Calculate the acute angle between the hour hand and the minute hand of a clock at 04:20.

Clock-angle questions need both hands measured from the same reference, usually the \(12\) mark, measured clockwise.

The minute hand makes a full \(360^{\circ}\) in \(60\) minutes, so it moves \(6^{\circ}\) per minute. At \(20\) minutes past:

Minute hand \(=20\times 6=120^{\circ}\) from 12 [M1]

The hour hand makes a full \(360^{\circ}\) in \(12\) hours, so it moves \(30^{\circ}\) per hour, which is \(0.5^{\circ}\) per minute. At 04:20 it has passed the \(4\) and crept a further \(20\) minutes towards the \(5\):

Hour hand \(=4\times 30+20\times 0.5=120+10=130^{\circ}\) from 12 [M1]

The angle between the hands is the difference:

\(130-120=10^{\circ}\) [A1]

The step that decides this question is the extra \(20\times 0.5=10^{\circ}\) of hour-hand drift. Treating the hour hand as sitting exactly on the \(4\) would give \(120-120=0^{\circ}\), suggesting the hands overlap at 04:20, which they do not. Since \(10^{\circ}\) is less than \(90^{\circ}\) it is already the acute angle, so no subtraction from \(360^{\circ}\) is needed.