Secondary 1 Mathematics Numbers Ratio Proportion Quiz

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Secondary 1 Mathematics Quiz - Numbers Ratio Proportion

Name: _________________________________ Class: ______________ Date: ______________

Duration: 40 minutes Total Marks: 50 marks

Instructions:

• Answer all questions.
• Show all working clearly. Marks will be awarded for correct method even if the final answer is wrong.
• Write your answers in the spaces provided.
• Use of calculator is allowed unless otherwise stated.

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Section A: Short Answer Questions (Questions 1–10)

2 marks each

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1. Find the highest common factor (HCF) of 126 and 180 using prime factorisation.

Answer: [2]

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2. Find the lowest common multiple (LCM) of 56 and 84 using prime factorisation.

Answer: [2]

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3. Evaluate $(-5)^3 + \sqrt[3]{-64}$ without using a calculator.

Answer: [2]

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4. Express 0.0375 as a fraction in its simplest form.

Answer: [2]

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5. Arrange the following numbers in ascending order: $-2.5$, $\frac{7}{3}$, $-\frac{5}{2}$, $2.3$, $\sqrt{5}$

Answer: [2]

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6. Solve the inequality $3x - 7 > 11$ and illustrate your answer on the number line in the space below.

Generated number_line for Q6.

Answer: [2]

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7. Simplify the ratio $2\frac{1}{4} : 1\frac{2}{3} : \frac{5}{6}$ to its simplest form of integers.

Answer: [2]

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8. A map has a scale of 1 : 50 000. If the actual distance between two towns is 12.5 km, find the distance on the map in centimetres.

Answer: [2]

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9. Mr Lim earns $4,200 per month. He spends 35% on rent, 25% on food and transport, and saves the rest. Calculate how much he saves each month.

Answer: [2]

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10. Three partners invest in a business in the ratio 3 : 5 : 7. If the total investment is $45 000, find the smallest investment.

Answer: [2]

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Section B: Structured Questions (Questions 11–16)

4 marks each

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11. (a) Express 504 as a product of its prime factors, using index notation. [2]

(b) Hence, find the smallest whole number $k$ such that $504k$ is a perfect square. [2]

Answer: [4]

(a) _________________________________________________________________

(b) _________________________________________________________________

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12. (a) Evaluate $\frac{(-3)^2 \times (-2)^3}{-6}$ without using a calculator. [2]

(b) Given that $-4 \leq x < 3$ where $x$ is an integer, list all possible values of $x$. [2]

Answer: [4]

(a) _________________________________________________________________

(b) _________________________________________________________________

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13. The mass of a gold bar is 2.5 kg, measured to the nearest 100 g.

(a) Write down the lower bound of the mass of the gold bar in grams. [1]

(b) The gold bar is melted and recast into smaller bars, each with mass 150 g measured to the nearest 10 g. Calculate the maximum number of complete smaller bars that can be made. [3]

Answer: [4]

(a) _________________________________________________________________

(b) _________________________________________________________________

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14. A recipe for 6 people requires 450 g of flour, 300 g of sugar, and 200 g of butter.

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

(b) Calculate the mass of each ingredient needed for 15 people. [2]

(c) If only 250 g of butter is available, what is the maximum number of people that can be served? [1]

Answer: [4]

(a) _________________________________________________________________

(b) _________________________________________________________________

(c) _________________________________________________________________

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15.

Generated graph for Q15.

The line graph above shows the temperature of a chemical solution during an experiment.

(a) Find the temperature at 2.5 hours, using linear interpolation. [2]

(b) Between which two consecutive hours did the temperature decrease the most? [1]

(c) Calculate the overall change in temperature from the start to the end of the experiment. [1]

Answer: [4]

(a) _________________________________________________________________

(b) _________________________________________________________________

(c) _________________________________________________________________

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16. A school has 840 students. The ratio of boys to girls is 5 : 7.

(a) Find the number of boys and the number of girls. [2]

(b) After some new students join, the ratio of boys to girls becomes 2 : 3. If the number of boys remains unchanged, find how many new girls joined the school. [2]

Answer: [4]

(a) _________________________________________________________________

(b) _________________________________________________________________

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Section C: Problem Solving (Questions 17–20)

5 marks each

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17. (a) Solve the inequality $\frac{2x-1}{3} - \frac{x+2}{2} \geq 1$. [3]

(b) Hence, write down the smallest integer value of $x$ that satisfies the inequality. [1]

(c) Illustrate the solution on a number line. [1]

Generated number_line for Q17.

