Reasoning Aptitude
CLOCK MCQs
Clocks
Total Questions : 146
| Page 6 of 15 pages
Answer: Option A. -> 126 minutes
Answer: (a)
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
Therefore, Required difference = (2 × 18 + 5 × 18) minutes = 126 minutes
Answer: (a)
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
Therefore, Required difference = (2 × 18 + 5 × 18) minutes = 126 minutes
Answer: Option C. -> 21.05
Answer: (c)
Clearly, time in London is 5 hrs 30 minutes behind 2.35 a.m. which is 9.05 p.m. or 21.05 hrs.
Answer: (c)
Clearly, time in London is 5 hrs 30 minutes behind 2.35 a.m. which is 9.05 p.m. or 21.05 hrs.
Answer: Option B. -> 12 noon, after 135 days
Answer: (b)
After 12 days, i.e., after 12 × 24 hours clock A will gain 48 minutes and will show 12:48 noon.
After 12 days, i.e., after 12 × 24 hours clock B will lose 16 minutes and will show 11:44 am.
The two clocks will show the same time after 135 days The time difference has to be 12 hours between then = 720 minutes.
A will gain 540 minutes in 135 days. B will lose 180 minutes in 135 days Total 720 minutes.
Further if we consider only time then the problem becomes simpler Total difference of minutes between the times shown by the clocks after 36 hours = $16/3$ minutes difference in 1 day = 12 × 60 minutes difference in $3/16$ × 12 × 60 = 135 days.
Answer: (b)
After 12 days, i.e., after 12 × 24 hours clock A will gain 48 minutes and will show 12:48 noon.
After 12 days, i.e., after 12 × 24 hours clock B will lose 16 minutes and will show 11:44 am.
The two clocks will show the same time after 135 days The time difference has to be 12 hours between then = 720 minutes.
A will gain 540 minutes in 135 days. B will lose 180 minutes in 135 days Total 720 minutes.
Further if we consider only time then the problem becomes simpler Total difference of minutes between the times shown by the clocks after 36 hours = $16/3$ minutes difference in 1 day = 12 × 60 minutes difference in $3/16$ × 12 × 60 = 135 days.
Answer: Option D. -> 1440 days
Answer: (d)
Clearly, the first watch will show the correct time when it has gained 12 hours
i.e., (12 × 60) = 720 min and the second watch will show the correct time when it has lost 720 min.
Time taken by first watch to gain 720 min = 720 days.
Time taken by second watch to gain 720 min = (720 ÷ 1 $1/2$) days = (720 × $2/3$ days) = 480 days.
So the first watch shows correct time after every 720 days and the second watch after every 480 days.
Therefore, Time after which both the clocks will together tell the correct time = L.C.M. of 720 and 480 = 1440 days.
Answer: (d)
Clearly, the first watch will show the correct time when it has gained 12 hours
i.e., (12 × 60) = 720 min and the second watch will show the correct time when it has lost 720 min.
Time taken by first watch to gain 720 min = 720 days.
Time taken by second watch to gain 720 min = (720 ÷ 1 $1/2$) days = (720 × $2/3$ days) = 480 days.
So the first watch shows correct time after every 720 days and the second watch after every 480 days.
Therefore, Time after which both the clocks will together tell the correct time = L.C.M. of 720 and 480 = 1440 days.
Answer: Option B. -> 8:45 AM
Answer: (b)
Clearly, Raveena left home 10 min before 8 : 40 am i.e., at 8 : 30 am
but it was 15 min earlier than usual,
so she usually leaves for the stop at (8 : 30 + 0 : 15) = 8 : 45 am.
Answer: (b)
Clearly, Raveena left home 10 min before 8 : 40 am i.e., at 8 : 30 am
but it was 15 min earlier than usual,
so she usually leaves for the stop at (8 : 30 + 0 : 15) = 8 : 45 am.
Answer: Option B. -> 9:15 AM
Answer: (b)
As next bus leaves at 9 : 35 am.
Therefore, Previous bus left at (9 : 35 - 0 : 30) = 9 : 05 am.
The time when enquiry clerk gave this information = 9 : 05 + 0 : 10 = 9 : 15 am.
Answer: (b)
As next bus leaves at 9 : 35 am.
Therefore, Previous bus left at (9 : 35 - 0 : 30) = 9 : 05 am.
The time when enquiry clerk gave this information = 9 : 05 + 0 : 10 = 9 : 15 am.
Question 57. There are 20 people working in an office. The first group of five works between 8 am and 2 pm. The second group of ten works between 10 am and 4 pm and the third group of five works between 12 noon and 6 pm. There are three computers in the office which all the employees frequently use. During which of the following hours, the computers are likely to be used the most?
Answer: Option C. -> 12 noon -2 : 00 pm
Answer: (c)
It is obvious that computers would be used most when all the three groups are working simultaneously and this happens during the period 12 noon to 2 pm.
Answer: (c)
It is obvious that computers would be used most when all the three groups are working simultaneously and this happens during the period 12 noon to 2 pm.
Answer: Option C. -> 16 : 10 h
Answer: (c)
Time of the last train leaving the station = (18 : 00 - 2 : 30) h = 15 : 30 h
But this happens 40 min before the announcement is made.
Therefore, Time of making announcement = (15 : 30 + 0 : 40) = 16 : 10 h.
Answer: (c)
Time of the last train leaving the station = (18 : 00 - 2 : 30) h = 15 : 30 h
But this happens 40 min before the announcement is made.
Therefore, Time of making announcement = (15 : 30 + 0 : 40) = 16 : 10 h.
Answer: Option C. -> 7:05 AM
Answer: (c)
Time of ringing last bell = (7 : 45- 0 : 45) = 7 : 00 am.
But it happened 5 min before the priest gave the information to the devotes.
Therefore, Time of giving information = 7 : 00 + 0 : 05 = 7 : 05 am.
Answer: (c)
Time of ringing last bell = (7 : 45- 0 : 45) = 7 : 00 am.
But it happened 5 min before the priest gave the information to the devotes.
Therefore, Time of giving information = 7 : 00 + 0 : 05 = 7 : 05 am.
Answer: Option D. -> 840°
Answer: (d)
Angle traced by the minute hand in 2 hrs 20 min,
i.e., 140 min = $(360/60 × 140)^o= 840°$
Answer: (d)
Angle traced by the minute hand in 2 hrs 20 min,
i.e., 140 min = $(360/60 × 140)^o= 840°$
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