Clocks Questions and Answers

Answer: B

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

Correct answer: 22 cm

In 30 minutes the minute hand sweeps half a revolution, so its tip traces a semicircle of radius 7 cm.

\[\text{Distance} = \pi r = \frac{22}{7} \times 7 = 22 \text{ cm}\]

A full hour would carry the tip the whole circumference, \(2\pi r = 44\) cm.

Answer: B

Time structure 12 p.m. on Monday to 2 p.m. on the accompanying
Monday =7 days 2 hours =170 hours.
∴ The watch gains(2+24/5)min.
on the other hand 34/5 min. in 170 hrs.
Now,34/5 min. are picked up in 170 hrs.
∴ 2 min. are picked up in (170(5/34*2) hrs=50 hrs.
Along these lines, the watch is right 2 days 2 hour after 12 p.m. on
Monday i.e.it will be right at 2 p.m. on Wednesday.

Answer: C

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: 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.