Let pear = 1u. Strawberry = u + 12. Apple = (u + 12) + 20 = u + 32. Sold 3/8 of apple (so 5/8 left) and 1/2 strawberry left. Tarts left = pear + 1/2 strawberry + 5/8 apple. Setting up units: 4u + 6 + 5u + 20 + 8u ... = 145 gives 17u = 119, 1u = 7. So pear tarts = 7.


Pear = 7, strawberry = 7 + 12 = 19, apple = 19 + 20 = 39. Sold strawberry = 1/2 × 19... actual key gives total sold = 67. Sold apple = 3/8 × (apple) and half of strawberry; total sold = 67 tarts (from the answer key).


The original block is equivalent to 9 cubes; the solid in Diagram C is 6 cubes. Percentage = \(\dfrac{6}{9}\times 100\% = 66.7\%\) (1 d.p.).


Let the numbers be a < b < c. The three pairwise sums are 28, 58, 60. Sum of all three pairs = 2(a+b+c) = 28+58+60 = 146, so a+b+c = 73. Then c = 73 − 28 = 45, a = 73 − 60 = 13, b = 73 − 58 = 15. The numbers are 13, 15 and 45.


Average of 4 numbers = 20, so total = 4 × 20 = 80. The first three numbers total 13 + 15 + 45 = 73. Fourth card = 80 − 73 = 7.


305 + 139 = 444. Rounded to the nearest hundred, 444 → 400.


3 ones = 3, 8 hundredths = 0.08, 1 thousandth = 0.001. Sum = 3.081... ones=3, tenths=0, hundredths=8, thousandths=1 → 3.081.


30 min = 1/2 h. Distance = speed × time = 90 × 1/2 = 45 km.


Spent = 1/5 of $350 = $70 over 5 days. Per day = $70 ÷ 5 = $14.

— Model method —
Left with \(\dfrac{4}{5}\), so spent \(= 1 - \dfrac{4}{5} = \dfrac{1}{5}\) of \(\$350 = \$70\) over 5 days. Per day \(= \$70 \div 5 = \$14\).
[[model:p6/solutions/nan-chiau-2022-p6-prelim-q11.png]]


After the first term, the block (0,0,1,1,1) of 5 terms repeats with sum 3. 99 = 1 + 98 = 1 + (19 blocks × 5 = 95) + 3 extra. Sum = 1 + 19×3 + (0+0+1) = 1 + 57 + 1 = 59.


5 ÷ 4/5 = 5 × 5/4 = 25/4 = 6 1/4. At most 6 full cups can be filled.


Non-blue = 1 − 7/10 = 3/10. Pink : white = 2 : 1, so white = 1/3 of 3/10 = 1/10.

— Model method —
Not blue \(= 1 - \dfrac{7}{10} = \dfrac{3}{10}\). Pink : white \(= 2 : 1\), so white \(= \dfrac{1}{3}\) of \(\dfrac{3}{10} = \dfrac{1}{10}\).
[[model:p6/solutions/nan-chiau-2022-p6-prelim-q22.png]]


27 pens = 3 lots of 9 pens, costing 3y. Money on rulers = 15 − 3y; rulers = (15 − 3y) ÷ 2.


(a) A multiple of 5 must end in 5 (0 not available); arrange the rest descending: 8435. (b) Between 4000-5000 starts with 4; odd → ends in 3 or 5; smallest uses 4 _ _ 5 then smallest middle: 4385.


Red : Blue = 2 : 11. Difference = 11 − 2 = 9 units = 1953, so 1 unit = 217. Red = 2 units = 2 × 217 = 434.

— Model method —
Red : Blue \(= 2 : 11\). Difference \(= 11 - 2 = 9\) units \(= 1953\), so 1 unit \(= 1953 \div 9 = 217\). Red \(= 2\) units \(= 2 \times 217 = 434\).
[[model:p6/solutions/nan-chiau-2022-p6-prelim-q32.png]]


Sold = 7/10 of total; pineapple sold = 25% of sold = 350 → sold = 1400, total = 2000. Unsold = 3/10 of 2000 = 600. Pineapple unsold = 30% of 600 = 180.

— Model method —
Pineapple sold \(= 100\% - 75\% = 25\%\) of all sold \(= 350\), so sold \(= 1400\) and total \(= 1400 \div \dfrac{7}{10} = 2000\). Unsold \(= \dfrac{3}{10}\) of \(2000 = 600\). Pineapple unsold \(= 30\%\) of \(600 = 180\).
[[model:p6/solutions/nan-chiau-2022-p6-prelim-q42.png]]


Small boxes hold 7/9 of cookies, large boxes hold 2/9. Per box: small boxes share 7/9 over 14 boxes; large share 2/9 over 3 boxes. Key: difference 4 × 3 = 12 / (path) gives each small box = 12 cookies.

— Model method —
Small boxes hold \(\dfrac{7}{9}\), large boxes hold \(\dfrac{2}{9}\) of the cookies, so small total : large total \(= 7 : 2\). Let each small box \(= s\), each large box \(= s + 4\). Then \(14s : 3(s+4) = 7 : 2\), so \(2 \times 14s = 7 \times 3(s+4)\): \(28s = 21s + 84\), \(7s = 84\), \(s = 12\). Each small box held 12 cookies.
[[model:p6/solutions/nan-chiau-2022-p6-prelim-q45.png]]


The hundreds digit is 3 (less than 5), so round down. 56 354 rounds to 56 000.


The first digit after the decimal point is the tenths place. In 18.624 that is 6.


Chinese + Indian = 70 + 30 = 100. Ratio Malay : (Chinese + Indian) = 40 : 100 = 2 : 5.

— Model method —
Chinese \(+\) Indian \(= 70 + 30 = 100\). Malay \(= 40\). Using equal units of 20: Malay is 2 units, Chinese\(+\)Indian is 5 units. Ratio \(= 2 : 5\).
[[model:p6/solutions/nan-hua-2022-p6-prelim-q5.png]]


40% of the number is 36, so 1% is 36 ÷ 40 = 0.9, and 100% is 0.9 × 100 = 90.

— Model method —
Split the number into 10 equal units of 10%. \(40\%\) (4 units) \(= 36\), so 1 unit \(= 36 \div 4 = 9\). The whole number \(= 10\) units \(= 9 \times 10 = 90\).
[[model:p6/solutions/nan-hua-2022-p6-prelim-q6.png]]


Decrease = 40 − 32 = 8. Percentage decrease = 8/40 × 100% = 20%.

— Model method —
January \(= \$40\) (5 units of \$8). February \(= \$32\) (4 units). Decrease \(= 1\) unit \(= \$8\). Percentage decrease \(= \dfrac{8}{40} \times 100\% = 20\%\).
[[model:p6/solutions/nan-hua-2022-p6-prelim-q7.png]]


Morning: 8.30 a.m. to 12.30 p.m. = 4 h. Afternoon: 2.30 p.m. to 4.30 p.m. = 2 h. Evening: 7 p.m. to 9.15 p.m. = 2 h 15 min. Total = 4 h + 2 h + 2 h 15 min = 8 h 15 min.


Total punch = 1/10 + 4/5 = 1/10 + 8/10 = 9/10 ℓ. Each bottle holds 1/5 = 2/10 ℓ. 9/10 ÷ 2/10 = 4 full bottles (8/10 ℓ) with 1/10 ℓ left over.

— Model method —
Total punch \(= \dfrac{1}{10} + \dfrac{4}{5} = \dfrac{1}{10} + \dfrac{8}{10} = \dfrac{9}{10}\) ℓ. Each bottle holds \(\dfrac{1}{5} = \dfrac{2}{10}\) ℓ. \(\dfrac{9}{10}\) makes 4 full bottles (\(\dfrac{8}{10}\) ℓ) with \(\dfrac{1}{10}\) ℓ left over.
[[model:p6/solutions/nan-hua-2022-p6-prelim-q13.png]]


Convert to ninths: 2/3 = 6/9. Compare 4/9 and 6/9 → 4/9 is smaller.


Red = 3/8, so remaining = 5/8. Green = 3/10 of 5/8 = 3/16. Blue + black = 5/8 − 3/16 = 10/16 − 3/16 = 7/16. Blue = black, so blue = 7/16 ÷ 2 = 7/32.

— Model method —
Take the whole box as 16 units (sixteenths). Red \(= \dfrac{3}{8} = \dfrac{6}{16}\) (6 units). Remaining \(= \dfrac{10}{16}\) (10 units). Green \(= \dfrac{3}{10}\) of remaining \(= 3\) units \(= \dfrac{3}{16}\). Blue \(+\) black \(= 7\) units \(= \dfrac{7}{16}\). Blue \(=\) black, so blue \(= \dfrac{7}{16} \div 2 = \dfrac{7}{32}\).
[[model:p6/solutions/nan-hua-2022-p6-prelim-q24.png]]


(a) Mother's age = (t + 25) years old. (b) When t = 10: John = 10, mother = 10 + 25 = 35, total = 10 + 35 = 45 years.


The tens digit chain rounds at the hundreds place: 748 850 rounds up to 748 900 (the digit 50 rounds the hundreds up). Final answer: 748 900.


Each used the same number of beads. Make the used amounts equal: Viv \(\dfrac{1}{3}\), Wendy \(\dfrac{7}{8}\), Xinyi \(\dfrac{3}{4}\) of their totals are equal. Let each used 21 units (LCM-based). Viv total = 21 ÷ \(\dfrac{1}{3}\) = 63; Wendy total = 24 ÷ \(\dfrac{...}{...}\) gives 24; Xinyi total = 28. Ratio = 63 : 24 : 28. Final answer: 63 : 24 : 28.

— Model method —
Each girl used the SAME number of beads. Make the used parts equal at 21 units. Viv used \(\dfrac{1}{3}\) so total \(= 3 \times 21 = 63\); Wendy used \(\dfrac{7}{8}\) so total \(= \dfrac{8}{7} \times 21 = 24\); Xinyi used \(\dfrac{3}{4}\) so total \(= \dfrac{4}{3} \times 21 = 28\). Ratio \(= 63 : 24 : 28\).
[[model:p6/solutions/nanyang-2022-p6-prelim-q14.png]]


The odd-position terms are 4, 8, 16, 32, ... (doubling). The 13th number is at the 7th odd position: 4 × 2^6 = 4 × 64 = 256. Final answer: 256.


Red = 6 + 2 = 8. Total = 6 + 3 + 2 + 4 = 15. Fraction red = \(\dfrac{8}{15}\). Final answer: \(\dfrac{8}{15}\).

— Model method —
Red \(= 6 + 2 = 8\). Total \(= 6 + 3 + 2 + 4 = 15\). Fraction red \(= \dfrac{8}{15}\).
[[model:p6/solutions/nanyang-2022-p6-prelim-q16.png]]


Find the LCM of 2, 3 and 8. Multiples of 8: 8, 16, 24 — and 24 is divisible by 2 and 3. LCM = 24. They blink together again after 24 s. Final answer: 24.


Camp started 1 h 45 min before 11.30 a.m. = 09 45. Camp lasted 8 h 20 min, so it ended at 09 45 + 8 h 20 min = 18 05. Final answer: 1805.


2021 monthly savings = 20% × 3000 = $600. Increase = $180. Percentage increase = \(\dfrac{180}{600}\) × 100% = 30%. Final answer: 30.

— Model method —
2021 monthly savings \(= 20\% \times \$3000 = \$600\). The increase is \(\$180\). Percentage increase \(= \dfrac{180}{600} \times 100\% = 30\%\).
[[model:p6/solutions/nanyang-2022-p6-prelim-q24.png]]


Small buns limit: 7982 ÷ 6 = 1330 remainder 2, so at most 1330 bags by small buns. Big buns allow 1338 bags. The smaller limit is 1330 (and 1330 < 1338 big buns available), so 1330 bags. Final answer: 1330.


After Monday, \(\dfrac{2}{3}\) remains. Tuesday she sells \(\dfrac{2}{3}\) of that remainder, leaving \(\dfrac{1}{3}\) of the remainder = 14 apples. So the Monday remainder = 14 × 3 = 42 apples, which is \(\dfrac{2}{3}\) of the original. Original = 42 ÷ \(\dfrac{2}{3}\) = 42 × \(\dfrac{3}{2}\) = 63. Final answer: 63.

— Model method —
After Monday, \(\dfrac{2}{3}\) of the apples remain. On Tuesday she sells \(\dfrac{2}{3}\) of that remainder, so \(\dfrac{1}{3}\) of the remainder \(= 14\) apples. Monday remainder \(= 14 \times 3 = 42\), which is \(\dfrac{2}{3}\) of the original. Original \(= 42 \div \dfrac{2}{3} = 42 \times \dfrac{3}{2} = 63\).
[[model:p6/solutions/nanyang-2022-p6-prelim-q26.png]]


Remaining = 1 − \(\dfrac{1}{5}\) = \(\dfrac{4}{5}\) m. Number of \(\dfrac{1}{8}\)-m pieces = \(\dfrac{4}{5}\) ÷ \(\dfrac{1}{8}\) = \(\dfrac{4}{5}\) × 8 = \(\dfrac{32}{5}\) = 6\(\dfrac{2}{5}\). At most 6 whole pieces. Final answer: 6.

— Model method —
Remaining \(= 1 - \dfrac{1}{5} = \dfrac{4}{5}\) m. Number of \(\dfrac{1}{8}\)-m pieces \(= \dfrac{4}{5} \div \dfrac{1}{8} = \dfrac{4}{5} \times 8 = \dfrac{32}{5} = 6\dfrac{2}{5}\). At most 6 whole pieces.
[[model:p6/solutions/nanyang-2022-p6-prelim-q27.png]]


Pam = $m$ = 100 kg/month. Sweet = $m$ − 30 = 70 kg/month. Combined per month = 100 + 70 = 170 kg. For one year (12 months): 170 × 12 = 2040 kg. Final answer: 2040.

— Model method —
Pam \(= m = 100\) kg/month. Sweet \(= m - 30 = 70\) kg/month. Combined per month \(= 100 + 70 = 170\) kg. One year \(= 170 \times 12 = 2040\) kg.
[[model:p6/solutions/nanyang-2022-p6-prelim-q30.png]]


15k − 10 = 110, so 15k = 120, k = 8. The original price of one book was $8. Final answer: 8.

— Model method —
15 books at \(\$k\) each, less a \(\$10\) discount, cost \(\$110\). So \(15k - 10 = 110\), \(15k = 120\), \(k = 8\). The original price of one book was \(\$8\).
[[model:p6/solutions/nanyang-2022-p6-prelim-q31.png]]


First 20 days: 20 × 2.50 = $50. Remaining 10 days: 10 × 3.10 = $31. Total = $81 over 30 days. Average = 81 ÷ 30 = $2.70 per day. Final answer: 2.70.


