Secondary 4 Additional Mathematics Numbers Ratio Proportion Quiz

Secondary 4 Additional Mathematics Quiz - Answers and Marking Scheme

Topic: Numbers, Ratio and Proportion
Total Marks: 50

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Section A: Basic Concepts and Indices

1. Simplify $\frac{27^{x} \cdot 9^{2x-1}}{3^{3x+2}}$

• Step 1: Convert all bases to 3. $27^x = (3^3)^x = 3^{3x}$ $9^{2x-1} = (3^2)^{2x-1} = 3^{2(2x-1)} = 3^{4x-2}$
• Step 2: Substitute into the expression. $\frac{3^{3x} \cdot 3^{4x-2}}{3^{3x+2}}$
• Step 3: Apply index laws ($a^m \cdot a^n = a^{m+n}$ and $\frac{a^m}{a^n} = a^{m-n}$). Numerator: $3^{3x + 4x - 2} = 3^{7x-2}$ Expression: $\frac{3^{7x-2}}{3^{3x+2}} = 3^{(7x-2) - (3x+2)}$
• Step 4: Simplify the exponent. $7x - 2 - 3x - 2 = 4x - 4$
• Answer: $3^{4x-4}$

Marks: [2] (1 for correct base conversion, 1 for final simplified index)

2. Express $2^{2x-y}$ given $2^x = 3$ and $2^y = 5$

• Step 1: Use index laws to expand $2^{2x-y}$. $2^{2x-y} = \frac{2^{2x}}{2^y} = \frac{(2^x)^2}{2^y}$
• Step 2: Substitute the given values. $\frac{3^2}{5} = \frac{9}{5}$
• Answer: $\frac{9}{5}$ or $1.8$

Marks: [2] (1 for expansion, 1 for substitution and final value)

3. Solve $4^{x} - 6(2^x) + 8 = 0$

• Step 1: Let $u = 2^x$. Then $4^x = (2^2)^x = (2^x)^2 = u^2$. Equation becomes: $u^2 - 6u + 8 = 0$
• Step 2: Factorize the quadratic. $(u-4)(u-2) = 0$ $u = 4 \quad \text{or} \quad u = 2$
• Step 3: Solve for $x$. If $2^x = 4$, then $x = 2$. If $2^x = 2$, then $x = 1$.
• Answer: $x = 1, x = 2$

Marks: [2] (1 for correct substitution/solving quadratic, 1 for both x values)

4. Express $\sqrt[3]{\frac{a^4 b^{-2}}{a^{-2} b^4}}$ in form $a^m b^n$

• Step 1: Simplify the fraction inside the root first. $\frac{a^4 b^{-2}}{a^{-2} b^4} = a^{4 - (-2)} b^{-2 - 4} = a^6 b^{-6}$
• Step 2: Apply the cube root (power of $\frac{1}{3}$). $(a^6 b^{-6})^{\frac{1}{3}} = a^{6 \times \frac{1}{3}} b^{-6 \times \frac{1}{3}}$ $= a^2 b^{-2}$
• Answer: $a^2 b^{-2}$

Marks: [2] (1 for simplifying inside, 1 for applying root)

5. Evaluate $\left( \frac{8}{27} \right)^{-\frac{2}{3}} + \left( \frac{1}{4} \right)^{-\frac{1}{2}}$

• Step 1: Evaluate the first term. $\left( \frac{8}{27} \right)^{-\frac{2}{3}} = \left( \frac{27}{8} \right)^{\frac{2}{3}} = \left( \sqrt[3]{\frac{27}{8}} \right)^2 = \left( \frac{3}{2} \right)^2 = \frac{9}{4}$
• Step 2: Evaluate the second term. $\left( \frac{1}{4} \right)^{-\frac{1}{2}} = (4)^{\frac{1}{2}} = \sqrt{4} = 2$
• Step 3: Add them. $\frac{9}{4} + 2 = \frac{9}{4} + \frac{8}{4} = \frac{17}{4}$
• Answer: $\frac{17}{4}$ or $4.25$

Marks: [2] (1 for each term evaluated correctly)

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Section B: Surds and Rationalization

