Answer Key

7.4 Decimals & Measurement

Pack A — Answers

#	Question	Answer
1	Write three different decimals between 0.3 and 0.4. Justify that each one lies in that range.	E.g. 0.31, 0.35, 0.39 — each is greater than 0.3 and less than 0.4
2	Round 4.372 to 1 decimal place and name the digit that decided the rounding. Justify.	4.4; the 2nd decimal digit (7) decided — it is 5 or more, so round up
3	The sum 4.37 + 2.85 should equal 7.22. A student writes 6.122. What error did they make? Give the correct answer.	Correct answer is 7.22; the student did not align the decimal points correctly
4	A pencil is 18 cm long. Express its length in mm AND in m, then explain the relationship between the units.	180 mm and 0.18 m; multiplying by 10 converts cm to mm, dividing by 100 converts cm to m
5	A recipe needs 3.67 g of flour. The cook has 8.52 g. How much flour is left over? Show your calculation with decimal points aligned.	4.85 g
6	Find 2.4 × 3 in two different ways: (i) directly, (ii) by treating 2.4 as a whole number, then adjusting. Show both methods give the same answer.	Both methods give 7.2
7	Without fully dividing, decide: will 8.4 ÷ 4 be greater or less than 2? Then calculate to check.	Greater than 2 (since 8 ÷ 4 = 2 and 8.4 > 8); exact answer: 2.1
8	Convert 4.7 cm to mm AND state what operation you performed and why.	47 mm; multiplied by 10 because 1 cm = 10 mm, so cm → mm requires × 10
9	Convert 3500 g to kg AND explain why you divide (not multiply) to convert g to kg.	3.5 kg; divide by 1 000 because kg is a larger unit (there are 1 000 g in each kg)
10	A rectangle has perimeter 9.4 cm. Give two different pairs of whole-number or one-decimal-place side lengths that could produce this perimeter. Justify each.	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)
11	Round 5.6847 to 2 decimal places.	5.68
12	Calculate 12.47 + 8.65 + 3.28.	24.40
13	Calculate 20.30 − 8.67.	11.63
14	Calculate 3.4 × 2.5.	8.5
15	Calculate 15.6 ÷ 1.2.	13
16	Convert 2.35 m into (a) cm and (b) mm.	(a) 235 cm, (b) 2 350 mm
17	A rectangle has area 22.4 cm² and width 3.2 cm. Find its length.	7 cm
18	A journey starts at 9:45 am and takes 2 hours and 35 minutes. What time does it finish?	12:20 pm
19	Convert 2.45 kg to g.	2 450 g
20	Shop A sells juice at £1.20 per bottle. Shop B sells the same juice in packs of 3 for £3.24. Which is better value? Show your working.	Shop B (£1.08 per bottle)
21	Substitute $x = 1.4$ into $3x + 2$.	6.2
22	Estimate $4.87 \times 3.2$ by rounding each factor to 1 decimal place, then calculate the exact answer.	Estimate: 4.9 × 3.2 = 15.68; exact: 15.584
23	A road is 2.4 km long. A second road is 850 m long. What is the total length in metres?	3 250 m
24	A rectangle has perimeter 24.6 cm. Its width is 3.4 cm. Find its length.	8.9 cm
25	Work out $(1.5)^2 + 3.2 \times 4$.	15.05
26	Work out −3.5 + 7.2.	3.7
27	Solve $5x = 8.5$.	$x = 1.7$
28	A train journey takes 1 hours and 45 minutes. How many seconds is that in total?	6 300 s
29	A rectangle has length 8.5 cm and width 4.2 cm. Find its area and its perimeter.	Area = 35.7 cm², Perimeter = 25.4 cm
30	A square has side 4.5 cm. Find its area. Then evaluate $\sqrt{\text{area}}$ and comment on the result.	Area = 20.25 cm²; $\sqrt{20.25} = 4.5$ — exact, equals the side length.
31	A rectangle has perimeter 20.4 cm. Its length is $(2x + 0.6)$ cm and its width is $x$ cm. Find $x$ and calculate the area.	$x = 3.2$ cm; area = 22.4 cm²
32	Substitute $t = 3.0$ into $\dfrac{t^2 - 1}{t + 2}$. Give your answer as a decimal.	1.6
33	The temperature rises from −6.4°C to 2.8°C over 4 hours. Find the rise and the rate of change in °C per hour.	Rise = 9.2°C; rate = 2.3°C/h
34	Write 196 as a product of prime factors in index form, then find $\sqrt{196}$.	$196 = 2^2 \times 7^2$; $\sqrt{196} = 14$
35	Simplify $5x + 3.2 - 2x + 1.7$, then evaluate it for $x = 0.5$.	$3x + 4.9$; value at $x = 0.5$: $6.4$
36	How many pieces of ribbon each 0.35 m long can be cut from a roll of 5.0 m? How much is left over? Give the leftover in cm.	14 pieces; 10 cm left over
37	The HCF of two numbers is 6 and their LCM is 90. One number is 18. Find the other.	30
38	Work out $\dfrac{3.6 \times (10 - 2^2)}{4}$. Give your answer as a decimal.	5.4
39	A tank holds 18.0 litres. Water drains at 2.5 litres/min and is added at 1.0 litres/min simultaneously. How long until the tank is empty?	12 minutes
40	A farmer's field is a rectangle. Its perimeter is 960 m. The length is 3 times the width. Find the area in m², then convert to hectares (1 ha = 10 000 m²).	Area = 43 200 m² = 4.32 ha