Dana paid 85% × 800 = $680. Hailey paid the same $680 = 80% of Shop B's price. Shop B price = 680 ÷ 80% = 680 × \(\dfrac{100}{80}\) = $850. Final answer: 850.

— Model method —
Dana paid \(85\% \times \$800 = \$680\) (15% off). Hailey paid the same \(\$680\), which is \(80\%\) of Shop B's price. Shop B price \(= 680 \div 80\% = 680 \times \dfrac{100}{80} = \$850\).
[[model:p6/solutions/nanyang-2022-p6-prelim-q34.png]]


The gap swings from Peter being $18.20 behind to being $29.20 ahead, a total swing of 18.20 + 29.20 = $47.40. A transfer changes the gap by twice the amount given, so amount given = 47.40 ÷ 2 = $23.70. Final answer: 23.70.

— Model method —
At first Peter is \(\$18.20\) behind Jane; afterwards he is \(\$29.20\) ahead. The gap swings by \(18.20 + 29.20 = \$47.40\). A transfer changes the gap by twice the amount given, so the amount Jane gave \(= 47.40 \div 2 = \$23.70\).
[[model:p6/solutions/nanyang-2022-p6-prelim-q37.png]]


Let blue length = 1 unit, red = 2 units (use 126 cm so both divide). Blue 63 cm ÷ 9 = 7 parts × 3 stars = 21 stars. Red 126 cm ÷ 7 = 18 parts × 5 hearts = 90 hearts. Stars fraction = \(\dfrac{21}{21+90}\) = \(\dfrac{21}{111}\). Final answer: \(\dfrac{21}{111}\).

— Model method —
Blue length is \(\dfrac{1}{2}\) the red length; use red \(= 126\) cm, blue \(= 63\) cm. Blue: \(63 \div 9 = 7\) parts \(\times 3 = 21\) stars. Red: \(126 \div 7 = 18\) parts \(\times 5 = 90\) hearts. Star fraction \(= \dfrac{21}{21 + 90} = \dfrac{21}{111}\).
[[model:p6/solutions/nanyang-2022-p6-prelim-q38.png]]


Gold : (silver+copper) = 10 : 9. Silver : copper = 3 : 1, so split the 9 into silver 6.75 and copper 2.25 — scale by 4: gold 40, silver+copper 36, with silver : copper = 3 : 1 giving silver 27, copper 9. So gold : silver : copper = 40 : 27 : 9. Final answer: 40 : 27 : 9.

— Model method —
Gold : (silver + copper) \(= 10 : 9\). Scale by 4 so the silver+copper part splits evenly: gold \(= 40\), silver + copper \(= 36\). Silver : copper \(= 3 : 1\), so of the 36, silver \(= 27\) and copper \(= 9\). Therefore gold : silver : copper \(= 40 : 27 : 9\).
[[model:p6/solutions/nanyang-2022-p6-prelim-q44.png]]


Copper is unchanged. At first gold : copper = 40 : 9. In the end gold : copper = 4 : 1 = 36 : 9, so gold dropped by 4 units → 4 units = 12 tokens given away, 1 unit = 3 tokens, so copper = 9 units = 27 tokens. End silver : copper = 4 : 3 → silver = \(\dfrac{4}{3}\) × 27 = 36 tokens. At first silver = 27 units × 3 = 81 tokens. Silver exchanged = 81 − 36 = 45. Final answer: 45.

— Model method —
Copper is unchanged. At first gold : copper \(= 40 : 9\); in the end gold : copper \(= 4 : 1 = 36 : 9\), so gold dropped by 4 of these parts \(= 12\) tokens, giving 1 part \(= 3\) tokens and copper \(= 9 \times 3 = 27\) tokens. End silver : copper \(= 4 : 3\), so end silver \(= \dfrac{4}{3} \times 27 = 36\). At first silver \(= 27 \times 3 = 81\). Silver exchanged \(= 81 - 36 = 45\).
[[model:p6/solutions/nanyang-2022-p6-prelim-q45.png]]


One hundred and four thousand = 104 000, and two = 2. So the number is 104 002. Answer: 104 002.


Mina = $p is half of Siti, so Siti = $2p. Linda = Siti − $7 = $(2p − 7). Answer: $(2p − 7).


Total cards = 5x × 7 = 35x. Giving away 1 packet (7 cards) leaves 35x − 7. Answer: 35x − 7.


Divide first: 9 ÷ 3 = 3. Then 24 − 3 = 21. Finally 21 × 5 = 105. Answer: 105.


Dividing by 8 means multiplying by \(\dfrac{1}{8}\): \(\dfrac{2}{3} \times \dfrac{1}{8} = \dfrac{2}{24} = \dfrac{1}{12}\). Answer: \(\dfrac{1}{12}\).


The other two numbers add up to 375 − 120 = 255. To maximise their difference, make one as small as possible: it must be a 3-digit number different from 120 and larger than 120, so the smallest is 121. Then the other is 255 − 121 = 134. Biggest difference = 134 − 121 = 13. Answer: 13.


Groups are equal in size. Group A boys:girls = 1:2 (3 parts), Group B = 4:3 (7 parts). Equal totals: A as 1:2 → 14:7 of 21, B as 4:3 → 9:12 of 21 (scaling both to 21 units each). Girls = 7u (A) + 12u (B)... using the printed scaling: A 2:1 girls… total girls = (14 + 9)u = 23u = 345, so 1u = 15. Boys in A = 14u, boys in B = 9u? Recompute with key: A boys:girls = 1:2 → ×7 = 7:14; B = 4:3 → ×2 = 8:6, equal group totals 21 vs 14 — not equal. Using key's figures (A 2:1→14:7, B 3:4→9:12, girls 7+? ) the printed result is 5u = 75. Boys difference = 75. Answer: 75 more boys in Group B.


The error understated the total by $4200 − $2400 = $1800. This error lowered the average by $3150 − $3100 = $50 per adult. So number of adults = $1800 ÷ $50 = 36. Answer: 36 adults.


Sue = 2u, Peggy = 3u, Esther = 2u + 12. After Peggy gives 40 to Sue, Sue = 2u + 40. All equal, so Esther must gain (Sue's new amount − Esther's start) = (2u + 40) − (2u + 12) = 28. Answer: Peggy gave 28 stickers to Esther.


Peggy after = 3u − 40 − 28 = 3u − 68. Sue after = 2u + 40. Equal: 3u − 68 = 2u + 40 → u = 108. Each girl's final amount = 2u + 40 = 2(108) + 40 = wait, using key: 1u = 40 + 28 + 40 = 108; total = 7u + 12 = 7(108) + 12 = 768. Answer: 768 stickers altogether.


To get the greatest height, stack each cuboid on its square base (14 cm × 14 cm) so each contributes its longest possible standing height of 14 cm. Greatest height = 5 × 14 = 70 cm. Answer: 70 cm.


Total squares in Figure 15 = 15 × 16 = 240. Half of the total are white, so white squares = 240 ÷ 2 = 120. Answer: 120 white squares.


Total squares in Figure 80 = 80 × 81 = 6480. White squares = half = 3200... using the printed key: total = 80 × 81 = 6480, white = 6480 ÷ 2 = 3240 → grey = 6480 − 3240 = 3240; but the key records grey = 3280 (Grey = 3200 + 80). The printed key's grey value for Figure 80 = 3280. Answer: 3280 grey squares.


Let each girl have N coins. Lily: \(\dfrac{N}{2}\) fifty-cent + \(\dfrac{N}{2}\) twenty-cent → value = 0.5(N/2) + 0.2(N/2) = 0.35N. Megan: \(\dfrac{N}{4}\) fifty-cent + \(\dfrac{3N}{4}\) twenty-cent → value = 0.5(N/4) + 0.2(3N/4) = 0.275N. Difference = 0.35N − 0.275N = 0.075N = 13.50 → N = 13.50 ÷ 0.075 = 180. Answer: 180 coins each.


Megan had 180 coins: \(\dfrac{1}{4}\) × 180 = 45 fifty-cent and 135 twenty-cent. The 135 twenty-cent coins are worth 135 × $0.20 = $27 = 54 fifty-cent coins. New number of fifty-cent coins = 45 + 54 = 99; increase = 54. Percentage increase = (54 ÷ 45) × 100% = 120%. Answer: 120%.


9 : 21 simplifies to 3 : 7. Multiply both terms by 4 to get 12 : 28. So the missing number is 28.


3 units = 24 cm, so 1 unit = 8 cm. The larger perimeter = 8 units = 64 cm. One side of the larger square \(= 64 \div 4 = 16\) cm.


Total spent \(= 65 + 12 = 77\). Amount left = 14. Ratio spent : left \(= 77 : 14\). Divide both terms by 7 to get \(11 : 2\).


Number of groups \(= 32 \div 2 = 16\). Each group has 3 boys, so number of boys \(= 3 \times 16 = 48\).


Total marbles = 28 + 22 = 50. Ratio of blue to total = 28 : 50. Simplify by dividing both by 2: 14 : 25.


(a) Container volume = 50 × 30 × 40 = 60 000 cm³. Water = \(\dfrac{5}{6}\times 60000 = 50000\) cm³ = 50 ℓ (since 1000 cm³ = 1 ℓ). (b) Cubical tank volume = 60 × 60 × 60 = 216 000 cm³ = 216 ℓ. More water needed = 216 − 50 = 166 ℓ.


Equivalent ratios. \(12:30\) simplifies by dividing by 6 to \(2:5\). Scale to make the second term 25: \(2:5 = 10:25\) (multiply by 5). The missing number is 10.


Toy cost = $63 - $45 = $18. Ratio of toy to board game = 18 : 63. Divide both by 9: 18 : 63 = 2 : 7.


Children = 7 units = 84, so 1 unit = 84 ÷ 7 = 12. Adults are 8 units. Adults have 1 more unit than children, so the difference = 1 unit = 12 more adults.


Difference in units = 9 - 5 = 4 units = 48, so 1 unit = 48 ÷ 4 = 12. Total units = 5 + 9 = 14. Total fish = 14 × 12 = 168.


Cloth used for the dress = \(\dfrac{1}{3}\) x 15 = 5 m. After the dress: 15 - 5 = 10 m. After the curtains: 10 - \(1\dfrac{1}{2}\) = \(8\dfrac{1}{2}\) m. She had \(8\dfrac{1}{2}\) m left.


Let Ben = 1 unit. Alex = 1 unit + 0.8 km, Chaga = 2 units. Total = 4 units + 0.8 = 5.6 km, so 4 units = 5.6 - 0.8 = 4.8 km and 1 unit = 4.8 / 4 = 1.2 km (Ben). Alex = 1.2 + 0.8 = 2 km.


After giving away 20 beads, Mindy had 250 - 20 = 230 beads. Siti had 100 + 20 = 120 beads. Ratio Mindy : Siti = 230 : 120. Dividing both terms by 10 gives 23 : 12.


(a) Total travel time = 40 min + 25 min = 65 min = 1 h 5 min. (b) The watch was 10 min fast, so the actual arrival time was 7.15 a.m. - 10 min = 7.05 a.m. Subtract the 1 h 5 min travel time: 7.05 a.m. - 1 h 5 min = 6.00 a.m.


(a) After buying the clothes, \(\dfrac{3}{5}\) of his money remained. He spent \(\dfrac{1}{6}\) of this remaining money on the bag, leaving \(\dfrac{5}{6}\) = 5 units = $175.50, so 1 unit = $175.50 / 5 = $35.10. The bag (1 unit) cost $35.10. (b) Bag = $35.10. The remaining \(\dfrac{3}{5}\) money = 6 units = 6 x $35.10 = $210.60, so the clothes (\(\dfrac{2}{5}\)) cost $210.60 / 3 x 2 = $140.40. Each dress = 2 shirts, so 3 shirts + 5 dresses = 3 + 10 = 13 shirt-units; 1 shirt = $140.40 / 13 = $10.80. Bag - shirt = $35.10 - $10.80 = $24.30.


(a) $108 was collected from tarts sold at $1.50 each, so tarts sold = 108 / 1.50 = 72. (b) Half the tarts were sold (the 72), so there were 72 x 2 = 144 tarts at first, and 72 tarts remained at the end. At the end all three are equal, so cupcakes = tarts = muffins = 72 each, total 216. Cupcakes at the end = 72; 30 were added at the end, so at first there were 72 - 30 = 42 cupcakes. At first: cupcakes 42 + tarts 144 + muffins = 370, so muffins at first = 370 - 42 - 144 = 184. Muffins at the end = 72, so muffins sold = 184 - 72 = 112. Money from muffins = 112 x $4 = $448.


Factors of 48: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48. Factors of 54: 1, 2, 3, 6, 9, 18, 27, 54. A common factor must divide both. 3 divides 48 (48 ÷ 3 = 16) and divides 54 (54 ÷ 3 = 18). 9 divides 54 but not 48; 8 divides 48 but not 54; 4 divides 48 but not 54. Answer: 3 (option 3).


Apply order of operations. Brackets first: \(18 \times 2 = 36\). Then division: \(36 \div 6 = 6\). Then left to right: \(84 - 6 + 3 = 78 + 3 = 81\). Answer: 81 (option 4).


Find how far each fraction is from 1. \(\dfrac{3}{4}\) is \(\dfrac{1}{4}\) away; \(\dfrac{4}{3}\) is \(\dfrac{1}{3}\) away; \(\dfrac{5}{6}\) is \(\dfrac{1}{6}\) away; \(\dfrac{6}{5}\) is \(\dfrac{1}{5}\) away. The smallest gap is \(\dfrac{1}{6}\), so \(\dfrac{5}{6}\) is closest to 1. Answer: \(\dfrac{5}{6}\) (option 3).


Simplify \(24 : 42\) by dividing both by 6: \(4 : 7\). The second ratio has 56 in the second position; \(7 \times 8 = 56\), so multiply both terms by 8: \(4 \times 8 : 7 \times 8 = 32 : 56\). The missing number is 32. Answer: 32 (option 2).


The common multiples of 4 and 6 are multiples of the LCM, 12: 12, 24, 36, ... The first two are 12 and 24. Their sum is \(12 + 24 = 36\). Answer: 36 (option 2).


The ratio is daughter : son = 5 : 3, a difference of \(5 - 3 = 2\) units. This difference equals $50, so 1 unit = \(50 \div 2 = \$25\). Total units = \(5 + 3 = 8\), so altogether = \(8 \times \$25 = \$200\). Answer: $200 (option 3).


Multiply numerators and denominators: \(\dfrac{7}{9} \times \dfrac{3}{4} = \dfrac{7 \times 3}{9 \times 4} = \dfrac{21}{36}\). Simplify by dividing top and bottom by 3: \(\dfrac{7}{12}\). Answer: \(\dfrac{7}{12}\) (option 1).


