Secondary 1 Mathematics Numbers Ratio Proportion Quiz

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