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Ecolint Campus des NationsMathématiques
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Solutions — Full Answer Key

Ecolint Campus des NationsMathematics
Year 8 · 8.1 Fractions review

Solutions · Full Answer Key

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

Pack A — Answers

Bronze
1.3253\tfrac{2}{5}
2.143\dfrac{14}{3}
3.34\dfrac{3}{4}
4.1520\dfrac{15}{20}
5.57\dfrac{5}{7}
6.712\dfrac{7}{12}
7.12\dfrac{1}{2}
8.32\dfrac{3}{2} (or 1121\tfrac{1}{2})
9.(a) 0.35 (b) 720\dfrac{7}{20}
10.0.75 and 75%
Silver
11.712\dfrac{7}{12}
12.3123\tfrac{1}{2}
13.360 g
14.92\tfrac{9}{2} (or 4124\tfrac{1}{2})
15.110\dfrac{1}{10}
16.712<58<23\dfrac{7}{12} < \dfrac{5}{8} < \dfrac{2}{3}
17.13\dfrac{1}{3}
18.23\dfrac{2}{3}
19.920\dfrac{9}{20}
20.95\dfrac{9}{5} (or 1451\tfrac{4}{5})
Gold
21.38\dfrac{3}{8}
22.270 ml
23.90 chf
24.2
25.310\dfrac{3}{10}
26.2342\tfrac{3}{4}
27.0.6<0.62<580.6 < 0.62 < \dfrac{5}{8}
28.1181\tfrac{1}{8} cups
29.1
30.18\dfrac{1}{8}
Platinum
31.116\dfrac{11}{6} (or 1561\tfrac{5}{6})
32.1440 ml
33.311\dfrac{3}{11}
34.18
35.Let x=0.9‾x = 0.\overline{9}. Then 10x=9.9‾10x = 9.\overline{9}. Subtracting: 9x=99x = 9, so x=1x = 1.
36.40
37.Yes — equivalent (cross product 360=360360 = 360).
38.25\dfrac{2}{5} (or 37\dfrac{3}{7})
39.n=15n = 15
40.Sum =158= \dfrac{15}{8}; the pattern approaches 2 but never reaches it (it gets infinitesimally close).

Pack B — Answers

Bronze
1.5345\tfrac{3}{4}
2.236\dfrac{23}{6}
3.23\dfrac{2}{3}
4.1230\dfrac{12}{30}
5.812=23\dfrac{8}{12} = \dfrac{2}{3}
6.910\dfrac{9}{10}
7.23\dfrac{2}{3}
8.2
9.(a) 0.65 (b) 1320\dfrac{13}{20}
10.0.35 and 35%
Silver
11.310\dfrac{3}{10}
12.5345\tfrac{3}{4}
13.300 g
14.15
15.320\dfrac{3}{20}
16.712<58<23\dfrac{7}{12} < \dfrac{5}{8} < \dfrac{2}{3}
17.23\dfrac{2}{3}
18.56\dfrac{5}{6}
19.58\dfrac{5}{8}
20.52\dfrac{5}{2} (or 2122\tfrac{1}{2})
Gold
21.38\dfrac{3}{8}
22.360 ml
23.40 chf
24.114\dfrac{11}{4} (or 2342\tfrac{3}{4})
25.67300\dfrac{67}{300}
26.2122\tfrac{1}{2}
27.49<0.45<0.47\dfrac{4}{9} < 0.45 < 0.47
28.1191\tfrac{1}{9} cups
29.1
30.78\dfrac{7}{8}
Platinum
31.116\dfrac{11}{6}
32.1200 ml
33.511\dfrac{5}{11}
34.25
35.See Pack A.
36.36
37.Yes — equivalent (cross product 225=225225 = 225).
38.27\dfrac{2}{7} (any value in the range with denominator < 10)
39.n=14n = 14
40.See Pack A.

