Secondary 4 Elementary Mathematics Numbers Ratio Proportion Quiz

Secondary 4 Elementary Mathematics Quiz - Numbers Ratio Proportion (Answer Key)

1.
$\frac{3^{2x} \times (3^2)^{x-1}}{(3^3)^x} = \frac{3^{2x} \times 3^{2x-2}}{3^{3x}}$
$= \frac{3^{4x-2}}{3^{3x}} = 3^{(4x-2)-3x} = 3^{x-2}$
Answer: $3^{x-2}$
[2 marks: 1 for converting bases, 1 for final simplified index]

2.
$(2^2)^{y+1} = (2^3)^{2y-3}$
$2^{2y+2} = 2^{6y-9}$
$2y + 2 = 6y - 9$
$11 = 4y \implies y = \frac{11}{4} = 2.75$
Answer: $y = 2.75$
[2 marks: 1 for equating indices, 1 for correct solution]

3.
$\left( \frac{27}{8} \right)^{-\frac{2}{3}} = \left( \frac{8}{27} \right)^{\frac{2}{3}} = \left( \sqrt[3]{\frac{8}{27}} \right)^2 = \left( \frac{2}{3} \right)^2 = \frac{4}{9}$
$5^0 = 1$
$\frac{4}{9} + 1 = \frac{13}{9} \text{ or } 1\frac{4}{9}$
Answer: $\frac{13}{9}$
[2 marks: 1 for evaluating index term, 1 for final addition]

4.
$\frac{1.67 \times 10^{-27}}{9.11 \times 10^{-31}} = \frac{1.67}{9.11} \times 10^{-27 - (-31)}$
$= 0.18331... \times 10^4 = 1833.1...$
Standard form: $1.83 \times 10^3$
Answer: $1.83 \times 10^3$
[2 marks: 1 for correct division, 1 for standard form and sig figs]

5.
$\frac{2.4 \times 10^5}{6.0 \times 10^{-3}} = \frac{2.4}{6.0} \times 10^{5 - (-3)}$
$= 0.4 \times 10^8 = 4.0 \times 10^7$
Answer: $4.0 \times 10^7$
[2 marks: 1 for calculation, 1 for standard form]

6.
Let shares be $3u, 5u, 2u$.
Difference between B and C: $5u - 2u = 3u$.
$3u = 450 \implies u = 150$.
Total sum = $3u + 5u + 2u = 10u$.
$10 \times 150 = 1500$.
Answer: \1500$
[3 marks: 1 for unit value, 1 for total units, 1 for final answer]

7.
Let boys = $7u$, girls = $4u$.
New boys = $7u - 12$, New girls = $4u + 12$.
Ratio $1:1 \implies 7u - 12 = 4u + 12$.
$3u = 24 \implies u = 8$.
Original boys = $7 \times 8 = 56$.
Answer: 56
[3 marks: 1 for equation, 1 for u, 1 for final answer]

8.
(a) $y = \frac{k}{x^2}$. When $x=2, y=12 \implies 12 = \frac{k}{4} \implies k=48$.
Equation: $y = \frac{48}{x^2}$
(b) When $x=6, y = \frac{48}{6^2} = \frac{48}{36} = \frac{4}{3}$ or $1.33$.
Answer: (a) $y = \frac{48}{x^2}$, (b) $1.33$
[3 marks: 2 for equation, 1 for value]

9.
Let original price = $P$.
After 20% increase: $1.20 P$.
After 15% decrease: $1.20 P \times (1 - 0.15) = 1.20 P \times 0.85$.
$1.20 \times 0.85 = 1.02$.
New price is $1.02 P$.
Percentage change = $(1.02 - 1) \times 100\% = 2\%$ increase.
Answer: $2\%$ increase
[3 marks: 1 for multipliers, 1 for product, 1 for percentage change]

10.
(a) $p = k \sqrt[3]{q}$. When $q=8, p=10 \implies 10 = k \sqrt[3]{8} = 2k \implies k=5$.
Equation: $p = 5 \sqrt[3]{q}$
(b) When $p=20, 20 = 5 \sqrt[3]{q} \implies 4 = \sqrt[3]{q} \implies q = 4^3 = 64$.
Answer: (a) $p = 5 \sqrt[3]{q}$, (b) 64
[4 marks: 2 for equation, 2 for value]

