Units, estimation, rounding and bounds – Year 10 GCSE maths worksheet answers to download as PDF

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1. Answer to Question 1

   1. \(3.5\times1000=\) 3500 m There are 1000 metres in a kilometre, so we multiply by 1000. So 3.5 km is 3500 m.
   2. \(2500\div1000=\) 2.5 litres There are 1000 ml in a litre, so we divide by 1000. So 2500 ml is 2.5 litres.
   3. \(0.8\times1000=\) 800 kg One tonne is 1000 kg, so we multiply by 1000. So 0.8 tonnes is 800 kg.
   4. \(1\text{ m}^2=100\times100=10\,000\text{ cm}^2\), so \(4.5\times10\,000=\) 45 000 cm² For areas the conversion factor is squared, so we multiply by 10 000, not 100.

2. Answer to Question 2

   1. 4.27 (2 d.p.) and 4.27 (3 s.f.) To 2 d.p., the third decimal digit 7 makes the 6 round up to 7. To 3 s.f. we keep 4, 2 and 6, and the next digit 7 rounds up.
   2. The first two significant figures are 4 and 5, next digit 6, so 0.0046 The zeros at the start are not significant.
   3. 25 500 (3 s.f.) and 25 500 (nearest 100) To 3 s.f. the fourth digit 8 makes the 4 round up to 5.
   4. 8.0 (1 d.p.) and 8.0 (2 s.f.). The zero shows the accuracy. Rounding 7.996 makes the 9s carry over, giving 8.0.

3. Answer to Question 3

   1. \(5\times20=\) 100 Round 4.8 to 5 and 19.7 to 20, then multiply. The estimate is 100.
   2. \(\dfrac{300}{0.5}=\) 600 Round 304 to 300 and 0.52 to 0.5. Dividing by 0.5 is the same as multiplying by 2, so the estimate is 600.
   3. \(\dfrac{60\times0.4}{0.2}=\) 120 Round 58.7 to 60, 0.41 to 0.4 and 0.19 to 0.2. Then \(60\times0.4=24\) and \(24\div0.2=120\). The estimate is 120.
   4. \(\sqrt{81}+\sqrt{25}=9+5=\) 14 82 is close to the square number 81 and 26 is close to 25, so the square roots are about 9 and 5. The estimate is 14.

4. Answer to Question 4

   1. \(7.25\le x\lt 7.35\) The value was rounded to 1 decimal place, so it can be up to 0.05 below or above 7.3.
   2. \(445\le x\lt 455\) Rounding to 2 significant figures means rounding to the nearest 10, so the value can be up to 5 below or above 450.
   3. \(11.5\le x\lt 12.5\) Rounding to the nearest whole number means the value can be up to 0.5 below or above 12.
   4. \(3.4\le x\lt 3.5\) (truncating cuts off the digits, so the value is at least 3.4) So the error interval for a truncated value starts at the value itself and goes up to the next tenth.

5. Answer to Question 5

   1. Half of 0.1 is 0.05: 8.35 and 8.45 cm The lower bound is \(8.4-0.05\) and the upper bound is \(8.4+0.05\).
   2. Half of 10 is 5: 245 and 255 g The lower bound is \(250-5\) and the upper bound is \(250+5\).
   3. 12.25 and 12.35 s The time was measured to the nearest 0.1 s, so the bounds are 0.05 s either side of 12.3 s.
   4. The accuracy is 0.1, half is 0.05: 2.45 and 2.55 litres

6. Answer to Question 6

   1. \(15\div60=0.25\), so 2.25 hours 15 minutes is a quarter of an hour, which is 0.25. So 2 hours 15 minutes is 2.25 hours.
   2. \(35\div5=7\) lots of 5 miles, \(7\times8=\) 56 km 35 miles is 7 lots of 5 miles, so it is 7 lots of 8 km. So 35 miles is about 56 km.
   3. \(3.5\times2.2=\) 7.7 lb Each kilogram is about 2.2 pounds, so we multiply by 2.2. So 3.5 kg is about 7.7 lb.
   4. \(0.4\times60=24\), so 1 hour 24 minutes Change the decimal part into minutes by multiplying by 60. So 1.4 hours is 1 hour 24 minutes.