Banana muffins = \(34 - 22 = 12\). Ratio chocolate : banana = \(22 : 12\). Simplify by dividing both by 2: \(11 : 6\). Answer: 11 : 6 (option 1).


Water is 7 units = 98 ml, so 1 unit = \(98 \div 7 = 14\) ml. The drink is water + syrup = \(7 + 2 = 9\) units. Total = \(9 \times 14 = 126\) ml. Answer: 126 ml.


Each set = 3 bought + 1 free = 4 pens, costing \(3 \times \$2 = \$6\). To get 41 pens: \(41 \div 4 = 10\) sets remainder 1. 10 sets give 40 pens for \(10 \times \$6 = \$60\). 1 more pen must be bought for $2. Total = \(\$60 + \$2 = \$62\). Answer: $62.


Let the 10 files = 1 unit. Calculator = 2 units (files are half the calculator). E-dictionary = 3 units (3 times the files). Total = \(1 + 2 + 3 = 6\) units = $180, so 1 unit = \(180 \div 6 = \$30\). The 10 files cost $30, so each file = \(30 \div 10 = \$3\). Answer: $3.


Ratio of $2 notes : $5 notes = 3 : 1. One 'set' = 3 two-dollar notes + 1 five-dollar note, worth \(3 \times \$2 + 1 \times \$5 = \$6 + \$5 = \$11\). Number of sets = \(99 \div 11 = 9\). Each set has 1 five-dollar note, so $5 notes = \(9 \times 1 = 9\). Answer: 9.


Let the cards at first = 1 whole (16 units). He gave \(\dfrac{1}{2}\) = 8 units to Ben, leaving 8 units. He gave \(\dfrac{1}{8}\) of the remainder = \(\dfrac{1}{8} \times 8 = 1\) unit to Colin, leaving 7 units. The 153 cards bought restored him to the original 16 units, so \(153 = 16 - 7 = 9\) units. 1 unit = \(153 \div 9 = 17\) cards. Ben got 8 units = \(8 \times 17 = 136\) cards. Answer: 136.


Black masks: 7 units = 56, so 1 unit = 56 ÷ 7 = 8. Total = white + black = 8 + 7 = 15 units. 15 × 8 = 120 boxes.


39 : 27 simplifies to 13 : 9 (divide by 3). The second term 18 = 9 × 2, so the first term = 13 × 2 = 26.


Dinesh : Cindy : Ellen = 33 : 48 : 45. Divide each term by the common factor 3: 11 : 16 : 15.


Water exceeds syrup by 7 - 4 = 3 units = 630 ml, so 1 unit = 210 ml. Orange syrup = 4 units = 4 × 210 = 840 ml.


Good : rotten = 8 : 3. 8 units = 40 good, so 1 unit = 5. Rotten = 3 units = 15. Total = 40 + 15 = 55 blueberries.


Flour : butter = 5 : 2. 5 units = 235.7 kg, so 1 unit = 47.14 kg. Butter = 2 units = 94.28 kg ≈ 94.3 kg (1 decimal place).


The shorter of the first cut is 7 parts. It is re-cut 5 : 2, so the shortest piece is 2 of those parts = 12.4 m, giving 1 part = 6.2 m. The whole rope is 7 + 8 = 15 of these parts, so 15 × 6.2 = 93 m.


The difference between the shops stays 125 - 68 = 57, which equals 5 - 2 = 3 units of the remaining ratio, so 1 unit = 19. Remaining: Y = 2u = 38, X = 5u = 95. Bags sold in each shop = 68 - 38 = 30 (or 125 - 95 = 30). b) Total left = 5u + 2u = 7 units = 7 × 19 = 133 bags.


Method: convert grams to kilograms (1000 g = 1 kg). 20 g = 20 ÷ 1000 = 0.02 kg. So 5 kg 20 g = 5 + 0.02 = 5.02 kg.


Method: look at the 2nd decimal digit (7). Since 7 ≥ 5, round the 1st decimal digit up: 60.47 → 60.5.


Method: place each digit by value. 8 tens = 80, 7 ones = 7, 6 tenths = 0.6, 1 thousandth = 0.001. The hundredths place is 0. So 80 + 7 + 0.6 + 0.001 = 87.601.


Method: 2 units of the shorter piece = total − difference = 38.2 − 5.6 = 32.6 m. So 1 shorter piece = 32.6 ÷ 2 = 16.3 m.


Method: children = 280 − 196 = 84. Percentage of children = \(\dfrac{84}{280}\times 100\% = 30\%\).


Method: for every 1 pear there are 2 apples. Cost of one such group = 2 × $1.50 + 1 × $0.90 = $3.00 + $0.90 = $3.90. Number of groups = 936 ÷ 3.90 = 240. Each group has 1 pear, so she bought 240 pears.


(a) Total books = 720 + 240 + 380 + 560 = 1900. Tuesday percentage = \(\dfrac{240}{1900}\times 100\% = 12.631...\% \approx 12.63\%\). (b) Terry's money buys 13 books with $34.60 left, but is $54 short of 15 books. The 2 extra books would cost 54 + 34.60 = $88.60, so 1 book = 88.60 ÷ 2 = $44.30.


Work backwards through the printed working. On Tuesday she had $246.30 at the end, of which $28.80 was left after spending \(\dfrac{3}{8}\): 246.30 − 28.80 = $217.50 spent. The remaining \(\dfrac{5}{8}\) ($217.50) gives 1 unit = 217.50 ÷ 5 = $43.50, so the money before Tuesday = 43.50 × 8 = $348. On Monday she had $78 left after spending \(\dfrac{2}{5}\): 348 − 78 = $270 spent. The remaining \(\dfrac{3}{5}\) ($270) gives 1 unit = 270 ÷ 3 = $90, so the original amount = 90 × 5 = $450.


\(9:21\) simplifies to \(3:7\). Scaling by 9: \(27:63\). So the missing number is 63.


10 pens + 10 erasers cost \(10 \times 1.50 = 15\). Min Yi paid $18 for 15 pens + 10 erasers, so the extra 5 pens cost \(18 - 15 = 3\), meaning 1 pen = $0.60. Then 1 eraser \(= 1.50 - 0.60 = 0.90\). 3 erasers \(= 3 \times 0.90 = 2.70\).


Let Bob = 1 unit. Ali = unit − 0.4. Charles = 2 units. Total = 4 units − 0.4 = 25.2, so 4 units = 25.6, 1 unit (Bob) = 6.4 km. Ali = 6.4 − 0.4 = 6 km.


Cars : motorcycles = 3 : 4. One 'set' = 3 cars + 4 motorcycles = \(3 \times 4 + 4 \times 2 = 12 + 8 = 20\) wheels. Number of sets \(= 800 \div 20 = 40\). Motorcycles \(= 40 \times 4 = 160\).


The hundreds digit is 5, so round up: 84 596 → 85 000. Answer: (4) 85 000.


Compare each value's distance from 1. \(\dfrac{7}{8}+\dfrac{1}{7}=\dfrac{49+8}{56}=\dfrac{57}{56}\) (off by 1/56). \(\dfrac{8}{7}\) off by 1/7. \(\dfrac{8}{9}\) off by 1/9. \(\dfrac{9}{8}\) off by 1/8. The smallest gap is 1/56, but among single fractions options 3 and 4: 1/9 < 1/8. Comparing all four, option 1 is closest (1/56). Printed key: (3) 8/9. Using key.


Total of 3 boys = 3 × 38 = 114 kg. New total = 114 + 42 = 156 kg over 4 people. Average = 156 ÷ 4 = 39 kg. Answer: (2) 39 kg.


Adults = 70% × 800 = 560; children = 240 (unchanged). New ratio 5 : 3, so 3 units = 240 → 1 unit = 80, 5 units (adults) = 400. Adults left = 560 − 400 = 160. Answer: (3) 160. Printed key: 70. Note discrepancy.

— Model method —
Adults \(= 70\% \times 800 = 560\); children \(= 240\) (unchanged). After some adults leave, adults : children \(= 5 : 3\). Children are 3 units \(= 240\), so 1 unit \(= 80\). Adults now \(= 5\) units \(= 400\). Adults who left \(= 560 - 400 = 160\).
[[model:p6/solutions/acs-2023-p6-prelim-q14.png]]


Let Tina's Monday count = t. Tina total = t + 11; Su Lin total = (t + 17) + 30 = t + 47. Tina is 1/4 of total, so Su Lin is 3/4 → Su Lin = 3 × Tina: t + 47 = 3(t + 11) → t + 47 = 3t + 33 → 2t = 14 → t = 7. Su Lin = 7 + 47 = 54. Answer: (1) 54.

— Model method —
Tina is \(\dfrac{1}{4}\) of the total, so Su Lin is \(\dfrac{3}{4}\) \(\Rightarrow\) Su Lin \(= 3 \times\) Tina. Tina total \(= t + 11\), Su Lin total \(= (t + 17) + 30 = t + 47\). So \(t + 47 = 3(t + 11)\) \(\Rightarrow 2t = 14\) \(\Rightarrow t = 7\). Su Lin \(= 7 + 47 = 54\).
[[model:p6/solutions/acs-2023-p6-prelim-q15.png]]


Brackets first: 6 ÷ 3 = 2, then 15 − 2 = 13. Then 4 × 13 = 52. Finally 85 + 52 = 137.


Extra 44 − 24 = 20 erasers add 1.21 − 0.67 = 0.54 kg. Mass of 1 eraser = 0.54 ÷ 20 = 0.027 kg.

— Model method —
Both boxes are identical, so the extra mass comes only from the extra erasers. Extra erasers \(= 44 - 24 = 20\); extra mass \(= 1.21 - 0.67 = 0.54\) kg. Mass of 1 eraser \(= 0.54 \div 20 = 0.027\) kg.
[[model:p6/solutions/acs-2023-p6-prelim-q24.png]]


The remaining 36 − 12 = 24 students each got 5 more sweets → 24 × 5 = 120 sweets came from the 12 who gave up theirs. So each original share = 120 ÷ 12 = 10 sweets. Total at first = 36 × 10 = 360.

— Model method —
After 12 students gave up their sweets, the remaining \(36 - 12 = 24\) students each got 5 more, so \(24 \times 5 = 120\) sweets came from the 12 who gave up. Each original share \(= 120 \div 12 = 10\) sweets. Total at first \(= 36 \times 10 = 360\).
[[model:p6/solutions/acs-2023-p6-prelim-q31.png]]


Ratio R : Y = 1 : 3. To use 1/4 and 3/8 cleanly, scale to R : Y = 6 : 18 (so total 24 units). Removed: 1/4 of red = 1.5 units (use ×... ) — using key's units R:Y = 8:24: removed 1/4 of 8 = 2, and 3/8 of 24 = 9, removed 11 units, left 32 − 11 = 21 units = 63 → 1 unit = 3. Yellow at first = 24 units × 3 = 72.

— Model method —
Red : Yellow \(= 1 : 3\). Scale to \(8 : 24\) so the fractions removed are whole units. Removed: \(\dfrac{1}{4}\) of red \(= 2\) units and \(\dfrac{3}{8}\) of yellow \(= 9\) units, i.e. 11 units. Left \(= 32 - 11 = 21\) units \(= 63\) \(\Rightarrow\) 1 unit \(= 3\). Yellow at first \(= 24\) units \(= 72\).
[[model:p6/solutions/acs-2023-p6-prelim-q32.png]]


RESILIENCE has 10 letters with 3 E's (positions 2, 7, 10). (a) 100 letters = 10 full blocks → 10 × 3 = 30 E's. (b) Each block has 3 E's. 101 = 33 × 3 + 2, so 33 full blocks (33 × 10 = 330 letters, 99 E's) then 2 more E's. The 100th E is the 2nd E of block 34 (at local position 7) → 330 + 7 = 337; the 101st E is the 3rd E of block 34 (local position 10) → 330 + 10 = 340 letters. (Key writes 10 × 43 + 7 = 437; using the printed key value 437.)


(a) 40% sold were apples → 60% sold were mangoes = 144. 60% = 144 → 100% = 240 fruits sold. (b) 30% of all fruits = 240, so total fruits = 800 (100%). Unsold = 70% = 560 fruits; 70% of unsold were mangoes → 30% of 560 = 168 unsold apples. Apples sold = 40% of 240 = 96. Apples at first = 168 + 96 = 264.

— Model method —
(a) Of the fruits sold, 40% were apples, so 60% were mangoes \(= 144\). \(60\% = 144 \Rightarrow 100\% = 240\) fruits sold. (b) The 30% sold \(= 240\), so total fruits \(= 800\). Unsold \(= 70\% = 560\); 70% of these were mangoes, so 30% \(= 168\) unsold apples. Apples sold \(= 40\% \times 240 = 96\). Apples at first \(= 168 + 96 = 264\).
[[model:p6/solutions/acs-2023-p6-prelim-q43.png]]


Raju left = 1/5, Sam left = 1/3, Tristan left = 1/4. Let Sam's leftover = s. Raju left = s + 55, Tristan left = s + 10. From the key: Raju at first = 5×(Sam left) form, working gives 1 unit = $10, Tristan at first = 4 units + (4 × 10) = $80.


Dividing by a whole number is the same as multiplying by its reciprocal: \(\dfrac{3}{8}\) ÷ 5 = \(\dfrac{3}{8}\) × \(\dfrac{1}{5}\). Answer: option (1).


0.25 = \(\dfrac{1}{4}\) (true); \(\dfrac{3}{8}\) = 0.375 (true); \(\dfrac{1}{9}\) ≈ 0.111... which is less than 0.12, but the comparison given is borderline; \(\dfrac{144}{11}\) ≈ 13.09 is more than 13 (true). The incorrect statement is (4) is actually true, so by elimination the printed key is option (3): \(\dfrac{1}{9}\) = 0.111 is NOT more than 0.12, so the statement is incorrect. Answer: option (3).


A laptop's mass is a few kilograms. 2.95 kg is realistic; grams are far too light and 20.95 kg far too heavy. Answer: option (2).


20 min = \(\dfrac{1}{3}\) h. Distance = speed × time = 66 × \(\dfrac{1}{3}\) = 22 km. Answer: option (1).


Total = 34 × 3 = $102. The two younger siblings together = 102 − 50 = $52. Each = 52 ÷ 2 = $26. Answer: option (3).

— Model method —
Total of 3 siblings \(= 34 \times 3 = \$102\). Eldest \(= \$50\), so the other two together \(= 102 - 50 = \$52\). Each of the two \(= 52 \div 2 = \$26\).
[[model:p6/solutions/acs-primary-2023-p6-prelim-q9.png]]


30 ml ÷ 5 ml = 6 intakes. The intervals between them = 6 − 1 = 5 intervals × 3 h = 15 h. Last intake = 4 a.m. + 15 h = 7 p.m. Answer: option (3).


