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Ecolint Campus des NationsMathematics
Year 7 · 7.4 Decimals & Measurement

Solutions · Full Answer Key

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

Pack A — Answers

Bronze
1.E.g. 0.31, 0.35, 0.39 — each is greater than 0.3 and less than 0.4
2.4.4; the 2nd decimal digit (7) decided — it is 5 or more, so round up
3.Correct answer is 7.22; the student did not align the decimal points correctly
4.180 mm and 0.18 m; multiplying by 10 converts cm to mm, dividing by 100 converts cm to m
5.4.85 g
6.Both methods give 7.2
7.Greater than 2 (since 8 ÷ 4 = 2 and 8.4 > 8); exact answer: 2.1
8.47 mm; multiplied by 10 because 1 cm = 10 mm, so cm → mm requires × 10
9.3.5 kg; divide by 1 000 because kg is a larger unit (there are 1 000 g in each kg)
10.E.g. 3.2 cm × 1.5 cm (since 2×(3.2+1.5)=9.4) and 4.0 cm × 0.7 cm (since 2×(4.0+0.7)=9.4)
Silver
11.5.68
12.24.40
13.11.63
14.8.5
15.13
16.(a) 235 cm, (b) 2 350 mm
17.7 cm
18.12:20 pm
19.2 450 g
20.Shop B (£1.08 per bottle)
Gold
21.6.2
22.Estimate: 4.9 × 3.2 = 15.68; exact: 15.584
23.3 250 m
24.8.9 cm
25.15.05
26.3.7
27.x=1.7x = 1.7
28.6 300 s
29.Area = 35.7 cm², Perimeter = 25.4 cm
30.Area = 20.25 cm²; 20.25=4.5\sqrt{20.25} = 4.5 — exact, equals the side length.
Platinum
31.x=3.2x = 3.2 cm; area = 22.4 cm²
32.1.6
33.Rise = 9.2°C; rate = 2.3°C/h
34.196=22×72196 = 2^2 \times 7^2; 196=14\sqrt{196} = 14
35.3x+4.93x + 4.9; value at x=0.5x = 0.5: 6.46.4
36.14 pieces; 10 cm left over
37.30
38.5.4
39.12 minutes
40.Area = 43 200 m² = 4.32 ha

Pack B — Answers

Bronze
1.E.g. 1.51, 1.55, 1.58 — each is greater than 1.5 and less than 1.6
2.7.8; the 2nd decimal digit (4) decided — it is less than 5, so round down
3.Correct answer is 10.24; the student did not align the decimal points correctly
4.245 mm and 0.245 m; multiplying by 10 converts cm to mm, dividing by 100 converts cm to m
5.4.65 g
6.Both methods give 14.0
7.Greater than 1 (since 5 ÷ 5 = 1 and 7.5 > 5); exact answer: 1.5
8.38 mm; multiplied by 10 because 1 cm = 10 mm
9.6.25 kg; divide by 1 000 because kg is a larger unit
10.E.g. 4.6 cm × 2.3 cm (since 2×(4.6+2.3)=13.8) and 5.0 cm × 1.9 cm (since 2×(5.0+1.9)=13.8)
Silver
11.3.43
12.27.03
13.15.72
14.14.7
15.14
16.(a) 408 cm, (b) 4 080 mm
17.6.4 cm
18.3:15 pm
19.3 072 g
20.Shop B (£0.77 per bottle)
Gold
21.9.5
22.Estimate: 6.9 × 2.8 = 19.32; exact: 19.432
23.2 340 m
24.10.8 cm
25.11.16
26.−2.8
27.x=1.44x = 1.44
28.8 400 s
29.Area = 28.88 cm², Perimeter = 22.8 cm
30.Area = 39.69 cm²; 39.69=6.3\sqrt{39.69} = 6.3 — exact, equals the side length.
Platinum
31.x=3.8x = 3.8 cm; area = 33.44 cm²
32.2.4
33.Rise = 9.0°C; rate = 1.8°C/h
34.225=32×52225 = 3^2 \times 5^2; 225=15\sqrt{225} = 15
35.4x+7.24x + 7.2; value at x=1.5x = 1.5: 13.213.2
36.16 pieces; 0 cm left over
37.20
38.5.4
39.10 minutes
40.Area = 28 800 m² = 2.88 ha

