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

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TuitionGoWhere Practice Paper — Mathematics Secondary 1

School: TuitionGoWhere Secondary School (AI) Subject: Mathematics Level: Secondary 1 (G3) Assessment: SA2 (End-of-Year Examination) Paper: Paper 1 (Calculator Allowed) Version: 2 of 5 Duration: 60 minutes Total Marks: 50

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Name: ___________________________ Class: _________ Date: ___________

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Instructions

1. Answer all questions in the spaces provided.
2. Show all working clearly. Marks are awarded for correct working, not just the final answer.
3. Do not use correction fluid or tape.
4. A calculator may be used where appropriate.
5. Give non-exact answers correct to 2 decimal places unless otherwise stated.
6. The number of marks available is shown in brackets [ ] at the end of each question or part-question.

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Section A: Short Answer Questions (20 marks)

Questions 1–10. Each question carries 2 marks.

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1. Express 360 as a product of its prime factors. Give your answer in index notation.

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___________________________________________________________________________ [2]

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

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___________________________________________________________________________ [2]

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3. Evaluate the following, giving your answer as a fraction in its simplest form.

$\frac{3}{4} - \frac{2}{5}$

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___________________________________________________________________________ [2]

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4. A recipe for 6 servings requires 450 g of flour. How much flour is needed for 10 servings?

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___________________________________________________________________________ [2]

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5. Write down the value of each of the following.

(a) $7^0$

(b) $\left(\frac{2}{3}\right)^{-1}$

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___________________________________________________________________________ [2]

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6. The ratio of boys to girls in a class is 5 : 4. There are 15 boys. How many students are in the class altogether?

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___________________________________________________________________________ [2]

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7. Round 47.628 to (a) 2 decimal places, (b) 1 significant figure.

(a) _______________________

(b) _______________________ [2]

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8. Simplify the ratio 48 : 72 to its simplest form.

___________________________________________________________________________ [2]

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9. Express 0.000073 in standard form.

___________________________________________________________________________ [2]

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10. Estimate the value of $\dfrac{4.92 \times 18.7}{0.48}$ by rounding each number to 1 significant figure.

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___________________________________________________________________________ [2]

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Section B: Structured Questions (20 marks)

Questions 11–15. Each question carries 4 marks.

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11. (a) Find the lowest common multiple (LCM) of 18 and 24. [2]

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(b) Hence, find the value of $\dfrac{5}{18} + \dfrac{7}{24}$. Give your answer as a mixed number in its simplest form. [2]

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___________________________________________________________________________ [4]

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12. A fruit seller has apples and oranges in the ratio 7 : 5. After selling 40 apples and buying 40 oranges, the ratio of apples to oranges becomes 1 : 1.

(a) How many apples did the fruit seller have at first? [3]

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(b) How many oranges did the fruit seller have at first? [1]

___________________________________________________________________________ [4]

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13. (a) Solve the inequality $-5x > 20$. [1]

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(b) Illustrate the solution to part (a) on the number line below. [2]

<---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|--->
  -7  -6  -5  -4  -3  -2  -1   0   1   2   3   4   5   6   7   8   9  10

(c) Write down the largest integer value of $x$ that satisfies the inequality. [1]

___________________________________________________________________________ [4]

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14. In a school, the ratio of students who wear spectacles to those who do not is 3 : 7. There are 48 more students who do not wear spectacles than those who do.

(a) How many students wear spectacles? [3]

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(b) What is the total number of students in the school? [1]

___________________________________________________________________________ [4]

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15. A car travels 240 km in 3 hours at a constant speed.

(a) At this speed, how far will the car travel in 5 hours? [2]

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(b) At this speed, how long will it take to travel 600 km? Give your answer in hours and minutes. [2]

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___________________________________________________________________________ [4]

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Section C: Problem-Solving Questions (10 marks)

Questions 16–20. Questions 16–19 carry 2 marks each. Question 20 carries 2 marks.

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16. The price of a laptop is $1200. During a sale, the price is reduced by 15%. Find the sale price of the laptop.

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___________________________________________________________________________ [2]

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17. Three friends, Ali, Bala, and Chris, share a sum of money in the ratio 2 : 3 : 5. If Chris receives $180 more than Ali, find the total sum of money shared.

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___________________________________________________________________________ [2]

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18. Given that $a : b = 3 : 4$ and $b : c = 6 : 5$, find the ratio $a : b : c$ in its simplest form.

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___________________________________________________________________________ [2]

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19. A map has a scale of 1 : 25 000. The distance between two towns on the map is 8.6 cm. Calculate the actual distance between the two towns in kilometres.

