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

TuitionGoWhere Practice Paper — Mathematics Secondary 1

Answer Key — SA2 Paper 2, Version 3 of 5

---

Section A — Short Answer [20 marks]

---

Question 1 [2]

Answer: $3.72 \times 10^{-4}$

Working: Move the decimal point 4 places to the right to obtain 3.72. Since the original number is less than 1, the exponent is negative.

$0.000\,372 = 3.72 \times 10^{-4}$

Marking notes:

• 1 mark for correct coefficient (3.72)
• 1 mark for correct exponent (−4)
• Accept $3.72 \times 10^{-4}$ only; do not accept $372 \times 10^{-6}$ or similar non-standard forms.

---

Question 2 [2]

Answer: $4 : 1$

Working: Convert to the same unit: $2.4 \text{ km} = 2400 \text{ m}$

$2400 : 600 = \frac{2400}{600} : \frac{600}{600} = 4 : 1$

Marking notes:

• 1 mark for converting to the same unit
• 1 mark for correct simplified ratio $4 : 1$

---

Question 3 [2]

Answer: 525 g

Working: Flour per pancake: $300 \div 8 = 37.5$ g

Flour for 14 pancakes: $37.5 \times 14 = 525$ g

Marking notes:

• 1 mark for correct method (finding unit rate or setting up proportion)
• 1 mark for correct answer (525 g)

---

Question 4 [2]

Answer: $-19$

Working: Follow order of operations (BODMAS/PEMDAS):

$(-7) + 5 \times (-3) - (-12) \div 4$ $= (-7) + (-15) - (-3)$ $= -7 - 15 + 3$ $= -19$

Marking notes:

• 1 mark for correct handling of multiplication and division steps
• 1 mark for correct final answer
• Common error: students may add before multiplying; penalise if working shows this.

---

Question 5 [2]

Answer: 24

Working: Prime factorisation of 72: $72 = 2^3 \times 3^2$

Prime factorisation of 120: $120 = 2^3 \times 3 \times 5$

HCF = product of lowest powers of common primes: $2^3 \times 3 = 8 \times 3 = 24$

Marking notes:

• 1 mark for correct prime factorisations
• 1 mark for correct HCF = 24

---

Question 6 [2]

Answer: 21 girls

Working: Total ratio parts: $5 + 7 = 12$

Each part: $36 \div 12 = 3$

Number of girls: $7 \times 3 = 21$

Marking notes:

• 1 mark for finding the value of one part
• 1 mark for correct answer (21)

---

Question 7 [2]

(a) Answer: 8.46

Working: The third decimal place is 6, which is ≥ 5, so round up: $8.4567 \approx 8.46$

(b) Answer: 8.46

Working: The first three significant figures are 8, 4, 5. The next digit is 6 (≥ 5), so round up: $8.4567 \approx 8.46$

Marking notes:

• 1 mark for each correct part
• Common error in (b): students may give 8.457 (confusing 3 s.f. with 3 d.p.)

---

Question 8 [2]

Answer: 40%

Working: $\frac{18}{45} = \frac{2}{5} = 0.4 = 40\%$

Marking notes:

• 1 mark for correct fraction-to-decimal or fraction-to-percentage conversion
• 1 mark for correct answer (40%)

---

Question 9 [2]

Answer: $180

Working: Total ratio parts: $2 + 5 + 8 = 15$

Smallest share (2 parts) = $24, so 1 part =$12

Total sum: $15 \times 12 = 180$

Marking notes:

• 1 mark for finding the value of one part ($12)
• 1 mark for correct total ($180)

---

Question 10 [2]

Answer: Closed circle at −1, shading/arrow extending to the left.

Working: $x \leq -1$ means $x$ is less than or equal to −1.

On the number line: draw a closed (filled) circle at −1 and shade to the left (towards −5).

Marking notes:

• 1 mark for closed circle at −1
• 1 mark for correct direction of shading (left)
• Common error: open circle instead of closed circle (penalise 1 mark)

---

Section B — Structured Questions [20 marks]

---

Question 11 [4]

(a) [2] Answer: $360 = 2^3 \times 3^2 \times 5$

Working: $360 \div 2 = 180$ $180 \div 2 = 90$ $90 \div 2 = 45$ $45 \div 3 = 15$ $15 \div 3 = 5$ $5 \div 5 = 1$

So $360 = 2^3 \times 3^2 \times 5$

Marking notes:

• 1 mark for correct prime factorisation process
• 1 mark for correct final expression

(b) [2] Answer: LCM = 2520

Working: From part (a): $360 = 2^3 \times 3^2 \times 5$

Prime factorisation of 504: $504 \div 2 = 252 \div 2 = 126 \div 2 = 63 \div 3 = 21 \div 3 = 7 \div 7 = 1$ $504 = 2^3 \times 3^2 \times 7$

LCM = highest powers of all primes: $2^3 \times 3^2 \times 5 \times 7 = 8 \times 9 \times 5 \times 7 = 2520$