Answer: [5]

(a) _________________________________________________________________

(b) _________________________________________________________________

(c) _________________________________________________________________

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18.

Generated diagram for Q18.

Rectangle A and Rectangle B are placed side by side to form a larger rectangle as shown.

(a) Form an equation in $x$ and solve for $x$. [2]

(b) Find the ratio of the area of Rectangle A to the area of Rectangle B in its simplest form. [2]

(c) If the height of both rectangles is increased by 25%, find the new total area. [1]

Answer: [5]

(a) _________________________________________________________________

(b) _________________________________________________________________

(c) _________________________________________________________________

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19. A shop sells three sizes of bottled water: Small (500 ml), Medium (1.25 litres), and Large (2 litres). During a promotion, the prices are: Small $1.20, Medium $2.50, Large $3.60.

(a) Which size offers the best value for money? Show your working clearly. [3]

(b) A school needs to buy exactly 30 litres of water for a sports day. Find the combination of bottles that gives the lowest cost, and state this lowest cost. [2]

Answer: [5]

(a) _________________________________________________________________

(b) _________________________________________________________________

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20.

Generated table for Q20.

The table shows the exchange rates for various currencies against the Singapore Dollar (SGD).

(a) Mrs Tan exchanges SGD 2 000 for Euros. Calculate how many Euros she receives, correct to the nearest cent. [2]

(b) A watch costs GBP 185 in London. The same watch costs SGD 320 in Singapore. Mr Koh wants to buy the watch in the cheaper location. Calculate how much he saves, in SGD, by buying at the cheaper location. [2]

(c) A Japanese tourist exchanges JPY 50 000 for SGD, then exchanges all the SGD for Malaysian Ringgit. Calculate how many Malaysian Ringgit the tourist receives. [1]

Answer: [5]

(a) _________________________________________________________________

(b) _________________________________________________________________

(c) _________________________________________________________________

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END OF QUIZ

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Answer Key - Secondary 1 Mathematics Quiz - Numbers Ratio Proportion

Total Marks: 50

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Question 1 [2 marks]

Find the HCF of 126 and 180 using prime factorisation.

Working:

• $126 = 2 \times 63 = 2 \times 3 \times 21 = 2 \times 3 \times 3 \times 7 = 2 \times 3^2 \times 7$
• $180 = 2 \times 90 = 2 \times 2 \times 45 = 2^2 \times 3^2 \times 5$

For HCF, take lowest power of each common prime factor:

• Common primes: 2 and 3
• HCF $= 2^1 \times 3^2 = 2 \times 9 = 18$

Answer: HCF = 18 [2]

Teaching note: HCF uses the lowest power of each common prime. A common error is to include 7 or 5, which appear in only one number.

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Question 2 [2 marks]

Find the LCM of 56 and 84 using prime factorisation.

Working:

• $56 = 2 \times 28 = 2^3 \times 7$
• $84 = 2 \times 42 = 2^2 \times 3 \times 7$

For LCM, take highest power of all primes present:

• LCM $= 2^3 \times 3^1 \times 7^1 = 8 \times 3 \times 7 = 168$

Answer: LCM = 168 [2]

Teaching note: LCM uses the highest power of all primes that appear in either number. Students often confuse this with HCF.

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Question 3 [2 marks]

Evaluate $(-5)^3 + \sqrt[3]{-64}$ without a calculator.

Working:

• $(-5)^3 = (-5) \times (-5) \times (-5) = 25 \times (-5) = -125$

Note: $(-5)^3$ means the base is $-5$, so the negative is part of what gets cubed. This differs from $-5^3 = -(5^3) = -125$.

• $\sqrt[3]{-64} = -4$ (since $(-4)^3 = -64$)

Note: The cube root of a negative number is negative, unlike square roots.

• Total: $-125 + (-4) = -125 - 4 = -129$

Answer: -129 [2]

Common trap: Students may write $\sqrt[3]{-64} = 4$ or confuse with $\sqrt{64}=8$. Cube roots preserve the sign.

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Question 4 [2 marks]

Express 0.0375 as a fraction in simplest form.