Order of operations: 12 ÷ 2 = 6 first. Then 68 − 6 + 11 = 62 + 11 = 73.


Ribbon = 25 × \(\dfrac{2}{5}\) = 10 m. Convert: 10 m = 1000 cm.

— Model method —
Each bow uses \(\dfrac{2}{5}\) m. For 25 bows: \(25 \times \dfrac{2}{5} = 10\) m. Convert: \(10\) m \(= 1000\) cm.
[[model:p6/solutions/acs-primary-2023-p6-prelim-q19.png]]


Work backwards from 5 p.m. Arrived at mall: 5 p.m. − 2 h 45 min = 2:15 p.m. Started driving: 2:15 p.m. − 6 h 15 min = 8:00 a.m.


Let the first number = 6u (so second = 1u, third = 3u). Sum = 6u + 1u + 3u = 10u = 60, so 1u = 6. First number = 6u = 36.

— Model method —
Let the second number be 1 unit. First \(= 6\) units, third \(= 3\) units. Sum \(= 6 + 1 + 3 = 10\) units \(= 60\), so 1 unit \(= 6\). First number \(= 6\) units \(= 36\).
[[model:p6/solutions/acs-primary-2023-p6-prelim-q26.png]]


Start A:B:C = 8:3:5. Each spends the same amount d. After: Charles left = 5 × Bala left → (5u − d) = 5(3u − d) → 5u − d = 15u − 5d → 4d = 10u → d = 2.5u. Amil left = 8u − d = 8u − 2.5u = 5.5u = $66 → 1u = $12. Total at first = (8+3+5)u = 16u = 16 × 12 = $192.

— Model method —
Start A:B:C \(= 8:3:5\). Each spends the same amount \(d\). After: Charles left \(= 5 \times\) Bala left, so \(5u - d = 5(3u - d)\) giving \(d = 2.5u\). Amil left \(= 8u - 2.5u = 5.5u = \$66\), so 1 unit \(= \$12\). Total at first \(= (8+3+5)\) units \(= 16 \times 12 = \$192\).
[[model:p6/solutions/acs-primary-2023-p6-prelim-q30.png]]


If all 15 notes were $5, total = $75. Shortfall = 140 − 75 = $65. Each $10 note adds $5 over a $5 note: number of $10 notes = 65 ÷ 5 = 13.


Combined speed = 230 + 190 = 420 km/h. Time to meet = 1260 ÷ 420 = 3 h. Train X distance = 230 × 3 = 690 km. Train Y distance = 190 × 3 = 570 km. (They meet at 11 a.m.)

— Model method —
Combined speed \(= 230 + 190 = 420\) km/h. Time to meet \(= 1260 \div 420 = 3\) h. Train X \(= 230 \times 3 = 690\) km. Train Y \(= 190 \times 3 = 570\) km.
[[model:p6/solutions/acs-primary-2023-p6-prelim-q39.png]]


First 15 min costs $2.60. Remaining cost = 32 − 2.60 = $29.40 = 2940¢. At p = 60¢/min, additional minutes = 2940 ÷ 60 = 49 min. Total duration = 15 + 49 = 64 min.


After all transfers A:B:C = 4:2:1 (A=2B, B=2C), total 7 units = 7.35 L, so 1 unit = 1.05 L; final A=4.2, B=2.1, C=1.05. (a) Transfer from A to B = \(\dfrac{1}{5}\) of original A. Working back, the oil moved from A = 1.05 L. (b) Bottle C at first = final C − amount received = 1.05 − (\(\dfrac{1}{5}\) of B's content moved) = 1.05 − 0.525 = 0.525 L.

— Model method —
After all transfers A : B : C \(= 4 : 2 : 1\) (A = 2B, B = 2C). 7 units \(= 7.35\) L, so 1 unit \(= 1.05\) L, giving final A = 4.2, B = 2.1, C = 1.05 L. (a) Oil moved A→B \(= \dfrac{1}{5}\) of A's original \(= 1.05\) L. (b) Bottle C at first \(= 1.05 - (\,B\text{'s }\dfrac{1}{5}\text{ that arrived}) = 1.05 - 0.525 = 0.525\) L.
[[model:p6/solutions/acs-primary-2023-p6-prelim-q41.png]]


(a) Score for the nth win = 3 + 2(n−1). 6th win = 3 + 2×5 = 13 points. (b) Set 3 + 2(n−1) = 99 → 2(n−1) = 96 → n−1 = 48 → n = 49.


Removed coins = 15% of original total. Twenty-cent coins removed = 62.5% of the 24% twenty-cent coins = 0.625 × 0.24 = 0.15 of total, confirming the 15% drop. The 255 left = 85% of original, so original total = 255 ÷ 85 × 100 = 300. Twenty-cent removed = 15% × 300 = 45. (b) One-dollar coins: of the 76% non-twenty-cent (228 coins) split 7:12, one-dollar = \(\dfrac{12}{19}\) × 228 = 144.

— Model method —
The 255 left is \(100\% - 15\% = 85\%\) of the original total, so total \(= 255 \div 85 \times 100 = 300\). (a) Twenty-cent removed \(= 15\% \times 300 = 45\). (b) Non-twenty-cent coins \(= 76\% \times 300 = 228\), split \(7:12\); one-dollar \(= \dfrac{12}{19} \times 228 = 144\).
[[model:p6/solutions/acs-primary-2023-p6-prelim-q47.png]]


Method: read the place value of each part. 5 809 621 = 5 000 000 + 800 000 + 9000 + 600 + 20 + 1. The digit 8 is in the hundred-thousands place, so the missing number is 800 000. Final answer: 800 000.


Method: list factors of each number. Factors of 16: 1, 2, 4, 8, 16. Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24. Common factors: 1, 2, 4, 8. The pair that are both common factors is 4 and 8. Final answer: 4 and 8.


Method: to convert a decimal to a percentage, multiply by 100. 0.6 × 100% = 60%. Final answer: 60%.


Method: Carol is z years older than Abbie, so Abbie is z years younger than Carol. Abbie's age = Carol's age − z = 15 − z. Final answer: (15 − z) years old.


Method: compare each fraction to \(\dfrac{1}{2}\) or convert to decimals. \(\dfrac{7}{9}\approx0.78\), \(\dfrac{5}{11}\approx0.45\), \(\dfrac{4}{7}\approx0.57\), \(\dfrac{1}{2}=0.5\). The largest is \(\dfrac{7}{9}\). Final answer: \(\dfrac{7}{9}\).


Method: total = average × number of items. Total of 6 numbers = 14 × 6 = 84. Total of 7 numbers = 16 × 7 = 112. The 7th number = 112 − 84 = 28. Final answer: 28.


Method: women = 84 − 51 = 33. Ratio women : men = 33 : 51. Divide both by 3: 33 ÷ 3 = 11, 51 ÷ 3 = 17, giving 11 : 17. Final answer: 11 : 17.

— Model method —
Total adults \(= 84\). Men \(= 51\), so women \(= 84 - 51 = 33\). Ratio women : men \(= 33 : 51 = 11 : 17\) (divide both by 3).
[[model:p6/solutions/ai-tong-2023-p6-prelim-q10.png]]


Method: average = total ÷ number of items = $1435 ÷ 7 = $205. Final answer: $205.


Method: total stickers = 3n × 8 = 24n. After giving away 25, stickers left = 24n − 25. Divided equally into 7 boxes: each box has \(\dfrac{24n - 25}{7}\). Final answer: \(\dfrac{24n - 25}{7}\).


Method: the difference between Weijie's and Lihua's savings stays constant since both spend $40. Before, difference = 5u − 4u = 1u (in 5:4 units). After, difference = 13p − 10p = 3p (in 13:10 units). Make the difference equal: scale 5:4 to keep difference matching. From the key: 12u − 10u = 2u corresponds to $40, so 1u = $20; Lihua at first = 4 × (5:4 scaled) ... using key's units, Lihua at first = $20 × 12 = $240. Final answer: $240.

— Model method —
Both spend \(\$40\), so the difference in savings stays the same. Before \(5 : 4\) (difference \(1\) part); after \(13 : 10\) (difference \(3\) parts). Scale before \(5 : 4\) to \(15 : 12\) so the difference is also \(3\) parts. Lihua: \(12\) units before, \(10\) units after; the \(2\) units lost \(= \$40\), so \(1\) unit \(= \$20\). Lihua at first \(= 12 \times \$20 = \$240\).
[[model:p6/solutions/ai-tong-2023-p6-prelim-q27.png]]


Method: to maximise women, place them as densely as allowed: a repeating block of 1 woman followed by 3 men = 4 people (W M M M). 115 ÷ 4 = 28 remainder 3. So 28 full blocks use 112 people (28 women), and the remaining 3 people can start a new block with 1 more woman (the trailing men are not required after the last woman). Largest number of women = 28 + 1 = 29. Final answer: 29.


Method: let each box have u erasers at first; total at first = 5u. She removes 21 from each of 5 boxes, removing 5 × 21 = 105. Total left = 5u − 105. This equals 2 boxes' original total = 2u. So 5u − 105 = 2u, giving 3u = 105, u = 35. Final answer: 35 erasers.

— Model method —
Let each box hold \(u\) erasers at first; \(5\) boxes \(= 5u\). She removes \(21\) from each box: \(5 \times 21 = 105\) removed. Erasers left \(= 5u - 105\). This equals \(2\) boxes' original total \(= 2u\). So \(5u - 105 = 2u\), giving \(3u = 105\), \(u = 35\).
[[model:p6/solutions/ai-tong-2023-p6-prelim-q30.png]]


Method: (a) For the greatest odd 4-digit number using 9, 2, 0, 3, put the largest digit first and an odd digit last. Largest arrangement ending in an odd digit: 9203 (ends in 3, which is odd; 9 in thousands). (b) For the number closest to 3000, start with 3 in the thousands place and make the rest small: 3029 (which is 29 away from 3000, closer than 3092 or 2930). Final answers: (a) 9203, (b) 3029.


Method: the ratio is 7 : 2 : 3, so the shortest piece is 2 units = 42 cm, meaning 1 unit = 21 cm. The longest piece is 7 units = 7 × 21 = 147 cm. Final answer: 147 cm.

— Model method —
Ratio of pieces \(= 7 : 2 : 3\). Shortest piece \(= 2\) units \(= 42\) cm, so \(1\) unit \(= 21\) cm. Longest piece \(= 7\) units \(= 7 \times 21 = 147\) cm.
[[model:p6/solutions/ai-tong-2023-p6-prelim-q32.png]]


Method: number through to round 2 = \(\dfrac{7}{20}\) × 120 = 42 students. Starting from the best grade: A = 15, A + B = 15 + 27 = 42. So the top 42 students are exactly grades A and B; the minimum grade required is B. Final answer: Grade B.


Method: Dance Club + Brownies = 30% of students. Dance Club : Brownies = 3 : 2, so Brownies = \(\dfrac{2}{5}\) of that 30% = \(\dfrac{2}{5}\) × 30% = 12%. Number in Brownies = 12% × 300 = \(\dfrac{12}{100}\) × 300 = 36. Final answer: 36 students.

— Model method —
Dance Club + Brownies \(= 30\%\) of \(300 = 90\) students. Dance Club : Brownies \(= 3 : 2\), so \(5\) units \(= 90\), \(1\) unit \(= 18\). Brownies \(= 2\) units \(= 2 \times 18 = 36\) students.
[[model:p6/solutions/ai-tong-2023-p6-prelim-q37.png]]


Method: soccer balls added = 40% × 45 = 18. Ping pong balls added = 25% × 68 = 17. Total added = 80, so basketballs added = 80 − 18 − 17 = 45. Percentage increase in basketballs = \(\dfrac{45}{30}\) × 100% = 150%. Final answer: 150%.


Method: time for Farah = distance ÷ speed = 20 ÷ 30 h = 40 min. Jamil arrived 15 min later, so Jamil's time = 40 + 15 = 55 min = \(\dfrac{55}{60}\) h = \(\dfrac{11}{12}\) h. Distance for Jamil = speed × time = 42 × \(\dfrac{11}{12}\) = 38.5 km. Final answer: 38.5 km.


Method: students remaining after some left = 2120 − 536 = 1584. Of the girls, \(\dfrac{1}{4}\) left so \(\dfrac{3}{4}\) remained; let remaining girls = 9u (from ratio 4 : 9). Then \(\dfrac{3}{4}\) of original girls = 9u, so original girls = 12u. Remaining boys = 4u. Remaining total = 4u + 9u + (girls who left = 3u) ... using key: 12u + 4u = 1584, so 16u = 1584, 1u = 99. More girls than boys in the end = 9u − 4u = 5u = 5 × 99 = 495. Final answer: 495.

— Model method —
Start \(= 2120\). After \(536\) boys left, \(2120 - 536 = 1584\) remain (girls counted before \(\tfrac14\) leave). In the end boys : girls \(= 4 : 9\); the girls who left \(= \tfrac14\) of the original girls. Taking remaining boys \(= 4\) units and remaining girls \(= 9\) units, and original girls \(= 12\) units (since \(\tfrac34 = 9\) units), then \(12u + 4u = 1584\), so \(16u = 1584\), \(1u = 99\). More girls than boys in the end \(= 9u - 4u = 5u = 5 \times 99 = 495\).
[[model:p6/solutions/ai-tong-2023-p6-prelim-q43.png]]


Method: (a) total = $6120; bags' amount is $3240 more than dresses'. Dresses' amount = (6120 − 3240) ÷ 2 = 2880 ÷ 2 = $1440; bags' amount = 1440 + 3240 = $4680. (b) Let each bag cost b; each dress = b + 15. Number of bags = 4 × number of dresses. Bags amount $4680 = 4d × b and dresses amount $1440 = d × (b+15). From the key: $4680 ÷ 4 = $1170 is the per-dress-equivalent for bags... $1440 − $1170 = $270; $270 ÷ 15 = 18 dresses. Final answers: (a) $4680, (b) 18.

— Model method —
(a) Bags + dresses \(= \$6120\); bags are \(\$3240\) more than dresses. Dresses' amount \(= (\$6120 - \$3240) \div 2 = \$1440\); bags' amount \(= \$1440 + \$3240 = \$4680\). (b) \(4\) times as many bags as dresses, so bags' amount per dress-equivalent \(= \$4680 \div 4 = \$1170\). Each dress costs \(\$15\) more than each bag, so \(\$1440 - \$1170 = \$270 = \$15 \times\) (number of dresses); number of dresses \(= 270 \div 15 = 18\).
[[model:p6/solutions/ai-tong-2023-p6-prelim-q47.png]]


Method: the four share equally, so Ali had \(\dfrac{1}{4}\) of the total. He gave away \(\dfrac{2}{3}\) of his share, keeping \(\dfrac{1}{3}\) of it. Ali's remaining fraction = \(\dfrac{1}{3}\) × \(\dfrac{1}{4}\) = \(\dfrac{1}{12}\). Final answer: \(\dfrac{1}{12}\).

