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Secondary 3 Additional Mathematics Numbers Ratio Proportion Quiz

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Questions

Secondary 3 Additional Mathematics Quiz - Numbers Ratio Proportion

Name: __________________________
Class: __________________________
Date: __________________________
Score: ________ / 50

Duration: 60 minutes
Total Marks: 50

Instructions:

Answer all questions.
Write your answers in the spaces provided.
Show all necessary working clearly. No marks will be given for correct answers without working.
Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question.
The use of an approved scientific calculator is expected, where appropriate.

Section A: Surds and Rationalisation (10 Marks)

1. Simplify the following expression, giving your answer in the form $a\sqrt{b}$ where $a$ and $b$ are integers and $b$ is as small as possible.
$\sqrt{75} - 2\sqrt{12} + \sqrt{48}$
[2]

2. Expand and simplify the following expression:
$(3\sqrt{2} - \sqrt{5})(2\sqrt{2} + 3\sqrt{5})$
[3]

3. Rationalise the denominator of the following fraction and simplify your answer:
$\frac{6}{\sqrt{12} - \sqrt{3}}$
[3]

4. Given that $x = \sqrt{3} + 1$ and $y = \sqrt{3} - 1$ , find the value of $x^2 + y^2$ .
[2]

Section B: Indices and Surd Equations (15 Marks)

5. Show that
$\frac{1}{\sqrt{5} + 2} + \frac{1}{\sqrt{5} - 2} = 2\sqrt{5}$
[4]

6. Solve the equation
$2^{2x+1} = 32$
[2]

7. Solve the equation
$3^{x^2 - 2x} = \frac{1}{27}$
[3]

8. Given that $4^x = 8^{y+1}$ , express $x$ in terms of $y$ .
[3]

9. Solve the equation involving surds:
$\sqrt{2x + 3} = x$
[3]

Section C: Ratio, Proportion and Variation I (15 Marks)

10. Solve the equation:
$\sqrt{x + 5} - \sqrt{x} = 1$
[3]

11. It is given that $y$ varies directly as the square of $x$ . When $x = 3$ , $y = 45$ .
(a) Find the equation connecting $y$ and $x$ .
(b) Find the value of $y$ when $x = 5$ .
[3]

12. It is given that $p$ varies inversely as the cube root of $q$ . When $q = 8$ , $p = 5$ .
(a) Find the equation connecting $p$ and $q$ .
(b) Find the value of $q$ when $p = 10$ .
[4]

13. The variable $z$ is such that $z = k \frac{x^2}{y}$ , where $k$ is a constant.
Given that $z = 12$ when $x = 4$ and $y = 2$ ,
(a) find the value of $k$ ,
(b) find the percentage change in $z$ if $x$ is increased by 10% and $y$ is decreased by 10%.
[5]

14. The resistance $R$ of a wire varies directly as its length $L$ and inversely as the square of its diameter $d$ .
A wire of length 100 m and diameter 2 mm has a resistance of 5 $\Omega$ .
(a) Find the formula for $R$ in terms of $L$ and $d$ .
(b) Calculate the resistance of a wire of length 150 m and diameter 3 mm.
[3]

Section D: Ratio, Proportion and Variation II (10 Marks)

15. The time $T$ taken for a journey varies partly as the distance $D$ and partly as the square of the distance $D$ .
When $D = 10$ km, $T = 25$ minutes.
When $D = 20$ km, $T = 70$ minutes.
(a) Express $T$ in terms of $D$ .
(b) Find the time taken when $D = 30$ km.
[4]

16. $A$ varies jointly as $B$ and the square of $C$ . If $A = 24$ when $B = 3$ and $C = 2$ , find $A$ when $B = 5$ and $C = 3$ .
[3]

17. The cost $C$ of producing a batch of items consists of a fixed cost and a variable cost which is proportional to the number of items $n$ .
Producing 100 items costs $500. Producing 200 items costs $900.
(a) Find the formula for $C$ in terms of $n$ .
(b) Find the cost of producing 150 items.
[3]

