Secondary 2 Mathematics Numbers Ratio Proportion Quiz

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Secondary 2 Mathematics Quiz - Numbers Ratio Proportion

Name: ___________________________
Class: ___________________________
Date: ___________________________
Score: ______ / 40

Duration: 45 minutes
Total Marks: 40

Instructions:

• Answer all questions.
• Write your answers in the spaces provided.
• Show all working clearly.
• Omission of essential working will result in loss of marks.
• Calculators may be used unless otherwise stated.
• Give answers correct to 3 significant figures where appropriate, unless exact values are required.

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Section A: Short Answer Questions (Questions 1–10, 2 marks each, Total 20 marks)

1. Express the ratio $48 : 72 : 96$ in its simplest form.
Answer: ___________________________ [2]

2. A sum of money is divided between Ali, Bala, and Charlie in the ratio $3 : 5 : 7$. If Bala receives $120 more than Ali, find the total sum of money. **Answer:**$___________________________ [2]

3. The scale of a map is $1 : 25\,000$. The distance between two towns on the map is $6.4$ cm. Find the actual distance between the two towns in kilometres.
Answer: ___________________________ km [2]

4. $y$ is inversely proportional to the square of $x$. When $x = 4$, $y = 9$. Find the value of $y$ when $x = 6$.
Answer: ___________________________ [2]

5. A car travels $180$ km using $15$ litres of petrol. How many litres of petrol are needed to travel $300$ km at the same rate?
Answer: ___________________________ litres [2]

6. The ratio of the number of boys to girls in a class is $4 : 5$. After $6$ boys join the class, the ratio becomes $5 : 5$. How many girls are in the class?
Answer: ___________________________ [2]

7. It takes $8$ workers $12$ days to complete a job. How many days will it take $6$ workers to complete the same job, assuming they work at the same rate?
Answer: ___________________________ days [2]

8. A recipe requires flour, sugar, and butter in the ratio $5 : 2 : 3$ by mass. If $400$ g of flour is used, find the total mass of the mixture.
Answer: ___________________________ g [2]

9. The price of a watch increased from $250 to$300. Express the increase as a percentage of the original price.
Answer: ___________________________ % [2]

10. $p$ is directly proportional to the cube root of $q$. When $q = 27$, $p = 12$. Find the value of $p$ when $q = 64$.
Answer: ___________________________ [2]

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Section B: Structured Questions (Questions 11–16, 3 marks each, Total 18 marks)

11. A map has a scale of $1 : 50\,000$. (a) The area of a lake on the map is $2.5$ cm². Calculate the actual area of the lake in km².
Answer: ___________________________ km² [2]

(b) The actual length of a river is $15$ km. Find its length on the map in centimetres.
Answer: ___________________________ cm [1]

12. The ratio of $A$ to $B$ is $3 : 7$ and the ratio of $B$ to $C$ is $5 : 9$. Find the ratio $A : B : C$ in its simplest form.
Answer: ___________________________ [3]

13. A factory produces widgets. The number of widgets produced is directly proportional to the number of machines and inversely proportional to the number of hours each machine operates per day. When $10$ machines operate for $8$ hours per day, $400$ widgets are produced. (a) Find an equation connecting the number of widgets $W$, the number of machines $m$, and the number of hours $h$.
Answer: ___________________________ [2]

(b) How many widgets are produced when $15$ machines operate for $6$ hours per day?
Answer: ___________________________ [1]

14. A sum of $4800 is divided among three children in the ratio of their ages. Their ages are$10$years,$12$years, and$14$years. How much does the youngest child receive? **Answer:**$___________________________ [3]

15. The time taken to fill a tank is inversely proportional to the number of taps used. $4$ taps take $3$ hours to fill the tank. (a) Find an equation connecting the time $T$ hours and the number of taps $n$.
Answer: ___________________________ [1]

(b) How long will it take $6$ taps to fill the tank?
Answer: ___________________________ hours [1]

(c) How many taps are needed to fill the tank in $1.5$ hours?
Answer: ___________________________ [1]

16. A car uses $8$ litres of petrol to travel $100$ km. (a) Find the petrol consumption rate in km per litre.
Answer: ___________________________ km/l [1]

(b) The car travels at a constant speed. After travelling $150$ km, the driver refills the tank with $15$ litres of petrol. How many more kilometres can the car travel before the petrol runs out?
Answer: ___________________________ km [2]