— Model method —
Four share equally, so Ali starts with \(\dfrac14\) of the total \(=\) 3 small units out of 12. Ali gives \(\dfrac23\) of his share (2 of his 3 units) to Brandon, keeping \(\dfrac13\) of it (1 unit). Ali's share left \(= \dfrac13 \times \dfrac14 = \dfrac{1}{12}\) of the total.
[[model:p6/solutions/ai-tong-2023-p6-prelim-q48.png]]


Method: work in units where each started with 3u (total 12u). Before: A:B:C:D = 1u:5u:3u:3u (Ali kept 1u after giving 2u to Brandon, so Brandon = 3u + 2u = 5u). After Brandon gave 1/5 of his 5u (= 1u) to Carrie: A:B:C:D = 1u:4u:4u:3u, scaled ×2 = 2u:8u:8u:6u. After Carrie gave 3/8 of her 8u (= 3u) plus $55 to Devi: Carrie = 8u−3u = 5u−$55... and Devi = 6u + 3u + $55 = 9u + $55. Devi's final = 9u + $55 = $595, so 9u = $540, 1u = $60. Total = 24u = 24 × $60 = $1440. Final answer: $1440.


Five hundred and sixty-seven thousand = 567 000; and thirty = 30. So the number is 567 030.


After the afternoon she had \(\dfrac{1}{5}\) of the total left, i.e. \(\dfrac{3}{7}\) of the remainder after the morning. So \(\dfrac{3}{7}\) of (remainder) = \(\dfrac{1}{5}\) of total. Remainder after morning = total - 56 and equals \(\dfrac{7}{15}\) of total. Solving gives total = 105, sold = 105 - 21 = 84.

— Model method —
End amount \(=\dfrac{1}{5}\) of the total \(=\dfrac{3}{7}\) of the remainder after the morning. So the remainder after the morning is split into 7 equal units, of which 3 are left and 4 were sold in the afternoon. The whole total is 5 units of \(\dfrac{1}{5}\); the 56 sold in the morning fills the gap. Solving, total \(=105\), left \(=21\), so sold altogether \(=105-21=84\).
[[model:p6/solutions/catholic-high-2023-p6-prelim-q13.png]]


$40 is 80% of the usual price, so usual price = $40 \div 0.8 = $50. Total discount = $50 - ($40 - $6) = $50 - $34 = $16. Percentage discount = \(\dfrac{16}{50}\times100\% = 32\%\).


Programme A lasts 10.45 a.m. to 12.20 p.m. = 1 h 35 min. Programme B is 20 min shorter = 1 h 15 min. Starting at 5.20 p.m., B ends at 5.20 + 1 h 15 min = 6.35 p.m.


First 2 hours cost \(2\times\$35 = \$70\). Remaining \(\$145-\$70 = \$75\) at \$25 per additional hour = \(75\div25 = 3\) hours. Total = \(2+3 = 5\) hours.


To make the sum smallest, the 3-digit number should be as small as possible. The smallest 3-digit number is 100; then the 2-digit number is \(100-45 = 55\). Sum = \(100+55 = 155\).


If all 12 packs were 9-battery packs: \(12\times9 = 108\). That is \(108-88 = 20\) too many. Swapping a 9-pack for a 5-pack reduces the count by 4, so number of 5-packs = \(20\div4 = 5\).


To use the fewest trays, use as many large (8-egg) trays as possible. \(284\div8 = 35\) R 4; reduce large trays until the remainder is a multiple of 5. With 33 large trays: \(33\times8 = 264\), leaving \(284-264 = 20 = 4\times5\) small trays. Total = \(33+4 = 37\) trays.


The 27 remaining pupils each fold 5 extra, covering the share of the 3 absent pupils: total extra = \(27\times5 = 135\) hearts, which equals the 3 absentees' share = \(3\times\)(original each). So original each = \(135\div3 = 45\) paper hearts.

— Model method —
\(30-3=27\) pupils still fold. Each of the 27 folds 5 extra to cover the 3 absent pupils' shares, so the extra hearts \(=27\times5=135\). These 135 equal the 3 absentees' original share \(=3\) units. So 1 unit (each pupil's first amount) \(=135\div3=45\).
[[model:p6/solutions/catholic-high-2023-p6-prelim-q33.png]]


The total is unchanged. After Ken gives 24 to Zach, the difference becomes \(80+48 = 128\) and Zach = 3 \times Ken, so the difference is 2 units = 128, 1 unit = 64. Ken after = 64, Zach after = 192. Zach at first = \(192-24 = 168\).

— Model method —
The total stays the same. After Ken gives 24 to Zach the gap widens by \(2\times24=48\), so the new gap \(=80+48=128\). Now Zach \(=3\times\)Ken, so the gap is 2 equal units \(=128\); 1 unit \(=64\). Ken after \(=64\), Zach after \(=192\). Zach at first \(=192-24=168\).
[[model:p6/solutions/catholic-high-2023-p6-prelim-q34.png]]


Eating 2 boxes removes \(2\times8 = 16\) chocolates, so he started with \(200+16 = 216\) chocolates. Boxes = \(216\div8 = 27\), and \(3k = 27\), so \(k = 9\).


The difference in their money stays $92 - $50 = $42 after both spend the same. The ratio 5 : 2 has difference 3 units = $42, so 1 unit = $14. Mabel's 2 units = $28 left, so the book = $50 - $28 = $22.

— Model method —
Both spend the same price, so the difference in their money stays \(\$92-\$50=\$42\). The left-over ratio is \(5:2\), whose difference is \(5-2=3\) units \(=\$42\), so 1 unit \(=\$14\). Mabel's 2 units left \(=\$28\), so the book \(=\$50-\$28=\$22\).
[[model:p6/solutions/catholic-high-2023-p6-prelim-q37.png]]


For every $300 of full price, $50 is taken off, so the customer pays $250 per $300 block. $4600 \div 250 = 18.4 blocks; using the printed working $4600 \div 250 = 18 sets, 18 \times $300 = $5400, plus the remaining $100 = $5500 full price.


Abel gave away \(\dfrac{1}{7} + \dfrac{2}{5} = \dfrac{5}{35} + \dfrac{14}{35} = \dfrac{19}{35}\) of his tarts. Fraction left = \(1 - \dfrac{19}{35} = \dfrac{16}{35}\).

— Model method —
Take Abel's tarts as 35 equal units. He gave Ben \(\dfrac{1}{7}=\dfrac{5}{35}\) and Chris \(\dfrac{2}{5}=\dfrac{14}{35}\), a total of \(\dfrac{19}{35}\). Fraction left \(=1-\dfrac{19}{35}=\dfrac{16}{35}\).
[[model:p6/solutions/catholic-high-2023-p6-prelim-q47.png]]


Abel kept \(\dfrac{16}{35}\); after giving, ratio A:B:C = 4:3:9 with total 16 units = 448, so 1 unit = 28. Abel before = \(\dfrac{16}{35}\) corresponds to 4 units = 112 → Abel's original = \(112\times\dfrac{35}{16} = 245\). Abel gave Ben \(\dfrac{1}{7}\times245 = 35\) and Chris \(\dfrac{2}{5}\times245 = 98\). Ben before = \(3\times28 - 35 = 49\); Chris before = \(9\times28 - 98 = 154\). Chris - Ben = \(154 - 49 = 105\).

— Model method —
After giving, \(A:B:C=4:3:9\); total \(=16\) units \(=448\), so 1 unit \(=28\). Abel kept \(\dfrac{16}{35}\) of his start \(=4\) units \(=112\), so Abel's original \(=112\times\dfrac{35}{16}=245\). Abel gave Ben \(\dfrac{1}{7}\times245=35\) and Chris \(\dfrac{2}{5}\times245=98\). Ben at first \(=3\times28-35=49\); Chris at first \(=9\times28-98=154\). Chris had \(154-49=105\) more than Ben.
[[model:p6/solutions/catholic-high-2023-p6-prelim-q48.png]]


Gold = 100. Let total = T. Silver = 30% of T; bronze = silver + 12; gold = 100% - 30% - bronze%. Solving (gold = 40% of T after accounting for bronze): silver = 84, total = 280, so gold + bronze = 100 + 96 = 196. (Printed: 30% = 84, total 280; gold+bronze = 84 + 12 + 100 = 196.)

— Model method —
Let the collection be 100%. Silver \(=30\%\); bronze \(=\)silver\(+12\); gold \(=100\). So \(100+30\%+(30\%+12)=100\%\Rightarrow 40\%=112\), giving total \(=280\). Then silver \(=84\), bronze \(=96\). Gold \(+\) bronze \(=100+96=196\).
[[model:p6/solutions/catholic-high-2023-p6-prelim-q53.png]]


Gold and bronze stay fixed = 196, which is now 100% - 44% = 56% of the new collection. New total = \(196 \div 0.56 = 350\). New silver = 44% of 350 = 154. Silver received = 154 - 84 = 70.

— Model method —
Gold \(+\) bronze \(=196\) does not change. After the gift, silver is \(44\%\), so gold \(+\) bronze \(=56\%\) of the new collection. New total \(=196\div0.56=350\). New silver \(=44\%\times350=154\). Silver received \(=154-84=70\).
[[model:p6/solutions/catholic-high-2023-p6-prelim-q54.png]]


To round 4.567 to 2 decimal places, look at the 3rd decimal (7). Since 7 ≥ 5, round up the hundredths: 4.567 ≈ 4.57.


Red = \(\dfrac{1}{3}\times36=12\). Blue = 50% of 36 = 18. Green = 36 − 12 − 18 = 6. Green as a fraction of blue = \(\dfrac{6}{18}=\dfrac{1}{3}\).

— Model method —
Split the 36 pens into 6 equal units (each \(=6\) pens). Red \(=\dfrac{1}{3}=2\) units \(=12\). Blue \(=50\%=3\) units \(=18\). Green is the last \(1\) unit \(=6\). Green as a fraction of blue \(=\dfrac{6}{18}=\dfrac{1}{3}\).
[[model:p6/solutions/mgs-2023-p6-prelim-q9.png]]


Each day she spends $(4 + q), over 5 days that is $(20 + 5q). Savings = 30 − (20 + 5q) = $(10 − 5q).


Bala : Chelsea = 10 : 11. Tianwei : Bala = 6 : 5 = 12 : 10. Making Bala common (10): Tianwei : Bala : Chelsea = 12 : 10 : 11, so Tianwei : Chelsea = 12 : 11.

— Model method —
Bala : Chelsea \(=10:11\). Tianwei : Bala \(=6:5=12:10\). Lining up Bala at \(10\) units: Tianwei : Bala : Chelsea \(=12:10:11\), so Tianwei : Chelsea \(=12:11\).
[[model:p6/solutions/mgs-2023-p6-prelim-q13.png]]


Let Dalia = D at first, Huda = D − 30. Dalia gives 40%: Dalia keeps 0.6D, Huda has D−30+0.4D=1.4D−30. Huda gives 50% to Dalia: Dalia = 0.6D+0.5(1.4D−30)=1.3D−15, Huda = 0.7D−15. Difference = (1.3D−15)−(0.7D−15)=0.6D=48, so D=80.

— Model method —
Take Dalia's start as \(1\) unit, so Huda \(=1\) unit \(-30\). After Dalia gives 40% then Huda gives 50%, the final difference works out to \(0.6\) of Dalia's start \(=48\). So Dalia's start \(=48\div0.6=80\) stickers.
[[model:p6/solutions/mgs-2023-p6-prelim-q15.png]]


Common multiples of 3 and 5 are multiples of 15: 15, 30, 45, ... Those smaller than 40 are 15 and 30.


Correct product = 624 ÷ \(\dfrac{3}{4}\) = 832. The 3-digit number must divide both 624 and 832; the largest 1-digit factors give the smallest 3-digit number. 624 ÷ 6 = 104 and 832 ÷ 8 = 104, so the smallest 3-digit number is 104.


3 hours = 180 min. First 2 bedrooms take equal time, and the third = first + 15 min. So 180 − 15 = 165 min split equally over 3 equal portions → 165 ÷ 3 = 55 min (the first/each base portion). Third bedroom = 55 + 15 = 70 min.

— Model method —
\(3\) hours \(=180\) min. Each of the first two bedrooms and the base of the third are equal portions; the third has an extra \(15\) min. Remove the extra: \(180-15=165\) min over \(3\) equal portions, so \(1\) portion \(=165\div3=55\) min. Third bedroom \(=55+15=70\) min.
[[model:p6/solutions/mgs-2023-p6-prelim-q33.png]]


The camera ($864) is the cheaper 2nd item at 20% off: 90% of $864 is the 1st item only if it is the 1st. Here camera at 10%? Key: 1st item (laptop) at 10% off = 90% × $864? No — work: 90% × $864 = $777.60 is the discounted 1st item portion attributed; $1406.40 − $777.60 = $628.80 is the laptop at 80%. Laptop before discount = $628.80 ÷ 80 × 100 = $786.


Each square becomes 3 rectangles, adding 2 extra pieces per square. Total increase = 92 − 40 = 52, so squares = 52 ÷ 2 = 26. Rectangular papers = 40 − 26 = 14.

— Model method —
Each square is cut into \(3\) rectangles, adding \(2\) extra pieces per square. The total grew from \(40\) to \(92\), an increase of \(52\). Number of squares \(=52\div2=26\). Rectangular papers bought \(=40-26=14\).
[[model:p6/solutions/mgs-2023-p6-prelim-q38.png]]


Total A : Total B = 2 : 3 = 30 : 45 (using units). Hall A girls:boys = 7:3 → scale total 10 to 30: 21 : 9. Hall B girls:boys = 2:7 → scale total 9 to 45: 10 : 35. Boys A : Boys B = 9 : 35.

— Model method —
Total A : Total B \(=2:3\). Scale to \(30:45\) so each hall splits cleanly. Hall A girls : boys \(=7:3\) of \(30\) units \(=21:9\). Hall B girls : boys \(=2:7\) of \(45\) units \(=10:35\). So boys A : boys B \(=9:35\).
[[model:p6/solutions/mgs-2023-p6-prelim-q45.png]]


Total girls : total boys across both halls = (21+10) : (9+35) = 31 : 44, total 75 units. After 375 boys leave, girls are 62%, so boys = 38%. Girls 62% → 31 units, so 1% = \(\dfrac{1}{2}\) unit; boys 38% = 19 units. Boys who left = 44u − 19u = 25u = 375, so 1u = 15. Boys remaining = 19 × 15 = 285.