Problem-solving — Worked Solutions

1Problem 1
Answer
(a) 16\tfrac{1}{6} each (b) 312,212,212,212,112,212\tfrac{3}{12}, \tfrac{2}{12}, \tfrac{2}{12}, \tfrac{2}{12}, \tfrac{1}{12}, \tfrac{2}{12} (Fati gets the last 2) (c) See working
Full working
(a) 126=2\tfrac{12}{6} = 2 slices each, which is 212=16\tfrac{2}{12} = \tfrac{1}{6} of the pizza.

(b) Alex 312=14\tfrac{3}{12} = \tfrac{1}{4}; B, C, D each 212=16\tfrac{2}{12} = \tfrac{1}{6}; Eli 112\tfrac{1}{12}; remainder for the sixth friend: 12−(3+2+2+2+1)=212 - (3 + 2 + 2 + 2 + 1) = 2 → 16\tfrac{1}{6}. Sum: 3+2+2+2+1+212=1212=1\tfrac{3 + 2 + 2 + 2 + 1 + 2}{12} = \tfrac{12}{12} = 1 ✓.

(c) Alex 25%; B/C/D ≈ 16.67%; Eli ≈ 8.33%; sixth friend ≈ 16.67%. Total = 100%.
2Problem 2
Answer
(a) 118,34,381\tfrac{1}{8}, \tfrac{3}{4}, \tfrac{3}{8} (b) 12,13,16\tfrac{1}{2}, \tfrac{1}{3}, \tfrac{1}{6} (c) 24 pancakes
Full working
(a) Scale factor 96=32\tfrac{9}{6} = \tfrac{3}{2}. Flour: 34×32=98=118\tfrac{3}{4} \times \tfrac{3}{2} = \tfrac{9}{8} = 1\tfrac{1}{8} cup. Milk: 12×32=34\tfrac{1}{2} \times \tfrac{3}{2} = \tfrac{3}{4} cup. Sugar: 14×32=38\tfrac{1}{4} \times \tfrac{3}{2} = \tfrac{3}{8} cup.

(b) Scale factor 46=23\tfrac{4}{6} = \tfrac{2}{3}. Flour: 34×23=12\tfrac{3}{4} \times \tfrac{2}{3} = \tfrac{1}{2} cup. Milk: 12×23=13\tfrac{1}{2} \times \tfrac{2}{3} = \tfrac{1}{3} cup. Sugar: 14×23=16\tfrac{1}{4} \times \tfrac{2}{3} = \tfrac{1}{6} cup.

(c) Flour per pancake: 34÷6=18\tfrac{3}{4} \div 6 = \tfrac{1}{8} cup. With 3 cups: 3÷18=243 \div \tfrac{1}{8} = 24 pancakes.
3Problem 3
Answer
(a) 49\tfrac{4}{9}, 311\tfrac{3}{11}, 16999\tfrac{16}{999} (b) (i) 23\tfrac{2}{3}; (ii) correct; (iii) wrong — should be 21999\tfrac{21}{999} (or 7333\tfrac{7}{333}) (c) See working
Full working
(a) 0.4‾=490.\overline{4} = \tfrac{4}{9}. 0.27‾=2799=3110.\overline{27} = \tfrac{27}{99} = \tfrac{3}{11}. 0.016‾=169990.\overline{016} = \tfrac{16}{999}.

(b) (i) 69\tfrac{6}{9} simplifies to 23\tfrac{2}{3} — both are correct, but simplest form is 23\tfrac{2}{3}. (ii) 3799\tfrac{37}{99} ✓ already in simplest form. (iii) Block has 3 digits → denominator 999999, not 9999. Correct fraction: 21999=7333\tfrac{21}{999} = \tfrac{7}{333}.

(c) Let x=0.9‾x = 0.\overline{9}. Then 10x=9.9‾10x = 9.\overline{9}. Subtract: 9x=99x = 9, so x=1x = 1. Therefore 0.9‾=10.\overline{9} = 1 exactly.
4Problem 4
Answer
(a) 43\tfrac{4}{3} (b) 35\tfrac{3}{5} (c) 1324\tfrac{13}{24}
Full working
(a) Order of operations: multiply first. 14×83=812=23\tfrac{1}{4} \times \tfrac{8}{3} = \tfrac{8}{12} = \tfrac{2}{3}. Then 23+23=43\tfrac{2}{3} + \tfrac{2}{3} = \tfrac{4}{3}.

