(a) Round $7.3864$ to 1 decimal place. [1]
(b) Round $0.04068$ to 2 significant figures. [1]
- (a) $7.4$B1
- (b) $0.041$B1
- Answer: (a) 7.4 (b) 0.041
- Examiner tip: Leading zeros are not significant: the first significant figure of 0.04068 is the 4.
Theatre tickets cost £19.75 each. A school buys 412 tickets.
Work out an estimate for the total cost. [2]
- for rounding both numbers sensibly, e.g. $20 \times 400$ (allow $20 \times 410$)M1
- £8000 (accept £8200 from $20\times410$); an exact answer of £8137 with no rounding scores 0A1
- Answer: £8000
- Examiner tip: An estimate must show the rounded numbers – an exact answer (£8137) with no rounding scores 0.
(a) Work out an estimate for the value of $\dfrac{6.92 \times 48.3}{0.512}$ [2]
(b) Is your answer to part (a) an overestimate or an underestimate? Give a reason for your answer. [1]
- (a) for at least two of the numbers correctly rounded to 1 significant figure, e.g. $\frac{7 \times 50}{0.5}$M1
- (a) $700$A1
- (b) Overestimate, with a reason, e.g. $6.92$ and $48.3$ were rounded up and $0.512$ was rounded down (dividing by a smaller number gives a bigger answer). Accept a reason based on their own rounding if consistent. Do not accept 'overestimate' with no reason.C1
- Answer: (a) 700 (b) overestimate
- Examiner tip: Dividing by 0.5 doubles the number – 7 × 50 ÷ 0.5 is 700, not 175.
(a) The length, $L$ cm, of a rod is 23 cm correct to the nearest centimetre. Write down the error interval for $L$. [2]
(b) A number, $x$, is truncated to 1 decimal place. The result is 4.7. Write down the error interval for $x$. [1]
- (a) for $22.5$ as the lower bound or $23.5$ as the upper boundB1
- (a) $22.5 \leq L < 23.5$ (fully correct with correct inequality signs)B1
- (b) $4.7 \leq x < 4.8$B1
- Answer: (a) $22.5 \le L < 23.5$ (b) $4.7 \le x < 4.8$
- Examiner tip: Truncation only chops digits off, so the lower bound is 4.7 itself – it is not ±0.05.
The diagram shows a rectangle. The length is 8.4 cm and the width is 5.2 cm, both correct to 1 decimal place.
(a) Work out the upper bound for the area of the rectangle. [2]
(b) Work out the lower bound for the perimeter of the rectangle. [2]
- (a) for $8.45 \times 5.25$ (both upper bounds used)M1
- (a) $44.3625$ cm² (accept $44.36$)A1
- (b) for $2 \times (8.35 + 5.15)$ or $8.35 + 8.35 + 5.15 + 5.15$M1
- (b) $27$ cmA1
- Answer: (a) 44.3625 cm² (b) 27 cm
- Examiner tip: Bounds of a measurement to 1 dp are ±0.05, not ±0.5 or ±0.1.
$$T = \frac{x}{y - z}$$
$x = 36$ correct to the nearest integer.
$y = 8.7$ correct to 1 decimal place.
$z = 2.1$ correct to 1 decimal place.
Work out the lower bound for the value of $T$. Give your answer correct to 3 significant figures. You must show your working. [3]
- for any correct bound: $35.5$, $36.5$, $8.65$, $8.75$, $2.05$ or $2.15$B1
- for $\frac{35.5}{8.75 - 2.05}$ (LB of $x$ divided by UB of $y$ minus LB of $z$)M1
- $5.30$ (accept $5.298…$)A1
- Answer: 5.30
- Examiner tip: To make y − z as large as possible use the UPPER bound of y and the LOWER bound of z.
Kofi runs a distance of 250 m, measured correct to the nearest metre. His time is 31.4 seconds, measured correct to the nearest 0.1 second.
