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

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

\[ \theta = \left| 300 - 137.5 \right| = 162.5^{\circ} \]

The reflex angle is what is left of the full turn:

\[ 360 - 162.5 = 197.5^{\circ} = 197\tfrac{1}{2}^{\circ} \]

A reflex angle is by definition greater than \( 180^{\circ} \), so the smaller 162.5 cannot be the answer.

Answer: D

The clock runs fast, so it reaches 11 am before the true time does.

From Sunday 1 am to Monday 11 am is 34 hours on the clock.

\[ \text{gain} = 20 \times \frac{34}{24} = \frac{85}{3} = 28\tfrac{1}{3} \text{ min} = 28 \text{ min } 20 \text{ s} \]

The true time is that much behind the clock:

\[ 11{:}00 - 28{:}20 = 10{:}31{:}40 \text{ am} \]

So the alarm sounded 31 minutes 40 seconds past 10 am, true time.

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

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: A

At 5 o'clock, the hands are 25 min. spaces apart. 
To be at right angles and that too between 5.30 and 6, the minute hand must gain 40 min
55 min. spaces are gained in 60 min.
40 min. spaces would be gained in (60/55 × 40) min. = 43 7/11 min. 
Required time = 43(7/11) min. past 5.

Answer: D

55 min. spaces are covered in 60 min.

60 min. spaces are covered in (60/55 × 60) min = 65 5/11 min
 Loss in 64 min = (65 5/11 - 64) = 16/11 min
Loss in 24 hrs= (16/11 × 1/64 × 24 × 60) min = 32 (8/11) min.

Answer: C

At 3 o’ clock, the hour hand is at 90 degrees from the vertical. In 40 minutes,
Hour hand = 90 + 40*(0.5) = 90 + 20 = 110 (The hour hand moves at 0.5 dpm)
Minute hand = 40*(6) = 240 (The minute hand moves at 6 dpm)
Difference or angle between the hands = 240 - 110 = 130 degrees.

Answer: C

At 7 o’ clock, the hour hand is at 210 degrees from the vertical.
In 20 minutes,
Hour hand = 210 + 20 x (0.5) = 220 (The hour hand moves at 0.5 dpm)
Minute hand = 20 x (6) = 120 (The minute hand moves at 6 dpm)
Difference or angle between the hands = 220 - 120 = 100 degree

Answer: A

In a correct clock the hands coincide every

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

Here they take 64 minutes, less than they should, so the watch is running fast - but the question asks how much it loses, and the conventional treatment measures the shortfall of the interval:

\[ \frac{720}{11} - 64 = \frac{16}{11} \text{ minutes per interval} \]

\[ \frac{16}{11} \times \frac{1440}{64} = \frac{23040}{704} = 32\tfrac{8}{11} \text{ minutes a day} \]

The 24-hour figure comes from the number of coincidence intervals in a day, \( \tfrac{1440}{64} = 22.5 \).

Answer: B

Correct answer: 130°

The minute hand moves \(6°\) per minute and the hour hand \(0.5°\) per minute.

\[\text{Minute hand} = 40 \times 6 = 240°\]

\[\text{Hour hand} = 3 \times 30 + 40 \times 0.5 = 90 + 20 = 110°\]

\[\text{Angle} = 240 - 110 = 130°\]

The hour hand has crept two-thirds of the way from 3 towards 4, which is what the extra \(20°\) represents.