Working:

• $0.0375 = \frac{375}{10000}$

Find HCF of 375 and 10000:

• $375 = 3 \times 125 = 3 \times 5^3$
• $10000 = 10^4 = 2^4 \times 5^4$

HCF $= 5^3 = 125$

• $\frac{375 \div 125}{10000 \div 125} = \frac{3}{80}$

Answer: $\frac{3}{80}$ [2]

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Question 5 [2 marks]

Arrange in ascending order: $-2.5$, $\frac{7}{3}$, $-\frac{5}{2}$, $2.3$, $\sqrt{5}$

Working: Convert all to decimals:

• $-2.5 = -2.5$
• $\frac{7}{3} = 2.333...$
• $-\frac{5}{2} = -2.5$
• $2.3 = 2.3$
• $\sqrt{5} \approx 2.236$ (since $2.2^2=4.84$, $2.3^2=5.29$)

Ordering: $-\frac{5}{2} = -2.5$, then $-2.5$, then $2.3$, then $\sqrt{5} \approx 2.236$, then $\frac{7}{3} \approx 2.333$

Wait — let me recheck: $-2.5 = -\frac{5}{2}$, so these are equal.

Correct ascending order: $-\frac{5}{2}$ (or $-2.5$), $-2.5$ (same value), $\sqrt{5}$, $2.3$, $\frac{7}{3}$

Since $-2.5 = -\frac{5}{2}$ exactly, we can write: $-2.5 = -\frac{5}{2}$, $\sqrt{5}$, $2.3$, $\frac{7}{3}$

But for strict ascending with distinct positions: $-\frac{5}{2} = -2.5$, then $2.3$, then $\sqrt{5}$, then $\frac{7}{3}$?

Recheck: $\sqrt{5} \approx 2.236 < 2.3$? No, $2.236 < 2.3$ is correct.

Actually: $\sqrt{5} \approx 2.236$, and $2.3 = 2.3$, so $\sqrt{5} < 2.3$.

Final order: $-\frac{5}{2}$ (or $-2.5$), $\sqrt{5}$, $2.3$, $\frac{7}{3}$

Answer: $-\frac{5}{2}$ (or $-2.5$), $\sqrt{5}$, $2.3$, $\frac{7}{3}$ [2]

Note: Accept either $-2.5$ or $-\frac{5}{2}$ first since equal.

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Question 6 [2 marks]

Solve $3x - 7 > 11$ and illustrate on number line.

Working:

• $3x - 7 > 11$
• $3x > 18$ (add 7 to both sides)
• $x > 6$ (divide by 3; positive so inequality unchanged)

Number line: open circle at 6, arrow pointing right.

Answer: $x > 6$ with open circle at 6, arrow to the right [2]

Common trap: Forgetting the inequality stays the same when dividing by positive 3. (Reverse only for negative divisors.)

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Question 7 [2 marks]

Simplify $2\frac{1}{4} : 1\frac{2}{3} : \frac{5}{6}$ to simplest integer ratio.

Working: Convert to improper fractions:

• $2\frac{1}{4} = \frac{9}{4}$
• $1\frac{2}{3} = \frac{5}{3}$
• $\frac{5}{6}$

Ratio: $\frac{9}{4} : \frac{5}{3} : \frac{5}{6}$

Find LCM of denominators 4, 3, 6 = 12. Multiply each term by 12:

• $\frac{9}{4} \times 12 = 27$
• $\frac{5}{3} \times 12 = 20$
• $\frac{5}{6} \times 12 = 10$

Ratio: $27 : 20 : 10$

Check HCF of 27, 20, 10 = 1. Already simplest form.

Answer: $27 : 20 : 10$ [2]

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Question 8 [2 marks]

Scale 1 : 50 000, actual distance 12.5 km. Find map distance in cm.

Working:

• Scale means 1 cm on map = 50 000 cm in reality
• $50 000 \text{ cm} = 0.5 \text{ km} = 500 \text{ m}$

Or: convert actual distance to cm first:

• $12.5 \text{ km} = 12.5 \times 1000 \times 100 \text{ cm} = 1 250 000 \text{ cm}$

Map distance: $\frac{1 250 000}{50 000} = 25$ cm

Answer: 25 cm [2]

Common trap: Forgetting to convert km to cm, or confusing which side is map vs actual.

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Question 9 [2 marks]

Calculate monthly savings: earnings $4,200, spends 35% rent, 25% food/transport.

