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Secondary 3 Elementary Mathematics Numbers Ratio Proportion Quiz
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Secondary 3 Elementary Mathematics Quiz - Numbers Ratio Proportion
Answer Key and Marking Scheme
Total Marks: 50
Section A: Indices and Standard Form (10 marks)
1. Simplify (\frac{a^7 \times a^{-3}}{a^2}) [2 marks]
Answer: (a^2)
Working:
(\frac{a^7 \times a^{-3}}{a^2} = \frac{a^{7+(-3)}}{a^2} = \frac{a^4}{a^2} = a^{4-2} = a^2)
Marking:
- M1: Correct application of index laws (adding indices in numerator, or equivalent)
- A1: Correct final answer (a^2)
2. Evaluate (\frac{4.5 \times 10^6 \times 2.0 \times 10^{-3}}{3.0 \times 10^2}) [2 marks]
Answer: (3.0 \times 10^1) or (3.0 \times 10)
Working:
(\frac{4.5 \times 2.0}{3.0} \times 10^{6+(-3)-2} = 3.0 \times 10^1)
Marking:
- M1: Correct separation of coefficients and powers of 10
- A1: Correct answer in standard form
3. Simplify (\left(\frac{27x^3}{8y^6}\right)^{\frac{2}{3}}) [2 marks]
Answer: (\frac{9x^2}{4y^4})
Working:
(\left(\frac{27x^3}{8y^6}\right)^{\frac{2}{3}} = \frac{27^{\frac{2}{3}} \cdot (x^3)^{\frac{2}{3}}}{8^{\frac{2}{3}} \cdot (y^6)^{\frac{2}{3}}} = \frac{9x^2}{4y^4})
(Since (27^{\frac{2}{3}} = (27^{\frac{1}{3}})^2 = 3^2 = 9) and (8^{\frac{2}{3}} = (8^{\frac{1}{3}})^2 = 2^2 = 4))
Marking:
- M1: Correct application of power to numerator and denominator
- A1: Correct simplified answer with positive indices
4. Express (0.0000528) in standard form. [1 mark]
Answer: (5.28 \times 10^{-5})
Marking:
- A1: Correct standard form (coefficient between 1 and 10, correct exponent)
5. Given (p = 3.6 \times 10^5) and (q = 1.2 \times 10^{-2}), find (\frac{p}{q}) in standard form. [3 marks]
Answer: (3.0 \times 10^7)
Working:
(\frac{p}{q} = \frac{3.6 \times 10^5}{1.2 \times 10^{-2}} = \frac{3.6}{1.2} \times 10^{5-(-2)} = 3.0 \times 10^7)
Marking:
- M1: Correct division of coefficients
- M1: Correct subtraction of exponents ((5 - (-2) = 7))
- A1: Correct answer in standard form
Section B: Ratio and Proportion (12 marks)
6. Ratio boys : girls = (5 : 7). 420 girls. Find total students. [2 marks]
Answer: 720
Working:
7 parts = 420 girls → 1 part = 60
Total parts = 5 + 7 = 12
Total students = 12 × 60 = 720
Marking:
- M1: Finding value of one part
- A1: Correct total
7. Ratio Ali : Ben : Chen = (3 : 5 : 7). Ben receives $45 more than Ali. Find total sum. [3 marks]
Answer: $337.50
Working:
Difference between Ben and Ali = 5 − 3 = 2 parts
2 parts = 22.50
Total parts = 3 + 5 + 7 = 15
Total sum = 15 × 337.50
Marking:
- M1: Identifying difference in parts
- M1: Finding value of one part
- A1: Correct total sum
8. Map scale (1 : 25,000). Distance on map = 8.4 cm. Find actual distance in km. [2 marks]
Answer: 2.1 km
Working:
Actual distance = 8.4 × 25 000 = 210 000 cm
= 2100 m = 2.1 km
Marking:
- M1: Correct multiplication and conversion to metres
- A1: Correct answer in kilometres
9. Side ratio (4 : 5 : 6). Perimeter = 45 cm. Find longest side. [2 marks]
Answer: 18 cm
Working:
Total parts = 4 + 5 + 6 = 15
1 part = 45 ÷ 15 = 3 cm
Longest side = 6 × 3 = 18 cm
Marking:
- M1: Finding value of one part
- A1: Correct longest side
10. Recipe for 8 people: 300 g flour. Flour for 12 people? [3 marks]
Answer: 450 g
Working:
Flour per person = 300 ÷ 8 = 37.5 g
Flour for 12 people = 37.5 × 12 = 450 g
Alternatively: (\frac{12}{8} \times 300 = 1.5 \times 300 = 450) g
Marking:
- M1: Correct method (unitary or scaling factor)
- M1: Correct multiplication
- A1: Correct answer with units
Section C: Direct and Inverse Proportion (13 marks)
