70, 70, 71, 72, 73, 74, 75, 76, 77, 78, 80.
- Arrange the data into a frequency distribution using class intervals of 10, beginning with the smallest observation. Include cumulative frequency and class midpoint.
- Determine the total number of observations and the total number of customers served, then calculate the mean.
- Use the management rule to state whether the average target is exceeded.
- Determine the number and percentage of staff who served at most 56 customers.
- Determine the number and percentage of staff who served at least 71 customers.
- Use the ordered data to determine the median and identify the median class.
- Explain how the frequency distribution, mean and percentages can help management make a decision.
(a) Frequency table: 31–40: f=6, cf=6, midpoint=35.5; 41–50: f=10, cf=16, midpoint=45.5; 51–60: f=12, cf=28, midpoint=55.5; 61–70: f=13, cf=41, midpoint=65.5; 71–80: f=9, cf=50, midpoint=75.5
(b) There are 50 staff members. Total customers served = 2868. Mean = 2868 ÷ 50 = 57.36.
(c) Since 57.36 > 56, the stated average target is exceeded.
(d) 23 staff served at most 56. Percentage = 23/50 × 100 = 46.0%.
(e) 9 staff served at least 71. Percentage = 9/50 × 100 = 18.0%.
(f) The 25th and 26th ordered observations are 58 and 59; median = 58.5. The median lies in the 51–60 class.
(g) The table shows concentration across ranges, the mean gives an overall average, and the percentages identify groups needing attention or further support.
ITEM 4
A class committee in Tourism Centre wants to choose one student representative from three candidates A, B and C. There are 142 students in the class.
17 students support A only, 31 support B only and 19 support C only. In addition, 12 support both A and B only, 15 support B and C only, 9 support A and C only, and 8 support all three.
Students who support none of the candidates are recorded separately. The teacher will select the candidate with the largest total support.
- Determine the number of students who support none of the three candidates.
- Determine the total number of supporters of A, B and C respectively.
- Identify the candidate with the greatest support and find the difference between the highest and second-highest totals.
- Determine the percentage of the whole class supporting the selected candidate.
- Determine the number and percentage of students who do not support the selected candidate.
- State whether the selected candidate has support from more than half the class and whether fewer than 50% oppose the candidate.
- Explain why the overlapping groups must be counted when making the final decision.
(a) Known regions total = 111; therefore none = 142 − 111 = 31 students.
(b) A = 46; B = 66; C = 51. Each total includes the relevant overlapping regions.
(c) Candidate B has the greatest support with 66 students. The second-highest total is 51, so the difference is 15.
(d) Support percentage = 66/142 × 100 = 46.48%.
(e) Non-supporters = 76. Percentage = 53.52%.
(f) Half the class is 71.0. Since 66 is greater than half, the selected candidate has majority support. The opposition percentage is 53.52%, which is below 50%.
(g) A student in an overlap supports more than one candidate. Ignoring these intersections would undercount candidates' totals and could change the result.
ITEM 5
A recreation facility in Tourism Centre is creating a paved common region where two circular lawns overlap. The radii of the circles are 2.05 m and 1.60 m.
The distance between the two intersection points is 1.80 m. The common region alone will be paved.
Each rectangular paver measures 6.0 cm by 10.0 cm. The supplier recommends buying an additional 8% above the calculated number because some pavers will be cut or damaged. Take π = 3.14.
- Determine the perpendicular distance from the centre of each circle to the common chord.
- Hence determine the distance between the centres of the two circles.
- Determine the central angle subtended by the common chord at each centre.
- Find the area of the common region by calculating the two circular segments.
- Convert the area of one paver into square metres and determine the minimum number of pavers required before the allowance.
- Allowing 8% extra, determine the number of pavers to purchase.
- Determine the total cost and explain why the final number must be rounded upward.
(a) Half-chord = 0.900 m. Distance for radius 2.05 m = √(2.05²−0.900²) = 1.842 m. Distance for radius 1.60 m = 1.323 m.
(b) Centre distance = 1.842 + 1.323 = 3.165 m.
(c) Central angles are approximately 52.08° and 68.46°.
(d) Using sector area minus triangle area for each segment, the common area is approximately 0.589 m².
(e) One paver area = 0.060 × 0.100 = 0.00600 m². Minimum before allowance = 0.589 ÷ 0.00600 ≈ 99 pavers.
(f) With 8% extra: 99 × (1 + 0.08) = 106.92, so 107 pavers should be purchased.
(g) Cost = 107 × Shs 225 = Shs 24,075. The result must be rounded upward because a fractional paver cannot be purchased and rounding downward would leave part of the region uncovered.
ITEM 6
A small manufacturing group in Tourism Centre produces open-topped ice-cream cones. One original cone has capacity 34 cm³ and height 4.75 cm.
The original cone costs Shs 55 to make and is sold for Shs 220. A new order requires cones with capacity 8 times the original capacity. The new cones are similar to the original cones.
The selling price is directly proportional to capacity, while production cost is directly proportional to the curved surface area of the cone. Use π = 3.14.
- Determine the radius and diameter of the original cone.
- Determine the slant height and curved surface area of the original cone.
- Determine the linear scale factor for the new cone, given that its capacity is 8 times the original.
- Determine the new height and diameter.
- Determine the curved surface area of one new cone and hence the total material area for an order of 250 cones.
- Determine the production cost of one new cone and the selling price of one new cone.
- Determine the profit per cone and the total profit on an order of 250 cones.
- Explain why capacity changes by a different factor from diameter.
(a) Using V = ⅓πr²h: r = √(3V/(πh)) = 2.615 cm. Diameter = 5.230 cm.
(b) Slant height l = √(r²+h²) = 5.422 cm. Curved surface area = πrl ≈ 44.525 cm².
(c) Volume factor = 8, so linear scale factor k = ∛8 = 2.000.
(d) New height = 4.75 × 2.000 = 9.500 cm. New diameter = 10.460 cm.
(e) New curved surface area = 178.099 cm² per cone. For 250 cones, total material area ≈ 4.452 m².
(f) Surface area scales as k², so new production cost = Shs 55 × 2.000² ≈ Shs 220. Selling price scales directly with capacity, so new selling price = Shs 220 × 8 = Shs 1,760.
(g) Profit per cone ≈ Shs 1,540. Total profit for 250 cones ≈ Shs 385,000.
(h) Capacity is a volume, so it changes with the cube of the linear scale factor, whereas diameter changes only linearly. Surface area changes with the square of the scale factor.
SET 13: COMMUNITY WATER PROJECT
ITEM 1
The local authority in Community Water Project has secured land for a community project. The project team has 20 acres available. The plan reserves 3/10 of the total land for the main buildings.
After the building area is removed, 1/5 of the land that remains is to be landscaped or planted. The rest is divided among staff facilities, a public waiting area and parking in the ratio 4:3:3.
The project engineer estimates that one vehicle needs 0.021 acres of parking space. However, 10% of the full parking capacity must remain free for emergency access and movement.
The landscaping must be cleared before construction. Two workers can clear half an acre in one day when working at the same rate. Each worker is paid Shs 16,000 for one day's work.
- Determine the area allocated to the main buildings.
- Determine the area remaining after the buildings and hence find the area reserved for landscaping or planting.
- Use the given ratio to determine the areas for staff facilities, the public waiting area and parking.
- Determine the theoretical maximum number of vehicles that the parking area can accommodate.
- After keeping 10% of the capacity free, determine the maximum number of ordinary vehicles that should be admitted at one time.
- Determine the number of workers required to clear the landscaping area in one day.
- Calculate the total amount required to pay those workers for the one-day job.
(a) Buildings = 20 × 3/10 = 6 acres.
(b) Remaining = 14 acres. Landscaping = 2.8 acres. Land still available after landscaping = 11.2 acres.
(c) Total ratio parts = 10. Staff = 4.48 acres; waiting area = 3.36 acres; parking = 3.36 acres.
(d) Parking capacity = 3.36 ÷ 0.021 = 160 vehicles.
(e) Emergency reserve = 10% × 160 = 16.0 vehicle spaces. Ordinary capacity = 90% × 160 = 144 whole vehicles.
(f) At 0.25 acre per worker-day, workers required = 2.8 ÷ 0.25 = 11.20, so 12 workers are required.
(g) Labour cost = 12 × Shs 16,000 = Shs 192,000.
ITEM 2
A small enterprise in Community Water Project uses two energy systems. Let S be the daily maintenance cost of the solar system and B the daily maintenance cost of the backup system.
Records show that S plus twice B is Shs 46,000. They also show that three times S exceeds twice B by Shs 10,000.
The enterprise produces two products, X and Y. Let x and y be the numbers produced per day. It must produce at least 20 units of X, X must be at least twice Y, total production cannot exceed 192, and the production budget is at most Shs 49,800.
Each X uses 3 units of a special input and each Y uses 4 units. At most 690 input units are available. Production costs are Shs 200 per X and Shs 300 per Y, while profits are Shs 500 and Shs 600 respectively.
- Form two simultaneous equations from the maintenance records and determine S and B.
- Define x and y and write all the production restrictions as inequalities, including non-negativity.
- Write the daily profit function P in terms of x and y.
- Determine the important feasible corner points and calculate the profit at each point.
- Identify the production plan that gives the greatest feasible profit.
- For that plan, calculate the total production cost and the total amount of special input used.
- Calculate the corresponding revenue and verify the profit.
(a) S + 2B = 46000; 3S − 2B = 10000. Adding gives 4S = 56000, so S = 14000. Then B = 16000. Thus solar maintenance is Shs 14,000 and backup maintenance is Shs 16,000.
(b) x ≥ 20; x ≥ 2y; x + y ≤ 192; 200x + 300y ≤ 49800; 3x + 4y ≤ 690; x ≥ 0, y ≥ 0.
(c) P = 500x + 600y.
(d) The key feasible corner points are (20,0), (20,10), (192,0) and (128,64). Their profits are Shs 10,000, Shs 16,000, Shs 96,000 and Shs 102,400 respectively.
(e) Therefore the maximum occurs at x = 128, y = 64, giving a maximum daily profit of Shs 102,400.
(f) Production cost = 200(128) + 300(64) = Shs 44,800. Input used = 3(128) + 4(64) = 640 units.
(g) Revenue = production cost + profit = Shs 44,800 + Shs 102,400 = Shs 147,200. Subtracting the cost from the revenue again gives Shs 102,400 profit.
ITEM 3
A management team in Community Water Project records the number of customers served by 50 staff members during a trial period. Management will continue the arrangement if the average number served exceeds 57.
Staff members who serve at most that threshold are also reviewed for additional support or possible reassignment.
The recorded values are listed below:
32, 34, 36, 38, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 61, 62, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 71, 71, 72, 73, 74, 75, 76, 77, 78, 79, 81.
- Arrange the data into a frequency distribution using class intervals of 10, beginning with the smallest observation. Include cumulative frequency and class midpoint.
- Determine the total number of observations and the total number of customers served, then calculate the mean.
- Use the management rule to state whether the average target is exceeded.
- Determine the number and percentage of staff who served at most 57 customers.
- Determine the number and percentage of staff who served at least 72 customers.
- Use the ordered data to determine the median and identify the median class.
- Explain how the frequency distribution, mean and percentages can help management make a decision.
(a) Frequency table: 32–41: f=6, cf=6, midpoint=36.5; 42–51: f=10, cf=16, midpoint=46.5; 52–61: f=12, cf=28, midpoint=56.5; 62–71: f=13, cf=41, midpoint=66.5; 72–81: f=9, cf=50, midpoint=76.5
(b) There are 50 staff members. Total customers served = 2918. Mean = 2918 ÷ 50 = 58.36.
