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Secondary 1 Mathematics Semestral Assessment 2 (End of Year) Paper 3

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TuitionGoWhere Practice Paper - Mathematics Secondary 1

TuitionGoWhere Secondary School (AI)


Subject:	Mathematics
Level:	Secondary 1 (G3)
Paper:	SA2 End-of-Year Practice
Version:	3 of 5
Duration:	1 hour 15 minutes
Total Marks:	60
Name:	_________________________
Class:	_________________________
Date:	_________________________

Instructions to Candidates

Write your name, class, and date in the spaces provided above.
This paper consists of THREE sections: Section A, Section B, and Section C.
Answer all questions.
Show all your working clearly. Marks will be awarded for correct method even if the final answer is wrong.
Write your answers in the spaces provided. If the space is insufficient, continue on the next page.
Non-exact numerical answers should be given correct to 3 significant figures, or 1 decimal place for angles in degrees, unless stated otherwise.
The use of calculators is allowed.
The number of marks available is shown in brackets [ ] at the end of each question or part question.

Section A: Short Answer Questions [20 marks]

Answer all questions. Show your working clearly.

1. Evaluate $-5 + 8 \times (-3) - (-7)$ . [2]

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2. Express 252 as a product of its prime factors, in index notation. [2]

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3. Find the highest common factor (HCF) of 84 and 126. [2]

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4. Find the lowest common multiple (LCM) of 18, 24, and 30. [2]

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5. Simplify $\frac{2}{5} - \frac{3}{4} \div \frac{9}{10}$ . [2]

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6. Write the following numbers in ascending order: $-\frac{3}{4}$ , $-0.7$ , $-\frac{2}{3}$ , $-0.65$ [2]

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7. Evaluate $\sqrt[3]{-64} + (-3)^2 - | -5 |$ . [2]

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8. Solve the inequality $-4x + 7 \geq 19$ and illustrate the solution on the number line in the space below. [2]

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<image_placeholder> id: Q8-fig1 type: number_line linked_question: Q8 description: A horizontal number line from -6 to 2 with evenly spaced tick marks at integer values, labeled appropriately labels: integers from -6 to 2 values: solution interval x ≤ -3 must_show: open or closed circle at -3, arrow pointing left, clear scale markings </image_placeholder>

9. The ratio of the number of apples to oranges in a basket is $5:8$ . If there are 24 more oranges than apples, how many fruits are there in the basket? [2]

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10. A map has a scale of $1 : 25\ 000$ . Find the actual distance, in kilometres, represented by 8.4 cm on the map. [2]

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Section B: Structured Questions [24 marks]

Answer all questions. Show your working clearly.

11. (a) Using a calculator, evaluate $\frac{\sqrt{7.29} \times 4.5^3}{1.8^2 - 0.96}$ . [2]

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(b) Express your answer in part (a) correct to 3 significant figures. [1]

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12. (a) Find the value of $(-2)^4 - (-2)^3$ . [1]

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(b) Evaluate $\frac{3}{8} + \left(-\frac{5}{6}\right) \times \frac{9}{10}$ . [2]

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13. A rectangular field has length $\frac{7}{8}$ m and width $\frac{4}{5}$ m.

(a) Find the perimeter of the field, giving your answer as a fraction in its simplest form. [2]

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(b) Find the area of the field, giving your answer as a fraction in its simplest form. [2]

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14. Simplify the following, leaving your answer in index notation where appropriate.

(a) $\frac{3^5 \times 3^7}{(3^2)^3}$ [2]

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(b) $\frac{2^4 \times 5^4}{10^3}$ [2]

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15. Three people, Ali, Beth, and Carl, share $480 in the ratio $2:3:5$ .

(a) Calculate the amount each person receives. [2]

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(b) Ali gives $\frac{1}{4}$ of his share to Beth. Find the new ratio of Ali's money to Beth's money to Carl's money. Give your ratio in its simplest form. [3]

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16. The temperatures recorded at a weather station over 5 days were: $-3.5^\circ\text{C}$ , $2.8^\circ\text{C}$ , $-1.2^\circ\text{C}$ , $0.6^\circ\text{C}$ , $-4.9^\circ\text{C}$ .

(a) Arrange the temperatures in ascending order. [1]

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(b) Find the difference between the highest and lowest temperatures. [2]

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Section C: Problem Solving [16 marks]

Answer all questions. Show your working clearly.