18. Given that $y$ is inversely proportional to $\sqrt{x}$ , and $y=10$ when $x=4$ .
Find the value of $x$ when $y=5$ .
[2]

19. The volume $V$ of a gas varies directly with its temperature $T$ (in Kelvin) and inversely with its pressure $P$ .
If $V = 200$ cm $^3$ when $T = 300$ K and $P = 100$ kPa, find the volume when $T = 350$ K and $P = 140$ kPa.
[3]

20. Simplify the expression fully:
$\frac{\sqrt{18} + \sqrt{8}}{\sqrt{2}}$
[2]

Answers

Secondary 3 Additional Mathematics Quiz - Numbers Ratio Proportion (Answer Key)

1. Simplify $\sqrt{75} - 2\sqrt{12} + \sqrt{48}$
$\sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3}$
$2\sqrt{12} = 2\sqrt{4 \times 3} = 2(2\sqrt{3}) = 4\sqrt{3}$
$\sqrt{48} = \sqrt{16 \times 3} = 4\sqrt{3}$
Expression $= 5\sqrt{3} - 4\sqrt{3} + 4\sqrt{3} = 5\sqrt{3}$
Answer: $5\sqrt{3}$
[2 marks: 1 for simplifying at least two terms correctly, 1 for final answer]

2. Expand $(3\sqrt{2} - \sqrt{5})(2\sqrt{2} + 3\sqrt{5})$
$= 3\sqrt{2}(2\sqrt{2}) + 3\sqrt{2}(3\sqrt{5}) - \sqrt{5}(2\sqrt{2}) - \sqrt{5}(3\sqrt{5})$
$= 6(2) + 9\sqrt{10} - 2\sqrt{10} - 3(5)$
$= 12 + 7\sqrt{10} - 15$
$= 7\sqrt{10} - 3$
Answer: $7\sqrt{10} - 3$
[3 marks: 1 for expansion, 1 for simplifying surds/integers, 1 for final answer]

3. Rationalise $\frac{6}{\sqrt{12} - \sqrt{3}}$
First simplify denominator: $\sqrt{12} - \sqrt{3} = 2\sqrt{3} - \sqrt{3} = \sqrt{3}$
Expression becomes $\frac{6}{\sqrt{3}}$
Rationalise: $\frac{6\sqrt{3}}{3} = 2\sqrt{3}$
Answer: $2\sqrt{3}$
[3 marks: 1 for simplifying denominator, 1 for rationalisation step, 1 for final answer]

4. Given $x = \sqrt{3} + 1, y = \sqrt{3} - 1$ , find $x^2 + y^2$
$x^2 = (\sqrt{3} + 1)^2 = 3 + 2\sqrt{3} + 1 = 4 + 2\sqrt{3}$
$y^2 = (\sqrt{3} - 1)^2 = 3 - 2\sqrt{3} + 1 = 4 - 2\sqrt{3}$
$x^2 + y^2 = (4 + 2\sqrt{3}) + (4 - 2\sqrt{3}) = 8$
Answer: 8
[2 marks: 1 for correct expansion of squares, 1 for final sum]

5. Show that $\frac{1}{\sqrt{5} + 2} + \frac{1}{\sqrt{5} - 2} = 2\sqrt{5}$
LHS $= \frac{1(\sqrt{5}-2)}{(\sqrt{5}+2)(\sqrt{5}-2)} + \frac{1(\sqrt{5}+2)}{(\sqrt{5}-2)(\sqrt{5}+2)}$
Denominator $= 5 - 4 = 1$
LHS $= (\sqrt{5} - 2) + (\sqrt{5} + 2)$
$= 2\sqrt{5}$
$= \text{RHS}$
[4 marks: 1 for correct conjugates, 1 for common denominator, 1 for numerator simplification, 1 for final conclusion]