(c) Since 58.36 > 57, the stated average target is exceeded.
(d) 23 staff served at most 57. Percentage = 23/50 × 100 = 46.0%.
(e) 9 staff served at least 72. Percentage = 9/50 × 100 = 18.0%.
(f) The 25th and 26th ordered observations are 59 and 60; median = 59.5. The median lies in the 52–61 class.
(g) The table shows concentration across ranges, the mean gives an overall average, and the percentages identify groups needing attention or further support.
ITEM 4
A class committee in Community Water Project wants to choose one student representative from three candidates A, B and C. There are 144 students in the class.
18 students support A only, 28 support B only and 14 support C only. In addition, 10 support both A and B only, 12 support B and C only, 10 support A and C only, and 6 support all three.
Students who support none of the candidates are recorded separately. The teacher will select the candidate with the largest total support.
- Determine the number of students who support none of the three candidates.
- Determine the total number of supporters of A, B and C respectively.
- Identify the candidate with the greatest support and find the difference between the highest and second-highest totals.
- Determine the percentage of the whole class supporting the selected candidate.
- Determine the number and percentage of students who do not support the selected candidate.
- State whether the selected candidate has support from more than half the class and whether fewer than 50% oppose the candidate.
- Explain why the overlapping groups must be counted when making the final decision.
(a) Known regions total = 98; therefore none = 144 − 98 = 46 students.
(b) A = 44; B = 56; C = 42. Each total includes the relevant overlapping regions.
(c) Candidate B has the greatest support with 56 students. The second-highest total is 44, so the difference is 12.
(d) Support percentage = 56/144 × 100 = 38.89%.
(e) Non-supporters = 88. Percentage = 61.11%.
(f) Half the class is 72.0. Since 56 is greater than half, the selected candidate has majority support. The opposition percentage is 61.11%, which is below 50%.
(g) A student in an overlap supports more than one candidate. Ignoring these intersections would undercount candidates' totals and could change the result.
ITEM 5
A recreation facility in Community Water Project is creating a paved common region where two circular lawns overlap. The radii of the circles are 2.10 m and 1.60 m.
The distance between the two intersection points is 1.85 m. The common region alone will be paved.
Each rectangular paver measures 4.0 cm by 8.0 cm. The supplier recommends buying an additional 5% above the calculated number because some pavers will be cut or damaged. Take π = 3.14.
- Determine the perpendicular distance from the centre of each circle to the common chord.
- Hence determine the distance between the centres of the two circles.
- Determine the central angle subtended by the common chord at each centre.
- Find the area of the common region by calculating the two circular segments.
- Convert the area of one paver into square metres and determine the minimum number of pavers required before the allowance.
- Allowing 5% extra, determine the number of pavers to purchase.
- Determine the total cost and explain why the final number must be rounded upward.
(a) Half-chord = 0.925 m. Distance for radius 2.10 m = √(2.10²−0.925²) = 1.885 m. Distance for radius 1.60 m = 1.306 m.
(b) Centre distance = 1.885 + 1.306 = 3.191 m.
(c) Central angles are approximately 52.27° and 70.64°.
(d) Using sector area minus triangle area for each segment, the common area is approximately 0.636 m².
(e) One paver area = 0.040 × 0.080 = 0.00320 m². Minimum before allowance = 0.636 ÷ 0.00320 ≈ 199 pavers.
(f) With 5% extra: 199 × (1 + 0.05) = 208.95, so 209 pavers should be purchased.
(g) Cost = 209 × Shs 250 = Shs 52,250. The result must be rounded upward because a fractional paver cannot be purchased and rounding downward would leave part of the region uncovered.
ITEM 6
A small manufacturing group in Community Water Project produces open-topped ice-cream cones. One original cone has capacity 36 cm³ and height 4.00 cm.
The original cone costs Shs 60 to make and is sold for Shs 240. A new order requires cones with capacity 27 times the original capacity. The new cones are similar to the original cones.
The selling price is directly proportional to capacity, while production cost is directly proportional to the curved surface area of the cone. Use π = 3.14.
- Determine the radius and diameter of the original cone.
- Determine the slant height and curved surface area of the original cone.
- Determine the linear scale factor for the new cone, given that its capacity is 27 times the original.
- Determine the new height and diameter.
- Determine the curved surface area of one new cone and hence the total material area for an order of 250 cones.
- Determine the production cost of one new cone and the selling price of one new cone.
- Determine the profit per cone and the total profit on an order of 250 cones.
- Explain why capacity changes by a different factor from diameter.
(a) Using V = ⅓πr²h: r = √(3V/(πh)) = 2.932 cm. Diameter = 5.865 cm.
(b) Slant height l = √(r²+h²) = 4.960 cm. Curved surface area = πrl ≈ 45.667 cm².
(c) Volume factor = 27, so linear scale factor k = ∛27 = 3.000.
(d) New height = 4.00 × 3.000 = 12.000 cm. New diameter = 17.594 cm.
(e) New curved surface area = 411.004 cm² per cone. For 250 cones, total material area ≈ 10.275 m².
(f) Surface area scales as k², so new production cost = Shs 60 × 3.000² ≈ Shs 540. Selling price scales directly with capacity, so new selling price = Shs 240 × 27 = Shs 6,480.
(g) Profit per cone ≈ Shs 5,940. Total profit for 250 cones ≈ Shs 1,485,000.
(h) Capacity is a volume, so it changes with the cube of the linear scale factor, whereas diameter changes only linearly. Surface area changes with the square of the scale factor.
SET 14: TEACHER RESOURCE CENTRE
ITEM 1
The local authority in Teacher Resource Centre has secured land for a community project. The project team has 18 acres available. The plan reserves 1/3 of the total land for the main buildings.
After the building area is removed, 1/4 of the land that remains is to be landscaped or planted. The rest is divided among staff facilities, a public waiting area and parking in the ratio 3:2:4.
The project engineer estimates that one vehicle needs 0.022 acres of parking space. However, 10% of the full parking capacity must remain free for emergency access and movement.
The landscaping must be cleared before construction. Two workers can clear half an acre in one day when working at the same rate. Each worker is paid Shs 19,000 for one day's work.
- Determine the area allocated to the main buildings.
- Determine the area remaining after the buildings and hence find the area reserved for landscaping or planting.
- Use the given ratio to determine the areas for staff facilities, the public waiting area and parking.
- Determine the theoretical maximum number of vehicles that the parking area can accommodate.
- After keeping 10% of the capacity free, determine the maximum number of ordinary vehicles that should be admitted at one time.
- Determine the number of workers required to clear the landscaping area in one day.
- Calculate the total amount required to pay those workers for the one-day job.
(a) Buildings = 18 × 1/3 = 6 acres.
(b) Remaining = 12 acres. Landscaping = 3 acres. Land still available after landscaping = 9 acres.
(c) Total ratio parts = 9. Staff = 3 acres; waiting area = 2 acres; parking = 4 acres.
(d) Parking capacity = 4 ÷ 0.022 = 180 vehicles.
(e) Emergency reserve = 10% × 180 = 18.0 vehicle spaces. Ordinary capacity = 90% × 180 = 162 whole vehicles.
(f) At 0.25 acre per worker-day, workers required = 3 ÷ 0.25 = 12.00, so 12 workers are required.
(g) Labour cost = 12 × Shs 19,000 = Shs 228,000.
ITEM 2
A small enterprise in Teacher Resource Centre uses two energy systems. Let S be the daily maintenance cost of the solar system and B the daily maintenance cost of the backup system.
Records show that S plus twice B is Shs 47,500. They also show that three times S exceeds twice B by Shs 10,500.
The enterprise produces two products, X and Y. Let x and y be the numbers produced per day. It must produce at least 20 units of X, X must be at least twice Y, total production cannot exceed 198, and the production budget is at most Shs 51,200.
Each X uses 3 units of a special input and each Y uses 4 units. At most 710 input units are available. Production costs are Shs 200 per X and Shs 300 per Y, while profits are Shs 500 and Shs 600 respectively.
- Form two simultaneous equations from the maintenance records and determine S and B.
- Define x and y and write all the production restrictions as inequalities, including non-negativity.
- Write the daily profit function P in terms of x and y.
- Determine the important feasible corner points and calculate the profit at each point.
- Identify the production plan that gives the greatest feasible profit.
- For that plan, calculate the total production cost and the total amount of special input used.
- Calculate the corresponding revenue and verify the profit.
(a) S + 2B = 47500; 3S − 2B = 10500. Adding gives 4S = 58000, so S = 14500. Then B = 16500. Thus solar maintenance is Shs 14,500 and backup maintenance is Shs 16,500.
(b) x ≥ 20; x ≥ 2y; x + y ≤ 198; 200x + 300y ≤ 51200; 3x + 4y ≤ 710; x ≥ 0, y ≥ 0.
(c) P = 500x + 600y.
(d) The key feasible corner points are (20,0), (20,10), (198,0) and (132,66). Their profits are Shs 10,000, Shs 16,000, Shs 99,000 and Shs 105,600 respectively.
(e) Therefore the maximum occurs at x = 132, y = 66, giving a maximum daily profit of Shs 105,600.
(f) Production cost = 200(132) + 300(66) = Shs 46,200. Input used = 3(132) + 4(66) = 660 units.
(g) Revenue = production cost + profit = Shs 46,200 + Shs 105,600 = Shs 151,800. Subtracting the cost from the revenue again gives Shs 105,600 profit.
ITEM 3
A management team in Teacher Resource Centre records the number of customers served by 50 staff members during a trial period. Management will continue the arrangement if the average number served exceeds 58.
Staff members who serve at most that threshold are also reviewed for additional support or possible reassignment.
The recorded values are listed below:
33, 35, 37, 39, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 62, 63, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 72, 72, 73, 74, 75, 76, 77, 78, 79, 80, 82.
- Arrange the data into a frequency distribution using class intervals of 10, beginning with the smallest observation. Include cumulative frequency and class midpoint.
- Determine the total number of observations and the total number of customers served, then calculate the mean.
- Use the management rule to state whether the average target is exceeded.
- Determine the number and percentage of staff who served at most 58 customers.
- Determine the number and percentage of staff who served at least 73 customers.
- Use the ordered data to determine the median and identify the median class.
- Explain how the frequency distribution, mean and percentages can help management make a decision.
(a) Frequency table: 33–42: f=6, cf=6, midpoint=37.5; 43–52: f=10, cf=16, midpoint=47.5; 53–62: f=12, cf=28, midpoint=57.5; 63–72: f=13, cf=41, midpoint=67.5; 73–82: f=9, cf=50, midpoint=77.5
(b) There are 50 staff members. Total customers served = 2968. Mean = 2968 ÷ 50 = 59.36.
(c) Since 59.36 > 58, the stated average target is exceeded.
(d) 23 staff served at most 58. Percentage = 23/50 × 100 = 46.0%.
(e) 9 staff served at least 73. Percentage = 9/50 × 100 = 18.0%.
(f) The 25th and 26th ordered observations are 60 and 61; median = 60.5. The median lies in the 53–62 class.
(g) The table shows concentration across ranges, the mean gives an overall average, and the percentages identify groups needing attention or further support.
ITEM 4
A class committee in Teacher Resource Centre wants to choose one student representative from three candidates A, B and C. There are 146 students in the class.
