Clocks Questions and Answers

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

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

Three minute-divisions is \( 3 \times 6 = 18^{\circ} \), with \( H = 5 \):

\[ \left| 150 - 5.5M \right| = 18 \]

\[ 5.5M = 132 \quad\Longrightarrow\quad M = 24 \]

The other root gives \( M = 30\tfrac{6}{11} \), on the far side of the coincidence at \( 27\tfrac{3}{11} \) minutes past 5.

The 24-minute answer is the exact one, which is what makes it the intended reading.

Answer: D

The hour hand sweeps the full \( 360^{\circ} \) in twelve hours, so it turns \( 30^{\circ} \) an hour.

\[ 6 \text{ hours} \times 30^{\circ} = 180^{\circ} \]

Only the hour hand is asked about; the minute hand would have turned \( 6 \times 360^{\circ} \) over the same period.

Answer: B

A mirror reverses the dial, so a reading of \( T \) appears as \( 12{:}00 - T \) (or \( 11{:}60 - T \), which is easier to subtract).

\[ 11{:}60 - 12{:}10 \]

Treating 12:10 as 0:10:

\[ 11{:}60 - 0{:}10 = 11{:}50 \]

So the mirror shows 11:50.

Answer: C

In each twelve-hour period the hands are at right angles 22 times - twice in most hours, but the crossings near 3 and 9 o'clock overlap so that two are lost.

\[ 22 \times 2 = 44 \text{ times in 24 hours} \]

The count is not 48, because between 2–3 and 3–4 (and again around 9) one right angle is shared rather than occurring twice.

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

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

\[ \theta = \left| 30(4) - 5.5(20) \right| = \left| 120 - 110 \right| = 10^{\circ} \]

The minute hand is at the 4, and the hour hand has drifted a third of the way past it - \( 20 \times 0.5 = 10^{\circ} \) - which is exactly the gap.

Answer: B

To be together between 9 and 10 o'clock, the minute hand must gain 45 min. space
55 min. spaces are gained in 60 min.
45 min spaces would be gained in (60/55 × 45) min. or 49 1/11 min. 
The hands are together at 49 1/11 min. past 9.

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: A

Correct answer: 22

The hands point in opposite directions once for every full \(360°\) the minute hand gains on the hour hand. That gain happens 11 times in 12 hours, so the hands are opposite 11 times in 12 hours and 22 times in 24 hours.

The same reasoning gives 22 coincidences in 24 hours, since each relative revolution produces exactly one of each.

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

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

"Same straight line but not together" means \( \theta = 180 \), with \( H = 7 \):

\[ \left| 210 - 5.5M \right| = 180 \]

\[ 5.5M = 30 \quad\Longrightarrow\quad M = \frac{60}{11} = 5\tfrac{5}{11} \]

The other root, \( 5.5M = 390 \), gives \( M = 70.9 \), beyond the hour.

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 \( \tfrac{1440}{720/11} = 22 \).