Clocks Questions and Answers

Answer: B

At 4 o' clock, the minute hand is 20 minute spaces behind the hour hand.
Now, when the two hands are at right angles, they are fifteen min spaces
apart. So, they are at right angles in following two cases.

Case I ? When the minute hand is 15 min spaces behind the hour hand.
In this case, min hand will have to gain (20-15) = 5 min spaces.
= (60/55)*5 = 60/11
∴ They are at right angle at 60/11 minutes past 4.
Case II ? When the minute hand is 15 minute spaces ahead of the hour hand.
In this case the minute hand will have to gain (20+15) = 35 min spaces.
= (60/55)*35 = 420/11 min
∴ They are at right angles at 420/11 min past 4.

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

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

\[ \theta = \left| 30(10) - 5.5(25) \right| = \left| 300 - 137.5 \right| \]

\[ = 162.5^{\circ} \]

This is the smaller of the two angles; the reflex angle on the other side is \( 360 - 162.5 = 197.5^{\circ} \).

Answer: A

At 3 O'clock, Minute hand is at 12 while the Hour hand is at 3. Again the minute hand has to sweep through ( 30 x 5 ) ie 150° for reaching the figure 5 to show 25 mins. 

Simultaneously the Hour hand will also rotate for 25 mins. Thus starting from the mark, 3 the hour hand will cover an angle = (25 x 30) / 60 = 12.5° 

Hence, Angle between Hour and the Minute hand = ( 60 - 12.5 ) = 47.5°

Answer: B

Correct answer: 11 p.m.

While the true time advances 1440 minutes, this clock advances only \(1440 - 16 = 1424\) minutes.

From 5 a.m. on the first day to 10 p.m. on the fourth is 3 days and 17 hours, or 89 hours, as shown by the clock. Scaling to true time:

\[89 \times \frac{1440}{1424} = 89 \times \frac{90}{89} = 90 \text{ hours}\]

Adding 90 hours to 5 a.m. gives 3 days and 18 hours later, which is 11 p.m. on the fourth day.

Answer: A

At h o' clock, the minutes hand is 5 h minute spaces
behind the hour hand.
The minute hand gains 55 min spaces in 60 mins.

∴ The minute hand will gain 5h minute spaces
in 60/55 * 5h = 60h/11 minutes.

∴ The two hands will be together between
'h' and 'h+1' o' clock at (60/11)h minutes h o' clock.

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

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

The hands are together when \( \theta = 0 \), with \( H = 7 \):

\[ 30(7) = 5.5M \quad\Longrightarrow\quad 210 = 5.5M \]

\[ M = \frac{210}{5.5} = \frac{420}{11} = 38\tfrac{2}{11} \]

So they meet at \( 38\tfrac{2}{11} \) minutes past 7. The hands coincide once in each hour except between 11 and 12, which is why there are 22 coincidences a day and not 24.

Answer: B

The hands of a clock point in opposite directions (in the same straight line) 11 times in every 12 hours. (Because between 5 and 7 they point in opposite directions at 6 o'clcok only).

So, in a day, the hands point in the opposite directions 22 times.

Answer: B

Correct answer: 100°

The hour hand covers \(360°\) in 12 hours, so it moves \(30°\) per hour, or \(0.5°\) per minute.

\[3 \times 30 + 20 \times 0.5 = 90 + 10 = 100°\]

The minute hand, by contrast, would sweep \(1200°\) in the same interval.

Answer: B

At 8 o’ clock, the hour hand is at 240 degrees from the vertical.
In 30 minutes,
Hour hand = 240 + 30 x (0.5) = 240 + 15 = 255 (The hour hand moves at 0.5 dpm)
Minute hand = 30 x (6) = 180 (The minute hand moves at 6 dpm)
Difference or angle between the hands = 255 - 180 = 75 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

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

\[ \left| 240 - 5.5M \right| = 180 \]

\[ 5.5M = 60 \quad\Longrightarrow\quad M = \frac{120}{11} \]

That is \( 10\tfrac{10}{11} \) minutes past 8. The other root, \( 5.5M = 420 \), gives \( M = 76.4 \), beyond the hour.