19 students support A only, 29 support B only and 15 support C only. In addition, 11 support both A and B only, 13 support B and C only, 11 support A and C only, and 7 support all three.
Students who support none of the candidates are recorded separately. The teacher will select the candidate with the largest total support.
- Determine the number of students who support none of the three candidates.
- Determine the total number of supporters of A, B and C respectively.
- Identify the candidate with the greatest support and find the difference between the highest and second-highest totals.
- Determine the percentage of the whole class supporting the selected candidate.
- Determine the number and percentage of students who do not support the selected candidate.
- State whether the selected candidate has support from more than half the class and whether fewer than 50% oppose the candidate.
- Explain why the overlapping groups must be counted when making the final decision.
(a) Known regions total = 105; therefore none = 146 − 105 = 41 students.
(b) A = 48; B = 60; C = 46. Each total includes the relevant overlapping regions.
(c) Candidate B has the greatest support with 60 students. The second-highest total is 48, so the difference is 12.
(d) Support percentage = 60/146 × 100 = 41.10%.
(e) Non-supporters = 86. Percentage = 58.90%.
(f) Half the class is 73.0. Since 60 is greater than half, the selected candidate has majority support. The opposition percentage is 58.90%, which is below 50%.
(g) A student in an overlap supports more than one candidate. Ignoring these intersections would undercount candidates' totals and could change the result.
ITEM 5
A recreation facility in Teacher Resource Centre is creating a paved common region where two circular lawns overlap. The radii of the circles are 2.15 m and 1.60 m.
The distance between the two intersection points is 1.90 m. The common region alone will be paved.
Each rectangular paver measures 5.0 cm by 10.0 cm. The supplier recommends buying an additional 6% above the calculated number because some pavers will be cut or damaged. Take π = 3.14.
- Determine the perpendicular distance from the centre of each circle to the common chord.
- Hence determine the distance between the centres of the two circles.
- Determine the central angle subtended by the common chord at each centre.
- Find the area of the common region by calculating the two circular segments.
- Convert the area of one paver into square metres and determine the minimum number of pavers required before the allowance.
- Allowing 6% extra, determine the number of pavers to purchase.
- Determine the total cost and explain why the final number must be rounded upward.
(a) Half-chord = 0.950 m. Distance for radius 2.15 m = √(2.15²−0.950²) = 1.929 m. Distance for radius 1.60 m = 1.287 m.
(b) Centre distance = 1.929 + 1.287 = 3.216 m.
(c) Central angles are approximately 52.45° and 72.85°.
(d) Using sector area minus triangle area for each segment, the common area is approximately 0.686 m².
(e) One paver area = 0.050 × 0.100 = 0.00500 m². Minimum before allowance = 0.686 ÷ 0.00500 ≈ 138 pavers.
(f) With 6% extra: 138 × (1 + 0.06) = 146.28, so 147 pavers should be purchased.
(g) Cost = 147 × Shs 275 = Shs 40,425. The result must be rounded upward because a fractional paver cannot be purchased and rounding downward would leave part of the region uncovered.
ITEM 6
A small manufacturing group in Teacher Resource Centre produces open-topped ice-cream cones. One original cone has capacity 38 cm³ and height 4.25 cm.
The original cone costs Shs 65 to make and is sold for Shs 260. A new order requires cones with capacity 4 times the original capacity. The new cones are similar to the original cones.
The selling price is directly proportional to capacity, while production cost is directly proportional to the curved surface area of the cone. Use π = 3.14.
- Determine the radius and diameter of the original cone.
- Determine the slant height and curved surface area of the original cone.
- Determine the linear scale factor for the new cone, given that its capacity is 4 times the original.
- Determine the new height and diameter.
- Determine the curved surface area of one new cone and hence the total material area for an order of 250 cones.
- Determine the production cost of one new cone and the selling price of one new cone.
- Determine the profit per cone and the total profit on an order of 250 cones.
- Explain why capacity changes by a different factor from diameter.
(a) Using V = ⅓πr²h: r = √(3V/(πh)) = 2.923 cm. Diameter = 5.846 cm.
(b) Slant height l = √(r²+h²) = 5.158 cm. Curved surface area = πrl ≈ 47.337 cm².
(c) Volume factor = 4, so linear scale factor k = ∛4 = 1.587.
(d) New height = 4.25 × 1.587 = 6.746 cm. New diameter = 9.279 cm.
(e) New curved surface area = 119.283 cm² per cone. For 250 cones, total material area ≈ 2.982 m².
(f) Surface area scales as k², so new production cost = Shs 65 × 1.587² ≈ Shs 163.79. Selling price scales directly with capacity, so new selling price = Shs 260 × 4 = Shs 1,040.
(g) Profit per cone ≈ Shs 876.21. Total profit for 250 cones ≈ Shs 219,052.57.
(h) Capacity is a volume, so it changes with the cube of the linear scale factor, whereas diameter changes only linearly. Surface area changes with the square of the scale factor.
SET 15: COMMUNITY HALL
ITEM 1
The local authority in Community Hall has secured land for a community project. The project team has 26 acres available. The plan reserves 7/13 of the total land for the main buildings.
After the building area is removed, 1/5 of the land that remains is to be landscaped or planted. The rest is divided among staff facilities, a public waiting area and parking in the ratio 2:3:4.
The project engineer estimates that one vehicle needs 0.018 acres of parking space. However, 10% of the full parking capacity must remain free for emergency access and movement.
The landscaping must be cleared before construction. Two workers can clear half an acre in one day when working at the same rate. Each worker is paid Shs 20,000 for one day's work.
- Determine the area allocated to the main buildings.
- Determine the area remaining after the buildings and hence find the area reserved for landscaping or planting.
- Use the given ratio to determine the areas for staff facilities, the public waiting area and parking.
- Determine the theoretical maximum number of vehicles that the parking area can accommodate.
- After keeping 10% of the capacity free, determine the maximum number of ordinary vehicles that should be admitted at one time.
- Determine the number of workers required to clear the landscaping area in one day.
- Calculate the total amount required to pay those workers for the one-day job.
(a) Buildings = 26 × 7/13 = 14 acres.
(b) Remaining = 12 acres. Landscaping = 2.4 acres. Land still available after landscaping = 9.6 acres.
(c) Total ratio parts = 9. Staff = 2.133 acres; waiting area = 3.2 acres; parking = 4.267 acres.
(d) Parking capacity = 4.267 ÷ 0.018 = 240 vehicles.
(e) Emergency reserve = 10% × 240 = 24.0 vehicle spaces. Ordinary capacity = 90% × 240 = 216 whole vehicles.
(f) At 0.25 acre per worker-day, workers required = 2.4 ÷ 0.25 = 9.60, so 10 workers are required.
(g) Labour cost = 10 × Shs 20,000 = Shs 200,000.
ITEM 2
A small enterprise in Community Hall uses two energy systems. Let S be the daily maintenance cost of the solar system and B the daily maintenance cost of the backup system.
Records show that S plus twice B is Shs 49,000. They also show that three times S exceeds twice B by Shs 11,000.
The enterprise produces two products, X and Y. Let x and y be the numbers produced per day. It must produce at least 20 units of X, X must be at least twice Y, total production cannot exceed 204, and the production budget is at most Shs 52,600.
Each X uses 3 units of a special input and each Y uses 4 units. At most 730 input units are available. Production costs are Shs 200 per X and Shs 300 per Y, while profits are Shs 500 and Shs 600 respectively.
- Form two simultaneous equations from the maintenance records and determine S and B.
- Define x and y and write all the production restrictions as inequalities, including non-negativity.
- Write the daily profit function P in terms of x and y.
- Determine the important feasible corner points and calculate the profit at each point.
- Identify the production plan that gives the greatest feasible profit.
- For that plan, calculate the total production cost and the total amount of special input used.
- Calculate the corresponding revenue and verify the profit.
(a) S + 2B = 49000; 3S − 2B = 11000. Adding gives 4S = 60000, so S = 15000. Then B = 17000. Thus solar maintenance is Shs 15,000 and backup maintenance is Shs 17,000.
(b) x ≥ 20; x ≥ 2y; x + y ≤ 204; 200x + 300y ≤ 52600; 3x + 4y ≤ 730; x ≥ 0, y ≥ 0.
(c) P = 500x + 600y.
(d) The key feasible corner points are (20,0), (20,10), (204,0) and (136,68). Their profits are Shs 10,000, Shs 16,000, Shs 102,000 and Shs 108,800 respectively.
(e) Therefore the maximum occurs at x = 136, y = 68, giving a maximum daily profit of Shs 108,800.
(f) Production cost = 200(136) + 300(68) = Shs 47,600. Input used = 3(136) + 4(68) = 680 units.
(g) Revenue = production cost + profit = Shs 47,600 + Shs 108,800 = Shs 156,400. Subtracting the cost from the revenue again gives Shs 108,800 profit.
ITEM 3
A management team in Community Hall records the number of customers served by 50 staff members during a trial period. Management will continue the arrangement if the average number served exceeds 59.
Staff members who serve at most that threshold are also reviewed for additional support or possible reassignment.
The recorded values are listed below:
34, 36, 38, 40, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 63, 64, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 73, 73, 74, 75, 76, 77, 78, 79, 80, 81, 83.
- Arrange the data into a frequency distribution using class intervals of 10, beginning with the smallest observation. Include cumulative frequency and class midpoint.
- Determine the total number of observations and the total number of customers served, then calculate the mean.
- Use the management rule to state whether the average target is exceeded.
- Determine the number and percentage of staff who served at most 59 customers.
- Determine the number and percentage of staff who served at least 74 customers.
- Use the ordered data to determine the median and identify the median class.
- Explain how the frequency distribution, mean and percentages can help management make a decision.
(a) Frequency table: 34–43: f=6, cf=6, midpoint=38.5; 44–53: f=10, cf=16, midpoint=48.5; 54–63: f=12, cf=28, midpoint=58.5; 64–73: f=13, cf=41, midpoint=68.5; 74–83: f=9, cf=50, midpoint=78.5
(b) There are 50 staff members. Total customers served = 3018. Mean = 3018 ÷ 50 = 60.36.
(c) Since 60.36 > 59, the stated average target is exceeded.
(d) 23 staff served at most 59. Percentage = 23/50 × 100 = 46.0%.
(e) 9 staff served at least 74. Percentage = 9/50 × 100 = 18.0%.
(f) The 25th and 26th ordered observations are 61 and 62; median = 61.5. The median lies in the 54–63 class.
(g) The table shows concentration across ranges, the mean gives an overall average, and the percentages identify groups needing attention or further support.
ITEM 4
A class committee in Community Hall wants to choose one student representative from three candidates A, B and C. There are 148 students in the class.
20 students support A only, 30 support B only and 16 support C only. In addition, 12 support both A and B only, 14 support B and C only, 12 support A and C only, and 8 support all three.
Students who support none of the candidates are recorded separately. The teacher will select the candidate with the largest total support.
- Determine the number of students who support none of the three candidates.
- Determine the total number of supporters of A, B and C respectively.
- Identify the candidate with the greatest support and find the difference between the highest and second-highest totals.
- Determine the percentage of the whole class supporting the selected candidate.
- Determine the number and percentage of students who do not support the selected candidate.
- State whether the selected candidate has support from more than half the class and whether fewer than 50% oppose the candidate.
- Explain why the overlapping groups must be counted when making the final decision.
(a) Known regions total = 112; therefore none = 148 − 112 = 36 students.
(b) A = 52; B = 64; C = 50. Each total includes the relevant overlapping regions.