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Section C: Problem Solving Questions (Questions 17–20, 4–5 marks each, Total 22 marks)

17. A rectangular field has length and breadth in the ratio $5 : 3$. The perimeter of the field is $320$ m. (a) Find the length and breadth of the field.
Answer: Length = __________ m, Breadth = __________ m [2]

(b) A path of uniform width $2$ m is constructed around the outside of the field. Find the area of the path.
Answer: ___________________________ m² [2]

18. Three friends, David, Ethan, and Fiona, share a sum of money. David receives $\frac{2}{5}$ of the total amount. Ethan receives $\frac{1}{3}$ of the remaining amount. Fiona receives the rest, which is $240. (a) Find the total sum of money. **Answer:**$___________________________ [2]

(b) Express the amounts received by David, Ethan, and Fiona as a ratio in its simplest form.
Answer: ___________________________ [2]

19. A paint mixture contains red, blue, and yellow paint in the ratio $7 : 4 : 3$ by volume. The mixture is made by mixing $2.8$ litres of red paint with some blue and yellow paint. (a) Find the total volume of the paint mixture.
Answer: ___________________________ litres [2]

(b) The mixture is poured into tins of capacity $500$ ml each. How many tins can be completely filled?
Answer: ___________________________ [2]

20. The cost $C$ of producing $x$ units of a product is given by $C = kx + \frac{m}{x}$, where $k$ and $m$ are constants. When $100$ units are produced, the cost is $500$. When $200$ units are produced, the cost is $800$. (a) Find the values of $k$ and $m$.
Answer: $k =$ __________, $m =$ __________ [3]

(b) Find the cost of producing $150$ units.
Answer: $___________________________ [1]

(c) Explain why the cost per unit decreases initially as production increases, but eventually increases.
Answer: _______________________________________________________________________________
_______________________________________________________________________________ [1]

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End of Quiz

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Secondary 2 Mathematics Quiz - Numbers Ratio Proportion (Answer Key)

Total Marks: 40

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Section A: Short Answer Questions (Questions 1–10, 2 marks each)

1. Express the ratio $48 : 72 : 96$ in its simplest form.

Answer: $2 : 3 : 4$ [2]

Working:

• Find HCF of 48, 72, 96: HCF = 24
• Divide each term by 24: $48 \div 24 = 2$, $72 \div 24 = 3$, $96 \div 24 = 4$
• Simplest form: $2 : 3 : 4$

Marking: 1 mark for correct HCF or partial simplification, 1 mark for final answer.

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2. A sum of money is divided between Ali, Bala, and Charlie in the ratio $3 : 5 : 7$. If Bala receives $120 more than Ali, find the total sum of money.

Answer: $900 [2]

Working:

• Difference in ratio units between Bala and Ali = $5 - 3 = 2$ units
• $2$ units = $120 \Rightarrow 1$ unit = $60
• Total units = $3 + 5 + 7 = 15$ units
• Total sum = $15 \times 60 = 900$

Marking: 1 mark for finding value of 1 unit, 1 mark for total sum.

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3. The scale of a map is $1 : 25\,000$. The distance between two towns on the map is $6.4$ cm. Find the actual distance between the two towns in kilometres.

Answer: $1.6$ km [2]

Working:

• Actual distance = $6.4 \times 25\,000 = 160\,000$ cm
• Convert to km: $160\,000 \div 100\,000 = 1.6$ km

Marking: 1 mark for correct multiplication, 1 mark for correct unit conversion to km.

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4. $y$ is inversely proportional to the square of $x$. When $x = 4$, $y = 9$. Find the value of $y$ when $x = 6$.

Answer: $4$ [2]

Working:

• $y = \frac{k}{x^2}$
• Substitute $x = 4$, $y = 9$: $9 = \frac{k}{16} \Rightarrow k = 144$
• Equation: $y = \frac{144}{x^2}$
• When $x = 6$: $y = \frac{144}{36} = 4$

Marking: 1 mark for finding $k = 144$, 1 mark for final answer $y = 4$.

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5. A car travels $180$ km using $15$ litres of petrol. How many litres of petrol are needed to travel $300$ km at the same rate?

Answer: $25$ litres [2]

Working:

• Rate = $180 \div 15 = 12$ km/litre
• Petrol needed = $300 \div 12 = 25$ litres
• Alternatively: $\frac{15}{180} = \frac{x}{300} \Rightarrow x = \frac{15 \times 300}{180} = 25$

Marking: 1 mark for finding rate or setting up proportion, 1 mark for final answer.

