Clock - Arithmetic Aptitude Questions and Answers

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Answer : Option D

Explanation :

Angle traced by the hour hand in 6 hours = $[ \frac {360 } { 12 }*6]^o$ =180°.

The reflex angle between the hands of a clock at 10.25 is:

A.

180°

B.

192 $\frac { 1^o } { 2 }$

C.

195°

D.

197 $\frac { 1^o } { 2 }$

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Answer : Option D

Explanation :

Angle traced by hour hand in $\frac { 125 } {1 2 }$ hrs=[ $\frac { 360 } {1 2 }$ x $\frac { 125 } {1 2 } ]^o$ =312 $\frac { 1^o } { 2 }$

Angle traced by minute hand in 25 min =[ $\frac { 360 } {1 2 }$ x $25 ]^o$ = $150^o$

Reflex angle = 360°- $[312\frac { 1 } { 2 } -150]^o$ -162 $\frac { 1^o } { 2 }$ =197 $\frac { 1^o } { 2 }$

A clock is started at noon. By 10 minutes past 5, the hour hand has turned through:

A.

145°

B.

150°

C.

155°

D.

160°

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Answer : Option C

Explanation :

Angle traced by hour hand in 12 hrs = 360°.

Angle traced by hour hand in 5 hrs 10 min. i.e., $\frac {3 1 } {6}$ hrs=[ $\frac { 360 } { 12 }$ x $\frac {3 1 } { 6 } ]^o$ == 155°.

A watch which gains 5 seconds in 3 minutes was set right at 7 a.m. In the afternoon of the same day, when the watch indicated quarter past 4 o'clock, the true time is:

A.

59 $\frac {7 } { 12 }$ min. past 3

B.

4 p.m.

C.

58 $\frac {7 } { 11 }$ min. past 3

D.

2 $\frac {3 } { 11 }$ min. past 3

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Answer : Option B

Explanation :

Time from 7 a.m. to 4.15 p.m. = 9 hrs 15 min. = $\frac { 37 } {4 }$ hrs.

3 min. 5 sec. of this clock = 3 min. of the correct clock.

$\frac { 37 } {720 }$ hrs of this clock = $\frac { 1 } { 20 }$ hrs of the correct clock

$\frac { 37 } {4 }$ hrs of this clock=[ $\frac { 1 } { 20 }$ x $\frac { 720 } {37}$ x $\frac { 37 } {4 }$ ]hrs of the correct clock

= 9 hrs of the correct clock.

The correct time is 9 hrs after 7 a.m. i.e., 4 p.m.

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.

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Answer : Option A

Explanation :

55 min. spaces are covered in 60 min.

60 min. spaces are covered in[ $\frac {60 } { 55 }$ x60]min=65 $\frac { 5 } { 11 }$ min

Loss in 64 min. =[65 $\frac { 5 } { 11 }$ -64]= $\frac { 16 } { 11 }$ min

Loss in 24 hrs =[ $\frac { 16 } { 11 }$ x $\frac { 1 } { 64 }$ x24x60]min=32 $\frac { 8} { 11}$ min

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 $\frac {5 } {11 }$

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Answer : Option D

Explanation :

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[ $\frac {60 } { 55 }$ x5]min=5 $\frac {5 } {11 }$ min

Required time =5 $\frac {5 } {11 }$ min.past 7.

At what time between 5.30 and 6 will the hands of a clock be at right angles?

A.

43 $\frac { 5 } { 11 }$ min. past 5

B.

43 $\frac { 7 } { 11 }$ min. past 5

C.

40 min. past 5

D.

45 min. past 5

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Answer : Option B

Explanation :

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 has to gain (25 + 15) = 40 min. spaces.

55 min. spaces are gained in 60 min.

40 min. spaces are gained in[ $\frac { 60 } { 55 }$ x40]min=43 $\frac { 7 } { 11 }$ min

Required time =43 $\frac { 7 } { 11 }$ min.past 5.

The angle between the minute hand and the hour hand of a clock when the time is 4.20, is:

A.

0°

B.

10°

C.

5°

D.

20°

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Answer : Option B

Explanation :

Angle traced by hour hand in $\frac { 13 } { 3 }$ hrs=[ $\frac { 360 } {1 2 }$ x $\frac { 13 } { 3 }]^o$ =130°.

Angle traced by min. hand in 20 min. =[ $\frac { 360 } {1 2 }$ x $20]^o$ = 120°.

Required angle = (130 - 120)° = 10°

At what angle the hands of a clock are inclined at 15 minutes past 5?

A.

58 $\frac { 1^o } { 2 }$

B.

64°

C.

67 $\frac { 1^o } { 2 }$

D.

72 $\frac { 1^o } { 2 }$

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Answer : Option C

Explanation :

Angle traced by hour hand in $\frac { 21 } { 4 }$ hrs=[ $\frac { 360 } { 12 }$ x $\frac { 21 } { 4 } ]^o$ =57 $\frac { 1^o } { 2 }$

Angle traced by min. hand in 15 min. =[ $\frac { 360 } { 60 }$ x $15]^o$ = 90°

Required angle =[157 $\frac { 1^o } { 2 }$ ]- 90° = 67 $\frac { 1^o } { 2 }$

At 3:40, the hour hand and the minute hand of a clock form an angle of:

A.

120°

B.

125°

C.

130°

D.

135°

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