Rounding, estimation, bounds & standard form – IGCSE Maths exam questions with mark scheme

1Grade 1–3FoundationCalculator4 marks

(a) Round 47 862 to the nearest hundred. [1]

(b) Round 0.006 384 9 to 3 significant figures. [1]

(d) Round 795 to 1 significant figure. [1]

(a) 47 900 caoB1
(b) 0.006 38 cao (do not accept 0.0064 or 0.006 39)B1
(c) 3.90 cao (must show the final zero; 3.9 is B0)B1
(d) 800 caoB1
Answer: (a) 47 900 (b) 0.006 38 (c) 3.90 (d) 800
Examiner tip: Leading zeros are never significant, and a rounded answer to 2 d.p. must show both decimal places even when the last one is 0.

2Grade 1–3Calculator3 marks

Work out an estimate for the value of

$$\frac{612\times4.87}{0.193}$$

Show your working clearly. [3]

for rounding at least two of the numbers to 1 significant figure, e.g. 600, 5, 0.2M1
for $\dfrac{3000}{0.2}$ or $600\times25$ or $3000\times5$ oe (all three rounded to 1 s.f.)M1
15 000 (accept answer from correct 1 s.f. rounding only; the exact value 15 442.69… gets no marks)A1
Answer: 15 000
Examiner tip: Round every number to 1 significant figure first, then remember that dividing by 0.2 is the same as multiplying by 5.

3Grade 4–5Calculator3 marks

The length, $L$ metres, of a wooden plank is 2.4 m correct to the nearest 0.1 m.

(a) Complete the error interval for $L$.

.................. $\leqslant L <$ ..................

[2]

The mass of a parcel is 650 g correct to the nearest 50 g.

(b) Write down the lower bound for the mass of the parcel. [1]

(a) 2.35 in the first boxB1
(a) 2.45 in the second box (accept $2.44\dot{9}$; B1 only for both values in the wrong boxes)B1
(b) 625 g (do not accept 645 or 649.5)B1
Answer: (a) $2.35\leqslant L<2.45$ (b) 625 g
Examiner tip: The bounds are half of the rounding unit either side: for ‘nearest 50 g’ that is ±25 g, not ±5 g.

Stuck on Rounding, estimation, bounds & standard form?

Get it sorted in a free 30-minute session

We work through your weakest questions on this topic on Zoom. I reply the same day.

Free 30-min session on this topic Send me 5 questions a day

4Grade 4–5Calculator4 marks

(a) Write 0.000 307 in standard form. [1]

(b) Write $5.2\times10^{4}$ as an ordinary number. [1]

(a) $3.07\times10^{-4}$B1
(b) 52 000B1
(c) for $24\times10^{3}$ or 24 000 or $3\times8=24$ and $10^{5}\times10^{-2}=10^{3}$ seenM1
(c) $2.4\times10^{4}$ caoA1
Answer: (a) $3.07\times10^{-4}$ (b) 52 000 (c) $2.4\times10^{4}$
Examiner tip: $24\times10^3$ is not in standard form: the number in front must be at least 1 and less than 10.

5Grade 4–5Calculator5 marks

The table shows the approximate areas of four countries.

Country	Area (km$^2$)
Algeria	$2.38\times10^{6}$
Egypt	$1.00\times10^{6}$
Malta	$3.16\times10^{2}$
Sri Lanka	$6.56\times10^{4}$

(a) Write the countries in order of area. Start with the smallest. [1]

(b) How many times larger is the area of Algeria than the area of Sri Lanka? Give your answer correct to the nearest whole number. [2]

(a) Malta, Sri Lanka, Egypt, AlgeriaB1
(b) for $\dfrac{2.38\times10^{6}}{6.56\times10^{4}}$ oe or 36.28…M1
(b) 36 caoA1
(c) for 1 000 000 + 65 600 or 1 065 600 oeM1
(c) $1.0656\times10^{6}$ (accept $1.07\times10^{6}$ if 1 065 600 seen)A1
Answer: (a) Malta, Sri Lanka, Egypt, Algeria (b) 36 (c) $1.0656\times10^{6}$ km$^2$
Examiner tip: When adding numbers in standard form, convert to ordinary numbers (or the same power of 10) first; you cannot just add the front numbers.

6Grade 6–7HigherCalculator3 marks

A rectangular rug has an area of 18.5 m$^2$, correct to 3 significant figures.

The width of the rug is 3.6 m, correct to 2 significant figures.

