Practice MCQ Questions and Answer on Clock

1.

A watch is 1 minute slow at 1 pm on Tuesday and 2 minutes fast at 1 pm on Thursday. When did it show the correct time = ?

• (A) 1 : 00 am on Wednesday
• (B) 5 : 00 am on Wednesday
• (C) 1 : 00 pm on Wednesday
• (D) 5 : 00 pm on Wednesday

Show Answer

 Time from 1 pm on Wednesday to 1 pm on Thursday = 48 hours So, the watch gains (1 + 2) minute or 3 minutes in 48 hours. Now, 3 minutes are gained in 48 hours So, 1 minute is gained in $$\left( {\frac{{48}}{3}} \right)$$   = 16 hours. Thus, the watch showed the correct time 16 hours after 1 pm on Tuesday, i.e., 5 am on Wednesday

2.

How many times do the hands of a clock coincide in a day?

• (A) 20
• (B) 21
• (C) 22
• (D) 24

Show Answer

 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). AM 12:00 1:05 2:11 3:16 4:22 5:27 6:33 7:38 8:44 9:49 10:55 PM 12:00 1:05 2:11 3:16 4:22 5:27 6:33 7:38 8:44 9:49 10:55 The hands overlap about every 65 minutes, not every 60 minutes. ∴ The hands coincide 22 times in a day.

3.

How much does a watch lose per day, if its hands coincide every 64 minutes?

• (A) $$32\frac{8}{{11}}$$ min.
• (B) $$36\frac{5}{{11}}$$ min.
• (C) 90 min.
• (D) 96 min.

Show Answer

 $$\eqalign{ & 55\,\min .\,{\text{spaces}}\,{\text{are}}\,{\text{covered}}\,{\text{in}}\,60\,\min \cr & 60\,\min .\,{\text{spaces}}\,{\text{are}}\,{\text{covered}}\,{\text{in}} \cr & = \left( {\frac{{60}}{{55}} \times 60} \right)\,\min . \cr & = 65\frac{5}{{11}}\,\min . \cr & {\text{Loss}}\,{\text{in}}\,64\,\min . \cr & = {65\frac{5}{{11}} - 64} = \frac{{16}}{{11}}\,\min . \cr & {\text{Loss}}\,{\text{in}}\,24\,hrs. \cr & = \left( {\frac{{16}}{{11}} \times \frac{1}{{64}} \times 24 \times 60} \right)\,\min. \cr & = 32\frac{8}{{11}}\,\min. \cr} $$

4.

An accurate clock shows 8 o'clock in the morning. Through how may degrees will the hour hand rotate when the clock shows 2 o'clock in the afternoon?

• (A) 144º
• (B) 150º
• (C) 168º
• (D) 180º

Show Answer

 $$\eqalign{ & {\text{Angle}}\,{\text{traced}}\,{\text{be}}\,{\text{the}}\,{\text{hour}}\,{\text{hand}}\,{\text{in}}\,{\text{6}}\,{\text{hours}} \cr & = {\left( {\frac{{360}}{{12}} \times 6} \right)^ \circ } = {180^ \circ } \cr} $$

5.

Between 5 and 6, a lady looked at her watch and mistaking the hour hand for the minute hand, she thought that the time was 57 minutes earlier than the correct time. The correct time was = ?

• (A) 12 minutes past 5
• (B) 24 minutes past 5
• (C) 36 minutes past 5
• (D) 48 minutes past 5

Show Answer

 Since the time read by the lady was 57 minutes earlier than the correct time, so the minute hand is (60 - 57) = 3 minutes spaces behind the hour hand. Now, at 5 o'clock, the minute hand is 25 minutes spaces behind the hour hand. To be 3 minutes spaces behind, it must gain (25 - 3) = 22 minutes spaces. 55 minutes spaces are gained in 60 minutes. 22 minutes spaces are gained in $$\left( {\frac{{60}}{{55}} \times 22} \right)$$   = 24 minutes Hence, the correct time was 24 minutes past 5.

6.

A watch which gains uniformly is 2 minutes low at noon on Monday and is 4 min. 48 sec fast at 2 p.m. on the following Monday. When was it correct?