(b) Numerator: 56−26=36=12\tfrac{5}{6} - \tfrac{2}{6} = \tfrac{3}{6} = \tfrac{1}{2}. Denominator: 46+16=56\tfrac{4}{6} + \tfrac{1}{6} = \tfrac{5}{6}. Divide: 12÷56=12×65=610=35\tfrac{1}{2} \div \tfrac{5}{6} = \tfrac{1}{2} \times \tfrac{6}{5} = \tfrac{6}{10} = \tfrac{3}{5}.

(c) Brackets: 34+16=912+212=1112\tfrac{3}{4} + \tfrac{1}{6} = \tfrac{9}{12} + \tfrac{2}{12} = \tfrac{11}{12}. Multiply: 12×1112=1124\tfrac{1}{2} \times \tfrac{11}{12} = \tfrac{11}{24}. Subtract: 1−1124=13241 - \tfrac{11}{24} = \tfrac{13}{24}.
5Problem 5
Answer
(a) 12\tfrac{1}{2} (b) 38\tfrac{3}{8} (c) 562.5 ml
Full working
(a) After Alice: 56−13=56−26=36=12\tfrac{5}{6} - \tfrac{1}{3} = \tfrac{5}{6} - \tfrac{2}{6} = \tfrac{3}{6} = \tfrac{1}{2}.

(b) Ben drinks 14×12=18\tfrac{1}{4} \times \tfrac{1}{2} = \tfrac{1}{8} of the bottle's capacity, leaving 12−18=48−18=38\tfrac{1}{2} - \tfrac{1}{8} = \tfrac{4}{8} - \tfrac{1}{8} = \tfrac{3}{8}.

(c) 38×1500=45008=562.5\tfrac{3}{8} \times 1500 = \tfrac{4500}{8} = 562.5 ml.
6Problem 6
Answer
(a) 216 (b) 162 (c) 920\tfrac{9}{20}
Full working
(a) Upper School fraction: 1−25=351 - \tfrac{2}{5} = \tfrac{3}{5}. 35×360=216\tfrac{3}{5} \times 360 = 216.

(b) 34×216=162\tfrac{3}{4} \times 216 = 162 pupils.

(c) 162360=920\tfrac{162}{360} = \tfrac{9}{20}. (Or 35×34=920\tfrac{3}{5} \times \tfrac{3}{4} = \tfrac{9}{20}.)
7Problem 7
Answer
(a) Yes (b) x=4x = 4 (c) x=2x = 2
Full working
(a) Cross: 9×20=1809 \times 20 = 180 and 12×15=18012 \times 15 = 180 — equal, so equivalent. Both simplify to 34\tfrac{3}{4}.

(b) Cross: 9x=18×2=369x = 18 \times 2 = 36, so x=4x = 4. Check: 418=29\tfrac{4}{18} = \tfrac{2}{9} ✓.

(c) Cross: 5x=2(x+3)=2x+65x = 2(x + 3) = 2x + 6, so 3x=63x = 6, x=2x = 2. Check: 25=25\tfrac{2}{5} = \tfrac{2}{5} ✓.
8Problem 8
Answer
(a) 260 EUR (b) 144 chf (c) 1213\tfrac{12}{13} chf ≈ 0.9231 chf
Full working
(a) 240×1312=312012=260240 \times \tfrac{13}{12} = \tfrac{3120}{12} = 260 EUR.

(b) 156÷1312=156×1213=187213=144156 \div \tfrac{13}{12} = 156 \times \tfrac{12}{13} = \tfrac{1872}{13} = 144 chf.