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6. The ratio of the number of boys to girls in a class is $4 : 5$. After $6$ boys join the class, the ratio becomes $5 : 5$. How many girls are in the class?

Answer: $30$ [2]

Working:

• Let boys = $4x$, girls = $5x$
• After 6 boys join: boys = $4x + 6$, girls = $5x$
• New ratio: $\frac{4x + 6}{5x} = \frac{5}{5} = 1$
• $4x + 6 = 5x \Rightarrow x = 6$
• Girls = $5x = 5 \times 6 = 30$

Marking: 1 mark for setting up equation, 1 mark for final answer.

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7. It takes $8$ workers $12$ days to complete a job. How many days will it take $6$ workers to complete the same job, assuming they work at the same rate?

Answer: $16$ days [2]

Working:

• Inverse proportion: workers $\times$ days = constant
• $8 \times 12 = 6 \times d \Rightarrow 96 = 6d \Rightarrow d = 16$
• Or: 1 worker takes $8 \times 12 = 96$ days; 6 workers take $96 \div 6 = 16$ days

Marking: 1 mark for correct method (inverse proportion), 1 mark for final answer.

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8. A recipe requires flour, sugar, and butter in the ratio $5 : 2 : 3$ by mass. If $400$ g of flour is used, find the total mass of the mixture.

Answer: $800$ g [2]

Working:

• Flour = $5$ units = $400$ g $\Rightarrow 1$ unit = $80$ g
• Total units = $5 + 2 + 3 = 10$ units
• Total mass = $10 \times 80 = 800$ g

Marking: 1 mark for finding value of 1 unit, 1 mark for total mass.

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9. The price of a watch increased from $250 to$300. Express the increase as a percentage of the original price.

Answer: $20\%$ [2]

Working:

• Increase = $300 - 250 = 50$
• Percentage increase = $\frac{50}{250} \times 100\% = 20\%$

Marking: 1 mark for finding increase, 1 mark for percentage calculation.

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10. $p$ is directly proportional to the cube root of $q$. When $q = 27$, $p = 12$. Find the value of $p$ when $q = 64$.

Answer: $16$ [2]

Working:

• $p = k\sqrt[3]{q}$
• When $q = 27$, $\sqrt[3]{27} = 3$: $12 = 3k \Rightarrow k = 4$
• Equation: $p = 4\sqrt[3]{q}$
• When $q = 64$, $\sqrt[3]{64} = 4$: $p = 4 \times 4 = 16$

Marking: 1 mark for finding $k = 4$, 1 mark for final answer $p = 16$.

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Section B: Structured Questions (Questions 11–16, 3 marks each)

11. A map has a scale of $1 : 50\,000$.

(a) The area of a lake on the map is $2.5$ cm². Calculate the actual area of the lake in km². Answer: $6.25$ km² [2]

Working:

• Area scale = $(1 : 50\,000)^2 = 1 : 2\,500\,000\,000$
• Actual area = $2.5 \times 2\,500\,000\,000 = 6\,250\,000\,000$ cm²
• Convert to km²: $1$ km² = $10^{10}$ cm²
• Actual area = $6\,250\,000\,000 \div 10^{10} = 0.625$ km²? Wait, let me recalculate.

Correction:

• $1$ cm on map = $50\,000$ cm = $0.5$ km
• $1$ cm² on map = $(0.5)^2 = 0.25$ km²
• Actual area = $2.5 \times 0.25 = 0.625$ km²

Wait, let me check again:

• $50\,000$ cm = $500$ m = $0.5$ km ✓
• $1$ cm² = $(0.5 \text{ km})^2 = 0.25 \text{ km}^2$ ✓
• $2.5 \text{ cm}^2 = 2.5 \times 0.25 = 0.625 \text{ km}^2$

Answer: $0.625$ km² [2]

(b) The actual length of a river is $15$ km. Find its length on the map in centimetres. Answer: $30$ cm [1]

Working:

• $15$ km = $1\,500\,000$ cm
• Map length = $1\,500\,000 \div 50\,000 = 30$ cm

Marking: (a) 1 mark for correct area scale or linear scale conversion, 1 mark for final answer in km². (b) 1 mark for correct answer.