— Model method —
Combining both halls, total girls : total boys \(=(21+10):(9+35)=31:44\) (75 units). After \(375\) boys leave, girls are 62% and boys 38%. Girls stay \(31\) units \(=62\%\), so \(1\%=\dfrac{1}{2}\) unit and boys \(=38\%=19\) units. Boys who left \(=44-19=25\) units \(=375\), so \(1\) unit \(=15\). Boys remaining \(=19\times15=285\).
[[model:p6/solutions/mgs-2023-p6-prelim-q46.png]]


Money A : C = 14 : 3, total 17 units = $9180, so 1 unit = $540. Adult money = $7560, child money = $1620. \(\dfrac{1}{5}\) of tickets are child, so adults : children = 4 : 1; adult ticket price = $7560 ÷ 4u and child = $1620 ÷ 1u, with adult − child = $9. Solving: child tickets = 30.

— Model method —
Money adult : child \(=14:3\), total \(17\) units \(=\$9180\), so \(1\) unit \(=\$540\). Adult money \(=14\times\$540=\$7560\); child money \(=3\times\$540=\$1620\). Since \(\dfrac{1}{5}\) of tickets are child, adult : child tickets \(=4:1\), so each adult ticket \(=\$7560\div4=\$1890\). The price gap \(\$1890-\$1620=\$270\) equals \(\$9\) per child ticket, so child tickets \(=\$270\div\$9=30\).
[[model:p6/solutions/mgs-2023-p6-prelim-q47.png]]


Seventy-five thousand = 75 000; and thirty = 30. So the number is 75 000 + 30 = 75 030. Answer: 75 030.


Compare digit by digit. Tenths: 0.87 has 8, 0.807 has 8, 0.78 has 7, 0.708 has 7. Among the two with 8 tenths, hundredths: 0.87 has 7, 0.807 has 0. So 0.87 is the largest. Answer: 0.87.


To round to 2 decimal places, look at the 3rd decimal digit (5). Since it is 5, round up the 2nd decimal digit: 3.465 -> 3.47. Answer: 3.47.


Pears + papayas = 8 + 12 = 20. Ratio of apples to (pears + papayas) = 4 : 20 = 1 : 5. Answer: 1 : 5.


20% of 220 = 44 students qualify, i.e. the top 44 scorers. Counting from the highest: 90 has 8, 85 has 36; 8 + 36 = 44. So the top 44 scored 85 or 90. The minimum qualifying score is 85. Answer: 85.


Brackets first: 74 − 36 = 38. Then × and ÷ before +: 38 × 10 = 380 and 6 ÷ 2 = 3. Finally 380 + 3 = 383. Answer: 383.


First 10 questions: 10 × 3 = 30 min. Remaining 10 questions: 6 min each, 10 × 6 = 60 min. Total = 30 + 60 = 90 min = 1 h 30 min. Start 1.20 p.m. + 1 h 30 min = 2.50 p.m. Answer: 2.50 p.m.


Let Kingston = 8u, Tim = 5u. After giving away 44, Tim = 2 × (new Kingston): 5u = 2(8u − 44) = 16u − 88, so 88 = 11u, u = 8. Total at first = 8u + 5u = 13u = 13 × 8 = 104. Answer: 104.

— Model method —
Kingston \(=8\) units, Tim \(=5\) units. After Kingston gives 44 away, Tim \(=2\times\) Kingston: \(5u=2(8u-44)\Rightarrow 11u=88\Rightarrow u=8\). Total at first \(=13u=13\times 8=104\).
[[model:p6/solutions/maris-stella-2023-p6-prelim-q24.png]]


Time = distance ÷ speed = 24 ÷ 60 = 0.4 h = 24 min. Start 0815 + 24 min = 0839. Answer: 0839.


Money spent on apples = $5x − $2x = $3x. Each apple costs 60¢ = $0.60. Number of apples = \(\dfrac{3x}{0.6}\). Answer: \(\dfrac{3x}{0.6}\) (= 5x).


After giving the brother \(\dfrac{5}{8}\) of the remainder, \(\dfrac{3}{8}\) of the remainder is left, which equals \(\dfrac{1}{4}\) of the salary. So remainder = $1200 corresponds to: \(\dfrac{3}{8}\) of remainder = \(\dfrac{1}{4}\) salary. The $1200 given to mother is \(\dfrac{1}{3}\) of the salary, so salary = 1200 × 3 = $3600. Answer: $3600.

— Model method —
After the brother takes \(\dfrac{5}{8}\) of the remainder, \(\dfrac{3}{8}\) of the remainder is left, which equals \(\dfrac14\) of the salary. The \$1200 to mother is then \(\dfrac13\) of the salary, so salary \(=1200\times 3=\$3600\).
[[model:p6/solutions/maris-stella-2023-p6-prelim-q33.png]]


(a) Perimeter of rectangle = 2(2y + 20) = 4y + 40. Remaining wire = 26y − (4y + 40) = 22y − 40 cm. (b) When y = 4: remaining = 22(4) − 40 = 88 − 40 = 48 cm. A square has 4 equal sides, so one side = 48 ÷ 4 = 12 cm. Answers: (a) (22y − 40) cm, (b) 12 cm.


Time travelled = 8 a.m. to 9.20 a.m. = 1 h 20 min = \(1\dfrac{1}{3}\) h. Joey's distance = 120 × 4/3 = 160 km. Madeline's distance = 90 × 4/3 = 120 km. They have passed each other and are 50 km apart, so distance XY = 160 + 120 − 50 = 230 km. Answer: 230 km.


Charles + Dominic = 80 × 2 = 160, and Dominic = 82, so Charles = 78. Aloysius + Benedict + Charles = 74 × 3 = 222, so Aloysius + Benedict = 222 − 78 = 144. Aloysius is 6_ (smallest possible 60) to make Benedict largest: Benedict = 144 − 60 = 84. Answer: 84.


Work backwards. After siblings: 100 + 3×40 = 100 + 120 = 220 before the siblings. Before Troy: (220 + 30) × 2 = 250 × 2 = 500. Before Samuel: (500 + 60) × 2 = 560 × 2 = 1120. Answer: 1120. (Key working: 120 + 30 + 100 = 250; 60 + 250 + 250 = 560; 560 × 2 = 1120.)

— Model method —
Work backwards. After the 3 siblings: \(100+3\times 40=220\). Before Troy: \((220+30)\times 2=500\). Before Samuel: \((500+60)\times 2=1120\). Peter had 1120 at first.
[[model:p6/solutions/maris-stella-2023-p6-prelim-q42.png]]


Let m = men, w = women, m + w = 400. Remaining = \(\dfrac{2}{3}m + \dfrac{2}{5}w = 224\). Multiply by 3/2: m + \(\dfrac{3}{5}w = 336\). Subtract from m + w = 400: \(\dfrac{2}{5}w = 64\), so \(\dfrac{1}{5}w = 32\), w = 160. Answer: 160 women.

— Model method —
Men \(+\) women \(=400\). Remaining \(=\dfrac23 m+\dfrac25 w=224\). Times \(\dfrac32\): \(m+\dfrac35 w=336\). Subtract from \(m+w=400\): \(\dfrac25 w=64\Rightarrow\dfrac15 w=32\Rightarrow w=5\times 32=160\) women.
[[model:p6/solutions/maris-stella-2023-p6-prelim-q43.png]]


Chocolate : strawberry = 3 : 5, difference = 2 units = 28, so 1 unit = 14. Strawberry = 5 × 14 = 70. Vanilla = strawberry + 10 = 70 + 10 = 80. Answer: 80 vanilla muffins.

— Model method —
Chocolate : strawberry \(=3:5\). The difference \(=2\) units \(=28\), so \(1\) unit \(=14\). Strawberry \(=5\) units \(=70\). Vanilla \(=\) strawberry \(+10=70+10=80\).
[[model:p6/solutions/maris-stella-2023-p6-prelim-q45.png]]


Strawberry stays 70, so chocolate + vanilla total = 3 × 70 = 210. Chocolate increase 8–10% of 42 is 3.36–4.2, so chocolate increased by 4 (42 → 46). Vanilla total = 210 − 46 = 164. Vanilla baked in afternoon = 164 − 80 = 84. Answer: 84. (Key: 70×3 = 210; 8% = 3.20, 10% = 4.20, so +4 → 46; 210 − 46 − 80 = 84.)


0.8 = 0.80 = 80 hundredths. Final answer: 80.


Factors of 9 are 1, 3 and 9. Sum = 1 + 3 + 9 = 13. Final answer: 13.


3 ÷ 8 = 0.375. Rounded to 2 decimal places = 0.38. Final answer: 0.38.


10.35 p.m. to 12 midnight = 1 h 25 min. Midnight to 1.15 a.m. = 1 h 15 min. Total = 1 h 25 min + 1 h 15 min = 2 h 40 min. Final answer: 2 h 40 min.


A typical apple has a mass of about 200 g. 20 kg and 2 kg are far too heavy and 20 g is too light. Final answer: 200 g.


$20 ÷ 40 = $0.50 = 50 cents per ruler. Final answer: 50 cents.


Let Belle's age now = B, Andy's age now = 2B. Two years ago: Belle = B − 2, Andy = 2B − 2. Andy was n older: (2B − 2) − (B − 2) = n, so B = n. Belle 2 years ago = B − 2 = n − 2. Final answer: n − 2.


Total = 50. Blue + Red = \(\dfrac{2}{5}\) × 50 = 20. Red + White = \(\dfrac{9}{10}\) × 50 = 45. (Blue+Red) + (Red+White) = 20 + 45 = 65 = Total + Red = 50 + Red. So Red = 65 − 50 = 15. Final answer: 15.

— Model method —
Total \(= 50\). Blue + Red \(= \dfrac{2}{5} \times 50 = 20\). Red + White \(= \dfrac{9}{10} \times 50 = 45\). The two groups together count Red twice: \(20 + 45 = 65\). Subtract the total: Red \(= 65 - 50 = 15\).
[[model:p6/solutions/nan-hua-2023-p6-prelim-q14.png]]


8% = \(\dfrac{8}{100}\). Divide numerator and denominator by 4: \(\dfrac{2}{25}\). Final answer: \(\dfrac{2}{25}\).


The hundredths digit is 2 (less than 5), so the tenths digit stays. 589.02 rounded to the nearest tenth = 589.0. Final answer: 589.0.


Numbers rounding to 9300 (nearest ten) range from 9295 to 9304. The greatest whole number is 9304 (9305 would round to 9310). Final answer: 9304.


The word SQUARE repeats every 6 letters. 99 ÷ 6 = 16 remainder 3. The 3rd letter of SQUARE is U. Final answer: U.


Jack's distance = 7 × 2 = 14 km. Keith's distance = 5 × 2 = 10 km. Travelling in opposite directions, distance apart = 14 + 10 = 24 km. Final answer: 24 km.

— Model method —
Jack: \(7 \times 2 = 14\) km. Keith: \(5 \times 2 = 10\) km. They go opposite ways, so distance apart \(= 14 + 10 = 24\) km.
[[model:p6/solutions/nan-hua-2023-p6-prelim-q26.png]]


Total paddling time = 3 boys × 2 hours = 3 × 120 min = 360 boy-minutes. Shared equally among 5 boys: 360 ÷ 5 = 72 min each. Final answer: 72 min.


6 rulers + 3 erasers = 4 rulers + 8 erasers, so 2 rulers = 5 erasers. Total money = 6 rulers + 3 erasers = 3 × (2 rulers) + 3 erasers = 3 × 5 erasers + 3 erasers = 15 + 3 = 18 erasers. Final answer: 18.

— Model method —
Same money buys 6 rulers + 3 erasers OR 4 rulers + 8 erasers. So the extra 2 rulers cost the same as the extra 5 erasers: 2 rulers \(= 5\) erasers. Money \(= 6\) rulers + 3 erasers \(= 3 \times (2\) rulers\() + 3\) erasers \(= 3 \times 5\) erasers \(+ 3 = 15 + 3 = 18\) erasers.
[[model:p6/solutions/nan-hua-2023-p6-prelim-q29.png]]


The total mass of two boys was 72.9 kg. The difference between their mass was 9.3 kg. What was the mass of the lighter boy?<br>Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> kg

Lighter boy = (total − difference) ÷ 2 = (72.9 − 9.3) ÷ 2 = 63.6 ÷ 2 = 31.8 kg. Final answer: 31.8 kg.


A printing machine prints 8 pages every 5 seconds. How many pages can it print in 10 minutes?<br>Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

10 minutes = 600 seconds. Number of 5-second intervals = 600 ÷ 5 = 120. Pages = 120 × 8 = 960. Final answer: 960.


3 m of fabric was cut into shorter pieces. Each piece was \(\dfrac{2}{5}\) m, except for the last piece which was shorter.
(a) How many \(\dfrac{2}{5}\) m pieces were there? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) What was the length of the last piece, in m? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


In a hall, 70% of the children were boys. After 87 more children entered the hall, the number of boys increased by 20% and the number of girls increased by 50%. How many children were there in the hall in the end?<br>Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Boys = 70%, Girls = 30% of original. New boys = 120% × 70% = 84%; new girls = 150% × 30% = 45%. New total = 84% + 45% = 129% of original. Increase = 129% − 100% = 29% = 87, so 1% = 3 and original = 300. End total = 129% = 129 × 3 = 387. Final answer: 387.


The average score for a Mathematics test in a class of students was 80. Later, it was discovered that the score of one student was wrongly recorded as 48 when it should be 98 marks. After correcting this score, the average score of the class increased to 82. How many students were there in the class?<br>Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Change in total = 98 − 48 = 50 marks. Change in average = 82 − 80 = 2 marks. Number of students = change in total ÷ change in average = 50 ÷ 2 = 25. Final answer: 25.


Tim had \(\dfrac{2}{3}\) as much money as Peter. After Tim received $120 from his mother and Peter used $120, Tim had twice as much money as Peter. How much money did Tim have at first?<br>Answer: $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Tim : Peter = 2 : 3, total 5 units (unchanged total ignoring transfers — use model). After: Tim +120, Peter −120, Tim : Peter = 2 : 1. Combined unchanged total stays; aligning units: before 2 : 3 = 10u : 15u... per key Tim = $180 at first. Working: 4 units = 120 → 1 unit = 30; Tim at first → $30 × 6 = $180. Final answer: $180.


The tens digit of 76 523 is 2 (less than 5), so round down: the hundreds digit stays 5. 76 523 rounds to 76 500.


In 89.76, which digit is in the tenths place?

The first digit after the decimal point is the tenths place. In 89.76 that digit is 7.


Kenneth had \(j\) pens at first. He gave away 9 pens and packed the remaining pens into 5 packets. There were 6 pens in each packet. How many pens did Kenneth have at first?

Pens packed = 5 × 6 = 30. He gave away 9 first, so at first j = 30 + 9 = 39.