Pack B — Answers

#	Question	Answer
1	Write three different decimals between 1.5 and 1.6. Justify that each one lies in that range.	E.g. 1.51, 1.55, 1.58 — each is greater than 1.5 and less than 1.6
2	Round 7.849 to 1 decimal place and name the digit that decided the rounding. Justify.	7.8; the 2nd decimal digit (4) decided — it is less than 5, so round down
3	The sum 6.48 + 3.76 should equal 10.24. A student writes 9.124. What error did they make? Give the correct answer.	Correct answer is 10.24; the student did not align the decimal points correctly
4	A pencil is 24.5 cm long. Express its length in mm AND in m, then explain the relationship between the units.	245 mm and 0.245 m; multiplying by 10 converts cm to mm, dividing by 100 converts cm to m
5	A recipe needs 2.39 g of flour. The cook has 7.04 g. How much flour is left over? Show your calculation with decimal points aligned.	4.65 g
6	Find 3.5 × 4 in two different ways: (i) directly, (ii) by treating 3.5 as a whole number, then adjusting. Show both methods give the same answer.	Both methods give 14.0
7	Without fully dividing, decide: will 7.5 ÷ 5 be greater or less than 2? Then calculate to check.	Greater than 1 (since 5 ÷ 5 = 1 and 7.5 > 5); exact answer: 1.5
8	Convert 3.8 cm to mm AND state what operation you performed and why.	38 mm; multiplied by 10 because 1 cm = 10 mm
9	Convert 6250 g to kg AND explain why you divide (not multiply) to convert g to kg.	6.25 kg; divide by 1 000 because kg is a larger unit
10	A rectangle has perimeter 13.8 cm. Give two different pairs of whole-number or one-decimal-place side lengths that could produce this perimeter. Justify each.	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)
11	Round 3.4251 to 2 decimal places.	3.43
12	Calculate 15.36 + 7.48 + 4.19.	27.03
13	Calculate 30.50 − 14.78.	15.72
14	Calculate 4.2 × 3.5.	14.7
15	Calculate 22.4 ÷ 1.6.	14
16	Convert 4.08 m into (a) cm and (b) mm.	(a) 408 cm, (b) 4 080 mm
17	A rectangle has area 28.8 cm² and width 4.5 cm. Find its length.	6.4 cm
18	A journey starts at 11:20 am and takes 3 hours and 55 minutes. What time does it finish?	3:15 pm
19	Convert 3.072 kg to g.	3 072 g
20	Shop A sells juice at £0.85 per bottle. Shop B sells the same juice in packs of 4 for £3.08. Which is better value? Show your working.	Shop B (£0.77 per bottle)
21	Substitute $x = 2.5$ into $3x + 2$.	9.5
22	Estimate $6.94 \times 2.8$ by rounding each factor to 1 decimal place, then calculate the exact answer.	Estimate: 6.9 × 2.8 = 19.32; exact: 19.432
