Edexcel IGCSE Math A 4.4 Measures Question Bank

Practise reading scales, estimating units, using clocks, measuring angles and applying compound measures such as speed and density.

Exam points

read values from instrument scales and choose sensible units for familiar measures
calculate time intervals from 12-hour or 24-hour times, including timetables
use bearings, angle measurements and formulae

Covered syllabus points

4.4.A Scales on a range of measuring instruments 4.4.B Time intervals in terms of the 24-hour and the 12-hour 4.4.C Make sensible estimates of a range of measures 4.4.D Angle measure 4.4.E Measure an angle to the nearest degree 4.4.F Average speed, distance and time 4.4.G Compound measure such as speed, density and pressure

[Maximum number: 1]

The scale drawing shows the position of a hall and the position of a library.

Figure for Question 4.4 Measures question 1 — Edexcel IGCSE Math A Higher

Scale: 1 cm represents 20 metres

A post box is 140 metres from the library on a bearing of $220^{\circ}$

Use your scale drawing to find

the bearing of the hall from the post box.

Accept bearing:
$330^\circ \text{ to } 334^\circ$

B1.

[Maximum number: 3]

Jalina left her home at 1000 to cycle to a park.
On her way to the park, she stopped at a friend's house and then continued her journey to the park.
Here is the distance-time graph for her journey to the park.

Figure for Question 4.4 Measures question 2 — Edexcel IGCSE Math A Higher

Work out Jalina's average cycling speed, in kilometres per hour, for the complete journey to the park and back.
Do not include the times when she was not cycling in your calculation.
Give your answer correct to 1 decimal place.
km/h

$1\text{ h }45\text{ m}+1\text{ h }30\text{ m}$
OR
$1+0.75+1.5=3.25\text{ h}$ $\frac{24\times2}{3.25}=14.8\ldots$

Final answer:
$14.8\text{ km/h}$

Marking guidance:
- M1 ft for total cycling time
- M1 ft dep for a full method
- A1 for 14.8, awrt 14.8

[Maximum number: 2]

The mass of a silver coin is 48.3 g
The density of silver is $10.5 \mathrm{~g} / \mathrm{cm}^{3}$
Work out the volume of the silver coin.
$\mathrm{cm}^{3}$

\rho=\frac{m}{V}

10.5=\frac{48.3}{V}

V=\frac{48.3}{10.5}=4.6

Final answer:

4.6\text{ cm}^3

M1 for substituting 10.5 and 48.3 correctly into a correct formula for density.
A1; allow $\frac{23}{5}$ or $4\frac35$ oe.

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Covered syllabus points

4.4 Measures question 1

4.4 Measures question 2

4.4 Measures question 3