O Level Additional Mathematics Numbers Ratio Proportion Quiz

O-Level Additional Mathematics Quiz - Numbers Ratio Proportion

ANSWER KEY AND MARKING SCHEME

Total Marks: 50

---

Section A: Short Answer (12 marks)

1. Express (\frac{7}{12}) as a decimal. [1 mark]

Answer: (0.583) (or (0.58333...))

Marking: A1 for correct decimal (accept (0.583) or (0.58\dot{3}))

---

2. Simplify the ratio (0.24 : 0.6 : 1.08), giving your answer in its simplest integer form. [2 marks]

Working: Multiply each term by 100: (24 : 60 : 108)
Divide by common factor 12: (2 : 5 : 9)

Answer: (2 : 5 : 9)

Marking: M1 for multiplying by 100 or equivalent scaling; A1 for (2 : 5 : 9)

---

3. A map is drawn to a scale of (1 : 25,000). Two towns are 8.4 cm apart on the map. Find the actual distance between the towns in kilometres. [2 marks]

Working: Actual distance = (8.4 \times 25,000 = 210,000) cm
(210,000 \div 100,000 = 2.1) km

Answer: (2.1) km

Marking: M1 for (8.4 \times 25,000) or equivalent; A1 for (2.1) km

---

4. The price of a laptop is increased by 15% to $1840. Find the original price of the laptop. [2 marks]

Working: Let original price = (x)
(x \times 1.15 = 1840)
(x = 1840 \div 1.15 = 1600)

Answer: $1600

Marking: M1 for (1840 \div 1.15) or equivalent; A1 for $1600

---

5. A sum of money is divided among Ali, Bala, and Chen in the ratio (3 : 5 : 7). If Chen receives $56 more than Ali, find the total sum of money. [3 marks]

Working: Let shares be (3k), (5k), (7k)
Chen (-) Ali (= 7k - 3k = 4k = 56)
(k = 14)
Total (= 3k + 5k + 7k = 15k = 15 \times 14 = 210)

Answer: $210

Marking: M1 for setting up (7k - 3k = 56); M1 for solving (k = 14); A1 for $210

---

Section B: Calculation and Problem Solving (26 marks)

6. The population of a town decreased by 8% in 2023 and then increased by 12% in 2024. At the end of 2024, the population was 51,520. Find the population at the start of 2023. [2 marks]

Working: Let start population = (P)
After 2023: (P \times 0.92)
After 2024: (P \times 0.92 \times 1.12 = 51,520)
(P \times 1.0304 = 51,520)
(P = 51,520 \div 1.0304 = 50,000)

Answer: (50,000)

Marking: M1 for (51,520 \div (0.92 \times 1.12)) or equivalent; A1 for (50,000)

---

7. (a) Expression for time for first journey. [1 mark]

Answer: (\frac{240}{x}) hours

Marking: A1 for (\frac{240}{x})

(b) Expression for time for return journey. [1 mark]

Answer: (\frac{240}{x + 10}) hours

Marking: A1 for (\frac{240}{x + 10})

(c) Form equation and show simplification. [3 marks]

Working: (\frac{240}{x} - \frac{240}{x + 10} = \frac{24}{60} = \frac{2}{5})
Multiply by (5x(x + 10)):
(5 \times 240(x + 10) - 5 \times 240x = 2x(x + 10))
(1200x + 12,000 - 1200x = 2x^2 + 20x)
(12,000 = 2x^2 + 20x)
(x^2 + 10x - 6000 = 0)

Marking: M1 for correct equation with 24 min = (\frac{2}{5}) hour; M1 for correct algebraic manipulation; A1 for showing simplification to given equation

(d) Solve the equation. [2 marks]

Working: (x^2 + 10x - 6000 = 0)
((x + 80)(x - 70) = 0) (or use quadratic formula)
(x = -80) or (x = 70)
Since speed > 0, (x = 70)

Answer: (70) km/h

Marking: M1 for factorisation or quadratic formula; A1 for (70) (reject (-80))

---

8. (a) Find equation connecting (y) and (x). [2 marks]

Working: (y \propto x^2 \implies y = kx^2)
When (x = 4), (y = 80): (80 = k(4^2) = 16k)
(k = 5)
(y = 5x^2)

Answer: (y = 5x^2)

Marking: M1 for (y = kx^2) and substituting; A1 for (y = 5x^2)

(b) Find (y) when (x = 7). [1 mark]

Working: (y = 5(7^2) = 5 \times 49 = 245)

Answer: (245)

Marking: A1 for 245

(c) Find (x) when (y = 20). [2 marks]

Working: (20 = 5x^2)
(x^2 = 4)
(x = 2) or (x = -2)

Answer: (x = 2) or (x = -2)

Marking: M1 for (x^2 = 4); A1 for both values (accept (x = \pm 2))

---

9. (a) Find equation connecting (p) and (q). [2 marks]