By considering bounds, work out his average speed to a suitable degree of accuracy. You must show all your working and give a reason for your final answer. [5]
- for any correct bound: $249.5$, $250.5$, $31.35$ or $31.45$B1
- for upper bound of speed $= \frac{250.5}{31.35}$M1
- for lower bound of speed $= \frac{249.5}{31.45}$M1
- for $7.990…$ and $7.933…$A1
- $8$ m/s, because both the upper and lower bounds round to 8 to 1 significant figure (but not to the same value to 2 sf)C1
- Answer: 8 m/s (UB 7.990…, LB 7.933…)
- Examiner tip: 'Suitable degree of accuracy' means the bounds must agree when rounded: 7.99 and 7.93 differ at 2 sf but both give 8 to 1 sf.
A total of 36 849 people attended a football match.
(a) Round 36 849 to the nearest hundred. [1]
(b) Round 36 849 to 2 significant figures. [1]
(c) Round $0.6097$ to 3 significant figures. [1]
- (a) $36\,800$B1
- (b) $37\,000$ (do not accept 37)B1
- (c) $0.610$ (must show the final zero; $0.61$ is B0)B1
- Answer: (a) 36 800 (b) 37 000 (c) 0.610
- Examiner tip: Rounding to significant figures keeps the size of the number: 36 849 to 2 s.f. is 37 000, not 37, and 0.6097 to 3 s.f. needs the trailing zero in 0.610.
Priya uses her calculator to work out $$\frac{297 \times 0.48}{6.1}$$ Her calculator shows the answer 233.7
Use estimation to decide whether her answer is sensible. You must show your working. [3]
- for rounding at least two of the numbers to 1 significant figure, e.g. $\dfrac{300 \times 0.5}{6}$ (allow $6.1 \approx 6$)M1
- for an estimate of $25$ (accept 24 to 25 from a correct method, e.g. $\frac{300\times0.5}{6}=25$ or $\frac{300\times0.48}{6}=24$)A1
- for a conclusion that her answer is not sensible, with a comparison, e.g. "the estimate is 25, so 233.7 is about 10 times too big" (dependent on M1)C1
- Answer: Estimate $\approx \frac{300\times0.5}{6} = 25$, so 233.7 is not sensible (about 10 times too big; the true value is 23.37…).
- Examiner tip: An estimate alone is not enough – you must state clearly whether her answer is sensible and compare it with your estimate.
(a) A number, $n$, is rounded to 2 significant figures. The result is 3800
Write down the error interval for $n$. [2]
(b) The mass, $m$ kg, of a sack of potatoes is 60 kg correct to the nearest 5 kg.
Write down the error interval for $m$. [2]
- (a) for $3750$ as the lower bound or $3850$ as the upper boundB1
- (a) $3750 \le n < 3850$ (accept $n \ge 3750$ and $n<3850$ written separately)B1
- (b) for $57.5$ as the lower bound or $62.5$ as the upper boundB1
- (b) $57.5 \le m < 62.5$ (both inequality signs correct)B1
- Answer: (a) $3750 \le n < 3850$ (b) $57.5 \le m < 62.5$
- Examiner tip: The bounds are half of the rounding unit either side: 2 s.f. of 3800 means nearest 100 (±50), and nearest 5 kg means ±2.5 kg – not ±0.5.
(a) Use your calculator to work out $$\frac{14.7 + 3.82}{2.6^2 - 1.95}$$ Write down all the figures on your calculator display. [2]
(b) Write your answer to part (a) correct to 2 significant figures. [1]
- (a) for $18.52$ or $4.81$ seenM1
- (a) $3.850311850\ldots$ (accept $3.8503118$ or better)A1
- (b) $3.9$ (ft their answer to (a) if it has more than 2 s.f.)B1
- Answer: (a) 3.850311850… (b) 3.9
- Examiner tip: Work out the top and the bottom separately (or use brackets) – typing it in one line without brackets divides only 3.82 by 2.6².