23	A road is 1.7 km long. A second road is 640 m long. What is the total length in metres?	2 340 m
24	A rectangle has perimeter 31.2 cm. Its width is 4.8 cm. Find its length.	10.8 cm
25	Work out $(2.4)^2 + 1.8 \times 3$.	11.16
26	Work out −8.4 + 5.6.	−2.8
27	Solve $5x = 7.2$.	$x = 1.44$
28	A train journey takes 2 hours and 20 minutes. How many seconds is that in total?	8 400 s
29	A rectangle has length 7.6 cm and width 3.8 cm. Find its area and its perimeter.	Area = 28.88 cm², Perimeter = 22.8 cm
30	A square has side 6.3 cm. Find its area. Then evaluate $\sqrt{\text{area}}$ and comment on the result.	Area = 39.69 cm²; $\sqrt{39.69} = 6.3$ — exact, equals the side length.
31	A rectangle has perimeter 25.2 cm. Its length is $(2x + 1.2)$ cm and its width is $x$ cm. Find $x$ and calculate the area.	$x = 3.8$ cm; area = 33.44 cm²
32	Substitute $t = 4.0$ into $\dfrac{t^2 - 4}{t + 1}$. Give your answer as a decimal.	2.4
33	The temperature rises from −3.5°C to 5.5°C over 5 hours. Find the rise and the rate of change in °C per hour.	Rise = 9.0°C; rate = 1.8°C/h
34	Write 225 as a product of prime factors in index form, then find $\sqrt{225}$.	$225 = 3^2 \times 5^2$; $\sqrt{225} = 15$
35	Simplify $7x + 4.8 - 3x + 2.4$, then evaluate it for $x = 1.5$.	$4x + 7.2$; value at $x = 1.5$: $13.2$
36	How many pieces of ribbon each 0.45 m long can be cut from a roll of 7.2 m? How much is left over? Give the leftover in cm.	16 pieces; 0 cm left over
37	The HCF of two numbers is 4 and their LCM is 60. One number is 12. Find the other.	20
38	Work out $\dfrac{5.4 \times (15 - 3^2)}{6}$. Give your answer as a decimal.	5.4
39	A tank holds 24.0 litres. Water drains at 3.2 litres/min and is added at 0.8 litres/min simultaneously. How long until the tank is empty?	10 minutes
40	A farmer's field is a rectangle. Its perimeter is 720 m. The length is 2 times the width. Find the area in m², then convert to hectares (1 ha = 10 000 m²).	Area = 28 800 m² = 2.88 ha

Problems — Worked Solutions

**The weighing problem.** You have a balance scale and four weights: 1 g, 3 g, 9 g, and 27 g. You can place weights on **either** side of the scale. (a) Can you measure exactly 5 g? Show how. (b) Can you measure exactly 11 g? Show how. (c) What is the heaviest whole-number mass (in grams) you **cannot** measure using these four weights? (d) How many different whole-number masses from 1 g to 40 g can you measure?

