Secondary 3 Elementary Mathematics Numbers Ratio Proportion Quiz

Answer Key: Secondary 3 Elementary Mathematics Quiz - Numbers Ratio Proportion

Total Marks: 45

Section A: Indices and Standard Form

1. Simplify $\frac{(2x^3)^4 \times 3x^{-2}}{6x^5}$

• Numerator: $(2^4 x^{12}) \times 3x^{-2} = 16x^{12} \times 3x^{-2} = 48x^{10}$
• Expression: $\frac{48x^{10}}{6x^5}$
• Divide coefficients: $48 \div 6 = 8$
• Divide indices: $x^{10-5} = x^5$
  Answer: $8x^5$
  [1 mark for numerator simplification, 1 mark for final answer]

2. Evaluate $27^{\frac{2}{3}} + 16^{-\frac{1}{4}}$

• $27^{\frac{2}{3}} = (\sqrt[3]{27})^2 = 3^2 = 9$
• $16^{-\frac{1}{4}} = \frac{1}{16^{\frac{1}{4}}} = \frac{1}{\sqrt[4]{16}} = \frac{1}{2}$
• Sum: $9 + 0.5 = 9.5$
  Answer: $9.5$ (or $\frac{19}{2}$)
  [1 mark for each term evaluated correctly]

3. Solve $5^{2x-1} = \frac{1}{125}$

• $\frac{1}{125} = \frac{1}{5^3} = 5^{-3}$
• Equate indices: $2x - 1 = -3$
• $2x = -2$
• $x = -1$
  Answer: $x = -1$
  [1 mark for converting RHS to base 5, 1 mark for solving linear eq]

4. Ratio of Proton mass to Electron mass

• $\frac{1.67 \times 10^{-27}}{9.11 \times 10^{-31}}$
• $= \frac{1.67}{9.11} \times 10^{-27 - (-31)}$
• $= 0.183315... \times 10^4$
• $= 1833.15...$
• Standard form: $1.83315... \times 10^3$
• Correct to 3 s.f.: $1.83 \times 10^3$
  Answer: $1.83 \times 10^3$
  [1 mark for division setup, 1 mark for power of 10, 1 mark for final s.f.]

5. Simplify $\left( \frac{a^{\frac{1}{2}} b^{-3}}{a^{-1} b^2} \right)^{\frac{2}{3}}$

• Inside bracket: $a^{\frac{1}{2} - (-1)} b^{-3 - 2} = a^{\frac{3}{2}} b^{-5}$
• Apply outer power: $(a^{\frac{3}{2}})^{\frac{2}{3}} (b^{-5})^{\frac{2}{3}}$
• $a^{\frac{3}{2} \times \frac{2}{3}} b^{-\frac{10}{3}}$
• $a^1 b^{-\frac{10}{3}}$
  Answer: $a b^{-\frac{10}{3}}$ (or $\frac{a}{b^{\frac{10}{3}}}$)
  [1 mark for simplifying inside, 1 mark for final indices]

Section B: Algebraic Fractions and Equations

6. $\frac{3}{x+2} - \frac{2}{x-1}$

• Common denominator: $(x+2)(x-1)$
• Numerator: $3(x-1) - 2(x+2)$
• $= 3x - 3 - 2x - 4$
• $= x - 7$
  Answer: $\frac{x-7}{(x+2)(x-1)}$
  [1 mark for common denom, 1 mark for expansion, 1 mark for final simplified numerator]

7. $\frac{x^2 - 9}{x^2 + 5x + 6} \div \frac{x-3}{x+4}$

• Factorise: $\frac{(x-3)(x+3)}{(x+2)(x+3)} \times \frac{x+4}{x-3}$
• Cancel $(x+3)$ and $(x-3)$:
• Remaining: $\frac{1}{x+2} \times \frac{x+4}{1}$
  Answer: $\frac{x+4}{x+2}$
  [1 mark for factorising, 1 mark for flipping second fraction, 1 mark for cancellation]

8. $\frac{2x}{x-3} = 5$

• $2x = 5(x-3)$
• $2x = 5x - 15$
• $15 = 3x$
• $x = 5$
  Answer: $x = 5$
  [1 mark for cross-multiplication, 1 mark for solution]

9. $\frac{3}{x+1} + \frac{2}{x-2} = 1$

• Multiply by $(x+1)(x-2)$:
• $3(x-2) + 2(x+1) = (x+1)(x-2)$
• $3x - 6 + 2x + 2 = x^2 - x - 2$
• $5x - 4 = x^2 - x - 2$
• $0 = x^2 - 6x + 2$
• Use quadratic formula: $x = \frac{6 \pm \sqrt{36 - 8}}{2} = \frac{6 \pm \sqrt{28}}{2}$
• $x = \frac{6 \pm 2\sqrt{7}}{2} = 3 \pm \sqrt{7}$
  Answer: $x = 3 + \sqrt{7}$ or $x = 3 - \sqrt{7}$ (approx $5.65, 0.35$)
  [1 mark for clearing denominators, 1 mark for forming quadratic, 1 mark for solving, 1 mark for both roots]

10. $y = \frac{2x+1}{x-3}$, express $x$ in terms of $y$.