11. (y \propto x), (y = 24) when (x = 6). [3 marks]
(a) Equation: (y = 4x)
(b) When (x = 10): (y = 40)
Working:
(a) (y = kx) → (24 = k \times 6) → (k = 4) → (y = 4x)
(b) (y = 4 \times 10 = 40)
Marking:
- M1: Finding (k)
- A1: Correct equation
- B1: Correct value for part (b)
12. (y \propto \frac{1}{x}), (y = 8) when (x = 5). [3 marks]
(a) Equation: (y = \frac{40}{x})
(b) When (y = 20): (x = 2)
Working:
(a) (y = \frac{k}{x}) → (8 = \frac{k}{5}) → (k = 40) → (y = \frac{40}{x})
(b) (20 = \frac{40}{x}) → (x = \frac{40}{20} = 2)
Marking:
- M1: Finding (k)
- A1: Correct equation
- B1: Correct value for part (b)
13. (y \propto x^2), (y = 75) when (x = 5). Find (y) when (x = 8). [3 marks]
Answer: (y = 192)
Working:
(y = kx^2) → (75 = k \times 25) → (k = 3)
When (x = 8): (y = 3 \times 64 = 192)
Marking:
- M1: Finding constant (k)
- M1: Substituting (x = 8)
- A1: Correct answer
14. (t \propto \frac{1}{v}). (t = 3) when (v = 80). Find (t) when (v = 60). [2 marks]
Answer: 4 hours
Working:
(t = \frac{k}{v}) → (3 = \frac{k}{80}) → (k = 240)
When (v = 60): (t = \frac{240}{60} = 4) hours
Marking:
- M1: Finding constant or using proportion method
- A1: Correct answer with units
15. ($C = a + bn). Cost for 200 pages = 18.50. Find cost for 500 pages. [2 marks]
Answer: $24.50
Working:
(12.50 = a + 200b) ... (1)
(18.50 = a + 350b) ... (2)
(2) − (1): (6.00 = 150b) → (b = 0.04)
From (1): (a = 12.50 - 200(0.04) = 12.50 - 8.00 = 4.50)
Cost for 500 pages = (4.50 + 500(0.04) = 4.50 + 20.00 = $24.50)
Marking:
- M1: Setting up and solving simultaneous equations for (a) and (b)
- A1: Correct final cost
Section D: Problem Solving with Ratio and Proportion (15 marks)
16. 8.5 L per 100 km. Petrol $2.35/L. Journey 420 km. Find cost. [3 marks]
Answer: 83.80)
Working:
Petrol used = (\frac{420}{100} \times 8.5 = 4.2 \times 8.5 = 35.7) litres
Cost = (35.7 \times 2.35 = $83.895 \approx $83.90) (to nearest cent)
(Allow 83.89 depending on rounding convention)
Marking:
- M1: Correct calculation of petrol used
- M1: Correct multiplication for cost
- A1: Correct answer (accept 83.89)
17. Scale (1 : 72). Model wingspan = 24 cm. Find actual wingspan in metres. [2 marks]
Answer: 17.28 m
Working:
Actual wingspan = 24 × 72 = 1728 cm = 17.28 m
Marking:
- M1: Correct multiplication and conversion
- A1: Correct answer in metres
18. Investment ratio (2 : 3 : 5). Total profit $48 000. Difference between largest and smallest shares. [3 marks]
Answer: $14 400
Working:
Total parts = 2 + 3 + 5 = 10
1 part = 4800
Largest share = 5 × 24 000
Smallest share = 2 × 9600
Difference = 9600 = $14 400
Marking:
- M1: Finding value of one part
- M1: Finding largest and smallest shares
- A1: Correct difference
19. (V \propto r^3). (V = 36\pi) when (r = 3). Find (V) when (r = 5) (in terms of (\pi)). [3 marks]
Answer: (\frac{500\pi}{3}) cm³ or (166\frac{2}{3}\pi) cm³
Working:
(V = kr^3) → (36\pi = k \times 27) → (k = \frac{36\pi}{27} = \frac{4\pi}{3})
When (r = 5): (V = \frac{4\pi}{3} \times 125 = \frac{500\pi}{3}) cm³
Marking:
- M1: Finding constant (k)
- M1: Substituting (r = 5)
- A1: Correct answer in terms of (\pi)
20. Machine: 240 components in 5 hours. [4 marks]
(a) Components in 8 hours: 384
(b) Time for 600 components: 12.5 hours (or 12 hours 30 minutes)
Working:
Rate = 240 ÷ 5 = 48 components per hour
(a) In 8 hours: 48 × 8 = 384 components
(b) Time = 600 ÷ 48 = 12.5 hours
Marking:
- M1: Finding rate (components per hour)
- A1: Correct answer for (a)
- M1: Correct method for (b)
- A1: Correct answer for (b) with units
END OF ANSWER KEY