What is N16: Upper and lower bounds?

A measured or rounded value stands for a range of possible true values. The lower and upper limits are usually half a rounding unit below and above the stated value, with the upper limit excluded.

How many marks are N16 questions worth?

In verified past-paper appearances, N16 questions on Edexcel GCSE Maths 1MA1 have been worth 2–5 marks.

Which past exam papers has N16 appeared in?

N16 has appeared in 10 verified past-paper questions indexed on this page, with direct links to the official question paper and mark scheme for each.

N16

Apply and interpret limits of accuracy, including upper and lower bounds

Upper and lower bounds

Worked answers, methods and verified real exam appearances for N16 on Edexcel GCSE Maths 1MA1.

Explanation

A measured or rounded value stands for a range of possible true values. The lower and upper limits are usually half a rounding unit below and above the stated value, with the upper limit excluded.
Use these limits to find a maximum possible error or to decide whether a claimed result is possible. For a sum, the smallest total uses all lower limits and the greatest possible total approaches all upper limits.
Higher-tier bounds calculations also choose numerator and denominator limits deliberately, but Foundation questions can require interpreting accuracy and maximum error.
Never treat rounded inputs as exact.
Examiners expect the chosen limits to be written before the calculation.

Worked example

Higher tier: Two lengths are $4.2$ cm and $3.7$ cm, each correct to the nearest $0.1$ cm. Could their exact total be less than $7.8$ cm?

1.Write the lower limits: the lengths are at least $4.15$ cm and $3.65$ cm.
2.Find the smallest possible total: $4.15+3.65=7.80$ cm.
3.Compare with $7.8$ cm: no exact total can be smaller.

Answer: No; the exact total is at least $7.80$ cm.

Common mistakes

Don't use the displayed rounded values as though they were exact.
Don't include an upper limit even though that endpoint rounds to the next displayed value.
Don't use upper limits when the question asks for the smallest possible total.

Exam tip

Write “lower” or “upper” beside each substituted value so the examiner can see why it gives the required extreme.

Worked practice

Tier 1 · Easy

A length is recorded as

12

cm to the nearest centimetre. Write down the maximum possible rounding error.

(1)

(Total for Question 1 is 1 mark)

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Mark scheme
1	$0.5$ cm	1	The rounding unit is $1$ cm, so the true length can differ from the recorded value by half of this: $1\div2=0.5$ cm.

Tier 3 · Hard

Higher only: A scale records the mass of each of

8

identical boxes as

2.4

kg to the nearest

0.1

kg. Could the exact total mass of the boxes be

20

kg? Justify your answer.

(3)

(Total for Question 2 is 3 marks)

Mark scheme

Mark scheme for question 2
Question	Answer	Mark	Mark scheme
2	No. The exact total must be less than $19.6$ kg.	3	A displayed mass of $2.4$ kg means one box has mass less than $2.45$ kg. Therefore $8$ boxes have total mass less than $8\times2.45=19.6$ kg. Since $20>19.6$ , a total of $20$ kg is impossible.

Tier 1 · Easy

Higher only: Each of

6

identical tubs has a mass of

260

g correct to the nearest

10

g. Work out the upper bound for their total mass.

(2)

(Total for Question 3 is 2 marks)

Mark scheme

Mark scheme for question 3
Question	Answer	Mark	Mark scheme
3	$1590$ g	2	The upper bound for one tub is $260+5=265$ g. The upper bound for the total mass is $6\times265=1590$ g.

Tier 2 · Standard

Higher only: A parcel has mass

7.6

kg correct to the nearest

0.1

kg. Could its exact mass be

7.66

kg? Give a reason.

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Answer	Mark	Mark scheme
4	No; the exact mass is less than $7.65$ kg.	2	The upper bound is $7.6+0.05=7.65$ kg and is not included. Since $7.66>7.65$ , the exact mass cannot be $7.66$ kg.

Tier 3 · Hard

Higher only: A laser measure records a length as

1.372

m to the nearest

0.001

m. A tape measure records the same length as

137

cm to the nearest centimetre. Which measure has the smaller maximum possible error? Give a reason.