11.
Scale $1 : 50,000$.
Area scale = $1^2 : 50,000^2 = 1 : 2,500,000,000$.
Map Area = $12 \text{ cm}^2$.
Actual Area = $12 \times 2,500,000,000 \text{ cm}^2 = 30,000,000,000 \text{ cm}^2$.
Convert to $\text{km}^2$: $1 \text{ km} = 100,000 \text{ cm} \implies 1 \text{ km}^2 = 10^{10} \text{ cm}^2$.
Actual Area = $\frac{3 \times 10^{10}}{10^{10}} = 3 \text{ km}^2$.
Answer: $3 \text{ km}^2$
[3 marks: 1 for area scale, 1 for calculation in cm², 1 for conversion]

12.
$A = P(1 + r)^n = 10000(1 + 0.035)^4$.
$A = 10000(1.035)^4 \approx 10000(1.14752)$.
$A \approx 11475.23$.
Answer: \11,475.23$
[3 marks: 1 for formula, 1 for substitution, 1 for correct rounding]

13.
(a) $T = \frac{k}{N}$. $8 = \frac{k}{15} \implies k = 120$.
Constant = 120 (worker-days).
(b) New time = 6 days. $6 = \frac{120}{N_{new}} \implies N_{new} = 20$.
Additional workers = $20 - 15 = 5$.
Answer: 5 workers
[3 marks: 1 for constant, 1 for new N, 1 for difference]

14.
$R \propto \frac{L}{d^2}$.
Answer: $R = \frac{kL}{d^2}$
[2 marks: 1 for direct/inverse structure, 1 for constant]

15.
Ratio Cement : Sand : Gravel = $1 : 2 : 4$.
Sand = 2 parts = $240 \text{ kg} \implies 1 \text{ part} = 120 \text{ kg}$.
(a) Cement (1 part) = $120 \text{ kg}$.
(b) Total parts = $1 + 2 + 4 = 7$. Total weight = $7 \times 120 = 840 \text{ kg}$.
Answer: (a) $120 \text{ kg}$, (b) $840 \text{ kg}$
[3 marks: 1 for unit weight, 1 for cement, 1 for total]

16.
Let distance one way = $d$. Total distance = $2d$.
Time AB = $\frac{d}{60}$. Time BA = $\frac{d}{90}$.
Total Time = $\frac{d}{60} + \frac{d}{90} = \frac{3d + 2d}{180} = \frac{5d}{180} = \frac{d}{36}$.
Average Speed = $\frac{\text{Total Distance}}{\text{Total Time}} = \frac{2d}{d/36} = 2d \times \frac{36}{d} = 72 \text{ km/h}$.
Answer: $72 \text{ km/h}$
[3 marks: 1 for times, 1 for total time, 1 for average speed formula]

17.
(a) $P_n = 50000(1.02)^n$.
(b) $n = 5$ (2025 - 2020).
$P_5 = 50000(1.02)^5 \approx 50000(1.10408) \approx 55204$.
Nearest hundred: $55,200$.
Answer: (a) $50000(1.02)^n$, (b) $55,200$
[3 marks: 1 for expression, 1 for calculation, 1 for rounding]

18.
Linear scale factor $k = \frac{15}{6} = 2.5$.
Volume scale factor = $k^3 = 2.5^3 = 15.625$.
Volume larger = $144 \times 15.625 = 2250 \text{ cm}^3$.
Answer: $2250 \text{ cm}^3$
[3 marks: 1 for linear SF, 1 for volume SF, 1 for final volume]

19.
Cost Price (CP) = \40$. Profit =$25%$of$40 = $10$. Selling Price (SP) =$40 + 10 = $50$. SP is$80%$of Marked Price (MP) (since 20$0.8 \times MP = 50 \implies MP = \frac{50}{0.8} = 62.5$. **Answer:**$$62.50$
[3 marks: 1 for SP, 1 for discount relation, 1 for MP]

20.
$x = \frac{k y^2}{z}$.
$10 = \frac{k (2^2)}{5} \implies 10 = \frac{4k}{5} \implies 50 = 4k \implies k = 12.5$.
New values: $y=6, z=15$.
$x = \frac{12.5 (6^2)}{15} = \frac{12.5 \times 36}{15}$.
$x = 12.5 \times 2.4 = 30$.
Answer: 30
[3 marks: 1 for k, 1 for substitution, 1 for final answer]