• (A) 2 p.m. on Tuesday
• (B) 2 p.m. on Wednesday
• (C) 3 p.m. on Thursday
• (D) 1 p.m. on Friday

Show Answer

 Time from 12 pm on Monday to 2 pm on the following Monday = 7 days 2 hours= 170 hours ∴ The watch gains \[\left( {2 + 4\frac{4}{5}} \right)\]   minutes or \[\frac{{34}}{5}\]   minutes in 170 hours. Now, $$\frac{{34}}{5}$$ minutes are gained in 170 hours ∴ 2 minutes are gained in \[\left( {170 \times \frac{5}{{34}} \times 2} \right)\]   hours = 50 hours ∴ Watch is correct 2 days 2 hours after 12 pm on Monday i.e., it will be correct at 2 pm on Wednesday.

7.

Imagine that your watch was correct at noon, but then it began to lose 30 minutes each hour. It now show 4 pm but it stopped 5 hours ago. What is the correct time now = ?

• (A) 9 : 30 pm
• (B) 11 pm
• (C) 1 am
• (D) 1 : 30 am

Show Answer

 The watch loses $$\frac{1}{2}$$ hour each hour. So, it must have take 8 hours to show 4 pm from 12 noon. Thus, it stopped at 8 pm. So, the correct time is 5 hours ahead of 8 pm, i,e., 1 am.

8.

At what time between 7 and 8 o'clock will the hands of a clock be in the same straight line but, not together?

• (A) 5 min. past 7
• (B) $$5\frac{2}{{11}}$$ min. past 7
• (C) $$5\frac{3}{{11}}$$ min. past 7
• (D) $$5\frac{5}{{11}}$$ min. past 7

Show Answer

 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 will have to gain only 5 min. spaces. 55 min. spaces are gained in 60 min. 5 min. spaces are gained in $$\eqalign{ & = \left( {\frac{{60}}{{55}} \times 5} \right){\kern 1pt} \min . \cr & = 5\frac{5}{{11}}{\kern 1pt} \min . \cr & \therefore {\text{Required time}} = 5\frac{5}{{11}}{\kern 1pt} \min .{\kern 1pt} \,{\text{past}}{\kern 1pt} 7 \cr} $$

9.

A mechanical grandfather clock is at present showing 7 hours 40 minutes 6 seconds. Assuming that it loses 4 seconds in every hour, what time will it show after exactly $$6\frac{1}{2}$$ hours ?

• (A) 14 hours 9 minutes 34 seconds
• (B) 14 hours 9 minutes 40 seconds
• (C) 14 hours 10 minutes 6 seconds
• (D) 14 hours 10 minutes 32 seconds

Show Answer

 $$\eqalign{ & {\text{Time lost in }}6\frac{1}{2}{\text{ hours}} \cr & = {\text{ }}\left( {6\frac{1}{2} \times 4} \right)\sec \cr & = 26\sec \cr} $$ Correct time after $${\text{6}}\frac{1}{2}$$ hours = 7 hours 40 minutes 6 seconds + 6 hours 30 minutes = 14 hours 10 minutes 6 seconds Time show by the clock = 14 hours 10 minutes 6 seconds - 26 sec = 14 hours 9 minutes 40 seconds

10.

Henry started a trip into the country between 8 am and 9 am when the hand of clock were together, He arrived at his destination between 2 pm and 3 pm when the hands of the clock were exactly 180° apart. How long did he travel ?

• (A) 6 hours
• (B) 7 hours
• (C) 9 hours
• (D) 11 hours

Show Answer

 To be together between 8 am and 9 am, the minute hand has to gain 40 minutes spaces. 55 minutes spaces are gained in 60 minutes. 40 minutes space are gained in $$\left( {\frac{{60}}{{55}} \times 40} \right)$$  minutes = $${\text{43}}\frac{7}{{11}}$$  minutes So, Henry started his trip at $${\text{43}}\frac{7}{{11}}$$  minutes past 8 am. Now, to be 180° apart, the hands must be 30 minutes spaces apart. At 2 pm, they are 10 minutes spaces apart. ∴ The minute hand will have to gain (10 + 30) = 40 minutes spaces. As calculate above, 40 minutes spaces are gained in $${\text{43}}\frac{7}{{11}}$$  minutes. So, Henry's trip ended at $${\text{43}}\frac{7}{{11}}$$  minutes past 2 pm ∴ Duration of travel = Duration from $${\text{43}}\frac{7}{{11}}$$  minutes past 8 am to $${\text{43}}\frac{7}{{11}}$$  minutes past 2 pm = 6 hours