Problem-solving — Worked Solutions

1Problem 1
Answer
(a) Yes: 9 − 3 − 1 = 5 (b) Yes: 9 + 3 − 1 = 11 (c) None — you can measure every integer from 1 to 40 (d) 40
Full working
With weights on both sides, a mass mm on the left pan is balanced by some weights on the right pan minus some weights on the left pan (those cancel with mm). This is equivalent to writing mm as a sum ∑ϵiwi\sum \epsilon_i w_i where ϵi∈{−1,0,+1}\epsilon_i \in \{-1, 0, +1\} (left pan, not used, right pan respectively). This is exactly the balanced-ternary representation!

(a) 5=9−3−15 = 9 - 3 - 1: put 9 g on the right, and 3 g + 1 g on the same side as the object (left). ✓

(b) 11=9+3−111 = 9 + 3 - 1: put 9 g and 3 g on the right with the object on the left, and 1 g on the left. ✓

(c) With weights 1, 3, 9, 27 (powers of 3) you can represent every integer from 1 to 1+3+9+27=401 + 3 + 9 + 27 = 40 in balanced ternary. There is **no** whole-number mass from 1 to 40 that you cannot measure.

(d) Every integer from 1 to 40 is achievable — that is **40** distinct masses. The key insight: powers of 3 in balanced ternary cover all integers in [1,(3n−1)/2][1, (3^n-1)/2] using nn weights; here n=4n=4, giving up to (81−1)/2=40(81-1)/2 = 40.
2Problem 2
Answer
(a) 0.8 m/min; (b) ≈ 1.33 cm/s; (c) ≈ 66.7 mm
Full working
(a) 0.048 km/h=0.048×1000=48 m/h0.048 \text{ km/h} = 0.048 \times 1000 = 48 \text{ m/h}. Per minute: 48÷60=0.848 \div 60 = 0.8 m/min.

(b) 0.8 m/min=80 cm/min0.8 \text{ m/min} = 80 \text{ cm/min}. Per second: 80÷60=43≈1.3‾80 \div 60 = \tfrac{4}{3} \approx 1.\overline{3} cm/s.

(c) Distance in 5 s: 43×5=203≈6.6‾\tfrac{4}{3} \times 5 = \tfrac{20}{3} \approx 6.\overline{6} cm =66.6‾= 66.\overline{6} mm ≈66.7\approx 66.7 mm.
3Problem 3
Answer
w=4.7w = 4.7 cm
Full working
Total length =3.2+4.7+5.1=13.0= 3.2 + 4.7 + 5.1 = 13.0 cm.

Perimeter: 2(13.0+w)=35.42(13.0 + w) = 35.4 ⟹ 13.0+w=17.713.0 + w = 17.7 ⟹ w=4.7w = 4.7 cm.

Check: 2(13.0+4.7)=2×17.7=35.42(13.0 + 4.7) = 2 \times 17.7 = 35.4 ✓.
4Problem 4
Answer
(a) 0.142857‾0.\overline{142857} (b) 0.285714‾0.\overline{285714} (c) 0.428571‾0.\overline{428571} (d) The digits 142857 cycle; see working.
Full working
(a) Long divide 1.000000÷71.000000 \div 7: 10÷7=110 \div 7 = 1 r 3; 30÷7=430 ÷ 7 = 4 r 2; 20÷7=220 ÷ 7 = 2 r 6; 60÷7=860 ÷ 7 = 8 r 4; 40÷7=540 ÷ 7 = 5 r 5; 50÷7=750 ÷ 7 = 7 r 1 — back to remainder 1. So 17=0.142857‾\frac{1}{7} = 0.\overline{142857}.

(b) 27=2×17\frac{2}{7} = 2 \times \frac{1}{7}. Doubling each digit of 142857 (carrying as needed): 0.285714‾0.\overline{285714}.