(c) Candidate B has the greatest support with 64 students. The second-highest total is 52, so the difference is 12.
(d) Support percentage = 64/148 × 100 = 43.24%.
(e) Non-supporters = 84. Percentage = 56.76%.
(f) Half the class is 74.0. Since 64 is greater than half, the selected candidate has majority support. The opposition percentage is 56.76%, which is below 50%.
(g) A student in an overlap supports more than one candidate. Ignoring these intersections would undercount candidates' totals and could change the result.
ITEM 5
A recreation facility in Community Hall is creating a paved common region where two circular lawns overlap. The radii of the circles are 2.20 m and 1.60 m.
The distance between the two intersection points is 1.60 m. The common region alone will be paved.
Each rectangular paver measures 6.0 cm by 8.0 cm. The supplier recommends buying an additional 7% above the calculated number because some pavers will be cut or damaged. Take π = 3.14.
- Determine the perpendicular distance from the centre of each circle to the common chord.
- Hence determine the distance between the centres of the two circles.
- Determine the central angle subtended by the common chord at each centre.
- Find the area of the common region by calculating the two circular segments.
- Convert the area of one paver into square metres and determine the minimum number of pavers required before the allowance.
- Allowing 7% extra, determine the number of pavers to purchase.
- Determine the total cost and explain why the final number must be rounded upward.
(a) Half-chord = 0.800 m. Distance for radius 2.20 m = √(2.20²−0.800²) = 2.049 m. Distance for radius 1.60 m = 1.386 m.
(b) Centre distance = 2.049 + 1.386 = 3.435 m.
(c) Central angles are approximately 42.65° and 60.00°.
(d) Using sector area minus triangle area for each segment, the common area is approximately 0.392 m².
(e) One paver area = 0.060 × 0.080 = 0.00480 m². Minimum before allowance = 0.392 ÷ 0.00480 ≈ 82 pavers.
(f) With 7% extra: 82 × (1 + 0.07) = 87.74, so 88 pavers should be purchased.
(g) Cost = 88 × Shs 300 = Shs 26,400. The result must be rounded upward because a fractional paver cannot be purchased and rounding downward would leave part of the region uncovered.
ITEM 6
A small manufacturing group in Community Hall produces open-topped ice-cream cones. One original cone has capacity 40 cm³ and height 4.50 cm.
The original cone costs Shs 70 to make and is sold for Shs 280. A new order requires cones with capacity 8 times the original capacity. The new cones are similar to the original cones.
The selling price is directly proportional to capacity, while production cost is directly proportional to the curved surface area of the cone. Use π = 3.14.
- Determine the radius and diameter of the original cone.
- Determine the slant height and curved surface area of the original cone.
- Determine the linear scale factor for the new cone, given that its capacity is 8 times the original.
- Determine the new height and diameter.
- Determine the curved surface area of one new cone and hence the total material area for an order of 250 cones.
- Determine the production cost of one new cone and the selling price of one new cone.
- Determine the profit per cone and the total profit on an order of 250 cones.
- Explain why capacity changes by a different factor from diameter.
(a) Using V = ⅓πr²h: r = √(3V/(πh)) = 2.914 cm. Diameter = 5.828 cm.
(b) Slant height l = √(r²+h²) = 5.361 cm. Curved surface area = πrl ≈ 49.058 cm².
(c) Volume factor = 8, so linear scale factor k = ∛8 = 2.000.
(d) New height = 4.50 × 2.000 = 9.000 cm. New diameter = 11.657 cm.
(e) New curved surface area = 196.233 cm² per cone. For 250 cones, total material area ≈ 4.906 m².
(f) Surface area scales as k², so new production cost = Shs 70 × 2.000² ≈ Shs 280. Selling price scales directly with capacity, so new selling price = Shs 280 × 8 = Shs 2,240.
(g) Profit per cone ≈ Shs 1,960. Total profit for 250 cones ≈ Shs 490,000.
(h) Capacity is a volume, so it changes with the cube of the linear scale factor, whereas diameter changes only linearly. Surface area changes with the square of the scale factor.
SET 16: PRODUCE MARKET
ITEM 1
The local authority in Produce Market has secured land for a community project. The project team has 30 acres available. The plan reserves 2/5 of the total land for the main buildings.
After the building area is removed, 1/6 of the land that remains is to be landscaped or planted. The rest is divided among staff facilities, a public waiting area and parking in the ratio 3:4:2.
The project engineer estimates that one vehicle needs 0.017 acres of parking space. However, 10% of the full parking capacity must remain free for emergency access and movement.
The landscaping must be cleared before construction. Two workers can clear half an acre in one day when working at the same rate. Each worker is paid Shs 18,000 for one day's work.
- Determine the area allocated to the main buildings.
- Determine the area remaining after the buildings and hence find the area reserved for landscaping or planting.
- Use the given ratio to determine the areas for staff facilities, the public waiting area and parking.
- Determine the theoretical maximum number of vehicles that the parking area can accommodate.
- After keeping 10% of the capacity free, determine the maximum number of ordinary vehicles that should be admitted at one time.
- Determine the number of workers required to clear the landscaping area in one day.
- Calculate the total amount required to pay those workers for the one-day job.
(a) Buildings = 30 × 2/5 = 12 acres.
(b) Remaining = 18 acres. Landscaping = 3 acres. Land still available after landscaping = 15 acres.
(c) Total ratio parts = 9. Staff = 5 acres; waiting area = 6.667 acres; parking = 3.333 acres.
(d) Parking capacity = 3.333 ÷ 0.017 = 200 vehicles.
(e) Emergency reserve = 10% × 200 = 20.0 vehicle spaces. Ordinary capacity = 90% × 200 = 180 whole vehicles.
(f) At 0.25 acre per worker-day, workers required = 3 ÷ 0.25 = 12.00, so 12 workers are required.
(g) Labour cost = 12 × Shs 18,000 = Shs 216,000.
ITEM 2
A small enterprise in Produce Market uses two energy systems. Let S be the daily maintenance cost of the solar system and B the daily maintenance cost of the backup system.
Records show that S plus twice B is Shs 50,500. They also show that three times S exceeds twice B by Shs 11,500.
The enterprise produces two products, X and Y. Let x and y be the numbers produced per day. It must produce at least 20 units of X, X must be at least twice Y, total production cannot exceed 210, and the production budget is at most Shs 54,000.
Each X uses 3 units of a special input and each Y uses 4 units. At most 750 input units are available. Production costs are Shs 200 per X and Shs 300 per Y, while profits are Shs 500 and Shs 600 respectively.
- Form two simultaneous equations from the maintenance records and determine S and B.
- Define x and y and write all the production restrictions as inequalities, including non-negativity.
- Write the daily profit function P in terms of x and y.
- Determine the important feasible corner points and calculate the profit at each point.
- Identify the production plan that gives the greatest feasible profit.
- For that plan, calculate the total production cost and the total amount of special input used.
- Calculate the corresponding revenue and verify the profit.
(a) S + 2B = 50500; 3S − 2B = 11500. Adding gives 4S = 62000, so S = 15500. Then B = 17500. Thus solar maintenance is Shs 15,500 and backup maintenance is Shs 17,500.
(b) x ≥ 20; x ≥ 2y; x + y ≤ 210; 200x + 300y ≤ 54000; 3x + 4y ≤ 750; x ≥ 0, y ≥ 0.
(c) P = 500x + 600y.
(d) The key feasible corner points are (20,0), (20,10), (210,0) and (140,70). Their profits are Shs 10,000, Shs 16,000, Shs 105,000 and Shs 112,000 respectively.
(e) Therefore the maximum occurs at x = 140, y = 70, giving a maximum daily profit of Shs 112,000.
(f) Production cost = 200(140) + 300(70) = Shs 49,000. Input used = 3(140) + 4(70) = 700 units.
(g) Revenue = production cost + profit = Shs 49,000 + Shs 112,000 = Shs 161,000. Subtracting the cost from the revenue again gives Shs 112,000 profit.
ITEM 3
A management team in Produce Market records the number of customers served by 50 staff members during a trial period. Management will continue the arrangement if the average number served exceeds 60.
Staff members who serve at most that threshold are also reviewed for additional support or possible reassignment.
The recorded values are listed below:
35, 37, 39, 41, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 64, 65, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 74, 74, 75, 76, 77, 78, 79, 80, 81, 82, 84.
- Arrange the data into a frequency distribution using class intervals of 10, beginning with the smallest observation. Include cumulative frequency and class midpoint.
- Determine the total number of observations and the total number of customers served, then calculate the mean.
- Use the management rule to state whether the average target is exceeded.
- Determine the number and percentage of staff who served at most 60 customers.
- Determine the number and percentage of staff who served at least 75 customers.
- Use the ordered data to determine the median and identify the median class.
- Explain how the frequency distribution, mean and percentages can help management make a decision.
(a) Frequency table: 35–44: f=6, cf=6, midpoint=39.5; 45–54: f=10, cf=16, midpoint=49.5; 55–64: f=12, cf=28, midpoint=59.5; 65–74: f=13, cf=41, midpoint=69.5; 75–84: f=9, cf=50, midpoint=79.5
(b) There are 50 staff members. Total customers served = 3068. Mean = 3068 ÷ 50 = 61.36.
(c) Since 61.36 > 60, the stated average target is exceeded.
(d) 23 staff served at most 60. Percentage = 23/50 × 100 = 46.0%.
(e) 9 staff served at least 75. Percentage = 9/50 × 100 = 18.0%.
(f) The 25th and 26th ordered observations are 62 and 63; median = 62.5. The median lies in the 55–64 class.
(g) The table shows concentration across ranges, the mean gives an overall average, and the percentages identify groups needing attention or further support.
ITEM 4
A class committee in Produce Market wants to choose one student representative from three candidates A, B and C. There are 150 students in the class.
16 students support A only, 31 support B only and 17 support C only. In addition, 10 support both A and B only, 15 support B and C only, 8 support A and C only, and 6 support all three.
Students who support none of the candidates are recorded separately. The teacher will select the candidate with the largest total support.
- Determine the number of students who support none of the three candidates.
- Determine the total number of supporters of A, B and C respectively.
- Identify the candidate with the greatest support and find the difference between the highest and second-highest totals.
- Determine the percentage of the whole class supporting the selected candidate.
- Determine the number and percentage of students who do not support the selected candidate.
- State whether the selected candidate has support from more than half the class and whether fewer than 50% oppose the candidate.
- Explain why the overlapping groups must be counted when making the final decision.
(a) Known regions total = 103; therefore none = 150 − 103 = 47 students.
(b) A = 40; B = 62; C = 46. Each total includes the relevant overlapping regions.
(c) Candidate B has the greatest support with 62 students. The second-highest total is 46, so the difference is 16.
(d) Support percentage = 62/150 × 100 = 41.33%.
(e) Non-supporters = 88. Percentage = 58.67%.
(f) Half the class is 75.0. Since 62 is greater than half, the selected candidate has majority support. The opposition percentage is 58.67%, which is below 50%.
(g) A student in an overlap supports more than one candidate. Ignoring these intersections would undercount candidates' totals and could change the result.
ITEM 5
A recreation facility in Produce Market is creating a paved common region where two circular lawns overlap. The radii of the circles are 2.00 m and 1.65 m.
The distance between the two intersection points is 1.65 m. The common region alone will be paved.
Each rectangular paver measures 4.0 cm by 10.0 cm. The supplier recommends buying an additional 8% above the calculated number because some pavers will be cut or damaged. Take π = 3.14.