6. Solve $2^{2x+1} = 32$
$32 = 2^5$
$2x + 1 = 5$
$2x = 4$
$x = 2$
Answer: $x = 2$
[2 marks: 1 for equating indices, 1 for solution]

7. Solve $3^{x^2 - 2x} = \frac{1}{27}$
$\frac{1}{27} = 3^{-3}$
$x^2 - 2x = -3$
$x^2 - 2x + 3 = 0$
Discriminant $\Delta = (-2)^2 - 4(1)(3) = 4 - 12 = -8 < 0$
No real solutions.
Answer: No real solution
[3 marks: 1 for converting RHS, 1 for quadratic equation, 1 for identifying no real roots]

8. Given $4^x = 8^{y+1}$ , express $x$ in terms of $y$
$(2^2)^x = (2^3)^{y+1}$
$2^{2x} = 2^{3(y+1)}$
$2x = 3(y+1)$
$2x = 3y + 3$
$x = \frac{3y + 3}{2}$
Answer: $x = \frac{3y + 3}{2}$
[3 marks: 1 for base conversion, 1 for equating indices, 1 for final expression]

9. Solve $\sqrt{2x + 3} = x$
Square both sides: $2x + 3 = x^2$
$x^2 - 2x - 3 = 0$
$(x - 3)(x + 1) = 0$
$x = 3$ or $x = -1$
Check: If $x = -1$ , LHS $= \sqrt{1} = 1$ , RHS $= -1$ . $1 \neq -1$ (Reject)
If $x = 3$ , LHS $= \sqrt{9} = 3$ , RHS $= 3$ . (Accept)
Answer: $x = 3$
[3 marks: 1 for solving quadratic, 1 for checking validity, 1 for final answer]

10. Solve $\sqrt{x + 5} - \sqrt{x} = 1$
$\sqrt{x + 5} = 1 + \sqrt{x}$
Square both sides: $x + 5 = 1 + 2\sqrt{x} + x$
$5 = 1 + 2\sqrt{x}$
$4 = 2\sqrt{x}$
$2 = \sqrt{x}$
Square again: $x = 4$
Check: $\sqrt{9} - \sqrt{4} = 3 - 2 = 1$ . (Valid)
Answer: $x = 4$
[3 marks: 1 for isolation and squaring, 1 for solving for $\sqrt{x}$ , 1 for final answer]

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

12. $p$ varies inversely as $\sqrt[3]{q}$ . $q=8, p=5$ .
(a) $p = \frac{k}{\sqrt[3]{q}}$
$5 = \frac{k}{\sqrt[3]{8}} \Rightarrow 5 = \frac{k}{2} \Rightarrow k = 10$
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) 1
[4 marks: 1 for k, 1 for equation, 1 for substitution, 1 for final value]

13. $z = k \frac{x^2}{y}$ . $z=12, x=4, y=2$ .
(a) $12 = k \frac{4^2}{2} = k \frac{16}{2} = 8k \Rightarrow k = 1.5$
(b) New $x = 1.1x$ , New $y = 0.9y$
New $z = 1.5 \frac{(1.1x)^2}{0.9y} = 1.5 \frac{1.21 x^2}{0.9 y} = \frac{1.21}{0.9} \left( 1.5 \frac{x^2}{y} \right)$
New $z = \frac{1.21}{0.9} z_{old} = 1.3444... z_{old}$
Percentage change $= (1.3444... - 1) \times 100\% = 34.4\%$ increase.
Answer: (a) $k=1.5$ , (b) 34.4% increase
[5 marks: 1 for k, 1 for setting up new ratio, 1 for calculation factor, 1 for percentage conversion, 1 for "increase"]