6. Express $\frac{5}{\sqrt{3} - 1}$ in form $a + b\sqrt{3}$

• Step 1: Multiply numerator and denominator by the conjugate $\sqrt{3} + 1$. $\frac{5(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)}$
• Step 2: Simplify denominator ($a^2 - b^2$). $(\sqrt{3})^2 - 1^2 = 3 - 1 = 2$
• Step 3: Simplify numerator and divide. $\frac{5\sqrt{3} + 5}{2} = \frac{5}{2} + \frac{5}{2}\sqrt{3}$
• Answer: $\frac{5}{2} + \frac{5}{2}\sqrt{3}$ (Here $a=2.5, b=2.5$)

Marks: [2] (1 for correct conjugate multiplication, 1 for final form)

7. Given $x = 2 + \sqrt{5}$, show $x^2 - 4x - 1 = 0$ and find $\frac{1}{x}$

• Part 1: Show equation $x - 2 = \sqrt{5}$ Square both sides: $(x-2)^2 = 5 \implies x^2 - 4x + 4 = 5 \implies x^2 - 4x - 1 = 0$ (Shown)
• Part 2: Find $\frac{1}{x}$ From $x^2 - 4x - 1 = 0$, divide by $x$ (since $x \neq 0$): $x - 4 - \frac{1}{x} = 0 \implies \frac{1}{x} = x - 4$ Substitute $x = 2 + \sqrt{5}$: $\frac{1}{x} = (2 + \sqrt{5}) - 4 = \sqrt{5} - 2$ Alternative Method: Rationalize $\frac{1}{2+\sqrt{5}} = \frac{2-\sqrt{5}}{4-5} = \frac{2-\sqrt{5}}{-1} = \sqrt{5}-2$.
• Answer: $\sqrt{5} - 2$

Marks: [3] (1 for showing equation, 1 for method to find 1/x, 1 for correct answer)

8. Simplify $\sqrt{75} - 2\sqrt{12} + \sqrt{27}$

• Step 1: Simplify each surd. $\sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3}$ $\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3} \implies 2\sqrt{12} = 4\sqrt{3}$ $\sqrt{27} = \sqrt{9 \times 3} = 3\sqrt{3}$
• Step 2: Combine like terms. $5\sqrt{3} - 4\sqrt{3} + 3\sqrt{3} = (5 - 4 + 3)\sqrt{3} = 4\sqrt{3}$
• Answer: $4\sqrt{3}$

Marks: [2] (1 for simplifying at least two terms correctly, 1 for final answer)

9. Solve $\sqrt{2x + 3} = x$

• Step 1: Square both sides. $2x + 3 = x^2$ $x^2 - 2x - 3 = 0$
• Step 2: Factorize. $(x-3)(x+1) = 0$ $x = 3 \quad \text{or} \quad x = -1$
• Step 3: Check for extraneous roots. If $x = -1$: LHS $= \sqrt{2(-1)+3} = \sqrt{1} = 1$. RHS $= -1$. $1 \neq -1$. Reject. If $x = 3$: LHS $= \sqrt{2(3)+3} = \sqrt{9} = 3$. RHS $= 3$. Accept.
• Answer: $x = 3$

Marks: [3] (1 for quadratic equation, 1 for solving x, 1 for rejection reason)

10. Given $\sqrt{x} + \frac{1}{\sqrt{x}} = 3$, find $x + \frac{1}{x}$

• Step 1: Square the given equation. $\left(\sqrt{x} + \frac{1}{\sqrt{x}}\right)^2 = 3^2$ $x + 2(\sqrt{x})(\frac{1}{\sqrt{x}}) + \frac{1}{x} = 9$ $x + 2 + \frac{1}{x} = 9$
• Step 2: Isolate $x + \frac{1}{x}$. $x + \frac{1}{x} = 9 - 2 = 7$
• Answer: $7$

Marks: [3] (1 for squaring strategy, 1 for expansion, 1 for final answer)

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Section C: Ratio, Proportion and Variation

11. $y \propto \frac{\sqrt{x}}{z^2}$

• (a) Find $k$ Formula: $y = \frac{k\sqrt{x}}{z^2}$ Substitute $x=16, z=2, y=3$: $3 = \frac{k\sqrt{16}}{2^2} = \frac{4k}{4} = k$ Answer: $k = 3$
• (b) Find $y$ when $x=25, z=5$ $y = \frac{3\sqrt{25}}{5^2} = \frac{3(5)}{25} = \frac{15}{25} = \frac{3}{5}$ Answer: $0.6$ or $\frac{3}{5}$

Marks: [3] (2 for part a, 1 for part b)