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___________________________________________________________________________ [2]

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20. The total mass of rice in three bags A, B, and C is 45 kg. The mass of rice in bag A to bag B is in the ratio 2 : 3, and the mass of rice in bag B to bag C is in the ratio 6 : 5. Find the mass of rice in bag C.

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___________________________________________________________________________ [2]

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End of Paper

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Total: 50 marks

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SA2 Practice Paper — Mathematics Secondary 1

Answer Key (Version 2 of 5)

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Section A

1. $360 = 2^3 \times 3^2 \times 5$

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 $360 = 2 \times 2 \times 2 \times 3 \times 3 \times 5 = 2^3 \times 3^2 \times 5$ [2]

Marking notes: Award 1 mark for correct prime factorisation (non-index form), 1 mark for correct index notation. Accept any correct method.

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2. $\text{HCF of 84 and 126} = 42$

Working: $84 = 2^2 \times 3 \times 7$ $126 = 2 \times 3^2 \times 7$

HCF = lowest powers of common primes $= 2^1 \times 3^1 \times 7^1 = 42$ [2]

Marking notes: Award 2 marks for correct answer with working. Award 1 mark for correct prime factorisations of both numbers even if HCF is wrong.

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3. $\dfrac{1}{20}$

Working: $\frac{3}{4} - \frac{2}{5} = \frac{15}{20} - \frac{8}{20} = \frac{7}{20}$ [2]

Marking notes: Award 1 mark for correct common denominator, 1 mark for correct final answer. Answer must be in simplest form.

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4. 750 g

Working: Flour per serving $= 450 \div 6 = 75$ g Flour for 10 servings $= 75 \times 10 = 750$ g [2]

Marking notes: Award 1 mark for finding unit amount, 1 mark for correct final answer with unit.

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5. (a) $7^0 = 1$ [1]

(b) $\left(\dfrac{2}{3}\right)^{-1} = \dfrac{3}{2}$ or $1\dfrac{1}{2}$ [1]

Marking notes: Each part worth 1 mark. Common error: students may write 0 for part (a).

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6. 27 students

Working: Ratio of boys : girls = 5 : 4 5 parts = 15, so 1 part = 3 Number of girls $= 4 \times 3 = 12$ Total students $= 15 + 12 = 27$ [2]

Marking notes: Award 1 mark for finding 1 part = 3, 1 mark for correct total.

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7. (a) 47.63 [1]

(b) 50 [1]

Marking notes: Part (a): rounding to 2 d.p. — the 8 causes the 2 to round up to 3. Part (b): 1 s.f. — the 7 causes the 4 to round up to 5.

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8. $2 : 3$

Working: $48 : 72 = \dfrac{48}{24} : \dfrac{72}{24} = 2 : 3$ [2]

Marking notes: Award 2 marks for correct simplified ratio. Award 1 mark for a correct attempt to divide by a common factor.

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9. $7.3 \times 10^{-5}$

Working: $0.000073 = 7.3 \times 10^{-5}$ [2]

Marking notes: Award 1 mark for correct coefficient (7.3), 1 mark for correct power of 10. Common error: writing $73 \times 10^{-6}$ — not standard form.

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10. 200

Working: $4.92 \approx 5$, $18.7 \approx 20$, $0.48 \approx 0.5$ $\frac{5 \times 20}{0.5} = \frac{100}{0.5} = 200$ [2]

Marking notes: Award 1 mark for correct rounding to 1 s.f., 1 mark for correct estimated answer.

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Section B

11. (a) LCM of 18 and 24 = 72 [2]

Working: $18 = 2 \times 3^2$ $24 = 2^3 \times 3$ LCM $= 2^3 \times 3^2 = 8 \times 9 = 72$

Marking notes: Award 2 marks for correct answer with working. Award 1 mark for correct prime factorisations.

(b) $1\dfrac{1}{72}$ [2]

Working: $\frac{5}{18} + \frac{7}{24} = \frac{20}{72} + \frac{21}{72} = \frac{41}{72}$ [2]

Marking notes: Award 1 mark for correct conversion to common denominator, 1 mark for correct final answer. Note: $\frac{41}{72}$ is already in simplest form. Common error: students may incorrectly add numerators and denominators.

