Secondary 4 Additional Mathematics Numbers Ratio Proportion Quiz

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

Total Marks: 40

Section A (10 marks)

1. $\frac{3}{\sqrt{5} - 2} = \frac{3(\sqrt{5} + 2)}{(\sqrt{5} - 2)(\sqrt{5} + 2)} = \frac{3\sqrt{5} + 6}{5 - 4} = 6 + 3\sqrt{5}$
Answer: $6 + 3\sqrt{5}$ [1]
Marking note: 1 mark for correct rationalisation and simplification. Common error: forgetting to multiply numerator by conjugate.

2. $x = 2 + \sqrt{3}$
$x^2 = (2 + \sqrt{3})^2 = 4 + 4\sqrt{3} + 3 = 7 + 4\sqrt{3}$
$\frac{1}{x} = \frac{1}{2 + \sqrt{3}} = \frac{2 - \sqrt{3}}{4 - 3} = 2 - \sqrt{3}$
$\frac{1}{x^2} = (2 - \sqrt{3})^2 = 4 - 4\sqrt{3} + 3 = 7 - 4\sqrt{3}$
$x^2 + \frac{1}{x^2} = (7 + 4\sqrt{3}) + (7 - 4\sqrt{3}) = 14$
Answer: $14$ (or $14 + 0\sqrt{3}$) [1]
Marking note: 1 mark for correct final answer. Key concept: conjugate pairs eliminate surds.

3. $\frac{4\sqrt{12} - 3\sqrt{27}}{\sqrt{3}} = \frac{4(2\sqrt{3}) - 3(3\sqrt{3})}{\sqrt{3}} = \frac{8\sqrt{3} - 9\sqrt{3}}{\sqrt{3}} = \frac{-\sqrt{3}}{\sqrt{3}} = -1$
Answer: $-1$ [1]
Marking note: 1 mark. Simplify surds first: $\sqrt{12}=2\sqrt{3}$, $\sqrt{27}=3\sqrt{3}$.

4. $2^{x+1} = 8^{x-2} = (2^3)^{x-2} = 2^{3x-6}$
Equate indices: $x + 1 = 3x - 6 \Rightarrow 2x = 7 \Rightarrow x = 3.5$
Answer: $3.5$ [1]
Marking note: 1 mark. Key step: express both sides with same base.

5. $\log_2 x = 3 \Rightarrow x = 2^3 = 8$
$\log_x 8 = \log_8 8 = 1$
Answer: $1$ [1]
Marking note: 1 mark. Alternatively: $\log_x 8 = \frac{\log_2 8}{\log_2 x} = \frac{3}{3} = 1$.

6. $\lg y = \lg k + n \lg x$
Gradient $= n = \frac{9 - 5}{4 - 2} = \frac{4}{2} = 2$
Answer: $2$ [1]
Marking note: 1 mark. Gradient of $\lg y$ vs $\lg x$ graph gives power $n$.

7. $\log_3 \frac{x+4}{x-2} = 1 \Rightarrow \frac{x+4}{x-2} = 3^1 = 3$
$x + 4 = 3x - 6 \Rightarrow 2x = 10 \Rightarrow x = 5$
Check: $x-2=3>0$, valid.
Answer: $5$ [1]
Marking note: 1 mark. Must check domain: $x > 2$.

8. $\frac{5 - 2\sqrt{6}}{\sqrt{3} - \sqrt{2}} = \frac{(5 - 2\sqrt{6})(\sqrt{3} + \sqrt{2})}{3 - 2} = 5\sqrt{3} + 5\sqrt{2} - 2\sqrt{18} - 2\sqrt{12}$
$= 5\sqrt{3} + 5\sqrt{2} - 6\sqrt{2} - 4\sqrt{3} = \sqrt{3} - \sqrt{2}$
Answer: $-\sqrt{2} + \sqrt{3}$ (or $\sqrt{3} - \sqrt{2}$) [1]
Marking note: 1 mark. Rationalise denominator, simplify $\sqrt{18}=3\sqrt{2}$, $\sqrt{12}=2\sqrt{3}$.

9. $\log_2 45 = \log_2 (9 \times 5) = \log_2 9 + \log_2 5 = 2\log_2 3 + \log_2 5 = 2a + b$
Answer: $2a + b$ [1]
Marking note: 1 mark. Use log laws: $\log(MN)=\log M+\log N$, $\log M^k=k\log M$.

