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

TuitionGoWhere Practice Paper - Mathematics Secondary 1

Answer Key and Marking Scheme

Version: 3 of 5

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Section A: Short Answer Questions [20 marks]

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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.

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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]

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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]

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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: $360$

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

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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.

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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).

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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 pending generation: number_line for Q8.

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.

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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

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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.

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Section B: Structured Questions [24 marks]

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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))

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

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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

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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: $10$

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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]

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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.

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Section C: Problem Solving [16 marks]

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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.

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

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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

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

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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

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End of Answer Key