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

With weights on both sides, a mass $m$ on the left pan is balanced by some weights on the right pan minus some weights on the left pan (those cancel with $m$). This is equivalent to writing $m$ as a sum $\sum \epsilon_i w_i$ where $\epsilon_i \in \{-1, 0, +1\}$ (left pan, not used, right pan respectively). This is exactly the balanced-ternary representation! (a) $5 = 9 - 3 - 1$: put 9 g on the right, and 3 g + 1 g on the same side as the object (left). ✓ (b) $11 = 9 + 3 - 1$: 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 $1 + 3 + 9 + 27 = 40$ 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 $[1, (3^n-1)/2]$ using $n$ weights; here $n=4$, giving up to $(81-1)/2 = 40$.

**Unit conversion chain.** A snail moves at 0.048 km/h. (a) Convert this speed to metres per minute. (b) Convert to centimetres per second. (c) How far (in mm) does the snail travel in 5 seconds?

Answer

(a) 0.8 m/min; (b) ≈ 1.33 cm/s; (c) ≈ 66.7 mm

(a) $0.048 \text{ km/h} = 0.048 \times 1000 = 48 \text{ m/h}$. Per minute: $48 \div 60 = 0.8$ m/min. (b) $0.8 \text{ m/min} = 80 \text{ cm/min}$. Per second: $80 \div 60 = \tfrac{4}{3} \approx 1.\overline{3}$ cm/s. (c) Distance in 5 s: $\tfrac{4}{3} \times 5 = \tfrac{20}{3} \approx 6.\overline{6}$ cm $= 66.\overline{6}$ mm $\approx 66.7$ mm.

**Perimeter and algebra.** Three rectangles are placed end to end along their lengths to form one large rectangle. Each small rectangle has the same width $w$ cm. Their lengths are 3.2 cm, 4.7 cm, and 5.1 cm. The total perimeter of the large rectangle is 35.4 cm. Find $w$.

Answer

$w = 4.7$ cm

Total length $= 3.2 + 4.7 + 5.1 = 13.0$ cm. Perimeter: $2(13.0 + w) = 35.4$ ⟹ $13.0 + w = 17.7$ ⟹ $w = 4.7$ cm. Check: $2(13.0 + 4.7) = 2 \times 17.7 = 35.4$ ✓.

**Repeating decimal mystery.** Without a calculator, work out the decimal expansions of the following fractions by doing the long division. Then explain the pattern. (a) $\dfrac{1}{7}$ (b) $\dfrac{2}{7}$ (c) $\dfrac{3}{7}$ (d) Without calculating, write down the decimal expansions of $\dfrac{4}{7}$, $\dfrac{5}{7}$, and $\dfrac{6}{7}$. Explain why they follow a cycle.

Answer

(a) $0.\overline{142857}$ (b) $0.\overline{285714}$ (c) $0.\overline{428571}$ (d) The digits 142857 cycle; see working.

(a) Long divide $1.000000 \div 7$: $10 \div 7 = 1$ r 3; $30 ÷ 7 = 4$ r 2; $20 ÷ 7 = 2$ r 6; $60 ÷ 7 = 8$ r 4; $40 ÷ 7 = 5$ r 5; $50 ÷ 7 = 7$ r 1 — back to remainder 1. So $\frac{1}{7} = 0.\overline{142857}$. (b) $\frac{2}{7} = 2 \times \frac{1}{7}$. Doubling each digit of 142857 (carrying as needed): $0.\overline{285714}$. (c) $\frac{3}{7} = 3 \times \frac{1}{7} = 0.\overline{428571}$. (d) The six digits 142857 form a cycle. Each fraction $\frac{n}{7}$ simply starts at a different point in the cycle: - $\frac{4}{7} = 0.\overline{571428}$ - $\frac{5}{7} = 0.\overline{714285}$ - $\frac{6}{7} = 0.\overline{857142}$ Why? Because when you divide by 7, the possible remainders are 1, 2, 3, 4, 5, 6 (in some order). Starting from remainder $n$ gives the decimal for $\frac{n}{7}$, which is just a rotation of the same six-digit block. The number 142857 is called a **cyclic number**.