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12. (a) 280 apples [3]

Working: Let the number of apples be $7x$ and oranges be $5x$. After selling 40 apples: apples $= 7x - 40$ After buying 40 oranges: oranges $= 5x + 40$ New ratio is 1 : 1, so: $7x - 40 = 5x + 40$ $2x = 80$ $x = 40$ Number of apples at first $= 7 \times 40 = 280$

Marking notes: Award 1 mark for setting up expressions, 1 mark for forming the equation, 1 mark for correct answer.

(b) 200 oranges [1]

Working: Number of oranges at first $= 5 \times 40 = 200$ [1]

Marking notes: Follow-through from part (a) accepted.

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13. (a) $x < -4$ [1]

Working: $-5x > 20$ $x < -4$ (inequality sign reversed when dividing by negative)

Marking notes: Common trap — students forget to reverse the inequality sign. Award 0 if answer is $x > -4$.

(b) Number line illustration [2]

Working: Open circle at $-4$, arrow/shading extending to the left (towards negative direction).

<---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|--->
  -7  -6  -5  -4  -3  -2  -1   0   1   2   3   4   5   6   7   8   9  10
              ==========================○→

Marking notes: Award 1 mark for open circle at -4, 1 mark for correct direction of shading. Common error: closed circle (should be open since it is strict inequality $x < -4$).

(c) $-5$ [1]

Working: The largest integer less than $-4$ is $-5$.

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14. (a) 36 students [3]

Working: Ratio of spectacles : no spectacles = 3 : 7 Difference in parts $= 7 - 3 = 4$ parts 4 parts = 48, so 1 part = 12 Students who wear spectacles $= 3 \times 12 = 36$

Marking notes: Award 1 mark for finding difference in parts (4), 1 mark for finding 1 part = 12, 1 mark for correct answer.

(b) 120 students [1]

Working: Total parts $= 3 + 7 = 10$ Total students $= 10 \times 12 = 120$ [1]

Marking notes: Follow-through accepted.

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15. (a) 400 km [2]

Working: Speed $= 240 \div 3 = 80$ km/h Distance in 5 hours $= 80 \times 5 = 400$ km

Marking notes: Award 1 mark for finding speed, 1 mark for correct distance.

(b) 7 hours 30 minutes [2]

Working: Time $= 600 \div 80 = 7.5$ hours $= 7$ hours $30$ minutes

Marking notes: Award 1 mark for correct time in hours (7.5 or equivalent), 1 mark for correct conversion to hours and minutes. Common error: writing 7 h 50 min (confusing 0.5 h with 50 min).

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Section C

16. $1020

Working: Discount = 15\% \times 1200 = 0.15 \times 1200 = \180$Sale price$= 1200 - 180 = $1020$

Alternative: Sale price = 85\% \times 1200 = 0.85 \times 1200 = \1020$ [2]

Marking notes: Award 1 mark for correct discount amount or correct percentage multiplier, 1 mark for correct final answer.

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17. $600

Working: Ratio Ali : Bala : Chris = 2 : 3 : 5 Difference between Chris and Ali $= 5 - 2 = 3$ parts 3 parts = $180, so 1 part = $60 Total $= 2 + 3 + 5 = 10$ parts = 10 \times 60 = \600$ [2]

Marking notes: Award 1 mark for finding 1 part = $60, 1 mark for correct total.

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18. $9 : 12 : 10$

Working: $a : b = 3 : 4 = 9 : 12$ (multiply by 3) $b : c = 6 : 5 = 12 : 10$ (multiply by 2) So $a : b : c = 9 : 12 : 10$ [2]

Marking notes: Award 1 mark for correctly making the $b$ values equal, 1 mark for correct combined ratio. Common error: students may write $3 : 4 : 5$ without matching the middle term.

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19. 2.15 km

Working: Actual distance $= 8.6 \times 25\,000 = 215\,000$ cm $215\,000$ cm $= 215\,000 \div 100\,000 = 2.15$ km [2]

Marking notes: Award 1 mark for correct multiplication (215 000 cm), 1 mark for correct conversion to km. Common error: forgetting to convert cm to km, or dividing by 100 instead of 100 000.

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20. 15 kg

Working: Bag A : Bag B = 2 : 3 = 4 : 6 Bag B : Bag C = 6 : 5 So Bag A : Bag B : Bag C = 4 : 6 : 5 Total parts $= 4 + 6 + 5 = 15$ parts 15 parts = 45 kg, so 1 part = 3 kg Mass of bag C $= 5 \times 3 = 15$ kg [2]

Marking notes: Award 1 mark for correctly combining the ratios (making B equal), 1 mark for correct answer. Common trap: students may add the ratios without matching the common term.

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End of Answer Key

Total: 50 marks