• $y(x-3) = 2x + 1$
• $xy - 3y = 2x + 1$
• $xy - 2x = 3y + 1$
• $x(y-2) = 3y + 1$
• $x = \frac{3y+1}{y-2}$
  Answer: $x = \frac{3y+1}{y-2}$
  [1 mark for cross multiply, 1 mark for grouping x terms, 1 mark for final answer]

11. $\frac{\frac{1}{x} + \frac{1}{y}}{\frac{1}{x} - \frac{1}{y}}$

• Numerator: $\frac{y+x}{xy}$
• Denominator: $\frac{y-x}{xy}$
• Divide: $\frac{x+y}{xy} \times \frac{xy}{y-x}$
• Cancel $xy$: $\frac{x+y}{y-x}$
  Answer: $\frac{x+y}{y-x}$
  [1 mark for combining numerator fractions, 1 mark for denominator, 1 mark for final simplification]

12. $\frac{kx^2 - 4}{x-2} = Ax + B$

• Since it simplifies, $x-2$ must be a factor of $kx^2 - 4$.
• Let $x=2$: $k(2)^2 - 4 = 0 \Rightarrow 4k = 4 \Rightarrow k=1$.
• Expression becomes $\frac{x^2 - 4}{x-2} = \frac{(x-2)(x+2)}{x-2} = x+2$.
• Compare to $Ax+B$: $A=1, B=2$.
  Answer: $k=1, A=1, B=2$
  [1 mark for finding k, 1 mark for simplifying, 1 mark for A and B]

Section C: Ratio, Proportion and Variation

13. $y$ varies directly as $x^2$. $x=3, y=45$.
(a) $y = kx^2 \Rightarrow 45 = k(3^2) \Rightarrow 45 = 9k \Rightarrow k=5$.
Equation: $y = 5x^2$
(b) When $x=5$: $y = 5(5^2) = 5(25) = 125$.
Answer: (a) $y=5x^2$, (b) $125$
[1 mark for k, 1 mark for eq, 1 mark for final value]

14. $P$ varies inversely as $\sqrt[3]{Q}$. $Q=8, P=5$.
(a) $P = \frac{k}{\sqrt[3]{Q}} \Rightarrow 5 = \frac{k}{\sqrt[3]{8}} \Rightarrow 5 = \frac{k}{2} \Rightarrow k=10$.
Equation: $P = \frac{10}{\sqrt[3]{Q}}$
(b) When $P=10$: $10 = \frac{10}{\sqrt[3]{Q}} \Rightarrow \sqrt[3]{Q} = 1 \Rightarrow Q = 1^3 = 1$.
Answer: (a) $P = \frac{10}{\sqrt[3]{Q}}$, (b) $Q=1$
[1 mark for k, 1 mark for eq, 1 mark for substitution, 1 mark for Q]

15. $T$ varies jointly as $W$ and $D^2$.
Answer: $T = k W D^2$
[1 mark for correct formula]

16. Ratio Cement : Sand : Gravel = $1 : 3 : 5$.

• Sand = 3 units = 240 kg.
• 1 unit = $240 \div 3 = 80$ kg.
• Total units = $1 + 3 + 5 = 9$ units.
• Total weight = $9 \times 80 = 720$ kg.
  Answer: $720$ kg
  [1 mark for unit value, 1 mark for total units, 1 mark for final answer]

17. Ratio $A : B = 5 : 3$. Difference = $5u - 3u = 2u$.

• 2u = \40 \Rightarrow u = $20$.
• Total sum = $5u + 3u = 8u$.
• 8 \times 20 = \160$. **Answer:**$$160$
  [1 mark for unit difference, 1 mark for unit value, 1 mark for total]

18. $R = \frac{kL}{d^2}$.

• New $L' = 2L$, New $d' = \frac{1}{2}d$.
• $R' = \frac{k(2L)}{(\frac{1}{2}d)^2} = \frac{2kL}{\frac{1}{4}d^2} = 2 \times 4 \times \frac{kL}{d^2} = 8R$.
• Resistance increases by a factor of 8.
  Answer: Factor of 8 (or 8 times)
  [1 mark for substitution, 1 mark for simplification, 1 mark for factor]

19. $y = k_1 x + \frac{k_2}{x}$.

• $x=1, y=5 \Rightarrow 5 = k_1(1) + k_2(1) \Rightarrow k_1 + k_2 = 5$ (Eq 1)
• $x=2, y=7 \Rightarrow 7 = k_1(2) + \frac{k_2}{2} \Rightarrow 14 = 4k_1 + k_2$ (Eq 2, multiplied by 2)
• Subtract Eq 1 from Eq 2: $(4k_1 + k_2) - (k_1 + k_2) = 14 - 5$
• $3k_1 = 9 \Rightarrow k_1 = 3$.
• Substitute into Eq 1: $3 + k_2 = 5 \Rightarrow k_2 = 2$.
• Equation: $y = 3x + \frac{2}{x}$.
• When $x=4$: $y = 3(4) + \frac{2}{4} = 12 + 0.5 = 12.5$.
  Answer: $12.5$
  [1 mark for setting up equations, 1 mark for solving constants, 1 mark for equation, 1 mark for final value]

20. Scale $1 : 50,000$. Area scale is square of linear scale.

• Linear scale: $1 \text{ cm} = 50,000 \text{ cm} = 0.5 \text{ km}$.
• Area scale: $1 \text{ cm}^2 = (0.5 \text{ km})^2 = 0.25 \text{ km}^2$.
• Map Area = $12 \text{ cm}^2$.
• Actual Area = $12 \times 0.25 = 3 \text{ km}^2$.
  Answer: $3 \text{ km}^2$
  [1 mark for area conversion factor, 1 mark for final calculation]