Calculate the lower bound for the length of the rug.
Show your working clearly and give your answer correct to 3 significant figures. [3]

for one correct bound seen: 18.45 or 18.55 or 3.55 or 3.65B1
for $\dfrac{18.45}{3.65}$M1
5.05 (from 5.0547…)A1
Answer: 5.05 m
Examiner tip: To make a quotient as small as possible, divide the lower bound of the top by the UPPER bound of the bottom.

7Grade 8–9HigherCalculator5 marks

A cyclist rides a distance of 1500 m, correct to the nearest 10 m.

She takes 263 seconds, correct to the nearest second.

By considering bounds, work out the average speed of the cyclist to a suitable degree of accuracy.
You must show all your working and give a reason for your answer. [5]

for at least two correct bounds seen from 1495, 1505, 262.5, 263.5B1
for upper bound of speed $\dfrac{1505}{262.5}$ (= 5.733…)M1
for lower bound of speed $\dfrac{1495}{263.5}$ (= 5.673…)M1
for 5.73(3…) and 5.67(3…) both correct (values given to at least 3 s.f.)A1
5.7 m/s with reason, e.g. both bounds round to 5.7 to 1 d.p. (or 2 s.f.) but do not agree to 2 d.p.C1
Answer: 5.7 m/s (both bounds 5.733… and 5.673… round to 5.7)
Examiner tip: A ‘suitable degree of accuracy’ is the most precise rounding on which the upper and lower bounds agree — you must state that they agree.

8Grade 8–9HigherCalculator4 marks

$x=4\times10^{2n}$ and $y=5\times10^{n-1}$, where $n$ is an integer.

(a) Find $xy$. Give your answer in standard form in terms of $n$. [2]

(b) Find $\dfrac{x}{y}$. Give your answer in standard form in terms of $n$. [2]

(a) for $20\times10^{3n-1}$ oe (multiplying 4×5 and adding the powers)M1
(a) $2\times10^{3n}$A1
(b) for $0.8\times10^{n+1}$ oe (dividing 4÷5 and subtracting the powers)M1
(b) $8\times10^{n}$A1
Answer: (a) $2\times10^{3n}$ (b) $8\times10^{n}$
Examiner tip: After multiplying or dividing the front numbers, adjust the power of 10 so the front number is between 1 and 10 (20 → 2 means add 1 to the power; 0.8 → 8 means subtract 1).

9Grade 6–7HigherCalculator4 marks

The diagram shows a right-angled triangle $ABC$.

$AB = 7.4$ cm, correct to 1 decimal place.
$BC = 4.6$ cm, correct to 1 decimal place.
Angle $ABC = 90^\circ$

(a) Write down the upper bound for the length of $BC$. [1]

(b) Calculate the lower bound for the size of angle $ACB$.
Give your answer correct to 1 decimal place. [3]

(a) 4.65 (accept 4.649…)B1
(b) for 7.35 seen (lower bound of $AB$)B1
(b) for $\tan ACB=\dfrac{7.35}{4.65}$ oeM1
(b) 57.7 (from 57.680…)A1
Answer: (a) 4.65 cm (b) $57.7^\circ$
Examiner tip: $\tan C=\dfrac{AB}{BC}$, so the smallest angle uses the smallest opposite side and the LARGEST adjacent side.

10Grade 6–7HigherCalculator4 marks

Priya says that $12\times10^{-8}$ is written in standard form.

(a) Explain why Priya is wrong. [1]

The diameter of a virus is $1.2\times10^{-7}$ metres.

Some of these viruses are placed in a straight line, touching each other. The line has a total length of 3 mm.

(b) Work out how many viruses are in the line. Give your answer in standard form. [3]

(a) e.g. 12 is not between 1 and 10 (the first number must be at least 1 and less than 10); accept it should be $1.2\times10^{-7}$C1
(b) for converting 3 mm to metres: 0.003 or $3\times10^{-3}$M1
(b) for $\dfrac{3\times10^{-3}}{1.2\times10^{-7}}$ oe (or $2.5\times10^{4}$ or 25 000 seen)M1
(b) $2.5\times10^{4}$ caoA1
Answer: (a) 12 is not between 1 and 10 (b) $2.5\times10^{4}$
Examiner tip: Always put both lengths in the same unit before dividing — most lost marks here come from leaving 3 mm as 3.

Stuck on this topic?

Book a free 30-minute session on WhatsApp and we'll find exactly which step is costing you the marks. Or build a timed paper on it.

Free 30-minute session