Practice MCQ Questions and Answer on Clock

11.

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

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

12.

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 $$\eqalign{ & = \left( {\frac{{60}}{{55}} \times 5} \right){\kern 1pt} \min . \cr & = 5\frac{5}{{11}}{\kern 1pt} \min . \cr & \therefore {\text{Required time}} = 5\frac{5}{{11}}{\kern 1pt} \min .{\kern 1pt} \,{\text{past}}{\kern 1pt} 7 \cr} $$

13.

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

14.

How many times are the hands of a clock at right angle in a day?

• (A) 22
• (B) 24
• (C) 44
• (D) 48

Show Answer

 In 12 hours, they are at right angles 22 times. ∴ In 24 hours, they are at right angles 44 times.

15.

There are two clocks, both set to show 10 pm on 21^st January 2010. One clock gains 2 minutes in an hour and the other clock loses 5 minutes in an hour. Then by how many minutes do the two clocks differ at 4 pm on 22^nd January 2010 ?

• (A) 126 minutes
• (B) 136 minutes
• (C) 96 minutes
• (D) 106 minutes

Show Answer

 One clock show 10 pm, on 21st January 2010 One clock gains = 2 minutes Other clock loses = 5 minutes Time period between 10 pm and 4 pm = 18 hours ∴ Required difference = (2 × 18 + 5 × 18 ) minutes = 126 minutes

16.

A clock is displaying correct time at 9 am on Monday. If the clock loses 12 minutes in 24 hours, then the actual time when the clock indicates 8 : 30 pm on Wednesday of the same week is = ?

• (A) 8 pm
• (B) 7 pm
• (C) 9 pm
• (D) 8 : 59 : 45 pm

Show Answer

 Time interval from 9 am on Monday to 8 : 30 pm on Wednesday. $$\eqalign{ & {\text{ = }}\left( {24 \times 2.5} \right) - {\text{0:30 hours }} \cr & {\text{ = 60}} - {\text{0}}{\text{:30 hours}} \cr & {\text{ = 59 hours 30 minutes}} \cr & = 59\frac{{30}}{{60}} \cr & = 59\frac{1}{2} \cr & = \frac{{119}}{2}{\text{ hours}} \cr & {\text{Also 24 hours}} - {\text{12 minutes}} \cr & = {\text{23 hours 48 minutes}} \cr & = 23 + \frac{{48}}{{60}} \cr & = 23\frac{4}{5} \cr & = \frac{{119}}{5}{\text{ hours}} \cr & \therefore \frac{{119}}{2}{\text{ hours of this clock}} \cr & = \frac{{24 \times 5}}{{119}} \times \frac{{119}}{2} \cr & = 60{\text{ hours}} \cr & \left( {60 - \frac{{119}}{2}} \right){\text{ hours}} \cr & {\text{ = }}\frac{{120 - 119}}{2}{\text{ hours}} \cr & {\text{ = }}\frac{1}{2}{\text{ hours}} \cr & = 30{\text{ minutes}} \cr} $$ Hence, the correct time is 30 minutes after 8:30 pm i.e., 9 pm

17.

In an accurate clock, in a period of 2 hours 20 minutes the minute hand will move over = ?

• (A) 140°
• (B) 320°
• (C) 520°
• (D) 840°

Show Answer

 Angle traced by the minute hand in 2 hours 20 minutes,i.e., $$\eqalign{ & = 140{\text{ minutes}} \cr & = {\left( {\frac{{360}}{{60}} \times 140} \right)^ \circ } \cr & = {840^ \circ } \cr} $$

18.

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.

19.

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

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

20.

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