10. $P = P_0 e^{kt}$. When $t=3$, $P=2P_0$.
$2P_0 = P_0 e^{3k} \Rightarrow 2 = e^{3k} \Rightarrow \ln 2 = 3k \Rightarrow k = \frac{\ln 2}{3} \approx 0.231$
Answer: $0.231$ [1]
Marking note: 1 mark. 3 s.f. required. Common error: using $\log_{10}$ instead of $\ln$.

Section B (18 marks)

11. (a) $\frac{\sqrt{50} + \sqrt{18}}{\sqrt{8}} = \frac{5\sqrt{2} + 3\sqrt{2}}{2\sqrt{2}} = \frac{8\sqrt{2}}{2\sqrt{2}} = 4$
Answer: $4$ [2]
Mark breakdown: 1 mark for simplifying each surd ($\sqrt{50}=5\sqrt{2}$, $\sqrt{18}=3\sqrt{2}$, $\sqrt{8}=2\sqrt{2}$), 1 mark for final simplification.

(b) From (a), $\sqrt{x} + \frac{2}{\sqrt{x}} = 4$
Let $u = \sqrt{x} > 0$. Then $u + \frac{2}{u} = 4 \Rightarrow u^2 + 2 = 4u \Rightarrow u^2 - 4u + 2 = 0$
$u = \frac{4 \pm \sqrt{16 - 8}}{2} = \frac{4 \pm \sqrt{8}}{2} = \frac{4 \pm 2\sqrt{2}}{2} = 2 \pm \sqrt{2}$
Both values positive. $x = u^2 = (2 \pm \sqrt{2})^2 = 4 \pm 4\sqrt{2} + 2 = 6 \pm 4\sqrt{2}$
Answer: $x = 6 + 4\sqrt{2}$ or $x = 6 - 4\sqrt{2}$ [2]
Mark breakdown: 1 mark for substitution and forming quadratic, 1 mark for solving and back-substitution. Common error: forgetting $x = u^2$.

12. (a) $3^{2x+1} = 5^{x-2}$
Take $\ln$ (or $\log$) both sides: $(2x+1)\ln 3 = (x-2)\ln 5$
$2x\ln 3 + \ln 3 = x\ln 5 - 2\ln 5$
$x(2\ln 3 - \ln 5) = -2\ln 5 - \ln 3$
$x = \frac{-2\ln 5 - \ln 3}{2\ln 3 - \ln 5} = \frac{2\ln 5 + \ln 3}{\ln 5 - 2\ln 3}$
Using calculator: $x \approx \frac{2(1.6094) + 1.0986}{1.6094 - 2(1.0986)} = \frac{4.3174}{-0.5878} \approx -7.345$
Answer: $-7.35$ (3 s.f.) [3]
Mark breakdown: 1 mark for taking logs, 1 mark for correct algebraic manipulation, 1 mark for correct 3 s.f. answer.

(b) Since $3^{2x+1}$ and $5^{x-2}$ are both increasing functions, and $3^{2x+1} = 5^{x-2}$ at $x \approx -7.35$, the inequality $3^{2x+1} > 5^{x-2}$ holds for $x > -7.35$.
Answer: $x > -7.35$ [1]
Marking note: 1 mark. Must recognise both sides are increasing functions of $x$.

13. (a) $\log_a \left(\frac{x^3}{y^2}\right) = 3\log_a x - 2\log_a y = 3(2) - 2(3) = 6 - 6 = 0$
Answer: $0$ [2]
Mark breakdown: 1 mark for applying log laws correctly, 1 mark for correct substitution and evaluation.

(b) $\log_2 (x+1) + \log_2 (x-1) = 3$
$\log_2 [(x+1)(x-1)] = 3 \Rightarrow \log_2 (x^2 - 1) = 3$
$x^2 - 1 = 2^3 = 8 \Rightarrow x^2 = 9 \Rightarrow x = \pm 3$
Domain: $x+1>0$ and $x-1>0 \Rightarrow x > 1$. So $x = 3$ only.
Answer: $x = 3$ [3]
Mark breakdown: 1 mark for combining logs, 1 mark for solving quadratic, 1 mark for domain check and rejecting $x=-3$.

14. $y = \frac{ax}{bx + 1} \Rightarrow \frac{1}{y} = \frac{bx + 1}{ax} = \frac{b}{a} + \frac{1}{a} \cdot \frac{1}{x}$
This is of the form $\frac{1}{y} = m \cdot \frac{1}{x} + c$ with gradient $m = \frac{1}{a}$ and intercept $c = \frac{b}{a}$.
Given gradient $= 2 \Rightarrow \frac{1}{a} = 2 \Rightarrow a = \frac{1}{2}$
Given intercept $= 3 \Rightarrow \frac{b}{a} = 3 \Rightarrow b = 3a = 3 \times \frac{1}{2} = \frac{3}{2}$
Answer: $a = \frac{1}{2}$, $b = \frac{3}{2}$ [4]
Mark breakdown: 1 mark for rearranging to linear form, 1 mark for identifying gradient and intercept expressions, 1 mark for finding $a$, 1 mark for finding $b$.