17. A recipe for 6 people requires 450 g of flour and 300 g of sugar.

(a) Find the ratio of flour to sugar in its simplest form. [1]

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(b) If the recipe is scaled up to serve 15 people, calculate the mass of flour needed. [2]

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(c) A chef has 2 kg of flour and 1.2 kg of sugar. What is the maximum number of people he can serve if he must use whole recipe quantities? [3]

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18. The mass of 8 identical metal cubes and 5 identical metal spheres is 3.7 kg. The mass of 4 identical metal cubes and 3 identical metal spheres is 1.9 kg. Let the mass of each cube be $c$ kg and the mass of each sphere be $s$ kg.

(a) Write down two equations in terms of $c$ and $s$ . [2]

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(b) By solving these equations simultaneously, find the mass of each cube and each sphere. [3]

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19. A shop sells three types of fruit juice: apple, orange, and mixed. The ratio of bottles sold was apple : orange : mixed = $4:5:3$ . The shop sold 36 bottles of mixed juice.

(a) How many bottles of apple juice were sold? [2]

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(b) The price of each bottle is: apple $2.40, orange $2.80, mixed $3.20. Calculate the total revenue from all juice sold. [3]

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20. A number $N$ is written as $2^3 \times 3^2 \times 5$ .

(a) Find the value of $N$ . [1]

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(b) Find the smallest positive integer $k$ such that $N \times k$ is a perfect square. [2]

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(d) Hence, find the smallest positive integer $p$ such that $\sqrt[3]{N \times p}$ is an integer and $\sqrt{N \times p}$ is also an integer. [3]

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END OF PAPER

Section A Total: 20 marks
Section B Total: 24 marks
Section C Total: 16 marks
Grand Total: 60 marks

Answers

TuitionGoWhere Practice Paper - Mathematics Secondary 1

Answer Key and Marking Scheme

Version: 3 of 5

Section A: Short Answer Questions [20 marks]

1. Evaluate $-5 + 8 \times (-3) - (-7)$ . [2]

Working: Using BODMAS/PEMDAS, multiplication comes before addition and subtraction.

First: $8 \times (-3) = -24$ [½]
Then: $-5 + (-24) - (-7)$ [½]
$= -5 - 24 + 7$ [½]
$= -29 + 7 = -22$ [½]

Answer: $-22$

Common mistake: Working left to right without priority: $-5 + 8 = 3$ , then $3 \times (-3) = -9$ , etc. Multiplication must be done first.

2. Express 252 as a product of its prime factors, in index notation. [2]

Working:

$252 \div 2 = 126$ [½]
$126 \div 2 = 63$
$63 \div 3 = 21$
$21 \div 3 = 7$
$7 \div 7 = 1$ [½ for complete factor tree or continuous division]

So $252 = 2 \times 2 \times 3 \times 3 \times 7$ [½]

Answer: $2^2 \times 3^2 \times 7$ [½]

3. Find the highest common factor (HCF) of 84 and 126. [2]

Working: Prime factorisation:

$84 = 2^2 \times 3 \times 7$ [½]
$126 = 2 \times 3^2 \times 7$ [½]

HCF uses the lowest power of each common prime factor:

Common primes: $2^1$ , $3^1$ , $7^1$ [½]

Answer: $\text{HCF} = 2 \times 3 \times 7 = 42$ [½]

4. Find the lowest common multiple (LCM) of 18, 24, and 30. [2]

Working: Prime factorisation:

$18 = 2 \times 3^2$ [½]
$24 = 2^3 \times 3$
$30 = 2 \times 3 \times 5$

LCM uses the highest power of each prime present:

$2^3$ , $3^2$ , $5^1$ [½]

LCM $= 8 \times 9 \times 5 = 360$ [1]

Answer: $360$

5. Simplify $\frac{2}{5} - \frac{3}{4} \div \frac{9}{10}$ . [2]

Working: Division first (BODMAS):

$\frac{3}{4} \div \frac{9}{10} = \frac{3}{4} \times \frac{10}{9}$ [½]
$= \frac{30}{36} = \frac{5}{6}$ [½]

Then subtraction:

$\frac{2}{5} - \frac{5}{6} = \frac{12}{30} - \frac{25}{30}$ [½]
$= \frac{-13}{30}$ [½]

Answer: $-\frac{13}{30}$

6. Write the following numbers in ascending order: $-\frac{3}{4}$ , $-0.7$ , $-\frac{2}{3}$ , $-0.65$ [2]

Working: Convert all to decimals:

$-\frac{3}{4} = -0.75$ [½]
$-\frac{2}{3} = -0.666...$ [½]

For negative numbers, the one with larger absolute value is smaller:

$-0.75 < -0.7 < -0.666... < -0.65$ [½]

Answer: $-\frac{3}{4}$ , $-0.7$ , $-\frac{2}{3}$ , $-0.65$ [½]

Concept: On the number line, numbers to the left are smaller. For negatives, closer to zero means larger.