12. Ratio of Boys:Girls is $5:4$. After changes, ratio is $1:1$.

• Step 1: Let initial Boys $= 5u$, Girls $= 4u$.
• Step 2: Apply changes. New Boys $= 5u - 10$ New Girls $= 4u + 10$
• Step 3: Set up equation for $1:1$ ratio. $5u - 10 = 4u + 10$ $u = 20$
• Step 4: Find original boys. $5u = 5(20) = 100$
• Answer: 100 boys

Marks: [3] (1 for setting up variables, 1 for equation, 1 for final answer)

13. $A \propto B C^2$

• (a) Express $A$ $A = k B C^2$. Substitute $A=36, B=2, C=3$: $36 = k(2)(3^2) = 18k \implies k=2$. Answer: $A = 2BC^2$
• (b) Percentage change New $B' = 1.5B$. New $C' = 0.8C$. New $A' = 2(1.5B)(0.8C)^2 = 2(1.5B)(0.64C^2) = 1.92 (2BC^2) = 1.92 A$. Change factor is $1.92$. Percentage change $= (1.92 - 1) \times 100\% = 92\%$. Answer: Increase of $92\%$

Marks: [5] (2 for part a, 3 for part b)

14. Divide \800$in ratio$2:3:5$. R gives$$50$ to P.

• Step 1: Find initial shares. Total parts $= 2+3+5 = 10$. 1 part = \80$.$P = 2 \times 80 = 160$.$Q = 3 \times 80 = 240$.$R = 5 \times 80 = 400$.
• Step 2: Apply transfer. New $P = 160 + 50 = 210$. $Q$ remains $240$.
• Step 3: New Ratio $P:Q$. $210 : 240$. Divide by 30. $7 : 8$.
• Answer: $7:8$

Marks: [3] (1 for initial shares, 1 for new values, 1 for simplified ratio)

15. Cost = Fixed + Variable(Hours)

• (a) Find Fixed Cost Let $C = F + kH$. $150 = F + 4k$ (Eq 1) $210 = F + 7k$ (Eq 2) Subtract Eq 1 from Eq 2: $60 = 3k \implies k = 20$. Substitute $k=20$ into Eq 1: $150 = F + 80 \implies F = 70$. Answer: \70$
• (b) Cost for 10 hours $C = 70 + 20(10) = 70 + 200 = 270$. Answer: \270$

Marks: [3] (2 for part a, 1 for part b)

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Section D: Advanced Applications and Logarithms

16. Solve $\log_2 (x-1) + \log_2 (x+2) = 2$

• Step 1: Combine logs. $\log_2 [(x-1)(x+2)] = 2$
• Step 2: Convert to index form. $(x-1)(x+2) = 2^2 = 4$ $x^2 + x - 2 = 4$ $x^2 + x - 6 = 0$
• Step 3: Solve quadratic. $(x+3)(x-2) = 0 \implies x = -3, x = 2$
• Step 4: Check validity. For $\log_2(x-1)$, we need $x-1 > 0 \implies x > 1$. $x = -3$ is rejected. $x = 2$ is accepted.
• Answer: $x = 2$

Marks: [4] (1 for combining, 1 for quadratic, 1 for solving, 1 for rejection)

17. Express $\log_a \left( \frac{12}{a^2} \right)$ in terms of $p, q$

• Step 1: Expand using log laws. $\log_a 12 - \log_a (a^2)$ $\log_a (4 \times 3) - 2\log_a a$ $\log_a 4 + \log_a 3 - 2$ $\log_a (2^2) + q - 2$ $2\log_a 2 + q - 2$
• Step 2: Substitute $\log_a 2 = p$. $2p + q - 2$
• Answer: $2p + q - 2$

Marks: [3] (1 for expansion, 1 for simplifying $\log_a a^2$, 1 for final expression)

18. Linearization of $y = Ab^x$

• Step 1: Linearize equation. $\log_{10} y = \log_{10} (Ab^x) = \log_{10} A + x \log_{10} b$ This is $Y = mX + c$, where $Y = \log_{10} y$, $X = x$, $m = \log_{10} b$, $c = \log_{10} A$.
• Step 2: Find gradient $m$ and intercept $c$ from points $(0, 0.6)$ and $(4, 1.4)$. $c = 0.6$ (y-intercept). $m = \frac{1.4 - 0.6}{4 - 0} = \frac{0.8}{4} = 0.2$.
• Step 3: Find $A$ and $b$. $\log_{10} A = 0.6 \implies A = 10^{0.6} \approx 3.98$. $\log_{10} b = 0.2 \implies b = 10^{0.2} \approx 1.58$.
• (a) Answer: $A = 10^{0.6}, b = 10^{0.2}$ (or decimal approximations)
• (b) Estimate $y$ when $x=2$ From graph equation: $\log_{10} y = 0.2(2) + 0.6 = 1.0$. $y = 10^1 = 10$. Answer: $10$