**Best buy with decimals.** Three brands of olive oil: | Brand | Size | Price | |---|---|---| | Alpha | 500 ml | £2.40 | | Beta | 750 ml | £3.45 | | Gamma | 1.2 litres | £5.16 | Which brand offers the best value for money? Show your method clearly.

Answer

Gamma (£0.43 per 100 ml)

Convert all prices to pence per 100 ml for a fair comparison. - **Alpha**: $240 \div 5 = 48\text{p}$ per 100 ml. - **Beta**: $345 \div 7.5 = 46\text{p}$ per 100 ml. - **Gamma**: $1.2 \text{ L} = 1200 \text{ ml}$; $516 \div 12 = 43\text{p}$ per 100 ml. Gamma is cheapest at 43p per 100 ml → **Gamma is best value**.

**Missing digits.** Each box represents a single decimal digit (0–9): $$3.\square\,7 + \square.4\,\square = 9.03$$ Find the missing digits. Show your working column by column.

Answer

3.57 + 5.46 = 9.03

Label the unknowns: $3.A7 + B.4C = 9.03$. **Hundredths column**: $7 + C$ must end in 3. So $7 + C = 13$ ⟹ $C = 6$ (carry 1 to tenths). **Tenths column**: $A + 4 + 1\text{ (carry)} $ must end in 0 (tenths digit of 9.03 is 0). So $A + 5 = 10$ ⟹ $A = 5$ (carry 1 to ones). **Ones column**: $3 + B + 1\text{ (carry)} = 9$ ⟹ $B = 5$. The missing digits are $A = 5$, $B = 5$, $C = 6$. Full sum: $3.57 + 5.46 = 9.03$ ✓.

**Directed decimals and temperature.** A freezer is at $-18.5$°C. The room is at $23.7$°C. (a) What is the difference in temperature between the room and the freezer? (b) After a power cut, the freezer warms up at 2.4°C per hour. How long until the inside reaches 0°C? Give your answer in hours and minutes. (c) How long until it reaches the room temperature?

Answer

(a) 42.2°C; (b) 7 h 42 min; (c) 17 h 35 min

(a) Difference $= 23.7 - (-18.5) = 23.7 + 18.5 = 42.2$°C. (b) Rise needed: $0 - (-18.5) = 18.5$°C. Time $= 18.5 \div 2.4 = 7.708\overline{3}$ h. $0.708\overline{3} \times 60 = 42.5$ min ≈ **7 h 42 min**. (c) Total rise needed: $42.2$°C. Time $= 42.2 \div 2.4 = 17.58\overline{3}$ h. $0.58\overline{3} \times 60 = 35$ min. So **17 h 35 min**.

**L-shape perimeter.** A shape is made by removing a rectangular notch from the top-right corner of a rectangle. Outer rectangle: 8.4 cm wide, 6.2 cm tall. Notch removed: 3.1 cm wide, 2.5 cm tall (from the top-right corner). Find the perimeter of the L-shaped figure.

Answer

29.2 cm

Trace the six sides of the L-shape clockwise from the bottom-left: 1. Bottom: $8.4$ cm 2. Right side (lower portion only): $6.2 - 2.5 = 3.7$ cm 3. Notch top (going left): $3.1$ cm 4. Notch left edge (going down — this is the inner vertical edge): $2.5$ cm 5. Remaining top (going left): $8.4 - 3.1 = 5.3$ cm 6. Left side (full height, going down): $6.2$ cm Perimeter $= 8.4 + 3.7 + 3.1 + 2.5 + 5.3 + 6.2 = 29.2$ cm ✓.

**Painted cube investigation.** A cube measuring $4 \times 4 \times 4$ is painted blue on all six faces, then cut into $1 \times 1 \times 1$ small cubes. (a) How many small cubes are there in total? (b) How many small cubes have **exactly 3** painted faces? (c) How many small cubes have **exactly 2** painted faces? (d) How many small cubes have **exactly 1** painted face? (e) How many small cubes have **0** painted faces? (f) Verify that your answers to (a)–(e) add up to the total from (a).

