Clocks Questions and Answers

Answer: C

In a correct clock the hands coincide every

\[ \frac{720}{11} = 65\tfrac{5}{11} \text{ minutes} \]

Here they coincide every 62 minutes, sooner than they should, so the clock is gaining.

\[ \text{gain per interval} = \frac{720}{11} - 62 = \frac{38}{11} \text{ min} \]

In 1440 minutes there are \( \tfrac{1440}{62} \) such intervals:

\[ \text{gain per day} = \frac{38}{11} \times \frac{1440}{62} = \frac{54720}{682} = 80\tfrac{80}{341} \text{ minutes} \]

Coinciding more often than every \( 65\tfrac{5}{11} \) minutes always means a fast clock; a slower interval, such as 64 minutes, would mean a loss.

Answer: D

The dial is divided into 60 minute-marks around \( 360^{\circ} \), so each mark spans

\[ \frac{360}{60} = 6^{\circ} \]

Being 25 minute-marks apart:

\[ 25 \times 6 = 150^{\circ} \]

The hands named do not matter here - the separation is given directly in minute-divisions.

Answer: B

The hour hand moves \( 0.5^{\circ} \) a minute and the minute hand \( 6^{\circ} \), so at \( H \) hours \( M \) minutes the angle between them is

\[ \theta = \left| 30H - 5.5M \right| \]

Eight minute-divisions is \( 8 \times 6 = 48^{\circ} \), so with \( H = 10 \):

\[ \left| 300 - 5.5M \right| = 48 \]

\[ 5.5M = 252 \quad\Longrightarrow\quad M = \frac{504}{11} = 45\tfrac{9}{11} \]

\[ 5.5M = 348 \quad\Longrightarrow\quad M = \frac{696}{11} = 63\tfrac{3}{11} \]

The second value exceeds 60, so it falls in the next hour: \( 3\tfrac{3}{11} \) minutes past 11.

The hands coincide at \( 54\tfrac{6}{11} \) minutes past 10, and the two answers sit on either side of that moment.

Answer: A

The hour hand moves \( 0.5^{\circ} \) a minute and the minute hand \( 6^{\circ} \), so at \( H \) hours \( M \) minutes the angle between them is

\[ \theta = \left| 30H - 5.5M \right| \]

Opposite directions means a straight line with \( \theta = 180 \), and \( H = 2 \):

\[ \left| 60 - 5.5M \right| = 180 \]

\[ 5.5M = 240 \quad\Longrightarrow\quad M = \frac{480}{11} = 43\tfrac{7}{11} \]

The other root, \( 5.5M = -120 \), is negative and so discarded.

Answer: B

The hour hand moves \( 0.5^{\circ} \) a minute and the minute hand \( 6^{\circ} \), so at \( H \) hours \( M \) minutes the angle between them is

\[ \theta = \left| 30H - 5.5M \right| \]

With \( H = 8 \) and \( \theta = 90 \):

\[ \left| 240 - 5.5M \right| = 90 \]

\[ 5.5M = 150 \quad\Longrightarrow\quad M = \frac{300}{11} = 27\tfrac{3}{11} \]

The other root gives \( M = 60 \), which is 9 o'clock exactly and so outside the hour asked about.

Answer: B

The hour hand moves \( 0.5^{\circ} \) a minute and the minute hand \( 6^{\circ} \), so at \( H \) hours \( M \) minutes the angle between them is

\[ \theta = \left| 30H - 5.5M \right| \]

\[ \theta = \left| 30(3) - 5.5(40) \right| = \left| 90 - 220 \right| = 130^{\circ} \]

Taking \( 30H \) as the hour hand's position from 12 automatically includes its drift during the 40 minutes, which is the step most often forgotten.

Answer: A

In twelve hours the hands coincide eleven times, not twelve: between 11 and 12 they meet only at 12 o'clock itself, so that hour contributes no separate coincidence.

\[ 11 \times 2 = 22 \text{ times in 24 hours} \]

Equivalently, the minute hand laps the hour hand once every \( \tfrac{720}{11} = 65\tfrac{5}{11} \) minutes, and \( \tfrac{1440}{720/11} = 22 \).

Answer: A

When the hands of the clock are in the same straight line but not together, they are 30 minute spaces apart.
At 7 o'clock, they are 25 min. spaces apart.
Minute hand must gain only 5 min. space.
55 min. spaces are gained in 60 min.
5 min. spaces would be gained in (60/55) x 5 = 5 (5/11) min
Required time = 5 (5/11) min past 7

Answer: D

At 4 o'clock, the hands of the watch are 20 min. spaces apart.
To be in opposite directions, they must be 30 min. spaces apart.
Minute hand must gain 50 min. spaces.
55 min. spaces are gained in 60 min.
50 min. spaces are gained in 60/55 x 50 = 54 (6/11) min
Required time = 54 6/11 min past 4

Answer: B

At 8 o’ clock, the hour hand is at 240 degrees from the vertical.
In 30 minutes,
Hour hand = 240 + 30 x (0.5) = 240 + 15 = 255 (The hour hand moves at 0.5 dpm)
Minute hand = 30 x (6) = 180 (The minute hand moves at 6 dpm)
Difference or angle between the hands = 255 - 180 = 75 degrees