(3)

(Total for Question 5 is 3 marks)

Mark scheme

Mark scheme for question 5
Question	Answer	Mark	Mark scheme
5	The laser measure; its maximum possible error is $0.5$ mm, compared with $5$ mm for the tape measure.	3	The laser measure has maximum error $0.001\div2=0.0005$ m, which is $0.5$ mm. The tape measure has maximum error $1\div2=0.5$ cm, which is $5$ mm. Therefore the laser measure has the smaller maximum possible error.

Tier 2 · Standard

A jar is marked as having mass

500

g. The maximum possible rounding error is

5

g. Write down the unit to which the mass was rounded and write the error interval for its exact mass

m

(3)

(Total for Question 6 is 3 marks)

Mark scheme

Mark scheme for question 6
Question	Answer	Mark	Mark scheme
6	Rounded to the nearest $10$ g; $495\leq m<505$	3	The rounding unit is twice the maximum possible error, so it is $2\times5=10$ g. The exact mass can be $5$ g below $500$ but must be less than $5$ g above it, giving $495\leq m<505$ .

Tier 3 · Hard

Higher only: An alloy bar has mass

1.8

kg correct to the nearest

0.1

kg. It is cut into

7

blocks of equal mass. Work out the upper bound for the mass of one block. Give your answer exactly in grams.

(4)

(Total for Question 7 is 4 marks)

Mark scheme

Mark scheme for question 7
Question	Answer	Mark	Mark scheme
7	$\dfrac{1850}{7}$ g (or $264\dfrac{2}{7}$ g)	4	The upper bound for the bar's mass is $1.85$ kg, which is $1850$ g. Dividing this upper bound equally between $7$ blocks gives $1850\div7=\dfrac{1850}{7}$ g for the upper bound of one block.

Tier 3 · Hard

Higher only: Two boards have recorded lengths of

8.4

m and

3.7

m, each correct to the nearest

0.1

m. A shelf design requires the first board to be at least

4.65

m longer than the second. Is this guaranteed? Justify your answer using bounds.

(4)

(Total for Question 8 is 4 marks)

Mark scheme

Mark scheme for question 8
Question	Answer	Mark	Mark scheme
8	No; the lower bound for the difference is $4.60$ m, which is $0.05$ m below $4.65$ m.	4	Use the lower bound of the first board and the upper bound of the second board. The lower bound for the difference is $8.35-3.75=4.60$ m. This is $0.05$ m less than the required $4.65$ m, so the required difference is not guaranteed.

Tier 3 · Hard

Higher only: A machine's output is recorded as

840

parts, correct to the nearest

10

parts. Its running time is recorded as

3.2

hours, correct to the nearest

0.1

hour. Could its exact output rate be greater than

270

parts per hour? You must show all your working.

(4)

(Total for Question 9 is 4 marks)

Mark scheme

Mark scheme for question 9
Question	Answer	Mark	Mark scheme
9	No; $270\times3.15=850.5>845$ .	4	The greatest possible output is less than $845$ parts and the least possible time is $3.15$ hours. A rate of $270$ parts per hour for $3.15$ hours would require $270\times3.15=850.5$ parts. Since $850.5>845$ , even the greatest possible output is too small, so the exact rate cannot be greater than $270$ parts per hour.

Verified exam appearances

Series	Paper	Question	Marks	Calculator	Tier	Links
2024-11	3F	Q23	2	Allowed	Foundation	QP MS
2024-11	2H	Q19	5	Allowed	Higher	QP MS
2022-11	2H	Q23	5	Allowed	Higher	QP MS
2022-06	3H	Q16	3	Allowed	Higher	QP MS
2019-11	2H	Q20	4	Allowed	Higher	QP MS
2024-06	3H	Q19	3	Allowed	Higher	QP MS
2023-06	2H	Q19	3	Allowed	Higher	QP MS
2019-06	3H	Q19	5	Allowed	Higher	QP MS
2021-11	2H	Q21	4	Allowed	Higher	QP MS
2023-11	3H	Q22	4	Allowed	Higher	QP MS

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