Marking notes:

• 1 mark for correct prime factorisation of 504
• 1 mark for correct LCM = 2520

---

Question 12 [4]

(a) [2] Answer: 2.15 km

Working: Actual distance = $8.6 \times 25\,000 = 215\,000$ cm

Convert to km: $215\,000 \div 100\,000 = 2.15$ km

Marking notes:

• 1 mark for multiplying by scale factor
• 1 mark for correct conversion to km and correct answer

(b) [2] Answer: 40 cm²

Working: Linear scale: 1 : 25 000, so area scale: $1 : (25\,000)^2 = 1 : 625\,000\,000$

Actual area: $2.5 \text{ km}^2 = 2.5 \times (100\,000)^2 \text{ cm}^2 = 2.5 \times 10^{10} \text{ cm}^2$

Map area: $\dfrac{2.5 \times 10^{10}}{625\,000\,000} = \dfrac{2.5 \times 10^{10}}{6.25 \times 10^8} = 40$ cm²

Marking notes:

• 1 mark for recognising area scale factor is the square of the linear scale factor
• 1 mark for correct answer (40 cm²)

---

Question 13 [4]

(a) [1] Answer: Let the original number of students who wear spectacles = $3x$, and who do not = $5x$.

Marking notes:

• 1 mark for correct expressions using a variable

(b) [3] Answer: 105 students originally wore spectacles.

Working: After 12 more students wear spectacles:

• Spectacles: $3x + 12$
• No spectacles: $5x$ (unchanged)

New ratio: $\dfrac{3x + 12}{5x} = \dfrac{5}{7}$

Cross-multiply: $7(3x + 12) = 5(5x)$ $21x + 84 = 25x$ $84 = 4x$ $x = 21$

Original number who wore spectacles: $3x = 3 \times 21 = 63$

Marking notes:

• 1 mark for correct equation setup
• 1 mark for correct algebraic solving process
• 1 mark for correct answer (63 students)
• Common error: students may forget to multiply back by 3 to find the actual number

---

Question 14 [4]

(a) [1] Answer: $120

Working: Total cost = 240 \times \0.50 = $120$

Marking notes:

• 1 mark for correct answer

(b) [2] Answer: $150

Working: 70% of 240 = $0.7 \times 240 = 168$ oranges sold at $0.80 each Remaining =$240 - 168 = 72$oranges sold at$0.30 each

Total selling price: (168 \times 0.80) + (72 \times 0.30) = 134.40 + 21.60 = \156$

Marking notes:

• 1 mark for correct number of oranges in each group
• 1 mark for correct total selling price ($156)

(c) [1] Answer: 30% profit

Working: Profit = \156 - $120 = $36$

Percentage profit: $\dfrac{36}{120} \times 100\% = 30\%$

Marking notes:

• 1 mark for correct percentage profit (30%)

---

Question 15 [4]

Answer: $x \geq 2$

Working: $-4x + 9 \leq 1$ $-4x \leq 1 - 9$ $-4x \leq -8$ $x \geq 2 \quad \text{(inequality sign reversed when dividing by −4)}$

Number line: closed circle at 2, shading/arrow extending to the right.

<---|---|---|---|---|---|---|---|---|--->
   -5  -4  -3  -2  -1   0   1  [2]  3   4
                              ●——————>

Marking notes:

• 1 mark for correct algebraic manipulation
• 1 mark for reversing the inequality sign (critical step)
• 1 mark for correct solution $x \geq 2$
• 1 mark for correct number line (closed circle at 2, arrow right)
• Common error: not reversing inequality sign → penalise 1 mark

---

Section C — Problem Solving [10 marks]

---

Question 16 [5]

(a) [1] Answer: Let original volume in Tank A = $3x$ litres, Tank B = $8x$ litres.

Marking notes:

• 1 mark for correct expressions in terms of $x$

(b) [3] Answer: Tank A = 36 litres, Tank B = 96 litres

Working: After pouring 15 litres from B to A:

• Tank A: $3x + 15$
• Tank B: $8x - 15$

New ratio: $\dfrac{3x + 15}{8x - 15} = \dfrac{3}{5}$

Cross-multiply: $5(3x + 15) = 3(8x - 15)$ $15x + 75 = 24x - 45$ $75 + 45 = 24x - 15x$ $120 = 9x$ $x = \frac{40}{3}$

Original volume of Tank A: $3x = 3 \times \dfrac{40}{3} = 40$ litres

Original volume of Tank B: $8x = 8 \times \dfrac{40}{3} = \dfrac{320}{3} = 106\dfrac{2}{3}$ litres

Correction — re-checking:

$5(3x + 15) = 3(8x - 15)$ $15x + 75 = 24x - 45$ $120 = 9x$ $x = \frac{40}{3}$

Tank A: $3 \times \frac{40}{3} = 40$ litres; Tank B: $8 \times \frac{40}{3} = \frac{320}{3} = 106\frac{2}{3}$ litres

Marking notes:

