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O Level Additional Mathematics Numbers Ratio Proportion Quiz
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O-Level Additional Mathematics Quiz - Numbers Ratio Proportion
Answer Key and Marking Scheme
Total Marks: 50
Section A: Direct Proportion and Variation (Questions 1–5)
1. ( y \propto x^2 \implies y = kx^2 ) (a) ( 45 = k(3^2) = 9k \implies k = 5 ) [M1] ( y = 5x^2 ) [A1] (b) When ( x = 5 ): ( y = 5(25) = 125 ) [A1]
2. ( d \propto t^2 \implies d = kt^2 ) (a) ( 78.4 = k(4^2) = 16k \implies k = 4.9 ) [M1] ( d = 4.9t^2 ) [A1] (b) When ( t = 7 ): ( d = 4.9(49) = 240.1 ) m [A1]
3. ( C = a + bn ), where ( a ) is the constant cost and ( b ) is the cost per page. (a) ( 12.50 = a + 200b ) ... (1) [M1] ( 18.50 = a + 350b ) ... (2) (2) – (1): ( 6.00 = 150b \implies b = 0.04 ) [M1] Substitute into (1): ( 12.50 = a + 200(0.04) = a + 8 \implies a = 4.50 ) ( C = 4.50 + 0.04n ) [A1] (b) When ( n = 500 ): ( C = 4.50 + 0.04(500) = 4.50 + 20 = $24.50 ) [A1]
4. ( V \propto \frac{T}{P} \implies V = \frac{kT}{P} ) (a) ( 45 = \frac{k(300)}{20} \implies 45 = 15k \implies k = 3 ) [M1] ( V = \frac{3T}{P} ) [A1] (b) When ( T = 360 ), ( P = 25 ): ( V = \frac{3(360)}{25} = \frac{1080}{25} = 43.2 ) [A1]
5. ( F \propto \frac{1}{d^2} \implies F = \frac{k}{d^2} ) (a) ( 15 = \frac{k}{8^2} = \frac{k}{64} \implies k = 960 ) [M1] ( F = \frac{960}{d^2} ) [A1] (b) When ( d = 4 ): ( F = \frac{960}{16} = 60 ) units [A1]
Section B: Ratio Problems (Questions 6–10)
6. Let shares be ( 3x, 5x, 7x ). (a) Chen – Ali = ( 7x - 3x = 4x = 84 \implies x = 21 ) [M1] Total = ( 3x + 5x + 7x = 15x = 15(21) = $315 ) [A1] (b) Bala receives ( 5x = 5(21) = $105 ) [A1]
7. Let original number of boys = ( 5x ), girls = ( 4x ). After changes: boys = ( 5x + 30 ), girls = ( 4x - 20 ). [M1] New ratio: ( \frac{5x + 30}{4x - 20} = \frac{7}{5} ) [M1] ( 5(5x + 30) = 7(4x - 20) ) ( 25x + 150 = 28x - 140 ) ( 290 = 3x \implies x = \frac{290}{3} ) Original students = ( 9x = 9 \times \frac{290}{3} = 870 ) [A1]
8. Let ( p = 2k, q = 3k, r = 5k ). ( p^2 + q^2 + r^2 = (2k)^2 + (3k)^2 + (5k)^2 = 4k^2 + 9k^2 + 25k^2 = 38k^2 ) [M1] ( 38k^2 = 1368 \implies k^2 = 36 \implies k = 6 ) (positive) [M1] ( p = 12, q = 18, r = 30 ) [A1]
9. Wire pieces: ( \frac{2}{10} \times 120 = 24 ) cm, ( \frac{3}{10} \times 120 = 36 ) cm, ( \frac{5}{10} \times 120 = 60 ) cm. [M1] Side of each square: ( 24 \div 4 = 6 ) cm, ( 36 \div 4 = 9 ) cm, ( 60 \div 4 = 15 ) cm. [M1] Areas: ( 6^2 = 36 ), ( 9^2 = 81 ), ( 15^2 = 225 ). Ratio of areas = ( 36 : 81 : 225 = 4 : 9 : 25 ) (dividing by 9). [A1]
10. Let length = ( 5x ), width = ( 3x ). (a) Perimeter = ( 2(5x + 3x) = 16x = 96 \implies x = 6 ) [M1] Length = ( 30 ) cm, width = ( 18 ) cm. [A1] (b) Area = ( 30 \times 18 = 540 ) cm(^2) [A1]
Section C: Rates and Proportional Reasoning (Questions 11–15)