7. Answer to Question 7

   1. \(50\text{ mm}\div500=\) 0.1 mm Change 5 cm into 50 mm, then divide by 500 sheets. One sheet is 0.1 mm thick.
   2. 4.5 cm and 5.5 cm (4.5 ≤ t < 5.5). The thickness was measured to the nearest centimetre, so it can be up to 0.5 cm below or above 5 cm.
   3. Least: \(45\div500=\) 0.090 mm. Greatest: \(55\div500=\) 0.110 mm. Divide each bound in millimetres by the 500 sheets.
   4. \(120\times5=600\) cm \(=\) 6 m. Each pack is about 5 cm thick, so the stack is about 6 m high.

8. Answer to Question 8

   1. \(100\div12.3=\) 8.13 m/s
   2. Time between 12.25 s and 12.35 s.
   3. Greatest: \(100\div12.25=8.163\) m/s. Least: \(100\div12.35=8.097\) m/s.
   4. To 2 s.f. the bounds are 8.2 and 8.1, which differ, so no. To 1 significant figure both bounds round to 8 m/s, so her speed is 8 m/s to 1 s.f.

9. Answer to Question 9

   1. \(4.2\times3.8=\) 15.96 m² Multiply the length by the width. The area is 15.96 m².
   2. Lower: \(4.15\times3.75=15.5625\). Upper: \(4.25\times3.85=16.3625\). So 15.56 m² and 16.36 m² (2 d.p.). For the lower bound multiply the two lower bounds; for the upper bound multiply the two upper bounds.
   3. \(16.3625\times18=\) £294.53 (nearest penny). The greatest cost uses the upper bound of the area.
   4. The greatest possible area is 16.3625 m², which is less than 16.4, so yes, it is certainly enough.

10. Answer to Question 10

    1. Round each number to 1 significant figure: 12 sections becomes 10, 480 people becomes 500, and 9 rows stays 9. Multiply: \(10\times500\times9=45\,000\). The estimate of the crowd is 45 000 people.
    2. Put the decimal point after the first digit: \(45\,000=4.5\times10\,000\). In standard form the estimate is \(4.5\times10^4\).
    3. Divide the estimate by the capacity: \(\dfrac{4.5\times10^4}{5.2\times10^4}=\dfrac{4.5}{5.2}=0.865\). Multiply by 100: about 87% of the seats might be filled.
    4. The accurate value is \(12\times480\times9=51\,840\). The estimate is an underestimate, because 12 was rounded down to 10 and this rounding has a bigger effect than rounding 480 up to 500.

11. Answer to Question 11

    1. \(2\times454=\) 908 g Multiply the number of pounds by 454 g. The flour is 908 g.
    2. \(6\times28.4=\) 170.4 g Multiply the number of ounces by 28.4 g. The butter is 170.4 g.
    3. \(10\times28.4=\) 284 ml \(=\) 0.284 litres Multiply by 28.4 ml, then divide by 1000 to change into litres. The milk is 284 ml.
    4. \(1000\div454=\) 2.2 lb Divide the grams by 454 to change into pounds. So 1 kg is about 2.2 lb.

12. Answer to Question 12

    1. 240 cm by 180 cm There are 100 cm in a metre, so multiply each length by 100.
    2. \(240\div15=16\) across and \(180\div15=12\) up, so \(16\times12=192\) tiles. Yes, they fit exactly.
    3. \(192\div20=9.6\), so 10 boxes (200 tiles). We round up because 9 boxes hold only 180 tiles, which is not enough.
    4. Cost \(=10\times24=\) £240. Left over \(=8\) tiles, which is \(\dfrac8{200}=\) 4%.

