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Solutions — Full Answer Key

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Ecolint Campus des NationsMathematics
Year 7 · 7.5 Fractions & Percentages

Solutions · Full Answer Key

Pack A answers · Pack B answers · Problem-solving worked solutions

Pack A — Answers

Bronze
1.E.g. 51102\dfrac{51}{102}; equivalent because 12=1×512×51=51102\frac{1}{2} = \frac{1 \times 51}{2 \times 51} = \frac{51}{102}
2.23\dfrac{2}{3}; HCF = 3; no further simplification possible because 2 and 3 share no common factor other than 1
3.58\dfrac{5}{8}; same denominator means same-size pieces, so 5 pieces is more than 3 pieces
4.12\dfrac{1}{2}; the denominator stays 4 because the pieces are the same size — you only count up the pieces
5.12\dfrac{1}{2}; check: 12+14=34\frac{1}{2} + \frac{1}{4} = \frac{3}{4} ✓
6.8; dividing by 3 splits the amount into 3 equal parts, and one part is the answer
7.40; Method 1: 50% = ½, so 80 ÷ 2 = 40. Method 2: 50% = 50/100, so 80 × 50 ÷ 100 = 40.
8.53\dfrac{5}{3}; the whole number 1 contributes 33\frac{3}{3}, plus 23\frac{2}{3} gives 53\frac{5}{3}
9.3 123\,\dfrac{1}{2}; check: 3×2+1=73 \times 2 + 1 = 7, so 72\frac{7}{2} ✓
10.0.5; 12\frac{1}{2} means 1 divided by 2, and 1÷2=0.51 \div 2 = 0.5
Silver
11.56\dfrac{5}{6}
12.512\dfrac{5}{12}
13.12\dfrac{1}{2}
14.1 121\,\dfrac{1}{2}
15.36
16.(a) 0.35 (b) 720\dfrac{7}{20}
17.16
18.3 343\,\dfrac{3}{4}
19.35<23<710\dfrac{3}{5} < \dfrac{2}{3} < \dfrac{7}{10}
20.£48
Gold
21.HCF = 6; 34\dfrac{3}{4}
22.−£12
23.4
24.2
25.£78
26.0.3‾0.\overline{3} (recurring)
27.−23<−12-\dfrac{2}{3} < -\dfrac{1}{2}
28.52\dfrac{5}{2} (or 2.5)
29.75 cm
30.120
Platinum
31.£54
32.£60
33.x=21x = 21
34.1716\dfrac{17}{16}
35.3 13243\,\dfrac{13}{24}
36.−34<−12<13<25-\dfrac{3}{4} < -\dfrac{1}{2} < \dfrac{1}{3} < \dfrac{2}{5}
37.13\dfrac{1}{3}
38.30% increase
39.16
40.Shop B (27.5p per 100g vs 30p per 100g)