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12. The ratio of $A$ to $B$ is $3 : 7$ and the ratio of $B$ to $C$ is $5 : 9$. Find the ratio $A : B : C$ in its simplest form.

Answer: $15 : 35 : 63$ [3]

Working:

• $A : B = 3 : 7 = 15 : 35$ (multiply by 5)
• $B : C = 5 : 9 = 35 : 63$ (multiply by 7)
• Make B the same: LCM of 7 and 5 is 35
• $A : B : C = 15 : 35 : 63$

Marking: 1 mark for finding LCM of B values (35), 1 mark for converting both ratios, 1 mark for final combined ratio.

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13. A factory produces widgets. The number of widgets produced is directly proportional to the number of machines and inversely proportional to the number of hours each machine operates per day. When $10$ machines operate for $8$ hours per day, $400$ widgets are produced.

(a) Find an equation connecting the number of widgets $W$, the number of machines $m$, and the number of hours $h$. Answer: $W = \frac{320m}{h}$ [2]

Working:

• $W \propto \frac{m}{h} \Rightarrow W = k\frac{m}{h}$
• Substitute $W = 400$, $m = 10$, $h = 8$: $400 = k\frac{10}{8} = \frac{10k}{8} = \frac{5k}{4}$
• $k = 400 \times \frac{4}{5} = 320$
• Equation: $W = \frac{320m}{h}$

(b) How many widgets are produced when $15$ machines operate for $6$ hours per day? Answer: $800$ [1]

Working:

• $W = \frac{320 \times 15}{6} = \frac{4800}{6} = 800$

Marking: (a) 1 mark for correct proportionality statement, 1 mark for finding $k$ and final equation. (b) 1 mark for correct substitution and answer.

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14. A sum of $4800 is divided among three children in the ratio of their ages. Their ages are$10$years,$12$years, and$14$ years. How much does the youngest child receive?

Answer: $1200$ [3]

Working:

• Ratio of ages = $10 : 12 : 14 = 5 : 6 : 7$ (divide by 2)
• Total units = $5 + 6 + 7 = 18$ units
• $18$ units = $4800 \Rightarrow 1$ unit = $4800 \div 18 = 266.67$? Wait.
• $4800 \div 18 = 266\frac{2}{3}$ — this gives non-integer. Let me check: $4800 \div 18 = 266.666...$
• Youngest (5 units) = $5 \times \frac{4800}{18} = \frac{24000}{18} = 1333.33$? That doesn't seem right for a school question.

Let me recalculate: $10 + 12 + 14 = 36$ total years.

• Youngest = $\frac{10}{36} \times 4800 = \frac{48000}{36} = 1333.33$ — still not clean.

Wait, maybe the question expects the ratio 10:12:14 simplified to 5:6:7, and total = 4800.

• $5+6+7 = 18$ parts
• $4800 \div 18 = 266\frac{2}{3}$ per part
• Youngest = $5 \times 266\frac{2}{3} = 1333\frac{1}{3}$

Hmm, this gives a recurring decimal. Let me adjust the numbers mentally — but as an answer key, I must work with the question as written.

• Actually, $4800 \div 18 = 266.666... = \frac{800}{3}$
• Youngest = $5 \times \frac{800}{3} = \frac{4000}{3} = 1333.33$

But wait — maybe I should use the unsimplified ratio 10:12:14?

• Total = 36 parts
• $4800 \div 36 = 133.333...$ per part
• Youngest = $10 \times 133.333... = 1333.33$ — same.

The answer is $1333.33$ or $1333\frac{1}{3}$. But this is unusual for a school question. Let me check if 4800 is divisible by 18... 18 × 266 = 4788, 18 × 267 = 4806. No.

Perhaps the question had a different total? But I must answer based on the question given. Answer: $1333.33$ (or $1333\frac{1}{3}$) [3]

Working:

• Ratio of ages = $10 : 12 : 14 = 5 : 6 : 7$
• Total parts = $5 + 6 + 7 = 18$
• Value of 1 part = $4800 \div 18 = \frac{800}{3} = 266\frac{2}{3}$
• Youngest child (5 parts) = $5 \times \frac{800}{3} = \frac{4000}{3} = 1333\frac{1}{3} \approx 1333.33$

Marking: 1 mark for correct ratio, 1 mark for total parts, 1 mark for final answer.

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15. The time taken to fill a tank is inversely proportional to the number of taps used. $4$ taps take $3$ hours to fill the tank.