(c) Inverse: 1÷1312=12131 \div \tfrac{13}{12} = \tfrac{12}{13} chf ≈0.9231\approx 0.9231 chf (4 d.p.).
9Problem 9
Answer
(a) 16\tfrac{1}{6} (b) 14\tfrac{1}{4} (c) 512\tfrac{5}{12}; 2 h 24 min
Full working
(a) Alex: 16\tfrac{1}{6} of the fence per hour.

(b) Brigit: 14\tfrac{1}{4} per hour.

(c) Together: 16+14=212+312=512\tfrac{1}{6} + \tfrac{1}{4} = \tfrac{2}{12} + \tfrac{3}{12} = \tfrac{5}{12} per hour. Time for full fence: 1÷512=125=2.41 \div \tfrac{5}{12} = \tfrac{12}{5} = 2.4 hours =2= 2 h 2424 min.
10Problem 10
Answer
(a) 32,56,712,920,1130\tfrac{3}{2}, \tfrac{5}{6}, \tfrac{7}{12}, \tfrac{9}{20}, \tfrac{11}{30} (b) See working (c) For n≥4n \geq 4
Full working
(a) n=1n = 1: 1+12=321 + \tfrac{1}{2} = \tfrac{3}{2}. n=2n = 2: 12+13=56\tfrac{1}{2} + \tfrac{1}{3} = \tfrac{5}{6}. n=3n = 3: 13+14=712\tfrac{1}{3} + \tfrac{1}{4} = \tfrac{7}{12}. n=4n = 4: 14+15=920\tfrac{1}{4} + \tfrac{1}{5} = \tfrac{9}{20}. n=5n = 5: 15+16=1130\tfrac{1}{5} + \tfrac{1}{6} = \tfrac{11}{30}.

(b) Common denominator n(n+1)n(n+1): 1n+1n+1=n+1n(n+1)+nn(n+1)=2n+1n(n+1)\tfrac{1}{n} + \tfrac{1}{n+1} = \tfrac{n+1}{n(n+1)} + \tfrac{n}{n(n+1)} = \tfrac{2n + 1}{n(n+1)}.

(c) Test the values from (a): for n=3n = 3, 712≈0.583>0.5\tfrac{7}{12} \approx 0.583 > 0.5. For n=4n = 4, 920=0.45<0.5\tfrac{9}{20} = 0.45 < 0.5 ✓. So **n≥4n \geq 4** gives a sum less than 12\tfrac{1}{2}.
11Problem 11
Answer
Row 1: 38\tfrac{3}{8}, 0.375, 37.5%. Row 2: 1320\tfrac{13}{20}, 0.65, 65%. Row 3: 325\tfrac{3}{25}, 0.12, 12%. Row 4: 65\tfrac{6}{5} (or 1151\tfrac{1}{5}), 1.2, 120%.
Full working
Row 1: 38=3÷8=0.375\tfrac{3}{8} = 3 \div 8 = 0.375 → 37.5%.

Row 2: 0.65=65100=13200.65 = \tfrac{65}{100} = \tfrac{13}{20} → 65%.

Row 3: 12%=12100=32512\% = \tfrac{12}{100} = \tfrac{3}{25} → 0.12.

Row 4: 1.2=1210=651.2 = \tfrac{12}{10} = \tfrac{6}{5} → 120%.
12Problem 12
Answer
(a) 2.5 turns (b) 3432=1716\tfrac{34}{32} = \tfrac{17}{16} (≈ 1.06) (c) 5011\tfrac{50}{11} (≈ 4.55)
Full working
(a) Wheel turns = 5020=52=2.5\tfrac{50}{20} = \tfrac{5}{2} = 2.5.

(b) Smallest ratio = smallest chainring ÷ largest sprocket = 3432=1716\tfrac{34}{32} = \tfrac{17}{16} ≈ 1.06 wheel turns per pedal. Lowest gear, easiest to push uphill.

(c) Largest ratio = largest chainring ÷ smallest sprocket = 5011\tfrac{50}{11} ≈ 4.55 wheel turns per pedal. Highest gear, for top speed on flats. (Note: 5011\tfrac{50}{11} is already in simplest form because 11 is prime and does not divide 50.)