Pack B — Answers

Bronze
1.E.g. 3451\dfrac{34}{51}; equivalent because 23=2×173×17=3451\frac{2}{3} = \frac{2 \times 17}{3 \times 17} = \frac{34}{51}
2.23\dfrac{2}{3}; HCF = 4; 2 and 3 share no common factor other than 1
3.710\dfrac{7}{10}; same denominator, so 7 pieces out of 10 is more than 4 pieces out of 10
4.35\dfrac{3}{5}; the denominator stays 5 because the piece-size (fifths) does not change
5.23\dfrac{2}{3}; check: 23+16=56\frac{2}{3} + \frac{1}{6} = \frac{5}{6} ✓
6.8; dividing by 4 splits the amount into 4 equal parts
7.30; Method 1: 50% = ½, so 60 ÷ 2 = 30. Method 2: 50% = 50/100, so 60 × 50 ÷ 100 = 30.
8.114\dfrac{11}{4}; each whole is 44\frac{4}{4}; two wholes give 84\frac{8}{4}, plus 34\frac{3}{4} gives 114\frac{11}{4}
9.2 142\,\dfrac{1}{4}; check: 2×4+1=92 \times 4 + 1 = 9, so 94\frac{9}{4} ✓
10.0.25; 14\frac{1}{4} means 1 divided by 4, and 1÷4=0.251 \div 4 = 0.25
Silver
11.712\dfrac{7}{12}
12.712\dfrac{7}{12}
13.12\dfrac{1}{2}
14.2 232\,\dfrac{2}{3}
15.24
16.(a) 0.60 (b) 35\dfrac{3}{5}
17.27
18.3 123\,\dfrac{1}{2}
19.38<512<12\dfrac{3}{8} < \dfrac{5}{12} < \dfrac{1}{2}
20.£69
Gold
21.HCF = 6; 57\dfrac{5}{7}
22.−£9
23.2 11122\,\dfrac{11}{12}
24.2 342\,\dfrac{3}{4}
25.£110
26.0.2‾0.\overline{2} (recurring)
27.−34<−58-\dfrac{3}{4} < -\dfrac{5}{8}
28.3
29.250 cm
30.180
Platinum
31.£81.60
32.£72
33.x=24x = 24
34.2536\dfrac{25}{36}
35.3 7123\,\dfrac{7}{12}
36.−56<−23<14<38-\dfrac{5}{6} < -\dfrac{2}{3} < \dfrac{1}{4} < \dfrac{3}{8}
37.711\dfrac{7}{11}
38.25% decrease
39.18
40.Shop B (33p per 100g vs 36p per 100g)

Problem-solving — Worked Solutions

1Problem 1
Answer
£54; total reduction is 32.5%, not 35%.
Full working
After 25% off: £80×0.75=£60\pounds 80 \times 0.75 = \pounds 60. After a further 10% off: £60×0.90=£54\pounds 60 \times 0.90 = \pounds 54. Total reduction from original: £80−£54=£26\pounds 80 - \pounds 54 = \pounds 26. As a percentage: 2680×100=32.5%\frac{26}{80} \times 100 = 32.5\%. The reductions are **not** simply added (35%) because the second 10% is taken off the already-reduced price, not the original £80. Sequential percentage changes multiply: 0.75×0.90=0.6750.75 \times 0.90 = 0.675, giving a single reduction of 1−0.675=32.5%1 - 0.675 = 32.5\%.
2Problem 2
Answer
101 (or 102 depending on rounding — see working)
Full working
38\frac{3}{8} of 360 = 360÷8×3=135360 \div 8 \times 3 = 135 students study French. 14\frac{1}{4} of 135 = 135÷4=33.75135 \div 4 = 33.75. Since we need a whole number of students, round to 34 who also study Spanish. French-only students: 135−34=101135 - 34 = \mathbf{101}. Note: the problem's fractions do not combine to give whole numbers with 360 students — a useful teaching point about mathematical modelling. If we round down (33 bilingual), French-only = 102.
3Problem 3
Answer
2 252\,\dfrac{2}{5} hours (2 hours 24 minutes)
Full working
In one hour, Pipe A fills 14\frac{1}{4} of the tank and Pipe B fills 16\frac{1}{6}. Together they fill 14+16\frac{1}{4} + \frac{1}{6} per hour. LCD = 12: 312+212=512\frac{3}{12} + \frac{2}{12} = \frac{5}{12} per hour. Time to fill whole tank = 1÷512=125=2251 \div \frac{5}{12} = \frac{12}{5} = 2\frac{2}{5} hours = **2 hours 24 minutes**.
4Problem 4
Answer
(a) 12+13\frac{1}{2} + \frac{1}{3} (b) 13+14\frac{1}{3} + \frac{1}{4} (c) 15+145\frac{1}{5} + \frac{1}{45} (d) Yes: 37=13+111+1231\frac{3}{7} = \frac{1}{3} + \frac{1}{11} + \frac{1}{231} (one way)
Full working
(a) 56=12+13\frac{5}{6} = \frac{1}{2} + \frac{1}{3}. Check: 36+26=56\frac{3}{6} + \frac{2}{6} = \frac{5}{6} ✓.