Answer

(a) 64 (b) 8 (c) 24 (d) 24 (e) 8 (f) 8+24+24+8 = 64 ✓

(a) $4 \times 4 \times 4 = \mathbf{64}$ 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 $4 - 2 = 2$ interior positions. A cube has 12 edges, so $12 \times 2 = \mathbf{24}$ cubes. (d) **1 painted face** — face cubes (not on any edge). Each face of the large cube has a $(4-2) \times (4-2) = 2 \times 2 = 4$ inner grid. 6 faces × 4 = $\mathbf{24}$ cubes. (e) **0 painted faces** — interior cubes. These form a $(4-2)^3 = 2^3 = \mathbf{8}$ cube inside. (f) $8 + 24 + 24 + 8 = 64$ ✓. *Extension for the curious:* For an $n \times n \times n$ cube: corners always = 8; edges = $12(n-2)$; faces = $6(n-2)^2$; interior = $(n-2)^3$.

**Rounding and bounds.** A plank of wood is measured as 2.4 m to the nearest 0.1 m. (a) Write the lower and upper bounds of the true length. (b) Planks are cut into shelves each exactly 0.8 m long. Using the upper bound, how many complete shelves can be cut? (c) Using the lower bound, how many complete shelves can be cut? (d) What does this tell you about the reliability of the answer to (b)?

Answer

(a) 2.35 m ≤ length < 2.45 m; (b) 3 shelves; (c) 2 shelves; (d) result is unreliable — it differs by 1 shelf

(a) Rounded to nearest 0.1 m: true length lies in $[2.35, 2.45)$. (b) Upper bound: $2.45 \div 0.8 = 3.0625$ → **3 complete shelves**. (c) Lower bound: $2.35 \div 0.8 = 2.9375$ → **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.

**Tiling a hall.** A school hall floor measures 24.5 m by 18.6 m. It is being tiled with square tiles each 0.5 m × 0.5 m, costing £3.80 each. A 10% wastage allowance must be added to the number of tiles ordered. Calculate the total cost of the tiles. Give your answer to the nearest pound.

Answer

£7 623

Floor area $= 24.5 \times 18.6$. $24 \times 18.6 = 446.4$; $0.5 \times 18.6 = 9.3$; total $= 455.7$ m². Each tile covers $0.5 \times 0.5 = 0.25$ m². Tiles needed (no wastage): $455.7 \div 0.25 = 1\,822.8$ → round up to $1\,823$. With 10% wastage: $1\,823 \times 1.1 = 2\,005.3$ → round up to $2\,006$ tiles. Cost: $2\,006 \times \pounds 3.80 = 2\,000 \times 3.80 + 6 \times 3.80 = \pounds 7\,600 + \pounds 22.80 = \pounds 7\,622.80 \approx \mathbf{\pounds 7\,623}$.

**Calendar arithmetic.** 1 January 2024 is a Monday. (a) What day of the week is 1 February 2024? (January has 31 days.) (b) What day of the week is 1 March 2024? (2024 is a leap year, so February has 29 days.) (c) What day of the week is 1 July 2024? (d) Explain why the day of the week advances by 1 for each non-leap year, but by 2 for each leap year.

Answer

(a) Thursday (b) Friday (c) Monday (d) 365 = 52×7 + 1 (advance 1); 366 = 52×7 + 2 (advance 2)

(a) From 1 January to 1 February = 31 days. $31 = 4 \times 7 + 3$. 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). $60 = 8 \times 7 + 4$. 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. $182 = 26 \times 7 + 0$. 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. $365 = 52 \times 7 + 1$, so the same date advances by 1 day of the week each non-leap year. A leap year has 366 days. $366 = 52 \times 7 + 2$, so it advances by 2 days. This explains why after a leap year, dates "skip" a day of the week.