13. Answer to Question 13

    1. \(4.5\le d\lt 5.5\) ml
    2. Least: \(4.5\times21=94.5\) ml. Greatest: \(5.5\times21=115.5\) ml. The total is 21 doses, because 3 doses a day for 7 days is \(3\times7=21\).
    3. No. The greatest total of 115.5 ml is more than 100 ml, so one bottle might not be enough.
    4. Spoons are made to measure exact standard amounts, so there is less error.

14. Answer to Question 14

    1. \(28\div8=3.5\), so 4 minibuses (round up: half a bus cannot be used).
    2. \(50\div12=4.17\), so 5 tins (round up to have enough).
    3. Money is paid to the nearest penny: £47.38.
    4. \(9.0\times10^6\) (the zero shows 2 significant figures).

15. Answer to Question 15

    1. \(6.45\le a\lt 6.55\) and \(2.35\le b\lt 2.45\). Each value was rounded to 1 decimal place, so the bounds are 0.05 either side.
    2. Upper bound of \(a+b=6.55+2.45=\) 9.00. Lower bound of \(a-b=6.45-2.45=\) 4.00. For the lower bound of a difference, subtract the upper bound of \(b\) from the lower bound of \(a\).
    3. Upper bound of \(\dfrac ab=\dfrac{\text{UB } a}{\text{LB } b}=\dfrac{6.55}{2.35}=\) 2.787. To make a fraction as large as possible, use the largest top and the smallest bottom.
    4. Lower bound of \(ab=6.45\times2.35=\) 15.1575. Multiply the two lower bounds to get the smallest possible product.

16. Answer to Question 16

    1. Distance 115 to 125 km. Time 1.45 to 1.55 h.
    2. Greatest \(=\dfrac{125}{1.45}=\) 86.2 km/h.
    3. Least \(=\dfrac{115}{1.55}=\) 74.2 km/h.
    4. The bounds round to 90 and 70 to 1 s.f., which are different, so the speed cannot be given to 1 s.f. with certainty.

17. Answer to Question 17

    1. Diameter 9.5 to 10.5 cm, radius 4.75 to 5.25 cm.
    2. Upper: \(\pi\times5.25^2=27.5625\pi=\) 86.6 cm². Lower: \(\pi\times4.75^2=22.5625\pi=\) 70.9 cm².
    3. The area could be anywhere from 70.9 to 86.6 cm², so the figure 78.5 (3 s.f.) suggests much more accuracy than is justified.
    4. 70.9 and 86.6 differ in the first digit (70 and 90 to 1 s.f.), so not even 1 significant figure can be quoted with certainty.

18. Answer to Question 18

    1. Rounded value 3.14 (\(3.14159\to3.14\)). Error interval \(3.135\le x\lt 3.145\).
    2. Truncated value 3.14. Error interval \(3.14\le x\lt 3.15\).
    3. Both give 3.14, which differs from \(\pi\) by 0.0016, so they are equally close here.
    4. Truncating 4.99 gives 4.9, whereas rounding gives 5.0. Truncating always cuts the number off and never rounds up.

19. Answer to Question 19

    1. \(|24.6-24.63|=\) 0.03 cm The absolute error is the difference between the measured value and the true value. The error is 0.03 cm.
    2. \(\dfrac{0.03}{24.63}\times100=\) 0.12% Divide the error by the true length and multiply by 100. The percentage error is 0.12%.
    3. 0.05 cm The rod was measured to the nearest 0.1 cm, so the measurement can be wrong by up to half of 0.1 cm, which is 0.05 cm.
    4. \(\dfrac{0.05}{24.6}\times100=\) 0.20% Divide the greatest error by the measured length and multiply by 100. The greatest percentage error is 0.20%.

20. Answer to Question 20

    1. 3.445 to 3.455 cm.
    2. Lower: \(3.445^2=11.868025\). Upper: \(3.455^2=11.937025\). So 11.8680 to 11.9370 cm². The bounds of the area are the squares of the bounds of the side.
    3. Both round to 11.9 to 3 s.f.: 3 significant figures. (To 4 s.f. they are 11.87 and 11.94.)
    4. The area is 11.9 cm² to 3 s.f.