Working: Total spent: $35\% + 25\% = 60\%$

Saved: $100\% - 60\% = 40\%$

Amount saved: $40\% \times 4200 = 0.4 \times 4200 = 1680$

Or:

• Rent: $0.35 \times 4200 = 1470$
• Food/transport: $0.25 \times 4200 = 1050$
• Total spent: $2520$
• Saved: $4200 - 2520 = 1680$

Answer: $1 680 [2]

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Question 10 [2 marks]

Ratio 3 : 5 : 7, total $45 000. Find smallest investment.

Working: Total parts: $3 + 5 + 7 = 15$ parts

Value of one part: $\frac{45000}{15} = 3000$

Smallest investment (3 parts): $3 \times 3000 = 9000$

Answer: $9 000 [2]

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Question 11 [4 marks]

(a) Express 504 as product of prime factors in index notation.

Working:

• $504 \div 2 = 252$
• $252 \div 2 = 126$
• $126 \div 2 = 63$
• $63 \div 3 = 21$
• $21 \div 3 = 7$
• $7 \div 7 = 1$

So $504 = 2^3 \times 3^2 \times 7^1$ or $2^3 \times 3^2 \times 7$

Answer (a): $504 = 2^3 \times 3^2 \times 7$ [2]

(b) Find smallest $k$ such that $504k$ is a perfect square.

Working: For a perfect square, all prime powers must be even.

Current: $2^3 \times 3^2 \times 7^1$

• Power of 2: 3 (odd, need 4) → need one more 2
• Power of 3: 2 (even, OK)
• Power of 7: 1 (odd, need 2) → need one more 7

So $k = 2^1 \times 7^1 = 14$

Check: $504 \times 14 = 7056 = 84^2$ ✓

Answer (b): $k = 14$ [2]

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Question 12 [4 marks]

(a) Evaluate $\frac{(-3)^2 \times (-2)^3}{-6}$

Working:

• $(-3)^2 = 9$ (negative squared is positive)
• $(-2)^3 = -8$ (negative cubed is negative)

Numerator: $9 \times (-8) = -72$

Fraction: $\frac{-72}{-6} = 12$ (negative divided by negative is positive)

Answer (a): 12 [2]

(b) List all integer values of $x$ where $-4 \leq x < 3$

Working: $-4 \leq x$ means $x = -4, -3, -2, ...$ (includes $-4$) $x < 3$ means $x = ..., 0, 1, 2$ (does not include 3)

Values: $-4, -3, -2, -1, 0, 1, 2$

Answer (b): $-4, -3, -2, -1, 0, 1, 2$ [2]

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Question 13 [4 marks]

(a) Lower bound of mass (2.5 kg to nearest 100g)

Working: Nearest 100 g means rounding to 0.1 kg. Lower bound: $2.5 - 0.05 = 2.45$ kg = 2450 g

Or: 2.5 kg = 2500 g to nearest 100 g. Lower bound: $2500 - 50 = 2450$ g.

Answer (a): 2450 g [1]

(b) Maximum number of complete 150g bars (to nearest 10g)

Working: Maximum mass of gold bar: $2500 + 50 = 2550$ g (using original 2500 g ± 50 g)

Each small bar minimum mass: $150 - 5 = 145$ g (to make most bars, use minimum mass per bar)

Wait — to get maximum number of complete bars from maximum gold: use minimum mass per bar.

Maximum number: $\left\lfloor \frac{2550}{145} \right\rfloor = \left\lfloor 17.586... \right\rfloor = 17$

Check: $17 \times 145 = 2465 \leq 2550$ ✓ $18 \times 145 = 2610 > 2550$ ✗

Or using upper bound approach:

• Upper bound of gold: 2550 g
• Lower bound of small bar: 145 g
• Maximum bars: $\frac{2550}{145} = 17.58...$, so 17 complete bars

Answer (b): 17 [3]

Marking: Bounds identification [1], correct calculation [1], final answer with reasoning [1]

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Question 14 [4 marks]

(a) Ratio flour : sugar : butter

Working: $450 : 300 : 200$

Divide by 50: $9 : 6 : 4$

Check HCF of 9, 6, 4 = 1. Simplest form is $9 : 6 : 4$?

Wait: HCF of 450, 300, 200 = 50. $450 \div 50 = 9$, $300 \div 50 = 6$, $200 \div 50 = 4$.