- Determine the perpendicular distance from the centre of each circle to the common chord.
- Hence determine the distance between the centres of the two circles.
- Determine the central angle subtended by the common chord at each centre.
- Find the area of the common region by calculating the two circular segments.
- Convert the area of one paver into square metres and determine the minimum number of pavers required before the allowance.
- Allowing 8% extra, determine the number of pavers to purchase.
- Determine the total cost and explain why the final number must be rounded upward.
(a) Half-chord = 0.825 m. Distance for radius 2.00 m = √(2.00²−0.825²) = 1.822 m. Distance for radius 1.65 m = 1.429 m.
(b) Centre distance = 1.822 + 1.429 = 3.251 m.
(c) Central angles are approximately 48.72° and 60.00°.
(d) Using sector area minus triangle area for each segment, the common area is approximately 0.443 m².
(e) One paver area = 0.040 × 0.100 = 0.00400 m². Minimum before allowance = 0.443 ÷ 0.00400 ≈ 111 pavers.
(f) With 8% extra: 111 × (1 + 0.08) = 119.88, so 120 pavers should be purchased.
(g) Cost = 120 × Shs 200 = Shs 24,000. The result must be rounded upward because a fractional paver cannot be purchased and rounding downward would leave part of the region uncovered.
ITEM 6
A small manufacturing group in Produce Market produces open-topped ice-cream cones. One original cone has capacity 32 cm³ and height 4.75 cm.
The original cone costs Shs 50 to make and is sold for Shs 200. A new order requires cones with capacity 27 times the original capacity. The new cones are similar to the original cones.
The selling price is directly proportional to capacity, while production cost is directly proportional to the curved surface area of the cone. Use π = 3.14.
- Determine the radius and diameter of the original cone.
- Determine the slant height and curved surface area of the original cone.
- Determine the linear scale factor for the new cone, given that its capacity is 27 times the original.
- Determine the new height and diameter.
- Determine the curved surface area of one new cone and hence the total material area for an order of 250 cones.
- Determine the production cost of one new cone and the selling price of one new cone.
- Determine the profit per cone and the total profit on an order of 250 cones.
- Explain why capacity changes by a different factor from diameter.
(a) Using V = ⅓πr²h: r = √(3V/(πh)) = 2.537 cm. Diameter = 5.074 cm.
(b) Slant height l = √(r²+h²) = 5.385 cm. Curved surface area = πrl ≈ 42.899 cm².
(c) Volume factor = 27, so linear scale factor k = ∛27 = 3.000.
(d) New height = 4.75 × 3.000 = 14.250 cm. New diameter = 15.222 cm.
(e) New curved surface area = 386.089 cm² per cone. For 250 cones, total material area ≈ 9.652 m².
(f) Surface area scales as k², so new production cost = Shs 50 × 3.000² ≈ Shs 450. Selling price scales directly with capacity, so new selling price = Shs 200 × 27 = Shs 5,400.
(g) Profit per cone ≈ Shs 4,950. Total profit for 250 cones ≈ Shs 1,237,500.
(h) Capacity is a volume, so it changes with the cube of the linear scale factor, whereas diameter changes only linearly. Surface area changes with the square of the scale factor.
SET 17: HEALTH OUTREACH BASE
ITEM 1
The local authority in Health Outreach Base has secured land for a community project. The project team has 24 acres available. The plan reserves 3/8 of the total land for the main buildings.
After the building area is removed, 1/4 of the land that remains is to be landscaped or planted. The rest is divided among staff facilities, a public waiting area and parking in the ratio 4:3:2.
The project engineer estimates that one vehicle needs 0.02 acres of parking space. However, 10% of the full parking capacity must remain free for emergency access and movement.
The landscaping must be cleared before construction. Two workers can clear half an acre in one day when working at the same rate. Each worker is paid Shs 21,000 for one day's work.
- Determine the area allocated to the main buildings.
- Determine the area remaining after the buildings and hence find the area reserved for landscaping or planting.
- Use the given ratio to determine the areas for staff facilities, the public waiting area and parking.
- Determine the theoretical maximum number of vehicles that the parking area can accommodate.
- After keeping 10% of the capacity free, determine the maximum number of ordinary vehicles that should be admitted at one time.
- Determine the number of workers required to clear the landscaping area in one day.
- Calculate the total amount required to pay those workers for the one-day job.
(a) Buildings = 24 × 3/8 = 9 acres.
(b) Remaining = 15 acres. Landscaping = 3.75 acres. Land still available after landscaping = 11.25 acres.
(c) Total ratio parts = 9. Staff = 5 acres; waiting area = 3.75 acres; parking = 2.5 acres.
(d) Parking capacity = 2.5 ÷ 0.02 = 125 vehicles.
(e) Emergency reserve = 10% × 125 = 12.5 vehicle spaces. Ordinary capacity = 90% × 125 = 112 whole vehicles.
(f) At 0.25 acre per worker-day, workers required = 3.75 ÷ 0.25 = 15.00, so 15 workers are required.
(g) Labour cost = 15 × Shs 21,000 = Shs 315,000.
ITEM 2
A small enterprise in Health Outreach Base uses two energy systems. Let S be the daily maintenance cost of the solar system and B the daily maintenance cost of the backup system.
Records show that S plus twice B is Shs 52,000. They also show that three times S exceeds twice B by Shs 12,000.
The enterprise produces two products, X and Y. Let x and y be the numbers produced per day. It must produce at least 20 units of X, X must be at least twice Y, total production cannot exceed 216, and the production budget is at most Shs 55,400.
Each X uses 3 units of a special input and each Y uses 4 units. At most 770 input units are available. Production costs are Shs 200 per X and Shs 300 per Y, while profits are Shs 500 and Shs 600 respectively.
- Form two simultaneous equations from the maintenance records and determine S and B.
- Define x and y and write all the production restrictions as inequalities, including non-negativity.
- Write the daily profit function P in terms of x and y.
- Determine the important feasible corner points and calculate the profit at each point.
- Identify the production plan that gives the greatest feasible profit.
- For that plan, calculate the total production cost and the total amount of special input used.
- Calculate the corresponding revenue and verify the profit.
(a) S + 2B = 52000; 3S − 2B = 12000. Adding gives 4S = 64000, so S = 16000. Then B = 18000. Thus solar maintenance is Shs 16,000 and backup maintenance is Shs 18,000.
(b) x ≥ 20; x ≥ 2y; x + y ≤ 216; 200x + 300y ≤ 55400; 3x + 4y ≤ 770; x ≥ 0, y ≥ 0.
(c) P = 500x + 600y.
(d) The key feasible corner points are (20,0), (20,10), (216,0) and (144,72). Their profits are Shs 10,000, Shs 16,000, Shs 108,000 and Shs 115,200 respectively.
(e) Therefore the maximum occurs at x = 144, y = 72, giving a maximum daily profit of Shs 115,200.
(f) Production cost = 200(144) + 300(72) = Shs 50,400. Input used = 3(144) + 4(72) = 720 units.
(g) Revenue = production cost + profit = Shs 50,400 + Shs 115,200 = Shs 165,600. Subtracting the cost from the revenue again gives Shs 115,200 profit.
ITEM 3
A management team in Health Outreach Base records the number of customers served by 50 staff members during a trial period. Management will continue the arrangement if the average number served exceeds 61.
Staff members who serve at most that threshold are also reviewed for additional support or possible reassignment.
The recorded values are listed below:
36, 38, 40, 42, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 65, 66, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 75, 75, 76, 77, 78, 79, 80, 81, 82, 83, 85.
- Arrange the data into a frequency distribution using class intervals of 10, beginning with the smallest observation. Include cumulative frequency and class midpoint.
- Determine the total number of observations and the total number of customers served, then calculate the mean.
- Use the management rule to state whether the average target is exceeded.
- Determine the number and percentage of staff who served at most 61 customers.
- Determine the number and percentage of staff who served at least 76 customers.
- Use the ordered data to determine the median and identify the median class.
- Explain how the frequency distribution, mean and percentages can help management make a decision.
(a) Frequency table: 36–45: f=6, cf=6, midpoint=40.5; 46–55: f=10, cf=16, midpoint=50.5; 56–65: f=12, cf=28, midpoint=60.5; 66–75: f=13, cf=41, midpoint=70.5; 76–85: f=9, cf=50, midpoint=80.5
(b) There are 50 staff members. Total customers served = 3118. Mean = 3118 ÷ 50 = 62.36.
(c) Since 62.36 > 61, the stated average target is exceeded.
(d) 23 staff served at most 61. Percentage = 23/50 × 100 = 46.0%.
(e) 9 staff served at least 76. Percentage = 9/50 × 100 = 18.0%.
(f) The 25th and 26th ordered observations are 63 and 64; median = 63.5. The median lies in the 56–65 class.
(g) The table shows concentration across ranges, the mean gives an overall average, and the percentages identify groups needing attention or further support.
ITEM 4
A class committee in Health Outreach Base wants to choose one student representative from three candidates A, B and C. There are 152 students in the class.
17 students support A only, 28 support B only and 18 support C only. In addition, 11 support both A and B only, 12 support B and C only, 9 support A and C only, and 7 support all three.
Students who support none of the candidates are recorded separately. The teacher will select the candidate with the largest total support.
- Determine the number of students who support none of the three candidates.
- Determine the total number of supporters of A, B and C respectively.
- Identify the candidate with the greatest support and find the difference between the highest and second-highest totals.
- Determine the percentage of the whole class supporting the selected candidate.
- Determine the number and percentage of students who do not support the selected candidate.
- State whether the selected candidate has support from more than half the class and whether fewer than 50% oppose the candidate.
- Explain why the overlapping groups must be counted when making the final decision.
(a) Known regions total = 102; therefore none = 152 − 102 = 50 students.
(b) A = 44; B = 58; C = 46. Each total includes the relevant overlapping regions.
(c) Candidate B has the greatest support with 58 students. The second-highest total is 46, so the difference is 12.
(d) Support percentage = 58/152 × 100 = 38.16%.
(e) Non-supporters = 94. Percentage = 61.84%.
(f) Half the class is 76.0. Since 58 is greater than half, the selected candidate has majority support. The opposition percentage is 61.84%, which is below 50%.
(g) A student in an overlap supports more than one candidate. Ignoring these intersections would undercount candidates' totals and could change the result.
ITEM 5
A recreation facility in Health Outreach Base is creating a paved common region where two circular lawns overlap. The radii of the circles are 2.05 m and 1.65 m.
The distance between the two intersection points is 1.70 m. The common region alone will be paved.
Each rectangular paver measures 5.0 cm by 8.0 cm. The supplier recommends buying an additional 5% above the calculated number because some pavers will be cut or damaged. Take π = 3.14.
- Determine the perpendicular distance from the centre of each circle to the common chord.
- Hence determine the distance between the centres of the two circles.
- Determine the central angle subtended by the common chord at each centre.
- Find the area of the common region by calculating the two circular segments.
- Convert the area of one paver into square metres and determine the minimum number of pavers required before the allowance.
- Allowing 5% extra, determine the number of pavers to purchase.
- Determine the total cost and explain why the final number must be rounded upward.
(a) Half-chord = 0.850 m. Distance for radius 2.05 m = √(2.05²−0.850²) = 1.865 m. Distance for radius 1.65 m = 1.414 m.
(b) Centre distance = 1.865 + 1.414 = 3.280 m.
(c) Central angles are approximately 48.99° and 62.02°.