7. Evaluate $\sqrt[3]{-64} + (-3)^2 - | -5 |$ . [2]

Working:

$\sqrt[3]{-64} = -4$ (since $(-4)^3 = -64$ ) [½]
$(-3)^2 = 9$ (note: brackets mean the negative is squared, giving positive) [½]
$| -5 | = 5$ [½]

So: $-4 + 9 - 5 = 0$ [½]

Answer: $0$

Common mistake: $\sqrt[3]{-64} = 4$ (forgetting cube root of negative is negative) or $(-3)^2 = -9$ (thinking only the 3 is squared).

8. Solve the inequality $-4x + 7 \geq 19$ and illustrate the solution on the number line. [2]

Working:

$-4x + 7 \geq 19$ [½]
$-4x \geq 12$ (subtract 7 from both sides) [½]
$x \leq -3$ (divide by negative 4, so reverse the inequality sign) [½]

Expected number line features:

Closed (filled) circle at $-3$ (since $\leq$ includes equality)
Arrow pointing to the left (towards smaller numbers)
Clear labeling with integers marked

<image_placeholder> id: Q8-fig1 type: number_line linked_question: Q8 description: Horizontal number line showing solution to x ≤ -3 labels: -6, -5, -4, -3, -2, -1, 0, 1, 2 values: closed circle at -3, shading/arrow extending left must_show: closed (filled) circle at -3, left-pointing arrow or shading, evenly spaced integer tick marks with labels </image_placeholder>

Answer: $x \leq -3$

Critical concept: When multiplying or dividing both sides of an inequality by a negative number, always reverse the inequality direction. This is a major exam trap.

9. The ratio of apples to oranges is $5:8$ . There are 24 more oranges than apples. How many fruits in the basket? [2]

Working:

Difference in ratio parts: $8 - 5 = 3$ parts [½]
These 3 parts represent 24 fruits
So 1 part = $24 \div 3 = 8$ fruits [½]
Total parts = $5 + 8 = 13$ parts
Total fruits = $13 \times 8 = 104$ [1]

Answer: $104$ fruits

10. Map scale $1 : 25\ 000$ . Find actual distance in km for 8.4 cm on map. [2]

Working:

Actual distance = $8.4 \times 25\ 000$ cm [½]
$= 210\ 000$ cm [½]
Convert to km: $210\ 000 \div 100\ 000 = 2.1$ km [1]

Answer: $2.1$ km

Method note: Scale $1:25\ 000$ means 1 cm on map = 25 000 cm actual = 0.25 km actual. So $8.4 \times 0.25 = 2.1$ km is an alternative valid method.

Section B: Structured Questions [24 marks]

11. (a) Using a calculator, evaluate $\frac{\sqrt{7.29} \times 4.5^3}{1.8^2 - 0.96}$ . [2]

Working:

$\sqrt{7.29} = 2.7$ [½]
$4.5^3 = 91.125$ [½]
$1.8^2 = 3.24$ ; so denominator $= 3.24 - 0.96 = 2.28$ [½]
Numerator: $2.7 \times 91.125 = 246.0375$
Final: $\frac{246.0375}{2.28} = 107.911...$ [½]

Accept calculator value: $107.9111842...$

(b) Express to 3 significant figures. [1]

Answer: $108$ (or $107.9$ if they truncated instead of rounding—award if correctly rounded to 3 s.f. from their (a))

12. (a) Find the value of $(-2)^4 - (-2)^3$ . [1]

Working:

$(-2)^4 = 16$ (even power, positive) [½]
$(-2)^3 = -8$ (odd power, negative) [½]
$16 - (-8) = 16 + 8 = 24$

Answer: $24$

(b) Evaluate $\frac{3}{8} + \left(-\frac{5}{6}\right) \times \frac{9}{10}$ . [2]

Working: Multiplication first:

$\left(-\frac{5}{6}\right) \times \frac{9}{10} = -\frac{45}{60} = -\frac{3}{4}$ [1]

Then addition:

$\frac{3}{8} + \left(-\frac{3}{4}\right) = \frac{3}{8} - \frac{6}{8} = -\frac{3}{8}$ [1]

Answer: $-\frac{3}{8}$

13. Rectangle: length $\frac{7}{8}$ m, width $\frac{4}{5}$ m.

(a) Perimeter [2]