(b) 712=13+14\frac{7}{12} = \frac{1}{3} + \frac{1}{4}. Check: 412+312=712\frac{4}{12} + \frac{3}{12} = \frac{7}{12} ✓.

(c) We need 1a+1b=29\frac{1}{a} + \frac{1}{b} = \frac{2}{9} with a<ba < b, a≠ba \neq b. Since 1a>19\frac{1}{a} > \frac{1}{9} (as it's the larger part), we need a<9a < 9. Also 1a<29\frac{1}{a} < \frac{2}{9} means a>92=4.5a > \frac{9}{2} = 4.5, so a≥5a \geq 5. Try a=5a = 5: 1b=29−15=1045−945=145\frac{1}{b} = \frac{2}{9} - \frac{1}{5} = \frac{10}{45} - \frac{9}{45} = \frac{1}{45}. So b=45b = 45. Check: 15+145=945+145=1045=29\frac{1}{5} + \frac{1}{45} = \frac{9}{45} + \frac{1}{45} = \frac{10}{45} = \frac{2}{9} ✓. Answer: 15+145\frac{1}{5} + \frac{1}{45}.

(d) 37\frac{3}{7}: try a=3a = 3 (largest unit fraction less than 37\frac{3}{7} since 13≈0.333<37≈0.429\frac{1}{3} \approx 0.333 < \frac{3}{7} \approx 0.429... actually 13<37\frac{1}{3} < \frac{3}{7} since 7<97 < 9). 37−13=921−721=221\frac{3}{7} - \frac{1}{3} = \frac{9}{21} - \frac{7}{21} = \frac{2}{21}. Now write 221\frac{2}{21} as a unit fraction sum: a=11a = 11 gives 1b=221−111=22231−21231=1231\frac{1}{b} = \frac{2}{21} - \frac{1}{11} = \frac{22}{231} - \frac{21}{231} = \frac{1}{231}. So 37=13+111+1231\frac{3}{7} = \frac{1}{3} + \frac{1}{11} + \frac{1}{231}. It is always possible (Fibonacci/Sylvester's sequence guarantees this).
5Problem 5
Answer
She is wrong. The final price is 96% of the original.
Full working
Take an example: original price £100. After 20% increase: £100×1.20=£120\pounds 100 \times 1.20 = \pounds 120. After 20% decrease: £120×0.80=£96\pounds 120 \times 0.80 = \pounds 96. The price is now **£96**, which is **less** than the original £100. Multiplying the multipliers: 1.20×0.80=0.961.20 \times 0.80 = 0.96, a 4% overall **decrease**. The 20% increase and 20% decrease do not cancel because the decrease is calculated on the higher (post-increase) price.
6Problem 6
Answer
(a) 15 km (b) 38\dfrac{3}{8} (c) 12 km/h
Full working
(a) 58\frac{5}{8} of 24 km: 24÷8×5=1524 \div 8 \times 5 = 15 km. (b) Remaining fraction: 1−58=381 - \frac{5}{8} = \frac{3}{8}. Remaining distance: 38×24=9\frac{3}{8} \times 24 = 9 km. (c) 45 minutes = 34\frac{3}{4} hour. Speed = distance ÷ time = 9÷34=9×43=129 \div \frac{3}{4} = 9 \times \frac{4}{3} = 12 km/h.
7Problem 7
Answer
(a) 30 (b) Many valid sets, e.g. {1, 2, 5, 8, 14}, {1, 3, 5, 7, 14} — see working for all (c) Yes — e.g. {2, 3, 5, 6, 14}.
Full working
(a) Mean = 6, five values: sum = 6×5=306 \times 5 = \mathbf{30}.

(b) The median is the 3rd value (when ordered), so the 3rd value = 5. We have: a<b<5<d<ea < b < 5 < d < e with a+b+5+d+e=30a + b + 5 + d + e = 30, so a+b+d+e=25a + b + d + e = 25.

Smallest = 1 (given): a=1a = 1. So b+d+e=24b + d + e = 24 with 1<b<51 < b < 5, 5<d<e5 < d < e, all different integers.

b∈{2,3,4}b \in \{2, 3, 4\}.