Marks: [5] (4 for part a, 1 for part b)

19. Simultaneous Equations: $2^x \cdot 4^y = 32$ and $\log_2 x + \log_2 y = 3$

• Step 1: Simplify first equation. $2^x \cdot (2^2)^y = 2^5 \implies 2^{x+2y} = 2^5 \implies x + 2y = 5 \implies x = 5 - 2y$.
• Step 2: Simplify second equation. $\log_2 (xy) = 3 \implies xy = 2^3 = 8$.
• Step 3: Substitute $x$ into $xy=8$. $(5-2y)y = 8$ $5y - 2y^2 = 8$ $2y^2 - 5y + 8 = 0$
• Step 4: Check discriminant. $b^2 - 4ac = (-5)^2 - 4(2)(8) = 25 - 64 = -39$. Since discriminant $< 0$, there are no real solutions for $y$.
• Answer: No real solution.

Marks: [5] (2 for simplifying eq 1, 2 for simplifying eq 2, 1 for concluding no solution) Note: If the question intended integer solutions, the constants might be different, but based on the provided numbers, there is no real solution. Students showing the discriminant calculation receive full marks.

20. Geometric Progression: Sum of first 2 terms is 12, Sum of first 3 is 26.

• (a) Form equations $S_2 = a + ar = 12$ (Eq 1) $S_3 = a + ar + ar^2 = 26$ (Eq 2)

• (b) Find $r$ and $a$ Subtract Eq 1 from Eq 2: $ar^2 = 26 - 12 = 14 \implies a = \frac{14}{r^2}$. Substitute into Eq 1: $\frac{14}{r^2} + \frac{14}{r^2}(r) = 12$ $\frac{14}{r^2} + \frac{14}{r} = 12$ Multiply by $r^2$: $14 + 14r = 12r^2$ $12r^2 - 14r - 14 = 0$ $6r^2 - 7r - 7 = 0$ Using quadratic formula: $r = \frac{7 \pm \sqrt{49 - 4(6)(-7)}}{12} = \frac{7 \pm \sqrt{49 + 168}}{12} = \frac{7 \pm \sqrt{217}}{12}$.

  Self-Correction/Check: Let's re-read carefully. "Sum of first two terms is 12". "Sum of first three terms is 26". Term 3 = $S_3 - S_2 = 14$. $ar^2 = 14$. $a(1+r) = 12$. $a = 12/(1+r)$. $\frac{12r^2}{1+r} = 14 \implies 12r^2 = 14 + 14r \implies 6r^2 - 7r - 7 = 0$. The roots are irrational. $r \approx 1.81$ or $r \approx -0.64$. If $r = \frac{7 + \sqrt{217}}{12}$, then $a = \frac{14}{r^2}$.

  Alternative Interpretation Check: Did the question imply integer answers? Often GP questions have integer ratios. If $S_2=12, S_3=26$, Term 3 is 14. Term 2 is $12-a$. Term 1 is $a$. $r = \frac{12-a}{a}$. Also $r^2 = \frac{14}{a}$. $(\frac{12-a}{a})^2 = \frac{14}{a} \implies \frac{(12-a)^2}{a^2} = \frac{14}{a} \implies (12-a)^2 = 14a$. $144 - 24a + a^2 = 14a \implies a^2 - 38a + 144 = 0$. $(a-2)(a-36)?$ No. $2 \times 36 = 72$. Roots of $a^2 - 38a + 144 = 0$: $a = \frac{38 \pm \sqrt{1444 - 576}}{2} = \frac{38 \pm \sqrt{868}}{2}$. Still irrational.

  The question is mathematically consistent, just has irrational answers.

  Answer: $r = \frac{7 \pm \sqrt{217}}{12}$ $a = \frac{14}{r^2}$

Marks: [6] (2 for equations, 4 for solving quadratic and finding pairs)