11. In 1 hour: Pipe A fills ( \frac{1}{6} ), Pipe B fills ( \frac{1}{8} ), Pipe C empties ( \frac{1}{12} ). [M1] Net fill in 1 hour = ( \frac{1}{6} + \frac{1}{8} - \frac{1}{12} = \frac{4}{24} + \frac{3}{24} - \frac{2}{24} = \frac{5}{24} ) [M1] Time to fill = ( \frac{24}{5} = 4.8 ) hours (or 4 hours 48 minutes). [A1]
12. Total work = ( 24 \times 45 \times 8 = 8640 ) worker-hours. [M1] Work done in first 15 days = ( 24 \times 15 \times 8 = 2880 ) worker-hours. Remaining work = ( 8640 - 2880 = 5760 ) worker-hours. Remaining workers = ( 24 - 8 = 16 ). [M1] Days needed = ( \frac{5760}{16 \times 8} = \frac{5760}{128} = 45 ) days. Total days = ( 15 + 45 = 60 ) days. Extra days = ( 60 - 45 = 15 ) days. [A1]
13. (a) Actual distance = ( 8.4 \times 50,000 = 420,000 ) cm = ( 4.2 ) km. [A1] (b) Area scale factor = ( (50,000)^2 = 2.5 \times 10^9 ). [M1] Map area = ( \frac{12.5 \times 10^{10}}{2.5 \times 10^9} = 5 ) cm(^2). [A1] (Or: ( 12.5 ) km(^2) = ( 12.5 \times 10^{10} ) cm(^2); map area = ( \frac{12.5 \times 10^{10}}{2.5 \times 10^9} = 5 ) cm(^2).)
14. Let distance = ( d ) km. Time from X to Y = ( \frac{d}{60} ) h; time from Y to X = ( \frac{d}{40} ) h. [M1] ( \frac{d}{60} + \frac{d}{40} = 5 ) [M1] Multiply by 120: ( 2d + 3d = 600 \implies 5d = 600 \implies d = 120 ) km. [A1]
15. Rate = ( \frac{480}{5} = 96 ) components per hour. (a) In 7 hours: ( 96 \times 7 = 672 ) components. [A1] (b) Time for 720 components: ( \frac{720}{96} = 7.5 ) hours. [A1]
Section D: Compound Variation and Advanced Problems (Questions 16–20)
16. ( D \propto \frac{W}{nh} \implies D = \frac{kW}{nh} ) (a) ( 15 = \frac{k(240)}{8 \times 10} = \frac{240k}{80} = 3k \implies k = 5 ) [M1] ( D = \frac{5W}{nh} ) [A1] (b) ( D = \frac{5(360)}{12 \times 8} = \frac{1800}{96} = 18.75 ) days. [M1, A1]
17. ( R \propto \frac{L}{d^2} \implies R = \frac{kL}{d^2} ) (a) ( 8 = \frac{k(50)}{2^2} = \frac{50k}{4} \implies 8 = 12.5k \implies k = 0.64 ) [M1] ( R = \frac{0.64L}{d^2} ) [A1] (b) ( R = \frac{0.64(80)}{4^2} = \frac{51.2}{16} = 3.2 ) ohms. [M1, A1]
18. ( I \propto \frac{S}{d^2} \implies I = \frac{kS}{d^2} ) (a) ( 8 = \frac{k(200)}{5^2} = \frac{200k}{25} = 8k \implies k = 1 ) [M1] ( I = \frac{S}{d^2} ) [A1] (b) ( I = \frac{450}{3^2} = \frac{450}{9} = 50 ) units. [A1]
19. ( T \propto \sqrt{L} \implies T = k\sqrt{L} ) (a) ( 1.6 = k\sqrt{64} = 8k \implies k = 0.2 ) [M1] ( T = 0.2\sqrt{L} ) [A1] (b) ( 2.4 = 0.2\sqrt{L} \implies \sqrt{L} = 12 \implies L = 144 ) cm. [A1]
20. ( C = a + \frac{b}{n} ), where ( a ) is the constant cost. (a) ( 850 = a + \frac{b}{200} ) ... (1) [M1] ( 580 = a + \frac{b}{500} ) ... (2) (1) – (2): ( 270 = b\left(\frac{1}{200} - \frac{1}{500}\right) = b\left(\frac{5 - 2}{1000}\right) = \frac{3b}{1000} ) [M1] ( b = 270 \times \frac{1000}{3} = 90,000 ) Substitute into (1): ( 850 = a + \frac{90,000}{200} = a + 450 \implies a = 400 ) ( C = 400 + \frac{90,000}{n} ) [A1] (b) When ( n = 1000 ): ( C = 400 + \frac{90,000}{1000} = 400 + 90 = $490 ) [A1]
END OF ANSWER KEY