- b=2b = 2: d+e=22d + e = 22, d≥6d \geq 6. Try d=6,e=16d = 6, e = 16; d=7,e=15d = 7, e = 15; d=8,e=14d = 8, e = 14; d=9,e=13d = 9, e = 13; d=10,e=12d = 10, e = 12; d=11,e=11d = 11, e = 11 (equal — invalid). Valid: {1,2,5,6,16}\{1,2,5,6,16\}, {1,2,5,7,15}\{1,2,5,7,15\}, {1,2,5,8,14}\{1,2,5,8,14\}, {1,2,5,9,13}\{1,2,5,9,13\}, {1,2,5,10,12}\{1,2,5,10,12\}.
- b=3b = 3: d+e=21d + e = 21, d≥6d \geq 6. Valid: {1,3,5,6,15}\{1,3,5,6,15\}, {1,3,5,7,14}\{1,3,5,7,14\}, {1,3,5,8,13}\{1,3,5,8,13\}, {1,3,5,9,12}\{1,3,5,9,12\}, {1,3,5,10,11}\{1,3,5,10,11\}.
- b=4b = 4: d+e=20d + e = 20, d≥6d \geq 6. Valid: {1,4,5,6,14}\{1,4,5,6,14\}, {1,4,5,7,13}\{1,4,5,7,13\}, {1,4,5,8,12}\{1,4,5,8,12\}, {1,4,5,9,11}\{1,4,5,9,11\}.

Many valid sets exist (the problem asks to "find all possible sets" — students can list them).

(c) If smallest = 2: a=2a = 2, so b+d+e=23b + d + e = 23, 2<b<52 < b < 5 means b∈{3,4}b \in \{3, 4\}.
- b=3b = 3: d+e=20d + e = 20, d≥6d \geq 6. Gives {2,3,5,6,14}\{2,3,5,6,14\} etc. — **valid** ✓.

So a valid set **does** still exist with smallest = 2. (The answer "No" above was incorrect — corrected here: **Yes**, e.g. {2,3,5,6,14}\{2,3,5,6,14\}.)
8Problem 8
Answer
(a) 126=2×32×7126 = 2 \times 3^2 \times 7; 210=2×3×5×7210 = 2 \times 3 \times 5 \times 7 (b) HCF = 42 (c) 35\dfrac{3}{5}
Full working
(a) 126÷2=63126 \div 2 = 63; 63÷3=2163 \div 3 = 21; 21÷3=721 \div 3 = 7; 7 prime. 126=2×32×7126 = 2 \times 3^2 \times 7. 210÷2=105210 \div 2 = 105; 105÷3=35105 \div 3 = 35; 35÷5=735 \div 5 = 7; 7 prime. 210=2×3×5×7210 = 2 \times 3 \times 5 \times 7. (b) HCF: take lowest powers of shared primes (21,31,712^1, 3^1, 7^1): 2×3×7=422 \times 3 \times 7 = 42. (c) 126210=126÷42210÷42=35\frac{126}{210} = \frac{126 \div 42}{210 \div 42} = \frac{3}{5}.
9Problem 9
Answer
5 145\,\dfrac{1}{4} cm
Full working
Perimeter = 2(l+w)=152(l + w) = 15, so l+w=7.5=712l + w = 7.5 = 7\frac{1}{2}. One side w=214w = 2\frac{1}{4}. Other side l=712−214l = 7\frac{1}{2} - 2\frac{1}{4}. Convert: 712=7247\frac{1}{2} = 7\frac{2}{4}. So l=724−214=514l = 7\frac{2}{4} - 2\frac{1}{4} = 5\frac{1}{4} cm. Check: 514+214=7125\frac{1}{4} + 2\frac{1}{4} = 7\frac{1}{2}; perimeter = 2×7.5=152 \times 7.5 = 15 ✓.
10Problem 10
Answer
(a) 01,14,13,12,23,34,11\frac{0}{1}, \frac{1}{4}, \frac{1}{3}, \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{1}{1} (b) bc−ad=1bc - ad = 1 always (c) 25\frac{2}{5}, yes (d) See working.
Full working
(a) Fractions with denominator ≤ 4, in order: 01, 14, 13, 12, 23, 34, 11\dfrac{0}{1},\ \dfrac{1}{4},\ \dfrac{1}{3},\ \dfrac{1}{2},\ \dfrac{2}{3},\ \dfrac{3}{4},\ \dfrac{1}{1}.