Working: (p \propto \frac{1}{\sqrt[3]{q}} \implies p = \frac{k}{\sqrt[3]{q}})
When (q = 8), (p = 15): (15 = \frac{k}{\sqrt[3]{8}} = \frac{k}{2})
(k = 30)
(p = \frac{30}{\sqrt[3]{q}})

Answer: (p = \frac{30}{\sqrt[3]{q}})

Marking: M1 for (p = \frac{k}{\sqrt[3]{q}}) and substituting; A1 for (p = \frac{30}{\sqrt[3]{q}})

(b) Find (p) when (q = 125). [1 mark]

Working: (p = \frac{30}{\sqrt[3]{125}} = \frac{30}{5} = 6)

Answer: (6)

Marking: A1 for 6

(c) Find percentage change in (p) when (q) increases by 72.8%. [3 marks]

Working: New (q = 1.728q)
New (p = \frac{30}{\sqrt[3]{1.728q}} = \frac{30}{\sqrt[3]{1.728} \times \sqrt[3]{q}} = \frac{30}{1.2 \times \sqrt[3]{q}} = \frac{p}{1.2})
Percentage change (= \frac{\frac{p}{1.2} - p}{p} \times 100% = \left(\frac{1}{1.2} - 1\right) \times 100%)
(= \left(\frac{5}{6} - 1\right) \times 100% = -\frac{1}{6} \times 100% = -16\frac{2}{3}%)
Decrease of (16.7%) (3 s.f.)

Answer: (16.7%) decrease (or (-16.7%))

Marking: M1 for (\sqrt[3]{1.728} = 1.2); M1 for correct ratio of new to old (p); A1 for (16.7%) decrease (accept (16\frac{2}{3}%))

---

10. (a) Form equation and find (k). [3 marks]

Working: Perimeter of square (= 4 \times 3k = 12k)
Perimeter of triangle (= 3 \times 4k = 12k)
Total wire (= 12k + 12k = 24k = 120)
(k = 5)

Answer: (k = 5)

Marking: M1 for perimeters (12k) and (12k); M1 for equation (24k = 120); A1 for (k = 5)

(b) Find length of each piece. [2 marks]

Working: Square wire (= 12k = 12 \times 5 = 60) cm
Triangle wire (= 12k = 60) cm

Answer: Square wire: (60) cm; Triangle wire: (60) cm

Marking: A1 for each correct length

---

Section C: Reasoning and Communication (12 marks)

11. (a) Components per machine per hour. [1 mark]

Working: 5 machines in 8 hours: (2400) components
1 machine in 8 hours: (2400 \div 5 = 480)
1 machine in 1 hour: (480 \div 8 = 60)

Answer: (60)

Marking: A1 for 60

(b) Time for 8 machines to produce 4800 components. [2 marks]

Working: Rate of 8 machines (= 8 \times 60 = 480) components per hour
Time (= 4800 \div 480 = 10) hours

Answer: (10) hours

Marking: M1 for rate of 8 machines or equivalent; A1 for 10 hours

---

12. (a) Find equation for cost per book. [4 marks]

Working: Let (C = a + \frac{b}{n}) where (a) is constant part and (\frac{b}{n}) is inverse variation part.
When (n = 500), (C = 8.40): (8.40 = a + \frac{b}{500}) ... (1)
When (n = 800), (C = 6.50): (6.50 = a + \frac{b}{800}) ... (2)
(1) (-) (2): (1.90 = b\left(\frac{1}{500} - \frac{1}{800}\right) = b\left(\frac{8 - 5}{4000}\right) = \frac{3b}{4000})
(b = 1.90 \times \frac{4000}{3} = \frac{7600}{3} = 2533\frac{1}{3})
From (1): (a = 8.40 - \frac{7600/3}{500} = 8.40 - \frac{7600}{1500} = 8.40 - 5.0666... = 3.333... = \frac{10}{3})
(C = \frac{10}{3} + \frac{7600}{3n} = \frac{10}{3} + \frac{7600}{3n})

Answer: (C = \frac{10}{3} + \frac{7600}{3n}) (or (C = 3.33 + \frac{2533.33}{n}))

Marking: M1 for (C = a + \frac{b}{n}); M1 for forming two equations; M1 for solving for (a) and (b); A1 for correct equation

(b) Cost per book when (n = 2000). [1 mark]

Working: (C = \frac{10}{3} + \frac{7600}{3 \times 2000} = \frac{10}{3} + \frac{7600}{6000} = \frac{10}{3} + \frac{19}{15} = \frac{50}{15} + \frac{19}{15} = \frac{69}{15} = 4.60)

Answer: $4.60

Marking: A1 for $4.60

(c) Explain why cost can never be less than a certain value. [2 marks]

Answer: As (n) increases, (\frac{7600}{3n}) approaches 0, so (C) approaches (\frac{10}{3} = 3.33). The cost per book can never be less than $3.33 (the constant part of the cost).