Working:

Perimeter $= 2 \times \left(\frac{7}{8} + \frac{4}{5}\right)$ [½]
$= 2 \times \left(\frac{35}{40} + \frac{32}{40}\right)$ [½]
$= 2 \times \frac{67}{40}$ [½]
$= \frac{67}{20}$ m or $3\frac{7}{20}$ m [½]

Answer: $\frac{67}{20}$ m (or $3.35$ m)

(b) Area [2]

Working:

Area $= \frac{7}{8} \times \frac{4}{5}$ [½]
$= \frac{28}{40}$ [½]
$= \frac{7}{10}$ m² [1]

Answer: $\frac{7}{10}$ m²

14. Simplify in index notation:

(a) $\frac{3^5 \times 3^7}{(3^2)^3}$ [2]

Working:

Numerator: $3^5 \times 3^7 = 3^{12}$ (add indices: $a^m \times a^n = a^{m+n}$ ) [½]
Denominator: $(3^2)^3 = 3^6$ (multiply indices: $(a^m)^n = a^{mn}$ ) [½]
Division: $3^{12} \div 3^6 = 3^6$ (subtract indices: $a^m \div a^n = a^{m-n}$ ) [1]

Answer: $3^6$ (or $729$ )

(b) $\frac{2^4 \times 5^4}{10^3}$ [2]

Working:

$2^4 \times 5^4 = (2 \times 5)^4 = 10^4$ (using $a^n \times b^n = (ab)^n$ ) [1]
$\frac{10^4}{10^3} = 10^1 = 10$ [1]

Answer: $10$

15. Ali, Beth, Carl share $480 in ratio $2:3:5$ .

(a) Amount each receives. [2]

Working:

Total parts = $2 + 3 + 5 = 10$ parts [½]
Value of 1 part = $480 \div 10 = \$ 48$ [½]
Ali: $2 \times \$ 48 = $96$ [½]
Beth: $3 \times \$ 48 = $144$
Carl: $5 \times \$ 48 = $240$ [½]

Answers: Ali $96, Beth $144, Carl $240

(b) Ali gives $\frac{1}{4}$ of his share to Beth. New ratio. [3]

Working:

Ali gives away: $\frac{1}{4} \times 96 = \$ 24$ [1]
Ali now has: $96 - 24 = \$ 72$ [½]
Beth now has: $144 + 24 = \$ 168$ [½]
Carl unchanged: $240

New amounts: $72 : 168 : 240$ [½]

Simplify by dividing by 24:

$72 \div 24 = 3$
$168 \div 24 = 7$
$240 \div 24 = 10$ [½]

Answer: $3 : 7 : 10$ [½]

16. Temperatures: $-3.5^\circ\text{C}$ , $2.8^\circ\text{C}$ , $-1.2^\circ\text{C}$ , $0.6^\circ\text{C}$ , $-4.9^\circ\text{C}$

(a) Ascending order. [1]

Answer: $-4.9^\circ\text{C}$ , $-3.5^\circ\text{C}$ , $-1.2^\circ\text{C}$ , $0.6^\circ\text{C}$ , $2.8^\circ\text{C}$

(b) Difference between highest and lowest. [2]

Working:

Highest: $2.8^\circ\text{C}$ [½]
Lowest: $-4.9^\circ\text{C}$ [½]
Difference = $2.8 - (-4.9)$ [½]
$= 2.8 + 4.9 = 7.7^\circ\text{C}$ [½]

Answer: $7.7^\circ\text{C}$

Concept: "Difference" always means the positive gap between two values. Subtracting a negative is equivalent to adding its absolute value.

Section C: Problem Solving [16 marks]

17. Recipe: 450 g flour, 300 g sugar for 6 people.

(a) Ratio flour : sugar in simplest form. [1]

Working:

$450 : 300 = 45 : 30 = 3 : 2$ [1]

Answer: $3 : 2$

(b) Flour for 15 people. [2]

Working:

Flour per person = $450 \div 6 = 75$ g [½]
For 15 people: $75 \times 15 = 1125$ g [½]

Or using ratio: $\frac{15}{6} = \frac{5}{2}$ , so $450 \times \frac{5}{2} = 1125$ g [1]

Answer: $1125$ g (or $1.125$ kg)

(c) Maximum people with 2 kg flour and 1.2 kg sugar. [3]

Working:

Flour per person: $450 \div 6 = 75$ g [½]
From flour: $2000 \div 75 = 26.666...$ So maximum 26 people from flour constraint [½]
Sugar per person: $300 \div 6 = 50$ g [½]
From sugar: $1200 \div 50 = 24$ people exactly [½]

The limiting ingredient is sugar. [½]

Answer: 24 people [½]

Key concept: Must check BOTH constraints and take the minimum. Whole number of people required.