(a) Find an equation connecting the time $T$ hours and the number of taps $n$. Answer: $T = \frac{12}{n}$ [1]

Working:

• $T \propto \frac{1}{n} \Rightarrow T = \frac{k}{n}$
• $3 = \frac{k}{4} \Rightarrow k = 12$
• $T = \frac{12}{n}$

(b) How long will it take $6$ taps to fill the tank? Answer: $2$ hours [1]

Working:

• $T = \frac{12}{6} = 2$ hours

(c) How many taps are needed to fill the tank in $1.5$ hours? Answer: $8$ [1]

Working:

• $1.5 = \frac{12}{n} \Rightarrow n = \frac{12}{1.5} = 8$

Marking: (a) 1 mark for equation. (b) 1 mark for answer. (c) 1 mark for answer.

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16. A car uses $8$ litres of petrol to travel $100$ km.

(a) Find the petrol consumption rate in km per litre. Answer: $12.5$ km/l [1]

Working:

• Rate = $100 \div 8 = 12.5$ km/l

(b) The car travels at a constant speed. After travelling $150$ km, the driver refills the tank with $15$ litres of petrol. How many more kilometres can the car travel before the petrol runs out? Answer: $187.5$ km [2]

Working:

• Petrol used for 150 km = $150 \div 12.5 = 12$ litres
• Petrol added = 15 litres
• Distance possible with 15 litres = $15 \times 12.5 = 187.5$ km

Marking: (a) 1 mark for rate. (b) 1 mark for finding petrol used or direct calculation, 1 mark for final answer.

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Section C: Problem Solving Questions (Questions 17–20)

17. A rectangular field has length and breadth in the ratio $5 : 3$. The perimeter of the field is $320$ m.

(a) Find the length and breadth of the field. Answer: Length = $100$ m, Breadth = $60$ m [2]

Working:

• Let length = $5x$, breadth = $3x$
• Perimeter = $2(5x + 3x) = 16x = 320$
• $x = 20$
• Length = $5 \times 20 = 100$ m, Breadth = $3 \times 20 = 60$ m

(b) A path of uniform width $2$ m is constructed around the outside of the field. Find the area of the path. Answer: $656$ m² [2]

Working:

• Outer length = $100 + 2(2) = 104$ m
• Outer breadth = $60 + 2(2) = 64$ m
• Outer area = $104 \times 64 = 6656$ m²
• Inner area = $100 \times 60 = 6000$ m²
• Path area = $6656 - 6000 = 656$ m²

Marking: (a) 1 mark for setting up equation, 1 mark for both dimensions. (b) 1 mark for outer dimensions, 1 mark for area calculation.

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18. Three friends, David, Ethan, and Fiona, share a sum of money. David receives $\frac{2}{5}$ of the total amount. Ethan receives $\frac{1}{3}$ of the remaining amount. Fiona receives the rest, which is $240.

(a) Find the total sum of money. Answer: $600$ [2]

Working:

• Let total = $T$
• David = $\frac{2}{5}T$
• Remaining = $T - \frac{2}{5}T = \frac{3}{5}T$
• Ethan = $\frac{1}{3} \times \frac{3}{5}T = \frac{1}{5}T$
• Fiona = Remaining after Ethan = $\frac{3}{5}T - \frac{1}{5}T = \frac{2}{5}T$
• Given Fiona = $240: \frac{2}{5}T = 240 \Rightarrow T = 240 \times \frac{5}{2} = 600$

(b) Express the amounts received by David, Ethan, and Fiona as a ratio in its simplest form. Answer: $2 : 1 : 2$ [2]

Working:

• David = $\frac{2}{5}T = \frac{2}{5} \times 600 = 240$
• Ethan = $\frac{1}{5}T = \frac{1}{5} \times 600 = 120$
• Fiona = $240$
• Ratio = $240 : 120 : 240 = 2 : 1 : 2$

Marking: (a) 1 mark for correct expression of remaining/Ethan/Fiona, 1 mark for total. (b) 1 mark for individual amounts, 1 mark for simplified ratio.

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19. A paint mixture contains red, blue, and yellow paint in the ratio $7 : 4 : 3$ by volume. The mixture is made by mixing $2.8$ litres of red paint with some blue and yellow paint.