Marking: A1 for identifying that the inverse variation term tends to 0; A1 for stating the minimum value is $3.33 (or (\frac{10}{3}))

---

13. (a) Express (V) in terms of (T) and (P). [3 marks]

Working: (V \propto \frac{T}{P} \implies V = k\frac{T}{P})
When (T = 300), (P = 1.0 \times 10^5), (V = 0.024):
(0.024 = k\frac{300}{1.0 \times 10^5})
(k = 0.024 \times \frac{1.0 \times 10^5}{300} = 0.024 \times \frac{1000}{3} = 8)
(V = \frac{8T}{P})

Answer: (V = \frac{8T}{P})

Marking: M1 for (V = k\frac{T}{P}); M1 for substituting and solving for (k); A1 for (V = \frac{8T}{P})

(b) Find volume when (T = 360) and (P = 1.2 \times 10^5). [1 mark]

Working: (V = \frac{8 \times 360}{1.2 \times 10^5} = \frac{2880}{120,000} = 0.024) m³

Answer: (0.024) m³

Marking: A1 for 0.024

(c) Describe what happens to pressure when volume is halved at constant temperature. [1 mark]

Answer: Since (V \propto \frac{1}{P}) at constant (T), halving (V) doubles (P). The pressure is doubled.

Marking: A1 for stating pressure doubles with correct justification (inverse proportion)

---

14. Given that (a : b = 3 : 4) and (b : c = 2 : 5), find the ratio (a : b : c) in its simplest form. [2 marks]

Working: (a : b = 3 : 4 = 3 \times 2 : 4 \times 2 = 6 : 8)
(b : c = 2 : 5 = 2 \times 4 : 5 \times 4 = 8 : 20)
Thus (a : b : c = 6 : 8 : 20 = 3 : 4 : 10)

Answer: (3 : 4 : 10)

Marking: M1 for making (b) the same in both ratios; A1 for (3 : 4 : 10)

---

15. A sum of $1200 is invested at a simple interest rate of 3.5$1410. [2 marks]

Working: Interest = (1410 - 1200 = 210)
Simple interest formula: (I = PRT)
(210 = 1200 \times 0.035 \times T)
(210 = 42T)
(T = 5)

Answer: (5) years

Marking: M1 for (210 = 1200 \times 0.035 \times T); A1 for 5

---

Section D: Advanced Application (0 marks)

16. The scale of a model car is (1 : 18). The model has a surface area of 450 cm². Find the surface area of the actual car in m². [2 marks]

Working: Area scale factor = (18^2 = 324)
Actual area = (450 \times 324 = 145,800) cm²
(145,800 \div 10,000 = 14.58) m²

Answer: (14.58) m²

Marking: M1 for multiplying by (18^2) or equivalent; A1 for 14.58

---

17. Three partners invest in a business in the ratio (2 : 3 : 5). The total profit is $15,000. If the profit is divided in the ratio of their investments, find the difference between the largest and smallest shares. [2 marks]

Working: Total parts = (2 + 3 + 5 = 10)
One part = (15,000 \div 10 = 1500)
Largest share = (5 \times 1500 = 7500)
Smallest share = (2 \times 1500 = 3000)
Difference = (7500 - 3000 = 4500)

Answer: $4500

Marking: M1 for finding value of one part; A1 for 4500

---

18. A solution is made by mixing acid and water in the ratio (3 : 7). How much pure acid must be added to 2 litres of the solution to change the ratio to (2 : 3)? [3 marks]

Working: In 2 L of solution, acid = (\frac{3}{10} \times 2 = 0.6) L, water = (1.4) L
Let (x) L of pure acid be added.
New acid = (0.6 + x), water remains (1.4)
New ratio: (\frac{0.6 + x}{1.4} = \frac{2}{3})
(3(0.6 + x) = 2 \times 1.4)
(1.8 + 3x = 2.8)
(3x = 1.0)
(x = \frac{1}{3}) L

Answer: (\frac{1}{3}) litre (or 0.333 L)

Marking: M1 for finding initial amounts; M1 for setting up proportion; A1 for (\frac{1}{3})

---

19. The time taken to paint a house varies inversely as the number of painters. If 6 painters take 10 days to paint the house, how many painters are needed to paint the house in 4 days? [2 marks]

Working: (T \propto \frac{1}{N} \implies T \times N = k)
(k = 6 \times 10 = 60)
For (T = 4): (4 \times N = 60 \implies N = 15)

Answer: (15) painters

Marking: M1 for (k = 60) or equivalent; A1 for 15

---

20. The cost of a banquet is partly constant and partly varies directly as the number of guests. For 50 guests, the cost is $2,500. For 80 guests, the cost is$3,400. Find the cost for 120 guests. [3 marks]

Working: Let (C = a + bn)
(2500 = a + 50b) ... (1)
(3400 = a + 80b) ... (2)
(2) (-) (1): (900 = 30b \implies b = 30)
From (1): (a = 2500 - 50 \times 30 = 1000)
(C = 1000 + 30n)
For (n = 120): (C = 1000 + 30 \times 120 = 1000 + 3600 = 4600)

Answer: $4600

Marking: M1 for setting up linear equations; M1 for solving for (a) and (b); A1 for 4600

---

END OF ANSWER KEY