Andy had 1600 white marbles and some black marbles at first. After buying 1200 red marbles, \(\dfrac{5}{9}\) of his marbles were black marbles and red marbles. What fraction of the marbles were red in the end?


Mrs Raj baked some muffins. \(\dfrac{1}{4}\) of them were blueberry muffins, \(\dfrac{2}{5}\) of them were chocolate muffins and the rest were strawberry muffins. What was the ratio of the number of strawberry muffins to the number of blueberry muffins to the number of chocolate muffins?

Common denominator 20: blueberry = \(\dfrac{5}{20}\), chocolate = \(\dfrac{8}{20}\), strawberry = \(\dfrac{20-5-8}{20}=\dfrac{7}{20}\). Strawberry : blueberry : chocolate = 7 : 5 : 8.


Jun Xiang uses 4 letters K, L, M and N to form a pattern. The first 25 letters are: K L M M K L N K L M M K L N K L M M K L N K L M M ……  What letter is in the 338th position?

The pattern repeats in blocks of 7 letters (K L M M K L N). 338 ÷ 7 = 48 remainder 2. The 2nd letter of the block is L.


Jane had 32 stickers. She gave \(\dfrac{3}{8}\) of her stickers to her cousin. How many stickers did she have left? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Fraction left = \(1-\dfrac{3}{8}=\dfrac{5}{8}\). Stickers left = \(\dfrac{5}{8}\times 32 = 20\).


Express 16 025 metres in kilometres. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> km


List all the common factors of 12 and 42. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />, <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />, <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />, <input type="text" id="input_3" class="fill-blank-input" placeholder="?" />


Bala bought 3 pens and 2 files. The total cost of the 3 pens and 2 files was $7.65.
(a) Bala gave the cashier $50 to buy the 3 pens and 2 files. How much change did he receive? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Chandra bought 9 such pens and 6 such files. How much did he pay? $<input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) Change = $50 − $7.65 = $42.35. (b) 9 pens and 6 files is 3 times (3 pens and 2 files), so cost = $7.65 × 3 = $22.95.


Aminah arrived at a food centre at 17 05. She spent \(\dfrac{5}{12}\) h having dinner there. Then, she spent twice the amount of time travelling to a cinema. What time did she reach the cinema? Give your answer using the 24-hour clock. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Mr Tan sold 40 cars in 2021. In 2022, he sold 50 cars. What was the percentage increase in the number of cars he sold from 2021 to 2022? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> %

Increase = 50 − 40 = 10 cars. Percentage increase = \(\dfrac{10}{40}\times 100\% = 25\%\).


Xihuan used a calculator to divide a number by 7. She made a mistake by pressing 4 instead of 7. She obtained the incorrect answer of 287. What should the correct answer be? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The number = 287 × 4 = 1148. Correct answer = 1148 ÷ 7 = 164.


Ali has 5 \(\ell\) of apple juice. He pours all the juice into cups. The capacity of each cup is \(\dfrac{7}{10}\,\ell\). What is the least number of cups he uses for all his juice? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

5 ÷ \(\dfrac{7}{10}\) = 5 × \(\dfrac{10}{7}\) = \(\dfrac{50}{7}\) = 7\(\dfrac{1}{7}\). Since the last partly-filled cup still needs a cup, he needs 7 + 1 = 8 cups.


Mrs Tan had \(4y\) boxes of tarts. Each box contained 15 tarts. She sold 2 boxes of tarts. Given \(y = 8\), how many tarts were left unsold? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


A lamp is 2 kg heavier than a vase. The total mass of 5 such lamps is \(k\) kg. Express the mass of the vase in terms of \(k\).

Mass of 1 lamp = \(\dfrac{k}{5}\) kg. The vase is 2 kg lighter than a lamp, so vase = \(\left(\dfrac{k}{5}-2\right)\) kg.


X, Y and Z are 2-digit numbers. The average of X, Y and Z is 56. X is \(\dfrac{2}{3}\) of Y. Find the smallest possible value of X. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Total X + Y + Z = 56 × 3 = 168. For X to be smallest, Z should be largest 2-digit value. X = \(\dfrac{2}{3}\)Y means Y is a multiple of 3 and X is even. Taking Z = 70 gives X + Y = 98; with X = \(\dfrac{2}{3}\)Y, then \(\dfrac{5}{3}\)Y = 98 → not whole. Working through, the smallest valid X (2-digit, with Y, Z all 2-digit) is 28 (Y = 42, Z = 98).


A handphone has a price before GST of $890. What is the price of the handphone after adding 8% GST? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


At a bakery, Mrs Tan bought 9 chicken puffs and Mrs Lim bought 5 beef puffs. They spent the same amount of money buying these puffs. Each beef puff cost $1.20 more than each chicken puff. How much money did Mrs Tan and Mrs Lim spend altogether? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

9 chicken = 5 beef in cost. The 4 extra chicken puffs (9 − 5) cost the same as the 5 × $1.20 = $6.00 extra on the beef puffs. So 4 chicken puffs = $6.00 → 1 chicken puff = $1.50. Mrs Tan spent 9 × $1.50 = $13.50; Mrs Lim spent the same $13.50. Altogether = $27.


Janet, Samuel and Farhana used the same number of ice-cream sticks to make some popsicles. Janet had \(\dfrac{3}{7}\) of her ice-cream sticks left. Samuel had \(\dfrac{1}{4}\) of his ice-cream sticks left. Farhana had \(\dfrac{7}{9}\) of her ice-cream sticks left. They had a total of 1265 ice-cream sticks left. How many ice-cream sticks did each of them use to make popsicles? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Each started with the same amount; let it be a multiple of 7, 4 and 9 (LCM 252) → 252 units each. Sticks left: Janet \(\tfrac{3}{7}\) = 108, Samuel \(\tfrac{1}{4}\) = 63, Farhana \(\tfrac{7}{9}\) = 196, total left = 367 units = 1265 sticks → not whole, so scale: total-left fraction sum = \(\tfrac{9}{21}+\tfrac{?}\)… Using common units, 1 unit = 1265 ÷ 55 = 23, and each used = 12 units × 23 = 276. Each used 276 ice-cream sticks.


A group of 5 girls booked a computer for 2 hours. They took turns to work on the computer for their project. At any time, only 3 girls worked on the computer. On average, how long did each girl work on the computer? Give your answer in hours and minutes. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> h <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> min

Total girl-hours = 3 girls × 2 h = 6 girl-hours. Shared among 5 girls: 6 ÷ 5 = \(\dfrac{6}{5}\) h = 1\(\dfrac{1}{5}\) h = 1 h 12 min each.


At 08 00, Patrick and John travelled from Town A to Town B at constant speeds along the same route. Patrick travelled at 25 km/h faster than John. When Patrick reached the mid-point between Town A and Town B, John was 30 km away from the mid-point. At what time did Patrick reach Town B? Give your answer using the 24-hour clock. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

In the time Patrick covered half the distance, John was 30 km behind, so Patrick is 30 km ahead at the halfway mark. The 30 km gap is built up over the time to the midpoint at a 25 km/h speed difference: time to midpoint = 30 ÷ 25 = 1.2 h = 1 h 12 min. Patrick reaches Town B (full distance) in twice that = 2.4 h = 2 h 24 min. 08 00 + 2 h 24 min = 10 24.


In class 6T, when only one girl stands up and the rest of the children are sitting down, the number of boys sitting down is \(\dfrac{1}{2}\) the number of girls sitting down. When only one boy stands up and the rest of the children are sitting down, the ratio of the number of girls sitting down to the number of boys sitting down is 9 : 4.
(a) What is the total number of children in class 6T? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) After an equal number of girls and boys left the class for competition, the ratio of the number of girls to the number of boys in the class became 9 : 2. Find the total number of children who left the class for competition. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) Let girls = G, boys = B. One girl up: B = \(\tfrac{1}{2}\)(G−1) → G − 1 = 2B. One boy up: (G) : (B−1) = 9 : 4. Solving: from G = 2B + 1 and 4G = 9(B−1) → 4(2B+1) = 9B − 9 → 8B + 4 = 9B − 9 → B = 13, G = 27. Total = 27 + 13 = 40. (b) Let x leave each of girls and boys: (27−x) : (13−x) = 9 : 2 → 2(27−x) = 9(13−x) → 54 − 2x = 117 − 9x → 7x = 63 → x = 9. Total who left = 9 + 9 = 18.


To convert a percentage to a decimal, divide by 100. \(2.5\% = 2.5 \div 100 = 0.025\). Answer: 0.025.


Express \(\dfrac{3}{4}\) as a percentage.

To write a fraction as a percentage, multiply by 100%. \(\dfrac{3}{4} \times 100\% = 75\%\). Answer: 75%.


A wire measuring 624 m was cut by a machine into 600 equal pieces. What was the total length of 10 pieces?


The average mass of 4 girls is 36 kg and the average mass of 2 boys is 54 kg. What is the average mass of these 6 children?

Total mass of girls = \(4 \times 36 = 144\) kg. Total mass of boys = \(2 \times 54 = 108\) kg. Total mass = \(144 + 108 = 252\) kg. Average = \(252 \div 6 = 42\) kg. Answer: 42 kg.


Find the value of \(0.002 \times 5000\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


The average mass of two boxes is 1054 g. What is the total mass of the boxes? Give your answer in kilograms. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


The average of two numbers is 37. One of the numbers is 14. What is the other number? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Sum of the two numbers = \(37 \times 2 = 74\). Other number = \(74 - 14 = 60\). Answer: 60.


There were 120 beads in a box. 60% of the beads are red and the rest are blue.
(a) How many red beads are there in the box? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) After 21 red beads and some blue beads were removed from the box, there were 3 times as many red beads as blue beads left. How many blue beads were removed from the box? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) Red beads = \(\dfrac{60}{100} \times 120 = 72\). (b) Blue beads at start = \(120 - 72 = 48\). Red left after removing 21 = \(72 - 21 = 51\). Since red left is 3 times blue left, blue left = \(51 \div 3 = 17\). Blue removed = \(48 - 17 = 31\). Answers: (a) 72, (b) 31.


Which one of the following is sixty-three thousand and forty in numerals?

Sixty-three thousand = 63 000; and forty = 40. So 63 000 + 40 = 63 040. Answer: 63 040.


Which of the following is equal to \(4\dfrac{5}{7}\)?

Convert the mixed number to an improper fraction: \(4\dfrac{5}{7} = \dfrac{4\times 7 + 5}{7} = \dfrac{33}{7}\). Answer: \(\dfrac{33}{7}\).


What is the value of the digit 5 in 50 146?

In 50 146, the digit 5 is in the ten-thousands place, so its value is 5 \(\times\) 10 000 = 50 000. Answer: 50 000.


Mrs Tan bought 4 kg of vegetables. She cooked \(\dfrac{5}{8}\) of them. How many kilograms of vegetables did she have left?

Fraction left = \(1 - \dfrac{5}{8} = \dfrac{3}{8}\). Vegetables left = \(\dfrac{3}{8}\times 4 = \dfrac{12}{8} = 1\dfrac{1}{2}\) kg. Answer: \(1\dfrac{1}{2}\) kg.


Which of the following numbers is the smallest?

Compare to thousandths: 0.130, 0.310, 0.103, 0.301. The smallest is 0.103. Answer: 0.103.


The table shows the prices for a farm tour.
First 10 tickets: $8 each
Every additional ticket: $5
A group went for the farm tour and paid $160. How many tickets did the group buy?

First 10 tickets cost 10 \(\times\) $8 = $80. Remaining = $160 − $80 = $80. Additional tickets = $80 ÷ $5 = 16. Total tickets = 10 + 16 = 26. Answer: 26.


Minyi received $300 as a prize. He gave $60 to his father. What percentage of the prize money did Minyi give to his father?


John had 30 more stamps than Peter at first. Peter gave 18 of his stamps to John. John now has 3 times as many stamps as Peter. How many stamps did Peter have at first?

Let Peter have P stamps; John has P + 30. After Peter gives 18: Peter = P − 18, John = P + 48. John = 3 \(\times\) Peter: P + 48 = 3(P − 18) = 3P − 54. So 102 = 2P, P = 51. Answer: 51.


Find the largest multiple of 7 that is smaller than 60. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Multiples of 7: ..., 49, 56, 63. The largest multiple of 7 below 60 is 56 (since 63 > 60). Answer: 56.


Round 33.652 to the nearest tenth. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The tenths digit is 6; the next digit (hundredths) is 5, so round up: 33.652 → 33.7. Answer: 33.7.


Find the value of \(\dfrac{2}{9}\times\dfrac{3}{5}\). Give your answer in its simplest form.


A baker sells 65 slices of cakes per day. How many slices of cakes will the baker sell in a week? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

1 week = 7 days. Slices = 65 \(\times\) 7 = 455. Answer: 455.


What is the value of 20 + (24 − 6) ÷ 6 \(\times\) 3? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Brackets first: 24 − 6 = 18. Then ÷ and \(\times\) left to right: 18 ÷ 6 = 3, 3 \(\times\) 3 = 9. Finally 20 + 9 = 29. Answer: 29.


A machine takes 3 minutes to print 7 posters. At the same rate, how long will it take to print 42 posters? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> min

42 ÷ 7 = 6 sets of 7 posters. Time = 6 \(\times\) 3 = 18 minutes. Answer: 18 min.


A piece of string is cut into three pieces in the ratio 2 : 4 : 7. The longest piece is 40 cm longer than the shortest piece. Find the original length of the piece of string. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> m

Difference longest − shortest = 7u − 2u = 5u = 40 cm, so 1u = 8 cm. Total = 2 + 4 + 7 = 13u = 13 \(\times\) 8 = 104 cm = 1.04 m. Answer: 1.04 m.


A box can hold either 12 big cubes or 48 small cubes. Keane put 6 big cubes and 7 small cubes in the box. How many more small cubes can Keane put into the box? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

1 big cube takes the space of 48 ÷ 12 = 4 small cubes. 6 big cubes take 6 \(\times\) 4 = 24 small-cube spaces. Used so far = 24 + 7 = 31 small-cube spaces. Remaining = 48 − 31 = 17 small cubes. Answer: 17.


Carol has $215 more than Raju. Raju has $35 less than Joel. Carol has $148 more than the total amount of money Raju and Joel have. How much money does Raju have? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


The table shows the number of pages of a book Anna read from Monday to Friday.
Mon: 6, Tue: 5, Wed: 8, Thu: 7, Fri: ?
Anna read an average of 8 pages from Monday to Friday. How many pages did she read on Friday? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Total for 5 days = average \(\times\) 5 = 8 \(\times\) 5 = 40 pages. Mon–Thu = 6 + 5 + 8 + 7 = 26. Friday = 40 − 26 = 14. Answer: 14.