(d) Using sector area minus triangle area for each segment, the common area is approximately 0.481 m².
(e) One paver area = 0.050 × 0.080 = 0.00400 m². Minimum before allowance = 0.481 ÷ 0.00400 ≈ 121 pavers.
(f) With 5% extra: 121 × (1 + 0.05) = 127.05, so 128 pavers should be purchased.
(g) Cost = 128 × Shs 225 = Shs 28,800. The result must be rounded upward because a fractional paver cannot be purchased and rounding downward would leave part of the region uncovered.
ITEM 6
A small manufacturing group in Health Outreach Base produces open-topped ice-cream cones. One original cone has capacity 34 cm³ and height 4.00 cm.
The original cone costs Shs 55 to make and is sold for Shs 220. A new order requires cones with capacity 8 times the original capacity. The new cones are similar to the original cones.
The selling price is directly proportional to capacity, while production cost is directly proportional to the curved surface area of the cone. Use π = 3.14.
- Determine the radius and diameter of the original cone.
- Determine the slant height and curved surface area of the original cone.
- Determine the linear scale factor for the new cone, given that its capacity is 8 times the original.
- Determine the new height and diameter.
- Determine the curved surface area of one new cone and hence the total material area for an order of 250 cones.
- Determine the production cost of one new cone and the selling price of one new cone.
- Determine the profit per cone and the total profit on an order of 250 cones.
- Explain why capacity changes by a different factor from diameter.
(a) Using V = ⅓πr²h: r = √(3V/(πh)) = 2.850 cm. Diameter = 5.699 cm.
(b) Slant height l = √(r²+h²) = 4.911 cm. Curved surface area = πrl ≈ 43.947 cm².
(c) Volume factor = 8, so linear scale factor k = ∛8 = 2.000.
(d) New height = 4.00 × 2.000 = 8.000 cm. New diameter = 11.399 cm.
(e) New curved surface area = 175.789 cm² per cone. For 250 cones, total material area ≈ 4.395 m².
(f) Surface area scales as k², so new production cost = Shs 55 × 2.000² ≈ Shs 220. Selling price scales directly with capacity, so new selling price = Shs 220 × 8 = Shs 1,760.
(g) Profit per cone ≈ Shs 1,540. Total profit for 250 cones ≈ Shs 385,000.
(h) Capacity is a volume, so it changes with the cube of the linear scale factor, whereas diameter changes only linearly. Surface area changes with the square of the scale factor.
SET 18: SKILLS WORKSHOP
ITEM 1
The local authority in Skills Workshop has secured land for a community project. The project team has 27 acres available. The plan reserves 1/3 of the total land for the main buildings.
After the building area is removed, 1/5 of the land that remains is to be landscaped or planted. The rest is divided among staff facilities, a public waiting area and parking in the ratio 3:2:4.
The project engineer estimates that one vehicle needs 0.02 acres of parking space. However, 10% of the full parking capacity must remain free for emergency access and movement.
The landscaping must be cleared before construction. Two workers can clear half an acre in one day when working at the same rate. Each worker is paid Shs 17,500 for one day's work.
- Determine the area allocated to the main buildings.
- Determine the area remaining after the buildings and hence find the area reserved for landscaping or planting.
- Use the given ratio to determine the areas for staff facilities, the public waiting area and parking.
- Determine the theoretical maximum number of vehicles that the parking area can accommodate.
- After keeping 10% of the capacity free, determine the maximum number of ordinary vehicles that should be admitted at one time.
- Determine the number of workers required to clear the landscaping area in one day.
- Calculate the total amount required to pay those workers for the one-day job.
(a) Buildings = 27 × 1/3 = 9 acres.
(b) Remaining = 18 acres. Landscaping = 3.6 acres. Land still available after landscaping = 14.4 acres.
(c) Total ratio parts = 9. Staff = 4.8 acres; waiting area = 3.2 acres; parking = 6.4 acres.
(d) Parking capacity = 6.4 ÷ 0.02 = 320 vehicles.
(e) Emergency reserve = 10% × 320 = 32.0 vehicle spaces. Ordinary capacity = 90% × 320 = 288 whole vehicles.
(f) At 0.25 acre per worker-day, workers required = 3.6 ÷ 0.25 = 14.40, so 15 workers are required.
(g) Labour cost = 15 × Shs 17,500 = Shs 262,500.
ITEM 2
A small enterprise in Skills Workshop uses two energy systems. Let S be the daily maintenance cost of the solar system and B the daily maintenance cost of the backup system.
Records show that S plus twice B is Shs 53,500. They also show that three times S exceeds twice B by Shs 12,500.
The enterprise produces two products, X and Y. Let x and y be the numbers produced per day. It must produce at least 20 units of X, X must be at least twice Y, total production cannot exceed 222, and the production budget is at most Shs 56,800.
Each X uses 3 units of a special input and each Y uses 4 units. At most 790 input units are available. Production costs are Shs 200 per X and Shs 300 per Y, while profits are Shs 500 and Shs 600 respectively.
- Form two simultaneous equations from the maintenance records and determine S and B.
- Define x and y and write all the production restrictions as inequalities, including non-negativity.
- Write the daily profit function P in terms of x and y.
- Determine the important feasible corner points and calculate the profit at each point.
- Identify the production plan that gives the greatest feasible profit.
- For that plan, calculate the total production cost and the total amount of special input used.
- Calculate the corresponding revenue and verify the profit.
(a) S + 2B = 53500; 3S − 2B = 12500. Adding gives 4S = 66000, so S = 16500. Then B = 18500. Thus solar maintenance is Shs 16,500 and backup maintenance is Shs 18,500.
(b) x ≥ 20; x ≥ 2y; x + y ≤ 222; 200x + 300y ≤ 56800; 3x + 4y ≤ 790; x ≥ 0, y ≥ 0.
(c) P = 500x + 600y.
(d) The key feasible corner points are (20,0), (20,10), (222,0) and (148,74). Their profits are Shs 10,000, Shs 16,000, Shs 111,000 and Shs 118,400 respectively.
(e) Therefore the maximum occurs at x = 148, y = 74, giving a maximum daily profit of Shs 118,400.
(f) Production cost = 200(148) + 300(74) = Shs 51,800. Input used = 3(148) + 4(74) = 740 units.
(g) Revenue = production cost + profit = Shs 51,800 + Shs 118,400 = Shs 170,200. Subtracting the cost from the revenue again gives Shs 118,400 profit.
ITEM 3
A management team in Skills Workshop records the number of customers served by 50 staff members during a trial period. Management will continue the arrangement if the average number served exceeds 62.
Staff members who serve at most that threshold are also reviewed for additional support or possible reassignment.
The recorded values are listed below:
37, 39, 41, 43, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 66, 67, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 76, 76, 77, 78, 79, 80, 81, 82, 83, 84, 86.
- Arrange the data into a frequency distribution using class intervals of 10, beginning with the smallest observation. Include cumulative frequency and class midpoint.
- Determine the total number of observations and the total number of customers served, then calculate the mean.
- Use the management rule to state whether the average target is exceeded.
- Determine the number and percentage of staff who served at most 62 customers.
- Determine the number and percentage of staff who served at least 77 customers.
- Use the ordered data to determine the median and identify the median class.
- Explain how the frequency distribution, mean and percentages can help management make a decision.
(a) Frequency table: 37–46: f=6, cf=6, midpoint=41.5; 47–56: f=10, cf=16, midpoint=51.5; 57–66: f=12, cf=28, midpoint=61.5; 67–76: f=13, cf=41, midpoint=71.5; 77–86: f=9, cf=50, midpoint=81.5
(b) There are 50 staff members. Total customers served = 3168. Mean = 3168 ÷ 50 = 63.36.
(c) Since 63.36 > 62, the stated average target is exceeded.
(d) 23 staff served at most 62. Percentage = 23/50 × 100 = 46.0%.
(e) 9 staff served at least 77. Percentage = 9/50 × 100 = 18.0%.
(f) The 25th and 26th ordered observations are 64 and 65; median = 64.5. The median lies in the 57–66 class.
(g) The table shows concentration across ranges, the mean gives an overall average, and the percentages identify groups needing attention or further support.
ITEM 4
A class committee in Skills Workshop wants to choose one student representative from three candidates A, B and C. There are 154 students in the class.
18 students support A only, 29 support B only and 19 support C only. In addition, 12 support both A and B only, 13 support B and C only, 10 support A and C only, and 8 support all three.
Students who support none of the candidates are recorded separately. The teacher will select the candidate with the largest total support.
- Determine the number of students who support none of the three candidates.
- Determine the total number of supporters of A, B and C respectively.
- Identify the candidate with the greatest support and find the difference between the highest and second-highest totals.
- Determine the percentage of the whole class supporting the selected candidate.
- Determine the number and percentage of students who do not support the selected candidate.
- State whether the selected candidate has support from more than half the class and whether fewer than 50% oppose the candidate.
- Explain why the overlapping groups must be counted when making the final decision.
(a) Known regions total = 109; therefore none = 154 − 109 = 45 students.
(b) A = 48; B = 62; C = 50. Each total includes the relevant overlapping regions.
(c) Candidate B has the greatest support with 62 students. The second-highest total is 50, so the difference is 12.
(d) Support percentage = 62/154 × 100 = 40.26%.
(e) Non-supporters = 92. Percentage = 59.74%.
(f) Half the class is 77.0. Since 62 is greater than half, the selected candidate has majority support. The opposition percentage is 59.74%, which is below 50%.
(g) A student in an overlap supports more than one candidate. Ignoring these intersections would undercount candidates' totals and could change the result.
ITEM 5
A recreation facility in Skills Workshop is creating a paved common region where two circular lawns overlap. The radii of the circles are 2.10 m and 1.65 m.
The distance between the two intersection points is 1.75 m. The common region alone will be paved.
Each rectangular paver measures 6.0 cm by 10.0 cm. The supplier recommends buying an additional 6% above the calculated number because some pavers will be cut or damaged. Take π = 3.14.
- Determine the perpendicular distance from the centre of each circle to the common chord.
- Hence determine the distance between the centres of the two circles.
- Determine the central angle subtended by the common chord at each centre.
- Find the area of the common region by calculating the two circular segments.
- Convert the area of one paver into square metres and determine the minimum number of pavers required before the allowance.
- Allowing 6% extra, determine the number of pavers to purchase.
- Determine the total cost and explain why the final number must be rounded upward.
(a) Half-chord = 0.875 m. Distance for radius 2.10 m = √(2.10²−0.875²) = 1.909 m. Distance for radius 1.65 m = 1.399 m.
(b) Centre distance = 1.909 + 1.399 = 3.308 m.
(c) Central angles are approximately 49.25° and 64.05°.
(d) Using sector area minus triangle area for each segment, the common area is approximately 0.521 m².
(e) One paver area = 0.060 × 0.100 = 0.00600 m². Minimum before allowance = 0.521 ÷ 0.00600 ≈ 87 pavers.
(f) With 6% extra: 87 × (1 + 0.06) = 92.22, so 93 pavers should be purchased.
(g) Cost = 93 × Shs 250 = Shs 23,250. The result must be rounded upward because a fractional paver cannot be purchased and rounding downward would leave part of the region uncovered.
ITEM 6
A small manufacturing group in Skills Workshop produces open-topped ice-cream cones. One original cone has capacity 36 cm³ and height 4.25 cm.
The original cone costs Shs 60 to make and is sold for Shs 240. A new order requires cones with capacity 4 times the original capacity. The new cones are similar to the original cones.