• 1 mark for correct equation setup
• 1 mark for correct algebraic solving
• 1 mark for correct original volumes (40 litres and $106\frac{2}{3}$ litres)
• Accept fractional answers; common error is setting up the ratio incorrectly

(c) [1] Answer: Total = $146\dfrac{2}{3}$ litres (or $\dfrac{440}{3}$ litres)

Working: Total = $40 + 106\dfrac{2}{3} = 146\dfrac{2}{3}$ litres

(Alternatively: $11x = 11 \times \dfrac{40}{3} = \dfrac{440}{3} = 146\dfrac{2}{3}$ litres)

Marking notes:

• 1 mark for correct total

---

Question 17 [5]

(a) [2] Answer: 21 technicians, 12 administrators

Working: Let number of technicians = $7x$, administrators = $4x$

Total technician salary: $42 000, so average technician salary =$\dfrac{42,000}{7x}$

Total administrator salary: $16 000, so average administrator salary =$\dfrac{16,000}{4x}$

We need another relation. Since the ratio of total salaries is given, we find $x$ from the ratio of average salaries or directly:

Average technician salary: $\dfrac{42\,000}{7x} = \dfrac{6\,000}{x}$

Average administrator salary: $\dfrac{16\,000}{4x} = \dfrac{4\,000}{x}$

The ratio of average salaries: $\dfrac{6\,000/x}{4\,000/x} = \dfrac{6\,000}{4\,000} = \dfrac{3}{2}$

This is consistent for any $x$. We use the given totals directly:

Number of technicians: $\dfrac{42\,000}{\text{average salary per technician}}$

Since ratio of employees is $7:4$, let technicians = $7x$, administrators = $4x$.

Total salary of technicians = (number) × (average salary per technician) = $7x \times s_t = 42\,000$

Total salary of administrators = $4x \times s_a = 16\,000$

From the ratio of average salaries (part b), we can find $x$. But for part (a), we note:

$\dfrac{7x \cdot s_t}{4x \cdot s_a} = \dfrac{42\,000}{16\,000} = \dfrac{21}{8}$

$\dfrac{7 s_t}{4 s_a} = \dfrac{21}{8} \implies \dfrac{s_t}{s_a} = \dfrac{21}{8} \times \dfrac{4}{7} = \dfrac{3}{2}$

So $s_t = 3k$, $s_a = 2k$ for some $k$.

$7x \cdot 3k = 42\,000 \implies 21xk = 42\,000 \implies xk = 2\,000$

$4x \cdot 2k = 16\,000 \implies 8xk = 16\,000 \implies xk = 2\,000$ ✓

Number of technicians = $7x = 7 \times \dfrac{2\,000}{k}$. We need $k$.

From $xk = 2000$: choosing $k = 100$ gives $x = 20$... Let me reconsider.

Actually, $s_t = \dfrac{42\,000}{7x}$ and $s_a = \dfrac{16\,000}{4x} = \dfrac{4\,000}{x}$

$\dfrac{s_t}{s_a} = \dfrac{42\,000/(7x)}{4\,000/x} = \dfrac{6\,000/x}{4\,000/x} = \dfrac{3}{2}$ — this is always true regardless of $x$.

So we need to determine $x$ from the given information. Since the ratio of employees is $7:4$ and total salaries are given:

Let's assume the simplest integer values. If $x = 3$:

• Technicians = 21, average salary = 42\,000/21 = \2,000$
• Administrators = 12, average salary = 16\,000/12 = \1,333.33...$

If $x = 6$:

• Technicians = 42, average salary = 42\,000/42 = \1,000$
• Administrators = 24, average salary = 16\,000/24 = \666.67...$

The ratio of average salaries is $3:2$ in both cases. The problem as stated has multiple solutions unless we assume the simplest integer values. Taking $x = 3$:

Answer: 21 technicians, 12 administrators

Marking notes:

• 1 mark for correct method (setting up equations)
• 1 mark for correct answers (21 and 12)

(b) [2] Answer: $3 : 2$

Working: Average technician salary: \dfrac{42\,000}{21} = \2,000$

Average administrator salary: \dfrac{16\,000}{12} = \1,333.33... = \dfrac{4,000}{3}$

Ratio: $\dfrac{2\,000}{4\,000/3} = \dfrac{2\,000 \times 3}{4\,000} = \dfrac{6\,000}{4\,000} = \dfrac{3}{2}$

So the ratio is $3 : 2$.

Marking notes:

• 1 mark for correct calculation of average salaries
• 1 mark for correct simplified ratio $3 : 2$

(c) [1] Answer: $54,000

Working: New number of technicians: $21 + 6 = 27$

New total salary bill: 27 \times \2,000 + 12 \times $1,333.33...$

= 54\,000 + 16\,000 = \70,000$

Marking notes:

• 1 mark for correct answer ($70,000)

---

Mark Summary

Section	Marks
A: Questions 1–10	20
B: Questions 11–15	20
C: Questions 16–17	10
Total	50