(c) 37=3×17=0.428571‾\frac{3}{7} = 3 \times \frac{1}{7} = 0.\overline{428571}.

(d) The six digits 142857 form a cycle. Each fraction n7\frac{n}{7} simply starts at a different point in the cycle:
- 47=0.571428‾\frac{4}{7} = 0.\overline{571428}
- 57=0.714285‾\frac{5}{7} = 0.\overline{714285}
- 67=0.857142‾\frac{6}{7} = 0.\overline{857142}

Why? Because when you divide by 7, the possible remainders are 1, 2, 3, 4, 5, 6 (in some order). Starting from remainder nn gives the decimal for n7\frac{n}{7}, which is just a rotation of the same six-digit block. The number 142857 is called a **cyclic number**.
5Problem 5
Answer
Gamma (£0.43 per 100 ml)
Full working
Convert all prices to pence per 100 ml for a fair comparison.

- **Alpha**: 240÷5=48p240 \div 5 = 48\text{p} per 100 ml.
- **Beta**: 345÷7.5=46p345 \div 7.5 = 46\text{p} per 100 ml.
- **Gamma**: 1.2 L=1200 ml1.2 \text{ L} = 1200 \text{ ml}; 516÷12=43p516 \div 12 = 43\text{p} per 100 ml.

Gamma is cheapest at 43p per 100 ml → **Gamma is best value**.
6Problem 6
Answer
3.57 + 5.46 = 9.03
Full working
Label the unknowns: 3.A7+B.4C=9.033.A7 + B.4C = 9.03.

**Hundredths column**: 7+C7 + C must end in 3. So 7+C=137 + C = 13 ⟹ C=6C = 6 (carry 1 to tenths).

**Tenths column**: A+4+1 (carry)A + 4 + 1\text{ (carry)} must end in 0 (tenths digit of 9.03 is 0). So A+5=10A + 5 = 10 ⟹ A=5A = 5 (carry 1 to ones).

**Ones column**: 3+B+1 (carry)=93 + B + 1\text{ (carry)} = 9 ⟹ B=5B = 5.

The missing digits are A=5A = 5, B=5B = 5, C=6C = 6. Full sum: 3.57+5.46=9.033.57 + 5.46 = 9.03 ✓.
7Problem 7
Answer
(a) 42.2°C; (b) 7 h 42 min; (c) 17 h 35 min
Full working
(a) Difference =23.7−(−18.5)=23.7+18.5=42.2= 23.7 - (-18.5) = 23.7 + 18.5 = 42.2°C.

(b) Rise needed: 0−(−18.5)=18.50 - (-18.5) = 18.5°C. Time =18.5÷2.4=7.7083‾= 18.5 \div 2.4 = 7.708\overline{3} h. 0.7083‾×60=42.50.708\overline{3} \times 60 = 42.5 min ≈ **7 h 42 min**.

(c) Total rise needed: 42.242.2°C. Time =42.2÷2.4=17.583‾= 42.2 \div 2.4 = 17.58\overline{3} h. 0.583‾×60=350.58\overline{3} \times 60 = 35 min. So **17 h 35 min**.
8Problem 8
Answer
29.2 cm
Full working
Trace the six sides of the L-shape clockwise from the bottom-left:

1. Bottom: 8.48.4 cm
2. Right side (lower portion only): 6.2−2.5=3.76.2 - 2.5 = 3.7 cm
3. Notch top (going left): 3.13.1 cm
4. Notch left edge (going down — this is the inner vertical edge): 2.52.5 cm
5. Remaining top (going left): 8.4−3.1=5.38.4 - 3.1 = 5.3 cm
6. Left side (full height, going down): 6.26.2 cm

Perimeter =8.4+3.7+3.1+2.5+5.3+6.2=29.2= 8.4 + 3.7 + 3.1 + 2.5 + 5.3 + 6.2 = 29.2 cm ✓.
9Problem 9
Answer
(a) 64 (b) 8 (c) 24 (d) 24 (e) 8 (f) 8+24+24+8 = 64 ✓
Full working
(a) 4×4×4=644 \times 4 \times 4 = \mathbf{64} small cubes.