The selling price is directly proportional to capacity, while production cost is directly proportional to the curved surface area of the cone. Use π = 3.14.
- Determine the radius and diameter of the original cone.
- Determine the slant height and curved surface area of the original cone.
- Determine the linear scale factor for the new cone, given that its capacity is 4 times the original.
- Determine the new height and diameter.
- Determine the curved surface area of one new cone and hence the total material area for an order of 250 cones.
- Determine the production cost of one new cone and the selling price of one new cone.
- Determine the profit per cone and the total profit on an order of 250 cones.
- Explain why capacity changes by a different factor from diameter.
(a) Using V = ⅓πr²h: r = √(3V/(πh)) = 2.845 cm. Diameter = 5.690 cm.
(b) Slant height l = √(r²+h²) = 5.114 cm. Curved surface area = πrl ≈ 45.684 cm².
(c) Volume factor = 4, so linear scale factor k = ∛4 = 1.587.
(d) New height = 4.25 × 1.587 = 6.746 cm. New diameter = 9.032 cm.
(e) New curved surface area = 115.116 cm² per cone. For 250 cones, total material area ≈ 2.878 m².
(f) Surface area scales as k², so new production cost = Shs 60 × 1.587² ≈ Shs 151.19. Selling price scales directly with capacity, so new selling price = Shs 240 × 4 = Shs 960.
(g) Profit per cone ≈ Shs 808.81. Total profit for 250 cones ≈ Shs 202,202.37.
(h) Capacity is a volume, so it changes with the cube of the linear scale factor, whereas diameter changes only linearly. Surface area changes with the square of the scale factor.
SET 19: ECO PARK
ITEM 1
The local authority in Eco Park has secured land for a community project. The project team has 25 acres available. The plan reserves 2/5 of the total land for the main buildings.
After the building area is removed, 1/4 of the land that remains is to be landscaped or planted. The rest is divided among staff facilities, a public waiting area and parking in the ratio 2:3:5.
The project engineer estimates that one vehicle needs 0.025 acres of parking space. However, 10% of the full parking capacity must remain free for emergency access and movement.
The landscaping must be cleared before construction. Two workers can clear half an acre in one day when working at the same rate. Each worker is paid Shs 16,000 for one day's work.
- Determine the area allocated to the main buildings.
- Determine the area remaining after the buildings and hence find the area reserved for landscaping or planting.
- Use the given ratio to determine the areas for staff facilities, the public waiting area and parking.
- Determine the theoretical maximum number of vehicles that the parking area can accommodate.
- After keeping 10% of the capacity free, determine the maximum number of ordinary vehicles that should be admitted at one time.
- Determine the number of workers required to clear the landscaping area in one day.
- Calculate the total amount required to pay those workers for the one-day job.
(a) Buildings = 25 × 2/5 = 10 acres.
(b) Remaining = 15 acres. Landscaping = 3.75 acres. Land still available after landscaping = 11.25 acres.
(c) Total ratio parts = 10. Staff = 2.25 acres; waiting area = 3.375 acres; parking = 5.625 acres.
(d) Parking capacity = 5.625 ÷ 0.025 = 225 vehicles.
(e) Emergency reserve = 10% × 225 = 22.5 vehicle spaces. Ordinary capacity = 90% × 225 = 202 whole vehicles.
(f) At 0.25 acre per worker-day, workers required = 3.75 ÷ 0.25 = 15.00, so 15 workers are required.
(g) Labour cost = 15 × Shs 16,000 = Shs 240,000.
ITEM 2
A small enterprise in Eco Park uses two energy systems. Let S be the daily maintenance cost of the solar system and B the daily maintenance cost of the backup system.
Records show that S plus twice B is Shs 55,000. They also show that three times S exceeds twice B by Shs 13,000.
The enterprise produces two products, X and Y. Let x and y be the numbers produced per day. It must produce at least 20 units of X, X must be at least twice Y, total production cannot exceed 228, and the production budget is at most Shs 58,200.
Each X uses 3 units of a special input and each Y uses 4 units. At most 810 input units are available. Production costs are Shs 200 per X and Shs 300 per Y, while profits are Shs 500 and Shs 600 respectively.
- Form two simultaneous equations from the maintenance records and determine S and B.
- Define x and y and write all the production restrictions as inequalities, including non-negativity.
- Write the daily profit function P in terms of x and y.
- Determine the important feasible corner points and calculate the profit at each point.
- Identify the production plan that gives the greatest feasible profit.
- For that plan, calculate the total production cost and the total amount of special input used.
- Calculate the corresponding revenue and verify the profit.
(a) S + 2B = 55000; 3S − 2B = 13000. Adding gives 4S = 68000, so S = 17000. Then B = 19000. Thus solar maintenance is Shs 17,000 and backup maintenance is Shs 19,000.
(b) x ≥ 20; x ≥ 2y; x + y ≤ 228; 200x + 300y ≤ 58200; 3x + 4y ≤ 810; x ≥ 0, y ≥ 0.
(c) P = 500x + 600y.
(d) The key feasible corner points are (20,0), (20,10), (228,0) and (152,76). Their profits are Shs 10,000, Shs 16,000, Shs 114,000 and Shs 121,600 respectively.
(e) Therefore the maximum occurs at x = 152, y = 76, giving a maximum daily profit of Shs 121,600.
(f) Production cost = 200(152) + 300(76) = Shs 53,200. Input used = 3(152) + 4(76) = 760 units.
(g) Revenue = production cost + profit = Shs 53,200 + Shs 121,600 = Shs 174,800. Subtracting the cost from the revenue again gives Shs 121,600 profit.
ITEM 3
A management team in Eco Park records the number of customers served by 50 staff members during a trial period. Management will continue the arrangement if the average number served exceeds 63.
Staff members who serve at most that threshold are also reviewed for additional support or possible reassignment.
The recorded values are listed below:
38, 40, 42, 44, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 67, 68, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 77, 77, 78, 79, 80, 81, 82, 83, 84, 85, 87.
- Arrange the data into a frequency distribution using class intervals of 10, beginning with the smallest observation. Include cumulative frequency and class midpoint.
- Determine the total number of observations and the total number of customers served, then calculate the mean.
- Use the management rule to state whether the average target is exceeded.
- Determine the number and percentage of staff who served at most 63 customers.
- Determine the number and percentage of staff who served at least 78 customers.
- Use the ordered data to determine the median and identify the median class.
- Explain how the frequency distribution, mean and percentages can help management make a decision.
(a) Frequency table: 38–47: f=6, cf=6, midpoint=42.5; 48–57: f=10, cf=16, midpoint=52.5; 58–67: f=12, cf=28, midpoint=62.5; 68–77: f=13, cf=41, midpoint=72.5; 78–87: f=9, cf=50, midpoint=82.5
(b) There are 50 staff members. Total customers served = 3218. Mean = 3218 ÷ 50 = 64.36.
(c) Since 64.36 > 63, the stated average target is exceeded.
(d) 23 staff served at most 63. Percentage = 23/50 × 100 = 46.0%.
(e) 9 staff served at least 78. Percentage = 9/50 × 100 = 18.0%.
(f) The 25th and 26th ordered observations are 65 and 66; median = 65.5. The median lies in the 58–67 class.
(g) The table shows concentration across ranges, the mean gives an overall average, and the percentages identify groups needing attention or further support.
ITEM 4
A class committee in Eco Park wants to choose one student representative from three candidates A, B and C. There are 156 students in the class.
19 students support A only, 30 support B only and 14 support C only. In addition, 10 support both A and B only, 14 support B and C only, 11 support A and C only, and 6 support all three.
Students who support none of the candidates are recorded separately. The teacher will select the candidate with the largest total support.
- Determine the number of students who support none of the three candidates.
- Determine the total number of supporters of A, B and C respectively.
- Identify the candidate with the greatest support and find the difference between the highest and second-highest totals.
- Determine the percentage of the whole class supporting the selected candidate.
- Determine the number and percentage of students who do not support the selected candidate.
- State whether the selected candidate has support from more than half the class and whether fewer than 50% oppose the candidate.
- Explain why the overlapping groups must be counted when making the final decision.
(a) Known regions total = 104; therefore none = 156 − 104 = 52 students.
(b) A = 46; B = 60; C = 45. Each total includes the relevant overlapping regions.
(c) Candidate B has the greatest support with 60 students. The second-highest total is 46, so the difference is 14.
(d) Support percentage = 60/156 × 100 = 38.46%.
(e) Non-supporters = 96. Percentage = 61.54%.
(f) Half the class is 78.0. Since 60 is greater than half, the selected candidate has majority support. The opposition percentage is 61.54%, which is below 50%.
(g) A student in an overlap supports more than one candidate. Ignoring these intersections would undercount candidates' totals and could change the result.
ITEM 5
A recreation facility in Eco Park is creating a paved common region where two circular lawns overlap. The radii of the circles are 2.15 m and 1.65 m.
The distance between the two intersection points is 1.80 m. The common region alone will be paved.
Each rectangular paver measures 4.0 cm by 8.0 cm. The supplier recommends buying an additional 7% above the calculated number because some pavers will be cut or damaged. Take π = 3.14.
- Determine the perpendicular distance from the centre of each circle to the common chord.
- Hence determine the distance between the centres of the two circles.
- Determine the central angle subtended by the common chord at each centre.
- Find the area of the common region by calculating the two circular segments.
- Convert the area of one paver into square metres and determine the minimum number of pavers required before the allowance.
- Allowing 7% extra, determine the number of pavers to purchase.
- Determine the total cost and explain why the final number must be rounded upward.
(a) Half-chord = 0.900 m. Distance for radius 2.15 m = √(2.15²−0.900²) = 1.953 m. Distance for radius 1.65 m = 1.383 m.
(b) Centre distance = 1.953 + 1.383 = 3.335 m.
(c) Central angles are approximately 49.49° and 66.11°.
(d) Using sector area minus triangle area for each segment, the common area is approximately 0.563 m².
(e) One paver area = 0.040 × 0.080 = 0.00320 m². Minimum before allowance = 0.563 ÷ 0.00320 ≈ 177 pavers.
(f) With 7% extra: 177 × (1 + 0.07) = 189.39, so 190 pavers should be purchased.
(g) Cost = 190 × Shs 275 = Shs 52,250. The result must be rounded upward because a fractional paver cannot be purchased and rounding downward would leave part of the region uncovered.
ITEM 6
A small manufacturing group in Eco Park produces open-topped ice-cream cones. One original cone has capacity 38 cm³ and height 4.50 cm.
The original cone costs Shs 65 to make and is sold for Shs 260. A new order requires cones with capacity 27 times the original capacity. The new cones are similar to the original cones.
The selling price is directly proportional to capacity, while production cost is directly proportional to the curved surface area of the cone. Use π = 3.14.
- Determine the radius and diameter of the original cone.
- Determine the slant height and curved surface area of the original cone.
- Determine the linear scale factor for the new cone, given that its capacity is 27 times the original.
- Determine the new height and diameter.
- Determine the curved surface area of one new cone and hence the total material area for an order of 250 cones.
- Determine the production cost of one new cone and the selling price of one new cone.
- Determine the profit per cone and the total profit on an order of 250 cones.
- Explain why capacity changes by a different factor from diameter.
(a) Using V = ⅓πr²h: r = √(3V/(πh)) = 2.840 cm. Diameter = 5.681 cm.
(b) Slant height l = √(r²+h²) = 5.321 cm. Curved surface area = πrl ≈ 47.462 cm².
