Unit 14 Section 1 : Rounding

There are three main ways to round numbers:

(i) to the nearest 10, 100, 1000, etc;
(ii) to a certain number of significant figures;
(iii) to a certain number of decimal places.

Note that a measured length such as '12 cm to the nearest cm' means that the actual length lies between 11.5 cm and 12.5 cm.

By convention, we normally round 0.5 up to the next whole number, so in fact,

11.5 cm ≤ actual length < 12.5 cm

We call 11.5 the lower bound and 12.5 the upper bound. We can also write the upper bound as

12.4999 . . . . or 12.49

where the dot above the 9 means that it is repeated indefinitely, or recurs.

It is important to see that 12.49 is not the upper bound, as, for example, the length could have been 12.498.

Example 1

A football match is watched by 56 742 people. Write this number correct to the nearest,

(a)10 000,
60 000
(b)1000,
57 000
(c)10.
56 740

Example 2

Write each of the following numbers correct to 3 significant figures:

(a)47 316
47 300
(b)303 971
304 000
(c)20.453
20.5
(d)0.004368
0.00437

Example 3

Write each of the following numbers correct to the number of decimal places stated:

(a)0.3741 to 2 d.p.
0.37
(b)3.8451 to 2 d.p.
3.85
(c)142.8315 to 1 d.p.
142.8
(d)0.000851 to 4 d.p.
0.0009

Example 4

State the upper and lower bounds for each of the following quantities and write an inequality for the actual value in each case.

(a)

4 mm to the nearest mm.

Upper bound = 4.5 mm
Lower bound = 3.5 mm

3.5 mm ≤ actual value < 4.5 mm

(b)

15 kg to the nearest kg.

Upper bound = 15.5 kg
Lower bound = 14.5 kg

14.5 kg ≤ actual value < 15.5 kg

(c)

4.56 m to the nearest cm.

Upper bound = 4.565 mm
Lower bound = 4.555 mm

4.555 m ≤ actual value < 4.565 m

Exercises

Question 1

Round each of the following numbers to the nearest 100:

(a)108
(b)199
(c)3471
(d)59
(e)33
(f)451
Question 2

Round the number 4 765 173 to:

(a)the nearest million,
(b)the nearest 10,
(c)the nearest 1000,
(d)the nearest 100.
Question 3

Write each of the following numbers correct to 3 significant figures:

(a)37 412
(b)84 563
(c)261.42
(d)0.3684
(e)0.002615
(f)0.0025713
(g)3.6213
(h)4.0071
(i)18.3071
Question 4

Write each of the following numbers correct to 2 decimal places:

(a)7.431
(b)8.269
(c)4.7135
(d)11.925
(e)24.8603
(f)44.0019
Question 5

Write each of the following numbers correct to the number of significant figures stated:

(a)6.475 to 2 s.f.
(b)1473 to 1 s.f.
(c)3681 to 2 s.f.
(d)571.32 to 4 s.f.
(e)16 001 to 3 s.f.
(f)148.25 to 3 s.f.
(g)16.999 to 3 s.f.
(h)38.9712 to 2 s.f.
(i)160.37 to 4 s.f.
Question 6

Write the number 183.9591 correct to:

(a)3 decimal places
(b)3 significant figures
(c)2 significant figures
(d)5 significant figures
(e)1 decimal place
(f)1 significant figure
Question 7

Barry rounds the number 374.49 to 375.

(a)

Is Barry correct?

Barry has rounded up instead of rounding off (rounding down).
(b)

Give the correct answer to the nearest whole number.

Question 8

Use a calculator to carry out the following calculations. In each case, give the answer correct to 2 significant figures.

(a)33 ÷ 4
(b)22 ÷ 0.7
(c)142 ÷ 0.8
(d)66 × 1.27
(e)3.25 × 1001
(f)62 × 47
(g)3.41 × 0.0092
(h)88 ÷ 0.007
(i)42 × 1.0952
Question 9

The mass of a ship is stated as 47 384 tonnes to the nearest tonne. Are the following statements correct?

(a)

The mass is less than 47 384.5 tonnes.

This statement is correct because 47 383.5 tonnes ≤ true mass < 47 384.5 tonnes.
(b)

The mass is greater than 47 384.4 tonnes.

This statement is incorrect; for example, the mass might have been 47 384.3 tonnes.
Question 10

Annie says that her rabbit has a mass of 1473 grams. Give this mass correct to:

(a)the nearest 100 grams, grams
(b)the nearest 10 grams, grams
(c)the nearest kg. kg
Question 11

At an athletics meeting, the discus throws are measured to the nearest centimetre.

(a)

Viv's best throw was measured as 35.42 m.
Could Viv's throw actually have been more than 35.42 m ?

The measured length of 35.42 m means that

35.415 m ≤ Viv's best throw < 35.425 m

so the throw could have been more than 35.42 m; for example, it might have been 35.423 m.

(b)

Chris won the hurdles race in a time of 14.6 seconds measured to the nearest tenth of a second.
Between what two values does Chris's time actually lie?

seconds ≤ actual value < seconds
Question 12

The table shows the lengths of some rivers to the nearest km.

(a)
(i)

Write the length of each river rounded to the nearest 100 km.

RiverLength in km to
the nearest km
Length in km to the
nearest 100 km
Severn354
Thames346
Trent297
Wye215
Dee113
(ii)

Which two rivers have the same length to the nearest 100 km?

and
(b)
(i)

Write the length of each river rounded to the nearest 10 km.

RiverLength in km to
the nearest km
Length in km to the
nearest 10 km
Severn354
Thames346
Trent297
Wye215
Dee113
(ii)

Which two rivers have the same length to the nearest 10 km?

and
(c)

There is another river which is not on the list.

It has a length of 200 km to the nearest 100 km, and a length of 150 km to the nearest 10 km.

Complete this sentence to give one possible length of the river to the nearest km.

The length of the river could be km.

150 km ≤ river length < 155 km
(d)

Two more rivers have different lengths to the nearest km.

They both have a length of 250 km to the nearest 10 km, but their lengths to the nearest 100 km are different.

Complete this sentence to give a possible length of each river to the nearest km.

The lengths of the rivers could be km and km.

250 km ≤ river length < 255 km
245 km ≤ river length < 250 km
Question 13

(a)

The length of an envelope is 21 cm to the nearest cm.

What is the smallest possible real length of the envelope?

cm
(b)

The interval on the number line below shows all the possible real lengths of the envelope.

Write down the correct numbers for the lower and upper limits of the interval, represented on the diagram by A and B.

A = cm

B = cm

(c)

The length of a card is 20.5 cm to the nearest tenth of a cm.

The interval on the number line below shows all the possible real lengths of the card.

Write down the numbers represented on the number line by C, D and E.

C = cm

D = cm

E = cm

(d)

Can you be sure the card will fit in the open envelope like this?

This is because the possible card length (20.55 cm) is larger than the smallest possible envelope length (20.5 cm).
For example, a card of length 20.53 cm would not fit in an envelope of length 20.52 cm.