15. (a) $\frac{7 + 4\sqrt{3}}{2 - \sqrt{3}} = \frac{(7 + 4\sqrt{3})(2 + \sqrt{3})}{4 - 3} = 14 + 7\sqrt{3} + 8\sqrt{3} + 12 = 26 + 15\sqrt{3}$
Answer: $26 + 15\sqrt{3}$ [2]
Mark breakdown: 1 mark for multiplying by conjugate, 1 mark for correct expansion and simplification.

(b) Area $= \text{length} \times \text{width} = (7 + 4\sqrt{3})(2 - \sqrt{3}) = 14 - 7\sqrt{3} + 8\sqrt{3} - 12 = 2 + \sqrt{3}$
Answer: $2 + \sqrt{3}$ [1]
Marking note: 1 mark. Direct multiplication (no rationalisation needed).

Section C (12 marks)

16. (a) For $y = Ab^x$, take $\lg$ (or $\ln$) both sides: $\lg y = \lg A + x \lg b$.
This is a linear equation in $x$ and $\lg y$ with gradient $\lg b$ and vertical intercept $\lg A$.
Plot $\lg y$ against $x$. If the points lie on a straight line, the relationship holds.
Then $b = 10^{\text{gradient}}$ and $A = 10^{\text{intercept}}$.
Answer: [2]
Mark breakdown: 1 mark for taking logs and showing linear form, 1 mark for explaining how to verify and find constants.

(b) From table:
$x=1, y=6.0 \Rightarrow \lg y = 0.778$
$x=2, y=18.0 \Rightarrow \lg y = 1.255$
$x=3, y=54.0 \Rightarrow \lg y = 1.732$
$x=4, y=162.0 \Rightarrow \lg y = 2.210$
$x=5, y=486.0 \Rightarrow \lg y = 2.687$

Plotting $\lg y$ vs $x$ gives a straight line.
Gradient $= \frac{2.687 - 0.778}{5 - 1} = \frac{1.909}{4} = 0.47725 \approx \lg 3$
$\Rightarrow \lg b = 0.477 \Rightarrow b = 10^{0.477} \approx 3$
Intercept (at $x=0$): $\lg A \approx 0.301 \Rightarrow A = 10^{0.301} \approx 2$
(Check: $y = 2 \times 3^x$ gives $y=6, 18, 54, 162, 486$ — exact match)
Answer: $A = 2$, $b = 3$ [4]
Mark breakdown: 1 mark for correct $\lg y$ values, 1 mark for plotting/drawing best-fit line, 1 mark for finding gradient and $b$, 1 mark for finding intercept and $A$.

(c) $y = 2 \times 3^{6.5} = 2 \times 3^6 \times 3^{0.5} = 2 \times 729 \times \sqrt{3} = 1458\sqrt{3} \approx 2525.4$
Answer: $2530$ (3 s.f.) or $1458\sqrt{3}$ [2]
Mark breakdown: 1 mark for correct substitution, 1 mark for correct evaluation (3 s.f.).

17. (a) $\log_4 (x^2 - 5x + 6) = 1 \Rightarrow x^2 - 5x + 6 = 4^1 = 4$
$x^2 - 5x + 2 = 0$
$x = \frac{5 \pm \sqrt{25 - 8}}{2} = \frac{5 \pm \sqrt{17}}{2}$
Domain: $x^2 - 5x + 6 > 0 \Rightarrow (x-2)(x-3) > 0 \Rightarrow x < 2$ or $x > 3$
Both solutions: $\frac{5 - \sqrt{17}}{2} \approx 0.44 < 2$ ✓, $\frac{5 + \sqrt{17}}{2} \approx 4.56 > 3$ ✓
Answer: $x = \frac{5 \pm \sqrt{17}}{2}$ [3]
Mark breakdown: 1 mark for removing log, 1 mark for solving quadratic, 1 mark for domain check.