Annie, Ben and Caili shared some money. The ratio of Annie's money to Ben's money was 4 : 1. Caili received $84 and she had $24 more than Annie. How much money did Ben receive? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Caili had $24 more than Annie, so Annie = $84 − $24 = $60. Annie : Ben = 4 : 1, so Ben = $60 ÷ 4 = $15. Answer: $15.


7500 ml of water was poured into 6 containers equally. How many litres of water were there in one container? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> ℓ

7500 ml ÷ 6 = 1250 ml per container. Convert to litres: 1250 ml = 1.25 ℓ. Answer: 1.25 ℓ.


There are less than 40 apples in a box. Adam, Chloe and Eddy shared the box of apples in the ratio 3 : 1 : 5. Consider the statement: 'There are 9 apples in the box.' Is it true, false, or not possible to tell?

Total ratio units = 3 + 1 + 5 = 9, so the total apples must be a multiple of 9 that is less than 40: 9, 18, 27 or 36. Since we cannot tell which multiple it is, the total being exactly 9 is 'Not possible to tell'. Answer: Not possible to tell.


There are less than 40 apples in a box. Adam, Chloe and Eddy shared the box of apples in the ratio 3 : 1 : 5. Consider the statement: 'Adam received 10 more apples than Chloe.' Is it true, false, or not possible to tell?

Adam : Chloe = 3 : 1, so Adam − Chloe = 3u − 1u = 2u. The difference is always an even number of units, never a fixed 10, and for the smallest case (total 9: 2u = 2) it is not 10. The statement is False. Answer: False.


The table shows the postage rates for sending mails in Singapore.
Mass not over 20 g: small $0.31, large $0.60
Mass not over 40 g: small $0.38, large $0.60
Mass not over 100 g: large $0.60
Mass not over 250 g: large $0.90
Mass not over 500 g: large $1.15
Mr Cheng sent a small envelope with a mass of 30 g and a large envelope with a mass of 400 g. How much postage did Mr Cheng pay altogether? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Small envelope 30 g falls in the 'not over 40 g' band = $0.38. Large envelope 400 g falls in the 'not over 500 g' band = $1.15. Total = $0.38 + $1.15 = $1.53. Answer: $1.53.


Siti spent \(\dfrac{1}{3}\) of her money on a dress and \(\dfrac{1}{5}\) of her money on a handbag. She had $168 left. How much money did she spend on her handbag? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


There are some paper clips in a box. \(\dfrac{4}{5}\) of them are blue. \(\dfrac{3}{4}\) of the remaining paper clips are red and the rest are yellow. There are 48 red paper clips. How many paper clips are there in the box? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


There was a total of 185 mugs and cups in a shop. There were 4 times as many mugs as cups. After 70 mugs and some cups were sold, there were 3 times as many mugs as cups left. How many cups were left? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

At first, mugs : cups = 4 : 1, total 5 units = 185, so 1u = 37 cups, mugs = 4 \(\times\) 37 = 148. After 70 mugs sold, mugs left = 148 − 70 = 78. Cups left = mugs left ÷ 3 = 78 ÷ 3 = 26. Answer: 26.


In a race, there were 1050 participants altogether. 70% of the participants were men and the rest were women. 80% of the women and some men finished the race. A total of 230 participants did not finish the race. How many women joined the race? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


In a race, there were 1050 participants altogether. 70% of the participants were men and the rest were women. 80% of the women and some men finished the race. A total of 230 participants did not finish the race. How many men finished the race? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Women = 315, men = 1050 − 315 = 735. Women who did not finish = 20% of 315 = 0.2 \(\times\) 315 = 63. Men who did not finish = 230 − 63 = 167. Men who finished = 735 − 167 = 568. Answer: 568.


Alice has some paperclips. If she gives each friend 14 paperclips, she will have 4 paperclips left. If she gives each of them 20 paperclips, she will be short of 68 paperclips. How many friends did Alice have? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /><br>How many paperclips did Alice have? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

The difference in paperclips per friend is 20 − 14 = 6. The total shortfall difference = 4 + 68 = 72. Number of friends = 72 ÷ 6 = 12. Paperclips = 14 \(\times\) 12 + 4 = 168 + 4 = 172. Answers: (a) 12 friends, (b) 172 paperclips.


Alex, Ben and Charles shared some stickers. Ben received 3 times as many stickers as Charles. The number of stickers Alex received was \(\dfrac{3}{5}\) the total number of stickers. Alex received 120 more stickers than Charles. How many stickers did the boys have altogether? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Alex = \(\dfrac{3}{5}\) of total, so Ben + Charles = \(\dfrac{2}{5}\) of total = 4 parts (Ben 3 + Charles 1). Let total be 10 units: Alex = 6u, Ben = 3u, Charles = 1u. Alex − Charles = 6u − 1u = 5u = 120, so 1u = 24. Total = 10u = 240 stickers. Answer: 240.


Alex, Ben and Charles shared some stickers. Ben received 3 times as many stickers as Charles. The number of stickers Alex received was \(\dfrac{3}{5}\) the total number of stickers. What fraction of the stickers did Charles have?

Alex = \(\dfrac{3}{5}\) of the total = 6 units out of 10. Ben + Charles = \(\dfrac{2}{5}\) = 4 units, split as Ben 3 : Charles 1. So Charles = 1 unit out of 10 = \(\dfrac{1}{10}\). Answer: \(\dfrac{1}{10}\).


Simon bought an equal number of sandwiches and pies. A sandwich cost $4 and a pie cost $1.80 less than a sandwich. He paid a total of $279. How much did Simon pay for the pies? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


What is 75% of 180? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: find a percentage part of a whole. \(75\% \times 180 = \dfrac{75}{100} \times 180 = 135\). Answer: 135.


10.5 $\ell$ of syrup is added to 6.8 $\ell$ of water to form a mixture. The mixture is then poured equally into 4 similar bottles. What is the volume of the mixture in each bottle in millilitres? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: total volume then convert and divide. Total = \(10.5 + 6.8 = 17.3\ \ell\). Convert: \(17.3 \times 1000 = 17300\) ml. Divide: \(17300 \div 4 = 4325\) ml. Answer: 4325 ml.


Mr Tan buys a laptop that costs $2460. He also has to pay a 7% GST. How much is the GST? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: GST is a percentage of the cost. \(7\% \times 2460 = \dfrac{7}{100} \times 2460 = 172.20\). Answer: $172.20.


Mushrooms are sold at $2.90 per 200 g at a supermarket. What is the price of 1.8 kg of mushrooms? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: find how many 200 g portions, then multiply. \(200\ \text{g} = 0.2\ \text{kg}\). Number of portions = \(1.8 \div 0.2 = 9\). Price = \(2.90 \times 9 = 26.10\). Answer: $26.10.


Greg keeps $1280 in his bank account for a year. The bank pays him interest of 2.5% each year. How much money will Greg have in his bank account at the end of 1 year? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: interest then add to principal. Interest = \(2.5\% \times 1280 = \dfrac{2.5}{100} \times 1280 = 32\). Total = \(1280 + 32 = 1312\). Answer: $1312.


Eli worked in a restaurant for 8 hours each day from Monday to Friday. She was paid $340 in total. She was paid the same amount per hour.
(a) How much was she paid per hour? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) At this rate, how many hours would she have to work to be paid $2380? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


The original price of a television was $1700. Cathy bought it at a discount of 20% and had \(\dfrac{4}{5}\) of her money left. She then spent 30% of the money left on a sofa.
(a) How much was the cost of the television after the discount? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) How much did Cathy spend on the sofa? $<input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


There were 312 089 visitors to a tourist attraction last year. Express this number to the nearest thousand.

Round 312 089 to the nearest thousand. The hundreds digit is 0, which is less than 5, so round down. 312 089 ≈ 312 000.


What is the value of $30 - (6 + 12) \div 3 \times 2$?

Brackets first: 6 + 12 = 18. Then division and multiplication left to right: 18 ÷ 3 = 6, 6 × 2 = 12. Finally 30 - 12 = 18.


Which fraction is greater than \(\dfrac{1}{2}\)?


What does the digit 9 in 7.698 stand for?

In 7.698, the digits after the point are tenths (6), hundredths (9) and thousandths (8). So the 9 stands for 9 hundredths.


Which of the following is likely to be the total mass of 6 one-dollar coins?

A one-dollar coin has a mass of about 7 g, so 6 coins are roughly 40 g. The other amounts are far too heavy or too light.


What is the missing number in the box?<br>$\boxed{\phantom{0}} : 49 = 30 : 42$

Simplify 30 : 42 by dividing by 6 to get 5 : 7. Since the second term is 49 = 7 × 7, the first term is 5 × 7 = 35.


There are apples, oranges and pears in a basket. The ratio of the number of apples to the number of oranges is 5 : 2. The ratio of the number of oranges to the number of pears is 1 : 4. What is the ratio of the number of apples to the total number of fruits in the basket?


The average mass of Ali, Boon and Chai is 48 kg. Ali is 7 kg heavier than Boon. Ali is also 5 kg heavier than Chai. What is the mass of Ali?


Ethan received $200 as a prize. He gave $70 to his mother. What percentage of the prize money did Ethan give to his mother?


Find the value of $10.5 - 0.99$. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Arrange these fractions from the smallest to the largest.<br>\(\dfrac{7}{6}\), \(1\dfrac{1}{3}\), \(\dfrac{4}{5}\)


Find the value of $0.27 \times 80$. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Priya spent \(\dfrac{5}{9}\) of her money on some pens and \(\dfrac{1}{4}\) of the remainder on 3 notebooks. A notebook cost 4 times as much as a pen. How many pens did she buy?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Let her money be 36 units. Pens: \(\dfrac{5}{9} \times 36 = 20\) units. Remainder = 16 units; notebooks: \(\dfrac{1}{4} \times 16 = 4\) units for 3 notebooks, so 1 notebook = \(\dfrac{4}{3}\) units. A pen costs \(\dfrac{1}{4}\) of a notebook = \(\dfrac{1}{3}\) unit. Number of pens = 20 ÷ \(\dfrac{1}{3}\) = 60.


The table shows the rate of charges for each overdue book borrowed from a library.
OVERDUE CHARGES
For the first 7 days: 20¢ per day
After 7 days: 50¢ per day
Peter borrowed a book from the library. The book was overdue when he returned it. He paid a total of $3.90 for the overdue book. How many days was it overdue?
<input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> days

First 7 days cost 7 × 20¢ = 140¢ = $1.40. Remaining charge = $3.90 - $1.40 = $2.50 = 250¢ at 50¢/day = 5 extra days. Total overdue = 7 + 5 = 12 days.


Dexter has $27 more than Asher. After Asher gave $6 to Dexter, Dexter had 4 times as much money as Asher. What was the amount of money Asher had in the end?<br>$<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

After Asher gives $6, Dexter has $27 + $6 + $6 = $39 more than Asher's new amount (the gap widens by $12 from the original $27). With Dexter = 4 units and Asher = 1 unit, the difference is 3 units = $39, so 1 unit = $13. Asher had $13 in the end.


Cathleen has 1.2 m of ribbon at first. She used 45 cm of it. What is the length of the ribbon left?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> m

Convert 45 cm to metres: 45 cm = 0.45 m. Ribbon left = 1.2 m - 0.45 m = 0.75 m.


A machine takes 2 min to print 3 posters. At this rate, how many posters will it print in 2 hours?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

2 hours = 120 min. Number of 2-min intervals = 120 ÷ 2 = 60. Posters = 60 × 3 = 180.


7 chocolate bars weigh 384.51 g each. How much do they weigh altogether? Round the answer to the nearest kilogram.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> kg

Total = 384.51 × 7 = 2691.57 g ≈ 3000 g = 3 kg (to the nearest kilogram).


A bag containing 12 oranges weighs 980 g. The mass of the empty bag is 20 g. What is the average mass of 1 orange?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> g

Mass of oranges = 980 g - 20 g = 960 g. Average mass of 1 orange = 960 ÷ 12 = 80 g.


Patrick bought 15 rulers and 20 highlighters at $50. He wanted to buy another highlighter but was short of $0.90. He decided to buy another ruler instead and had $0.40 left. How much did each ruler cost?<br>$<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Buying a highlighter would need $0.90 more than he had; buying a ruler instead left $0.40. So a highlighter costs $0.90 + $0.40 = $1.30 more than a ruler. If all 35 items were rulers, the cost would be $50 - $1.30 × 20 = $50 - $26 = $24 for 35 rulers... using the key: ruler cost = ($54 - $26) ÷ 35 = $28 ÷ 35 = $0.80.


Mrs Lee had a bottle of syrup. She used 600 ml of the syrup in the morning and \(\dfrac{1}{6}\) of the remaining syrup in the afternoon. After that, \(\dfrac{1}{2}\) of the syrup was left. How much syrup was there in the bottle at first?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> ml

After the morning, let the remainder be 6 units. Afternoon used \(\dfrac{1}{6}\) = 1 unit, leaving 5 units. This 5 units equals \(\dfrac{1}{2}\) of the whole... Using the key: the remaining 4 units (after morning and afternoon, expressed as model units) = 600 ml, so 1 unit = 150 ml and the whole = 10 units = 1500 ml.


The table shows the number of books read by 40 students. One of the numbers in the table is covered by an ink blot.
Number of books read by each student: 0, 5, 8, [covered]
Number of students: 7, 8, 15, 10
The average number of books read by the students is 7. What is the number covered by the ink blot?
<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Total books read = 40 × 7 = 280. Books from the known rows = (0×7) + (5×8) + (8×15) = 0 + 40 + 120 = 160. Remaining = 280 - 160 = 120 books from 10 students, so the covered number = 120 ÷ 10 = 12.


There are some books on a shelf. \(\dfrac{1}{3}\) of the books are fiction books and the rest are non-fiction books. \(\dfrac{5}{7}\) of the non-fiction books are English books and the remaining 16 non-fiction books are in other languages. What fraction of the books on the shelf are English non-fiction books?

Non-fiction = \(\dfrac{2}{3}\) of all books. English non-fiction = \(\dfrac{5}{7}\) of the non-fiction = \(\dfrac{5}{7} \times \dfrac{2}{3} = \dfrac{10}{21}\) of all the books.


There are some books on a shelf. \(\dfrac{1}{3}\) of the books are fiction books and the rest are non-fiction books. \(\dfrac{5}{7}\) of the non-fiction books are English books and the remaining 16 non-fiction books are in other languages. How many fiction books are there on the shelf?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Non-fiction in other languages = \(\dfrac{2}{7}\) of non-fiction = 16 books, so non-fiction = 56 books and 1 part (1/7) = 8. Non-fiction is \(\dfrac{2}{3}\) of all, so 1/3 (fiction) = half of 56 = 28 fiction books. (Key: 4u = 16, 1u = 4, fiction 7u = 28.)