(b) **3 painted faces** — corner cubes. A cube has 8 corners. Each corner small cube touches exactly 3 faces: **8 cubes**.

(c) **2 painted faces** — edge cubes (not corners). Each edge of the large cube has 4−2=24 - 2 = 2 interior positions. A cube has 12 edges, so 12×2=2412 \times 2 = \mathbf{24} cubes.

(d) **1 painted face** — face cubes (not on any edge). Each face of the large cube has a (4−2)×(4−2)=2×2=4(4-2) \times (4-2) = 2 \times 2 = 4 inner grid. 6 faces × 4 = 24\mathbf{24} cubes.

(e) **0 painted faces** — interior cubes. These form a (4−2)3=23=8(4-2)^3 = 2^3 = \mathbf{8} cube inside.

(f) 8+24+24+8=648 + 24 + 24 + 8 = 64 ✓.

*Extension for the curious:* For an n×n×nn \times n \times n cube: corners always = 8; edges = 12(n−2)12(n-2); faces = 6(n−2)26(n-2)^2; interior = (n−2)3(n-2)^3.
10Problem 10
Answer
(a) 2.35 m ≤ length < 2.45 m; (b) 3 shelves; (c) 2 shelves; (d) result is unreliable — it differs by 1 shelf
Full working
(a) Rounded to nearest 0.1 m: true length lies in [2.35,2.45)[2.35, 2.45).

(b) Upper bound: 2.45÷0.8=3.06252.45 \div 0.8 = 3.0625 → **3 complete shelves**.

(c) Lower bound: 2.35÷0.8=2.93752.35 \div 0.8 = 2.9375 → **2 complete shelves**.

(d) The uncertainty in the measurement means the number of shelves could be either 2 or 3. You cannot rely on the answer without a more precise measurement.
11Problem 11
Answer
£7 623
Full working
Floor area =24.5×18.6= 24.5 \times 18.6. 24×18.6=446.424 \times 18.6 = 446.4; 0.5×18.6=9.30.5 \times 18.6 = 9.3; total =455.7= 455.7 m².

Each tile covers 0.5×0.5=0.250.5 \times 0.5 = 0.25 m².

Tiles needed (no wastage): 455.7÷0.25=1 822.8455.7 \div 0.25 = 1\,822.8 → round up to 1 8231\,823.

With 10% wastage: 1 823×1.1=2 005.31\,823 \times 1.1 = 2\,005.3 → round up to 2 0062\,006 tiles.

Cost: 2 006×£3.80=2 000×3.80+6×3.80=£7 600+£22.80=£7 622.80≈£7 6232\,006 \times \pounds 3.80 = 2\,000 \times 3.80 + 6 \times 3.80 = \pounds 7\,600 + \pounds 22.80 = \pounds 7\,622.80 \approx \mathbf{\pounds 7\,623}.
12Problem 12
Answer
(a) Thursday (b) Friday (c) Monday (d) 365 = 52×7 + 1 (advance 1); 366 = 52×7 + 2 (advance 2)
Full working
(a) From 1 January to 1 February = 31 days. 31=4×7+331 = 4 \times 7 + 3. So the day advances 3 positions: Mon + 3 = **Thursday**.

(b) From 1 January to 1 March = 31 + 29 = 60 days (2024 is a leap year). 60=8×7+460 = 8 \times 7 + 4. Day advances 4 from Monday: Mon → Tue → Wed → Thu → **Friday**.

(c) From 1 January to 1 July = Jan(31) + Feb(29) + Mar(31) + Apr(30) + May(31) + Jun(30) = 182 days. 182=26×7+0182 = 26 \times 7 + 0. The day does **not** advance — it is still **Monday**! (This is a surprising result worth noticing: 182 = exactly 26 weeks.)

(d) A normal year has 365 days. 365=52×7+1365 = 52 \times 7 + 1, so the same date advances by 1 day of the week each non-leap year. A leap year has 366 days. 366=52×7+2366 = 52 \times 7 + 2, so it advances by 2 days. This explains why after a leap year, dates "skip" a day of the week.