(b) $\log_4 (x^2 - 5x + 6) < 1 \Rightarrow 0 < x^2 - 5x + 6 < 4$
Solve $x^2 - 5x + 6 > 0 \Rightarrow x < 2$ or $x > 3$
Solve $x^2 - 5x + 6 < 4 \Rightarrow x^2 - 5x + 2 < 0 \Rightarrow \frac{5 - \sqrt{17}}{2} < x < \frac{5 + \sqrt{17}}{2}$
Intersection: $\frac{5 - \sqrt{17}}{2} < x < 2$ or $3 < x < \frac{5 + \sqrt{17}}{2}$
Answer: $\frac{5 - \sqrt{17}}{2} < x < 2$ or $3 < x < \frac{5 + \sqrt{17}}{2}$ [2]
Mark breakdown: 1 mark for setting up compound inequality with domain, 1 mark for correct final intervals.

18. (a) Sum of first 4 terms: $S_4 = \frac{a(1-r^4)}{1-r} = 30$
Sum to infinity: $S_\infty = \frac{a}{1-r} = 32$
Answer: $\frac{a(1-r^4)}{1-r} = 30$ and $\frac{a}{1-r} = 32$ [2]
Mark breakdown: 1 mark for each correct equation.

(b) From $S_\infty = 32$: $a = 32(1-r)$
Substitute into $S_4$: $\frac{32(1-r)(1-r^4)}{1-r} = 30 \Rightarrow 32(1-r^4) = 30$
$1 - r^4 = \frac{30}{32} = \frac{15}{16} \Rightarrow r^4 = \frac{1}{16} \Rightarrow r = \frac{1}{2}$ (since $0<r<1$)
$a = 32(1 - \frac{1}{2}) = 16$
Answer: $a = 16$, $r = \frac{1}{2}$ [3]
Mark breakdown: 1 mark for substituting $a$, 1 mark for solving $r^4 = 1/16$, 1 mark for finding $a$.

(c) $S_n = \frac{16(1 - 0.5^n)}{1 - 0.5} = 32(1 - 0.5^n)$
Need $S_n > 31.5 \Rightarrow 32(1 - 0.5^n) > 31.5 \Rightarrow 1 - 0.5^n > \frac{31.5}{32} = 0.984375$
$0.5^n < 0.015625 = \frac{1}{64} = 0.5^6$
Since $0.5^n$ is decreasing, $n > 6$. Least integer $n = 7$.
Answer: $n = 7$ [2]
Mark breakdown: 1 mark for setting up inequality, 1 mark for solving and finding least $n$.

19. (a) $y = \frac{kx^2}{x+1} \Rightarrow y = k \left(\frac{x^2}{x+1}\right)$
Graph of $y$ vs $\frac{x^2}{x+1}$ is straight line through origin with gradient $k$.
Passes through $(4, 12)$: $k = \frac{12}{4} = 3$
Answer: $k = 3$ [2]
Mark breakdown: 1 mark for identifying gradient $= k$, 1 mark for correct calculation.

(b) $y = 27 = \frac{3x^2}{x+1} \Rightarrow 27(x+1) = 3x^2 \Rightarrow 27x + 27 = 3x^2$
$3x^2 - 27x - 27 = 0 \Rightarrow x^2 - 9x - 9 = 0$
$x = \frac{9 \pm \sqrt{81 + 36}}{2} = \frac{9 \pm \sqrt{117}}{2} = \frac{9 \pm 3\sqrt{13}}{2}$
Since $x+1$ in denominator, $x \neq -1$. Both solutions valid.
Answer: $x = \frac{9 \pm 3\sqrt{13}}{2}$ [3]
Mark breakdown: 1 mark for substituting $y=27$ and $k=3$, 1 mark for forming quadratic, 1 mark for solving.

20. (a) Let $2^x = 3^y = 6^z = k$.
Then $x = \log_2 k$, $y = \log_3 k$, $z = \log_6 k$.
$\frac{1}{z} = \log_k 6 = \log_k (2 \times 3) = \log_k 2 + \log_k 3 = \frac{1}{x} + \frac{1}{y}$
$\Rightarrow z = \frac{1}{\frac{1}{x} + \frac{1}{y}} = \frac{xy}{x + y}$
Answer: $z = \frac{xy}{x + y}$ [3]
Mark breakdown: 1 mark for setting common value $k$, 1 mark for expressing reciprocals, 1 mark for deriving $z = \frac{xy}{x+y}$.

(b) $x = 2$, $y = 3 \Rightarrow z = \frac{2 \times 3}{2 + 3} = \frac{6}{5} = 1.2$
Answer: $1.20$ (3 s.f.) [1]
Marking note: 1 mark. Exact fraction $\frac{6}{5}$ also accepted.

End of Answer Key