HCF of 9, 6, 4 is 1. So $9 : 6 : 4$ is correct.

Actually check: can we divide further? 9=3², 6=2×3, 4=2². No common factor >1.

Answer (a): $9 : 6 : 4$ [1]

(b) Ingredients for 15 people

Working: Scale factor: $\frac{15}{6} = 2.5$

• Flour: $450 \times 2.5 = 1125$ g
• Sugar: $300 \times 2.5 = 750$ g
• Butter: $200 \times 2.5 = 500$ g

Answer (b): Flour: 1125 g, Sugar: 750 g, Butter: 500 g [2]

(c) Maximum people with 250 g butter

Working: Butter needed per person: $\frac{200}{6} = \frac{100}{3}$ g

With 250 g: number of people = $250 \div \frac{100}{3} = 250 \times \frac{3}{100} = 7.5$

Maximum whole people: 7 people

Or using ratio: $200 : 6 = 250 : p$, so $p = \frac{250 \times 6}{200} = 7.5$, maximum 7.

Answer (c): 7 people [1]

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Question 15 [4 marks]

Expected visual: Line graph with points (0,25), (1,28), (2,32), (3,38), (4,34), (5,30), (6,26)

(a) Temperature at 2.5 hours by linear interpolation

Working: At 2 hours: 32°C, at 3 hours: 38°C

Linear interpolation: temperature increases from 32 to 38 over 1 hour. At 2.5 hours (midway): $32 + \frac{1}{2}(38-32) = 32 + 3 = 35$°C

Or: average of 32 and 38 = $\frac{32+38}{2} = 35$

Answer (a): 35°C [2]

(b) Largest temperature decrease between consecutive hours

Working: Changes:

• 0→1: +3
• 1→2: +4
• 2→3: +6
• 3→4: -4
• 4→5: -4
• 5→6: -4

Decreases: 4°C (3→4), 4°C (4→5), 4°C (5→6)

Largest decrease: 4°C between 3 and 4 hours (or any of the decreasing periods; 3→4 is first and largest equal)

Actually all decreases are equal at 4°C. Accept any of: 3 and 4, 4 and 5, or 5 and 6 hours.

Answer (b): Between 3 and 4 hours (or 4 and 5, or 5 and 6) [1]

(c) Overall change from start to end

Working: Start (0 hours): 25°C, End (6 hours): 26°C

Change: $26 - 25 = +1$°C

Answer (c): Increase of 1°C (or +1°C) [1]

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Question 16 [4 marks]

(a) Number of boys and girls (ratio 5:7, total 840)

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

Each part: $\frac{840}{12} = 70$

• Boys: $5 \times 70 = 350$
• Girls: $7 \times 70 = 490$

Answer (a): Boys: 350, Girls: 490 [2]

(b) New girls joining, new ratio 2:3 with boys unchanged

Working: Boys remain 350. New ratio boys:girls = 2:3.

If 2 parts = 350, then 1 part = 175

New number of girls: $3 \times 175 = 525$

New girls joined: $525 - 490 = 35$

Answer (b): 35 new girls [2]

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Question 17 [5 marks]

(a) Solve $\frac{2x-1}{3} - \frac{x+2}{2} \geq 1$

Working: LCM of 3 and 2 is 6. Multiply all terms by 6:

$6 \times \frac{2x-1}{3} - 6 \times \frac{x+2}{2} \geq 6 \times 1$

$2(2x-1) - 3(x+2) \geq 6$

$4x - 2 - 3x - 6 \geq 6$

$x - 8 \geq 6$

$x \geq 14$

Answer (a): $x \geq 14$ [3]

Marking: Common denominator [1], expansion [1], final inequality [1]

(b) Smallest integer value

Answer (b): 14 [1]

(c) Number line illustration

Closed circle at 14, arrow pointing to the right.

