Clocks Questions and Answers

Answer: B

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

From 8 am Sunday to 8 pm the following Sunday is

\[ 7 \times 24 + 12 = 180 \text{ hours} \]

\[ \text{total gain} = 5 + 5\tfrac{4}{5} = 10.8 \text{ minutes} \]

\[ \text{rate} = \frac{10.8}{180} = 0.06 \text{ min per hour} \]

Starting 5 minutes slow, it becomes correct after

\[ \frac{5}{0.06} = 83\tfrac{1}{3} \text{ h} = 83 \text{ h } 20 \text{ min} \]

\[ 8 \text{ a.m. Sunday} + 83 \text{ h } 20 \text{ min} = 7{:}20 \text{ p.m. on Wednesday} \]

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: A

Time from 5am on a day to 10 pm on fourth day= 89 hours
Presently 23hrs. 44min. of this clock= 24 hours of right clock.
∴ 356/15 hrs of this clock= (24*15/356*89) hrs. Of right clock.
= 90 hrs of right clock.
Along these lines, the right time is 11 pm.

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

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

\[ \theta = \left| 30(8) - 5.5(30) \right| = \left| 240 - 165 \right| = 75^{\circ} \]

At half past the hour the hour hand has moved \( 15^{\circ} \) beyond the 8, which is why the answer is not the round \( 90^{\circ} \) between the 6 and the 8.

Answer: C

Correct answer: 120°

The minute hand completes \(360°\) in 60 minutes, so it turns \(6°\) each minute.

\[20 \times 6 = 120°\]

This is a third of the dial, matching the fact that 20 minutes is a third of an hour.

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.