14. $R \propto \frac{L}{d^2} \Rightarrow R = \frac{kL}{d^2}$ .
$L=100, d=2, R=5$ .
(a) $5 = \frac{k(100)}{2^2} = \frac{100k}{4} = 25k \Rightarrow k = 0.2$
Formula: $R = \frac{0.2L}{d^2}$
(b) $L=150, d=3$ .
$R = \frac{0.2(150)}{3^2} = \frac{30}{9} = \frac{10}{3} = 3.33 \, \Omega$
Answer: (a) $R = \frac{0.2L}{d^2}$ , (b) $3.33 \, \Omega$
[3 marks: 1 for formula/constants, 1 for substitution, 1 for final answer]

15. $T = k_1 D + k_2 D^2$ .
$D=10, T=25 \Rightarrow 25 = 10k_1 + 100k_2$ (Eq 1)
$D=20, T=70 \Rightarrow 70 = 20k_1 + 400k_2$ (Eq 2)
Divide Eq 2 by 10: $7 = 2k_1 + 40k_2$
From Eq 1: $2.5 = k_1 + 10k_2 \Rightarrow k_1 = 2.5 - 10k_2$
Sub into modified Eq 2: $7 = 2(2.5 - 10k_2) + 40k_2$
$7 = 5 - 20k_2 + 40k_2$
$2 = 20k_2 \Rightarrow k_2 = 0.1$
$k_1 = 2.5 - 10(0.1) = 1.5$
(a) $T = 1.5D + 0.1D^2$
(b) $D=30$ : $T = 1.5(30) + 0.1(30^2) = 45 + 0.1(900) = 45 + 90 = 135$ minutes.
Answer: (a) $T = 1.5D + 0.1D^2$ , (b) 135 minutes
[4 marks: 1 for setting up equations, 1 for solving constants, 1 for equation, 1 for final calculation]

16. $A = k B C^2$ .
$24 = k(3)(2^2) = 12k \Rightarrow k = 2$ .
Formula: $A = 2BC^2$ .
When $B=5, C=3$ :
$A = 2(5)(3^2) = 10(9) = 90$ .
Answer: 90
[3 marks: 1 for k, 1 for formula, 1 for final answer]

17. $C = a + bn$ .
$500 = a + 100b$ (1)
$900 = a + 200b$ (2)
(2) - (1): $400 = 100b \Rightarrow b = 4$ .
$a = 500 - 100(4) = 100$ .
(a) $C = 100 + 4n$
(b) $n=150 \Rightarrow C = 100 + 4(150) = 100 + 600 = 700$ .
Answer: (a) $C = 100 + 4n$ , (b) $700
[3 marks: 1 for constants, 1 for formula, 1 for final calculation]

18. $y = \frac{k}{\sqrt{x}}$ .
$10 = \frac{k}{\sqrt{4}} = \frac{k}{2} \Rightarrow k = 20$ .
Equation: $y = \frac{20}{\sqrt{x}}$ .
When $y=5$ : $5 = \frac{20}{\sqrt{x}} \Rightarrow \sqrt{x} = 4 \Rightarrow x = 16$ .
Answer: 16
[2 marks: 1 for constant/equation, 1 for final answer]

19. $V = \frac{kT}{P}$ .
$200 = \frac{k(300)}{100} = 3k \Rightarrow k = \frac{200}{3}$ .
New V: $V = \frac{\frac{200}{3}(350)}{140} = \frac{200 \times 350}{3 \times 140}$ .
$350/140 = 2.5 = 5/2$ .
$V = \frac{200 \times 5}{3 \times 2} = \frac{1000}{6} = 166.66...$
Answer: 167 cm $^3$ (3 s.f.)
[3 marks: 1 for constant, 1 for substitution, 1 for final answer]

20. Simplify $\frac{\sqrt{18} + \sqrt{8}}{\sqrt{2}}$
$\sqrt{18} = 3\sqrt{2}$ , $\sqrt{8} = 2\sqrt{2}$ .
Numerator $= 3\sqrt{2} + 2\sqrt{2} = 5\sqrt{2}$ .
Expression $= \frac{5\sqrt{2}}{\sqrt{2}} = 5$ .
Answer: 5
[2 marks: 1 for simplifying surds, 1 for final division]