Clocks Questions and Answers

Answer: D

Correct answer: 54 (6/11) min past 10

At 10 o’clock the hour hand leads the minute hand by \(300°\), and the minute hand gains \(5.5°\) per minute.

\[t = \frac{300}{5.5} = \frac{600}{11} = 54\tfrac{6}{11}\ \text{minutes}\]

So they coincide at \(54\tfrac{6}{11}\) minutes past 10, the last coincidence before the hands meet again at 12.

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: 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: A

The speed of hour hand is 0.5 dpm (Degree per Minute)
And the speed of minute hand is 6 dpm.
So, 300 degree traverse by minute hand in 300/6 = 50 Minutes.
Thus, angle moved by the hour hand in 50 minutes is 50/0.5 = 25° degrees.

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: D

The clock gains 20 seconds in every 3 hours, so first count the hours.

From Thursday 2:00 am to Friday 6:30 pm is

\[ 22 + 18.5 = 40.5 \text{ hours} \]

\[ \text{gain} = 20 \times \frac{40.5}{3} = 20 \times 13.5 = 270 \text{ s} = 4 \text{ min } 30 \text{ s} \]

The clock therefore reads ahead of the true 6:30 pm:

\[ 6{:}30{:}00 + 4{:}30 = 6{:}34{:}30 \text{ pm} \]

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: B

Time from 5 am On a day to 3 pm on 8th day = 178 hours
Now 23 hrs 44 min. of this clock = 24 hours of correct clock
365/15 hrs of this clock = 24 hours of correct clock
178 hrs of this clock = (24 × 15/356 × 89) hrs of correct clock
= 180 hrs of correct clock.
So, the correct time is 5 pm.

Answer: C

In 12 hours, the hands coincide or are in opposite direction 22 times.

 In 24 hours, the hands coincide or are in opposite direction 44 times a day.

Answer: B

The hands of a clock coincide 11 times in every 12 hours (Since between 11 and 1, they coincide only once, i.e., at 12 o'clock).

The hands overlap about every 65 minutes, not every 60 minutes.
Therefore, The hands coincide 22 times in a day.