(b) Check adjacent pairs: 01\frac{0}{1} and 14\frac{1}{4}: 1×1−0×4=11 \times 1 - 0 \times 4 = 1. 14\frac{1}{4} and 13\frac{1}{3}: 1×4−1×3=11 \times 4 - 1 \times 3 = 1. 13\frac{1}{3} and 12\frac{1}{2}: 1×3−1×2=11 \times 3 - 1 \times 2 = 1. The cross-product bc−ad=1bc - ad = \mathbf{1} for every adjacent pair. This is a remarkable property of Farey sequences.

(c) Mediant of 13\frac{1}{3} and 12\frac{1}{2}: 1+13+2=25\frac{1+1}{3+2} = \frac{2}{5}. Is 13<25<12\frac{1}{3} < \frac{2}{5} < \frac{1}{2}? 13≈0.333\frac{1}{3} \approx 0.333, 25=0.4\frac{2}{5} = 0.4, 12=0.5\frac{1}{2} = 0.5. Yes, 25\frac{2}{5} lies strictly between them.

(d) If ab<cd\frac{a}{b} < \frac{c}{d} are adjacent Farey fractions, their mediant is a+cb+d\frac{a+c}{b+d}. To show ab<a+cb+d\frac{a}{b} < \frac{a+c}{b+d}: cross-multiply: a(b+d)<b(a+c)⇔ab+ad<ab+bc⇔ad<bc⇔bc−ad>0a(b+d) < b(a+c) \Leftrightarrow ab + ad < ab + bc \Leftrightarrow ad < bc \Leftrightarrow bc - ad > 0, which holds since bc−ad=1>0bc - ad = 1 > 0. Similarly a+cb+d<cd\frac{a+c}{b+d} < \frac{c}{d} since d(a+c)<c(b+d)⇔cd−ad<bc+cd−dc⇔d(a+c) < c(b+d) \Leftrightarrow cd - ad < bc + cd - dc \Leftrightarrow same condition. So the mediant always lies strictly between adjacent Farey fractions.
11Problem 11
Answer
(a) 1141\dfrac{1}{4} kg (b) 56\dfrac{5}{6} (c) 16.6‾%16.\overline{6}\%
Full working
(a) Flour per biscuit: 34÷12=348=116\frac{3}{4} \div 12 = \frac{3}{48} = \frac{1}{16} kg. For 20 biscuits: 116×20=2016=54=114\frac{1}{16} \times 20 = \frac{20}{16} = \frac{5}{4} = 1\frac{1}{4} kg. (b) Fraction of bag: 5/43/2=54×23=1012=56\frac{5/4}{3/2} = \frac{5}{4} \times \frac{2}{3} = \frac{10}{12} = \frac{5}{6}. (c) Fraction remaining: 1−56=161 - \frac{5}{6} = \frac{1}{6}. As a percentage: 16×100=16.6‾%\frac{1}{6} \times 100 = 16.\overline{6}\%.
12Problem 12
Answer
£280.80
Full working
Savings: 14\frac{1}{4} of £960 = £240. Remainder after saving: £960−£240=£720\pounds 960 - \pounds 240 = \pounds 720. Rent: 35% of £720 = £720×0.35=£252\pounds 720 \times 0.35 = \pounds 252. After rent: £720−£252=£468\pounds 720 - \pounds 252 = \pounds 468. Food and bills: 25\frac{2}{5} of £468 = £468÷5×2=£93.60×2=£187.20\pounds 468 \div 5 \times 2 = \pounds 93.60 \times 2 = \pounds 187.20. Discretionary: £468−£187.20=£280.80\pounds 468 - \pounds 187.20 = \mathbf{\pounds 280.80}.