(c) Volume factor = 27, so linear scale factor k = ∛27 = 3.000.
(d) New height = 4.50 × 3.000 = 13.500 cm. New diameter = 17.042 cm.
(e) New curved surface area = 427.154 cm² per cone. For 250 cones, total material area ≈ 10.679 m².
(f) Surface area scales as k², so new production cost = Shs 65 × 3.000² ≈ Shs 585. Selling price scales directly with capacity, so new selling price = Shs 260 × 27 = Shs 7,020.
(g) Profit per cone ≈ Shs 6,435. Total profit for 250 cones ≈ Shs 1,608,750.
(h) Capacity is a volume, so it changes with the cube of the linear scale factor, whereas diameter changes only linearly. Surface area changes with the square of the scale factor.
SET 20: INNOVATION HUB
ITEM 1
The local authority in Innovation Hub has secured land for a community project. The project team has 28 acres available. The plan reserves 3/7 of the total land for the main buildings.
After the building area is removed, 1/5 of the land that remains is to be landscaped or planted. The rest is divided among staff facilities, a public waiting area and parking in the ratio 3:2:4.
The project engineer estimates that one vehicle needs 0.024 acres of parking space. However, 10% of the full parking capacity must remain free for emergency access and movement.
The landscaping must be cleared before construction. Two workers can clear half an acre in one day when working at the same rate. Each worker is paid Shs 19,500 for one day's work.
- Determine the area allocated to the main buildings.
- Determine the area remaining after the buildings and hence find the area reserved for landscaping or planting.
- Use the given ratio to determine the areas for staff facilities, the public waiting area and parking.
- Determine the theoretical maximum number of vehicles that the parking area can accommodate.
- After keeping 10% of the capacity free, determine the maximum number of ordinary vehicles that should be admitted at one time.
- Determine the number of workers required to clear the landscaping area in one day.
- Calculate the total amount required to pay those workers for the one-day job.
(a) Buildings = 28 × 3/7 = 12 acres.
(b) Remaining = 16 acres. Landscaping = 3.2 acres. Land still available after landscaping = 12.8 acres.
(c) Total ratio parts = 9. Staff = 4.267 acres; waiting area = 2.844 acres; parking = 5.689 acres.
(d) Parking capacity = 5.689 ÷ 0.024 = 240 vehicles.
(e) Emergency reserve = 10% × 240 = 24.0 vehicle spaces. Ordinary capacity = 90% × 240 = 216 whole vehicles.
(f) At 0.25 acre per worker-day, workers required = 3.2 ÷ 0.25 = 12.80, so 13 workers are required.
(g) Labour cost = 13 × Shs 19,500 = Shs 253,500.
ITEM 2
A small enterprise in Innovation Hub uses two energy systems. Let S be the daily maintenance cost of the solar system and B the daily maintenance cost of the backup system.
Records show that S plus twice B is Shs 56,500. They also show that three times S exceeds twice B by Shs 13,500.
The enterprise produces two products, X and Y. Let x and y be the numbers produced per day. It must produce at least 20 units of X, X must be at least twice Y, total production cannot exceed 234, and the production budget is at most Shs 59,600.
Each X uses 3 units of a special input and each Y uses 4 units. At most 830 input units are available. Production costs are Shs 200 per X and Shs 300 per Y, while profits are Shs 500 and Shs 600 respectively.
- Form two simultaneous equations from the maintenance records and determine S and B.
- Define x and y and write all the production restrictions as inequalities, including non-negativity.
- Write the daily profit function P in terms of x and y.
- Determine the important feasible corner points and calculate the profit at each point.
- Identify the production plan that gives the greatest feasible profit.
- For that plan, calculate the total production cost and the total amount of special input used.
- Calculate the corresponding revenue and verify the profit.
(a) S + 2B = 56500; 3S − 2B = 13500. Adding gives 4S = 70000, so S = 17500. Then B = 19500. Thus solar maintenance is Shs 17,500 and backup maintenance is Shs 19,500.
(b) x ≥ 20; x ≥ 2y; x + y ≤ 234; 200x + 300y ≤ 59600; 3x + 4y ≤ 830; x ≥ 0, y ≥ 0.
(c) P = 500x + 600y.
(d) The key feasible corner points are (20,0), (20,10), (234,0) and (156,78). Their profits are Shs 10,000, Shs 16,000, Shs 117,000 and Shs 124,800 respectively.
(e) Therefore the maximum occurs at x = 156, y = 78, giving a maximum daily profit of Shs 124,800.
(f) Production cost = 200(156) + 300(78) = Shs 54,600. Input used = 3(156) + 4(78) = 780 units.
(g) Revenue = production cost + profit = Shs 54,600 + Shs 124,800 = Shs 179,400. Subtracting the cost from the revenue again gives Shs 124,800 profit.
ITEM 3
A management team in Innovation Hub records the number of customers served by 50 staff members during a trial period. Management will continue the arrangement if the average number served exceeds 64.
Staff members who serve at most that threshold are also reviewed for additional support or possible reassignment.
The recorded values are listed below:
39, 41, 43, 45, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 68, 69, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 78, 78, 79, 80, 81, 82, 83, 84, 85, 86, 88.
- Arrange the data into a frequency distribution using class intervals of 10, beginning with the smallest observation. Include cumulative frequency and class midpoint.
- Determine the total number of observations and the total number of customers served, then calculate the mean.
- Use the management rule to state whether the average target is exceeded.
- Determine the number and percentage of staff who served at most 64 customers.
- Determine the number and percentage of staff who served at least 79 customers.
- Use the ordered data to determine the median and identify the median class.
- Explain how the frequency distribution, mean and percentages can help management make a decision.
(a) Frequency table: 39–48: f=6, cf=6, midpoint=43.5; 49–58: f=10, cf=16, midpoint=53.5; 59–68: f=12, cf=28, midpoint=63.5; 69–78: f=13, cf=41, midpoint=73.5; 79–88: f=9, cf=50, midpoint=83.5
(b) There are 50 staff members. Total customers served = 3268. Mean = 3268 ÷ 50 = 65.36.
(c) Since 65.36 > 64, the stated average target is exceeded.
(d) 23 staff served at most 64. Percentage = 23/50 × 100 = 46.0%.
(e) 9 staff served at least 79. Percentage = 9/50 × 100 = 18.0%.
(f) The 25th and 26th ordered observations are 66 and 67; median = 66.5. The median lies in the 59–68 class.
(g) The table shows concentration across ranges, the mean gives an overall average, and the percentages identify groups needing attention or further support.
ITEM 4
A class committee in Innovation Hub wants to choose one student representative from three candidates A, B and C. There are 158 students in the class.
20 students support A only, 31 support B only and 15 support C only. In addition, 11 support both A and B only, 15 support B and C only, 12 support A and C only, and 7 support all three.
Students who support none of the candidates are recorded separately. The teacher will select the candidate with the largest total support.
- Determine the number of students who support none of the three candidates.
- Determine the total number of supporters of A, B and C respectively.
- Identify the candidate with the greatest support and find the difference between the highest and second-highest totals.
- Determine the percentage of the whole class supporting the selected candidate.
- Determine the number and percentage of students who do not support the selected candidate.
- State whether the selected candidate has support from more than half the class and whether fewer than 50% oppose the candidate.
- Explain why the overlapping groups must be counted when making the final decision.
(a) Known regions total = 111; therefore none = 158 − 111 = 47 students.
(b) A = 50; B = 64; C = 49. Each total includes the relevant overlapping regions.
(c) Candidate B has the greatest support with 64 students. The second-highest total is 50, so the difference is 14.
(d) Support percentage = 64/158 × 100 = 40.51%.
(e) Non-supporters = 94. Percentage = 59.49%.
(f) Half the class is 79.0. Since 64 is greater than half, the selected candidate has majority support. The opposition percentage is 59.49%, which is below 50%.
(g) A student in an overlap supports more than one candidate. Ignoring these intersections would undercount candidates' totals and could change the result.
ITEM 5
A recreation facility in Innovation Hub is creating a paved common region where two circular lawns overlap. The radii of the circles are 2.20 m and 1.65 m.
The distance between the two intersection points is 1.85 m. The common region alone will be paved.
Each rectangular paver measures 5.0 cm by 10.0 cm. The supplier recommends buying an additional 8% above the calculated number because some pavers will be cut or damaged. Take π = 3.14.
- Determine the perpendicular distance from the centre of each circle to the common chord.
- Hence determine the distance between the centres of the two circles.
- Determine the central angle subtended by the common chord at each centre.
- Find the area of the common region by calculating the two circular segments.
- Convert the area of one paver into square metres and determine the minimum number of pavers required before the allowance.
- Allowing 8% extra, determine the number of pavers to purchase.
- Determine the total cost and explain why the final number must be rounded upward.
(a) Half-chord = 0.925 m. Distance for radius 2.20 m = √(2.20²−0.925²) = 1.996 m. Distance for radius 1.65 m = 1.366 m.
(b) Centre distance = 1.996 + 1.366 = 3.362 m.
(c) Central angles are approximately 49.73° and 68.20°.
(d) Using sector area minus triangle area for each segment, the common area is approximately 0.608 m².
(e) One paver area = 0.050 × 0.100 = 0.00500 m². Minimum before allowance = 0.608 ÷ 0.00500 ≈ 122 pavers.
(f) With 8% extra: 122 × (1 + 0.08) = 131.76, so 132 pavers should be purchased.
(g) Cost = 132 × Shs 300 = Shs 39,600. The result must be rounded upward because a fractional paver cannot be purchased and rounding downward would leave part of the region uncovered.
ITEM 6
A small manufacturing group in Innovation Hub produces open-topped ice-cream cones. One original cone has capacity 40 cm³ and height 4.75 cm.
The original cone costs Shs 70 to make and is sold for Shs 280. A new order requires cones with capacity 8 times the original capacity. The new cones are similar to the original cones.
The selling price is directly proportional to capacity, while production cost is directly proportional to the curved surface area of the cone. Use π = 3.14.
- Determine the radius and diameter of the original cone.
- Determine the slant height and curved surface area of the original cone.
- Determine the linear scale factor for the new cone, given that its capacity is 8 times the original.
- Determine the new height and diameter.
- Determine the curved surface area of one new cone and hence the total material area for an order of 250 cones.
- Determine the production cost of one new cone and the selling price of one new cone.
- Determine the profit per cone and the total profit on an order of 250 cones.
- Explain why capacity changes by a different factor from diameter.
(a) Using V = ⅓πr²h: r = √(3V/(πh)) = 2.836 cm. Diameter = 5.673 cm.
(b) Slant height l = √(r²+h²) = 5.532 cm. Curved surface area = πrl ≈ 49.275 cm².
(c) Volume factor = 8, so linear scale factor k = ∛8 = 2.000.
(d) New height = 4.75 × 2.000 = 9.500 cm. New diameter = 11.346 cm.
(e) New curved surface area = 197.100 cm² per cone. For 250 cones, total material area ≈ 4.928 m².
(f) Surface area scales as k², so new production cost = Shs 70 × 2.000² ≈ Shs 280. Selling price scales directly with capacity, so new selling price = Shs 280 × 8 = Shs 2,240.
(g) Profit per cone ≈ Shs 1,960. Total profit for 250 cones ≈ Shs 490,000.
(h) Capacity is a volume, so it changes with the cube of the linear scale factor, whereas diameter changes only linearly. Surface area changes with the square of the scale factor.
this is a real thing lets roll
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