Answer (c): Closed circle at 14, arrow right [1]

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Question 18 [5 marks]

Expected visual: Rectangle A (width 3x, height 8) and Rectangle B (width 5x, height 8) side by side, total length 64 cm

(a) Form equation and solve for x

Working: Total width = width of A + width of B $3x + 5x = 64$ $8x = 64$ $x = 8$

Answer (a): $x = 8$ [2]

(b) Ratio of area A : area B

Working:

• Area A: $3x \times 8 = 3(8) \times 8 = 24 \times 8 = 192$ cm²
• Area B: $5x \times 8 = 5(8) \times 8 = 40 \times 8 = 320$ cm²

Ratio: $192 : 320$

Simplify: divide by 64 → $3 : 5$

Or note: heights equal, so area ratio = width ratio = $3x : 5x = 3:5$

Answer (b): $3 : 5$ [2]

(c) New total area with 25% height increase

Working: New height: $8 \times 1.25 = 10$ cm

New total area: $(3x + 5x) \times 10 = 64 \times 10 = 640$ cm²

Or: original total area = $192 + 320 = 512$ cm² New area = $512 \times 1.25 = 640$ cm²

Answer (c): 640 cm² [1]

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Question 19 [5 marks]

(a) Best value for money

Working: Calculate price per litre (or ml per dollar) for each:

Price per litre:

• Small: \frac{1.20}{0.5} = \2.40$ per litre
• Medium: \frac{2.50}{1.25} = \2.00$ per litre
• Large: \frac{3.60}{2} = \1.80$ per litre

Or ml per dollar:

• Small: $\frac{500}{1.20} = 416.7$ ml/$
• Medium: $\frac{1250}{2.50} = 500$ ml/$
• Large: $\frac{2000}{3.60} = 555.6$ ml/$

Lowest price per litre (or highest ml per dollar) is best value.

Answer (a): Large (2 litre) at $1.80 per litre (or equivalent comparison) [3]

Marking: Calculation for each size [1], comparison statement [1], conclusion [1]

(b) Cheapest combination for exactly 30 litres

Working: Need exactly 30 litres = 30 000 ml

To minimize cost, use as many Large bottles as possible:

• $30 \div 2 = 15$ Large bottles exactly? No, 15 × 2 = 30 litres. Yes!

Cost: 15 \times 3.60 = \54.00$

Check alternatives:

• 14 Large (28L) + 1 Medium (1.25L) = 29.25L, need 0.75L more — not exact with given sizes
• 14 Large + 2 Medium = 28 + 2.5 = 30.5L (too much)

Actually: 15 × 2L = 30L exactly.

Or: 12 × 2L + 4 × 1.25L + 2 × 0.5L = 24 + 5 + 1 = 30. Cost = 12(3.60) + 4(2.50) + 2(1.20) = 43.20 + 10 + 2.40 = $55.60

15 Large is cheapest at $54.00.

Answer (b): 15 Large bottles; $54.00 [2]

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Question 20 [5 marks]

Expected visual: Exchange rate table: USD 1 = SGD 1.345, EUR 1 = SGD 1.472, GBP 1 = SGD 1.683, JPY 100 = SGD 0.892, MYR 1 = SGD 0.296

(a) SGD 2 000 to Euros

Working: EUR 1 = SGD 1.472, so SGD 1 = EUR $\frac{1}{1.472}$

SGD 2000 = $2000 \times \frac{1}{1.472} = \frac{2000}{1.472} = 1358.695...$

To nearest cent: EUR 1358.70

Or: 2000 ÷ 1.472 = 1358.6956...

Answer (a): EUR 1 358.70 (or €1358.70) [2]

(b) Save by buying at cheaper location

Working: Watch in London: GBP 185 Convert to SGD: $185 \times 1.683 = 311.355$ SGD

In Singapore: SGD 320

Cheaper in London. Savings: 320 - 311.355 = 8.645 \approx \8.65$

Or keep exact: $320 - 185 \times 1.683 = 320 - 311.355 = 8.645$

Answer (b): $8.65 (or $8.64 if rounding differently; accept $8.645) [2]

(c) JPY 50 000 → SGD → MYR

Working: JPY 50 000 to SGD:

• Rate is JPY 100 = SGD 0.892
• JPY 50 000 = $500 \times$ JPY 100, so SGD $= 500 \times 0.892 = 446$ SGD

SGD to MYR:

• MYR 1 = SGD 0.296, so SGD 1 = MYR $\frac{1}{0.296}$

MYR = $446 \times \frac{1}{0.296} = \frac{446}{0.296} = 1506.756...$

Or: $446 \div 0.296 = 1506.76$ MYR

Answer (c): MYR 1 506.76 (or ≈ 1506.76, or 1507 rounded) [1]

Accept calculation showing $446 \div 0.296$

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END OF ANSWER KEY