18. 8 cubes + 5 spheres = 3.7 kg; 4 cubes + 3 spheres = 1.9 kg.

(a) Two equations. [2]

Answers:

$8c + 5s = 3.7$ [1]
$4c + 3s = 1.9$ [1]

(b) Solve simultaneously. [3]

Working: Elimination method:

Multiply second equation by 2: $8c + 6s = 3.8$ [½]
Subtract first equation: $(8c + 6s) - (8c + 5s) = 3.8 - 3.7$ [½]
So $s = 0.1$ kg [½]

Substitute back:

$4c + 3(0.1) = 1.9$ [½]
$4c + 0.3 = 1.9$
$4c = 1.6$
$c = 0.4$ kg [½]

Answers: Each cube = $0.4$ kg, each sphere = $0.1$ kg [½]

Verification: $8(0.4) + 5(0.1) = 3.2 + 0.5 = 3.7$ ✓

19. Apple : orange : mixed = $4:5:3$ . Mixed = 36 bottles.

(a) Apple bottles sold. [2]

Working:

3 parts (mixed) = 36 [½]
1 part = $36 \div 3 = 12$ bottles [½]
Apple (4 parts) = $4 \times 12 = 48$ bottles [1]

Answer: 48 bottles

(b) Total revenue. [3]

Working:

Orange: 5 parts = $5 \times 12 = 60$ bottles [½]
Total bottles: $48 + 60 + 36 = 144$ bottles (or use parts: $4+5+3=12$ parts = 144) [½]

Revenue calculation:

Apple: $48 \times \$ 2.40 = $115.20$ [½]
Orange: $60 \times \$ 2.80 = $168.00$ [½]
Mixed: $36 \times \$ 3.20 = $115.20$ [½]

Total: $115.20 + 168.00 + 115.20 = \$ 398.40$ [½]

Answer: $398.40

20. $N = 2^3 \times 3^2 \times 5$

(a) Value of $N$ . [1]

Working:

$N = 8 \times 9 \times 5 = 360$

Answer: $360$

(b) Smallest $k$ such that $N \times k$ is perfect square. [2]

Working: For perfect square, all prime exponents must be even:

Current: $2^3$ , $3^2$ , $5^1$ [½]
Need: $2^4$ (need one more 2), $3^2$ (already even), $5^2$ (need one more 5) [½]

So $k = 2^1 \times 5^1 = 10$ [1]

Answer: $k = 10$

(c) Smallest $m$ such that $N \times m$ is perfect cube. [2]

Working: For perfect cube, all prime exponents must be multiples of 3:

Current: $2^3$ (good), $3^2$ (need one more 3), $5^1$ (need two more 5s) [½]
Need: $3^3$ , $5^3$ [½]

So $m = 3^1 \times 5^2 = 3 \times 25 = 75$ [1]

Answer: $m = 75$

(d) Smallest $p$ such that both $\sqrt[3]{N \times p}$ and $\sqrt{N \times p}$ are integers. [3]

Working: Need $N \times p$ to be both a perfect square AND perfect cube, i.e., a perfect sixth power (LCM of 2 and 3 is 6). [1]

Exponents in $N \times p$ must be multiples of 6:

For 2: need exponent $\geq 3$ and multiple of 6, so need $2^3$ (to make $2^6$ ) [½]
For 3: need exponent $\geq 2$ and multiple of 6, so need $3^4$ (to make $3^6$ ) [½]
For 5: need exponent $\geq 1$ and multiple of 6, so need $5^5$ (to make $5^6$ ) [½]

So $p = 2^3 \times 3^4 \times 5^5$ [½]

Calculate: $8 \times 81 \times 3125 = 8 \times 253125 = 2\ 025\ 000$

Answer: $p = 2\ 025\ 000$ (or $2^3 \times 3^4 \times 5^5$ )

Verification: $N \times p = 2^6 \times 3^6 \times 5^6 = (2 \times 3 \times 5)^6 = 30^6 = (30^3)^2 = (30^2)^3 = 27\ 000^2 = 900^3$

Mark Allocation Summary

Section	Marks
Section A (Questions 1-10)	20
Section B (Questions 11-16)	24
Section C (Questions 17-20)	16
Grand Total	60

End of Answer Key