(a) Work out an estimate for the value of $$\frac{\sqrt{99.2} \times 3.08^2}{0.0487}$$ [2]
(b) Use your answer to part (a) to write down an estimate for the value of $$\frac{\sqrt{99.2} \times 3.08^2}{4.87}$$ [1]
- (a) for rounding at least two of the values correctly, e.g. $\sqrt{100}$ or $10$, $3^2$ or $9$, $0.05$M1
- (a) $1800$ (from $\frac{10 \times 9}{0.05}$; accept $1800$ only from correct rounding)A1
- (b) $18$ (ft their (a) $\div 100$)B1
- Answer: (a) 1800 (b) 18
- Examiner tip: Dividing by 0.05 is the same as multiplying by 20 – many students divide by 20 or multiply by 0.05 and lose the accuracy mark.
Sam is loading a goods lift. The maximum safe load of the lift is 285 kg, correct to the nearest 5 kg.
The table shows the masses of four boxes and of Sam. Each mass is correct to the nearest kilogram.
| Item | Box A | Box B | Box C | Box D | Sam |
|---|---|---|---|---|---|
| Mass (kg) | 48 | 52 | 61 | 39 | 82 |
Sam says, "The total mass is 282 kg, which is less than 285 kg, so the lift is definitely safe."
Use bounds to show that Sam could be wrong. You must show your working. [4]
- for one correct bound of a mass in the table, e.g. $48.5$ or $82.5$, or for $5 \times 0.5$ or $2.5$ seenB1
- for the upper bound of the total mass: $282 + 2.5 = 284.5$ kg (or $48.5+52.5+61.5+39.5+82.5$)M1
- for the lower bound of the safe load: $282.5$ kgM1
- for a correct conclusion with both values, e.g. "upper bound of the total mass 284.5 kg is greater than the lower bound of the safe load 282.5 kg, so the load could exceed the limit"C1
- Answer: Upper bound of total = 284.5 kg; lower bound of safe load = 282.5 kg; 284.5 > 282.5, so the lift could be overloaded.
- Examiner tip: Compare the worst cases: the largest the load could be against the smallest the limit could be – using 285 kg or the rounded total gets no conclusion mark.
The diagram shows a right-angled triangle $ABC$.
$BC = 7.3$ cm, correct to 1 decimal place.
$AC = 12.6$ cm, correct to 1 decimal place.
Work out the upper bound for the size of angle $x$. Give your answer correct to 1 decimal place. You must show your working. [4]
- for $7.35$ or $12.55$ seen (a correct bound used in the right place)B1
- for $\sin x = \dfrac{7.35}{12.55}$ (upper bound of $BC$ ÷ lower bound of $AC$)M1
- for $x = \sin^{-1}\left(\dfrac{7.35}{12.55}\right)$ or $\sin^{-1}(0.5856\ldots)$M1
- $35.8^\circ$ (awrt $35.8$; accept $35.85$)A1
- Answer: $35.8^\circ$ ($\sin^{-1}(7.35/12.55) = 35.849\ldots^\circ$)
- Examiner tip: To make $\sin x$ as large as possible you need the biggest opposite side divided by the smallest hypotenuse – using 7.35 ÷ 12.65 gives the wrong bound.
$$R = \sqrt{a - b}$$
$a = 9.6$, correct to 1 decimal place.
$b = 4.3$, truncated to 1 decimal place.
By considering bounds, work out the value of $R$ to a suitable degree of accuracy. You must show all your working and give a reason for your final answer. [5]
- for $9.55$ or $9.65$ seen, and $4.4$ seen as the upper bound of $b$ (or $4.3 \le b < 4.4$)B1
- for upper bound $\sqrt{9.65 - 4.3}$ ($=\sqrt{5.35}$)M1
- for lower bound $\sqrt{9.55 - 4.4}$ ($=\sqrt{5.15}$)M1
- for $2.313\ldots$ and $2.269\ldots$ (both correct)A1
- $R = 2.3$ with reason, e.g. "both bounds round to 2.3 correct to 2 s.f." (dependent on both M marks)C1
- Answer: UB = 2.313…, LB = 2.269…; both round to 2.3 (2 s.f.), so $R = 2.3$
- Examiner tip: A truncated value can only be bigger than its recorded value, so $b$ lies between 4.3 and 4.4 – not between 4.25 and 4.35.
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