(a) Find the total volume of the paint mixture. Answer: $5.6$ litres [2]

Working:

• Red = $7$ units = $2.8$ litres $\Rightarrow 1$ unit = $0.4$ litres
• Total units = $7 + 4 + 3 = 14$ units
• Total volume = $14 \times 0.4 = 5.6$ litres

(b) The mixture is poured into tins of capacity $500$ ml each. How many tins can be completely filled? Answer: $11$ [2]

Working:

• Total volume = $5.6$ litres = $5600$ ml
• Number of tins = $5600 \div 500 = 11.2$
• Completely filled tins = $11$

Marking: (a) 1 mark for value of 1 unit, 1 mark for total volume. (b) 1 mark for unit conversion, 1 mark for integer answer (rounding down).

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20. The cost $C$ of producing $x$ units of a product is given by $C = kx + \frac{m}{x}$, where $k$ and $m$ are constants. When $100$ units are produced, the cost is $500$. When $200$ units are produced, the cost is $800$.

(a) Find the values of $k$ and $m$. Answer: $k = 3.5$, $m = 15000$ [3]

Working:

• Equation 1: $500 = 100k + \frac{m}{100} \Rightarrow 500 = 100k + 0.01m$
• Equation 2: $800 = 200k + \frac{m}{200} \Rightarrow 800 = 200k + 0.005m$
• Multiply Eq 1 by 100: $50000 = 10000k + m$
• Multiply Eq 2 by 200: $160000 = 40000k + m$
• Subtract: $110000 = 30000k \Rightarrow k = \frac{11}{3} \approx 3.667$? Wait.

Let me recalculate carefully:

• $500 = 100k + \frac{m}{100}$ ... (1)
• $800 = 200k + \frac{m}{200}$ ... (2)

Multiply (1) by 100: $50000 = 10000k + m$ ... (1a) Multiply (2) by 200: $160000 = 40000k + m$ ... (2a)

Subtract (1a) from (2a): $110000 = 30000k \Rightarrow k = \frac{11}{3} = 3\frac{2}{3}$

Substitute into (1a): $50000 = 10000(\frac{11}{3}) + m = \frac{110000}{3} + m$ $m = 50000 - \frac{110000}{3} = \frac{150000 - 110000}{3} = \frac{40000}{3} \approx 13333.33$

Hmm, these are not nice numbers. Let me check the question setup again. The question says: "When 100 units are produced, the cost is 500. When 200 units are produced, the cost is 800."

$C = kx + \frac{m}{x}$

$500 = 100k + \frac{m}{100}$ $800 = 200k + \frac{m}{200}$

Multiply first by 100: $50000 = 10000k + m$ Multiply second by 200: $160000 = 40000k + m$

Subtract: $110000 = 30000k \Rightarrow k = \frac{11}{3}$ $m = 50000 - 10000(\frac{11}{3}) = 50000 - \frac{110000}{3} = \frac{40000}{3}$

These are the correct mathematical answers. I'll present them as fractions.

Answer: $k = \frac{11}{3}$ (or $3\frac{2}{3}$), $m = \frac{40000}{3}$ (or $13333\frac{1}{3}$) [3]

(b) Find the cost of producing $150$ units. Answer: $\frac{11}{3} \times 150 + \frac{40000/3}{150} = 550 + \frac{40000}{450} = 550 + \frac{800}{9} = 550 + 88\frac{8}{9} = 638\frac{8}{9} \approx 638.89$ [1]

Working:

• $C = \frac{11}{3}(150) + \frac{40000/3}{150} = 550 + \frac{40000}{450} = 550 + \frac{800}{9} = \frac{4950 + 800}{9} = \frac{5750}{9} = 638\frac{8}{9}$

(c) Explain why the cost per unit decreases initially as production increases, but eventually increases. Answer: The cost per unit is $\frac{C}{x} = k + \frac{m}{x^2}$. The term $\frac{m}{x^2}$ decreases as $x$ increases (spreading fixed costs), but the $kx$ term represents variable costs that increase linearly. Initially, the decreasing $\frac{m}{x^2}$ dominates, lowering average cost. Eventually, the linear $kx$ term dominates, causing average cost to rise. [1]

Marking: (a) 1 mark for setting up two equations, 1 mark for solving simultaneously, 1 mark for correct $k$ and $m$. (b) 1 mark for correct substitution and answer. (c) 1 mark for correct explanation referencing the two components of the cost function.

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End of Answer Key