Reasoning Aptitude
CLOCK MCQs
Clocks
Total Questions : 146
| Page 4 of 15 pages
Answer: Option C. -> 18
Answer: (c)
Time period from 1 O' clock afternoon to 10 O' clock night = (10 - 1)h = 9 h
As we know, hands are at right angle 2 times in an hour.
Hence, in 9 h they will be at right angle 9 × 2 = 18 times
Answer: (c)
Time period from 1 O' clock afternoon to 10 O' clock night = (10 - 1)h = 9 h
As we know, hands are at right angle 2 times in an hour.
Hence, in 9 h they will be at right angle 9 × 2 = 18 times
Answer: Option C. -> 21$ 9/11 $min past 4
Answer: (c)
At 4 O' clock, hour hand is at 4 and minute hand is at 12.
To be together with hour hand minute hand will have to gain 20 min. As 55 min are gained by minute hand in 60 min.
Therefore, 20 min will be gained in $(60/55 × 20)$ min = $(60 × 4/11)$ min = $240 / 11$ min
= 21 $9/11$ min Hence, the hands will be together at 21 $9/11$ min past 4.
Answer: (c)
At 4 O' clock, hour hand is at 4 and minute hand is at 12.
To be together with hour hand minute hand will have to gain 20 min. As 55 min are gained by minute hand in 60 min.
Therefore, 20 min will be gained in $(60/55 × 20)$ min = $(60 × 4/11)$ min = $240 / 11$ min
= 21 $9/11$ min Hence, the hands will be together at 21 $9/11$ min past 4.
Answer: Option A. -> 49$1/11$min past 6 and 16$4/11$min past 6
Answer: (a)
Answer: (a)
Answer: Option D. -> 44
Answer: (d)
We know that the hands of a clock are at right angle twice in every hour but between 2 and 4 O' clock there is a common position at 3 O' clock and also between 8 and 10 O' clock there is common position at 9 O' clock.
So, they are at right angles 22 times in 12 h and therefore, in 24 h or in a day they are at right angle 44 times.
Answer: (d)
We know that the hands of a clock are at right angle twice in every hour but between 2 and 4 O' clock there is a common position at 3 O' clock and also between 8 and 10 O' clock there is common position at 9 O' clock.
So, they are at right angles 22 times in 12 h and therefore, in 24 h or in a day they are at right angle 44 times.
Answer: Option B. -> 22
Answer: (b)
The hands of a clock are in the same straight line (but opposite in direction) 11 times in every 12 h, because between 5 and 7 they point in opposite direction at 6 O' clock only.
Therefore, in a day (24 h) the hands points in the opposite direction (2 × 11) = 22 times.
Answer: (b)
The hands of a clock are in the same straight line (but opposite in direction) 11 times in every 12 h, because between 5 and 7 they point in opposite direction at 6 O' clock only.
Therefore, in a day (24 h) the hands points in the opposite direction (2 × 11) = 22 times.
Answer: Option B. -> 22
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' clock only).
So, in a day, the hands point in the opposite directions 22 times.
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' clock only).
So, in a day, the hands point in the opposite directions 22 times.
Answer: Option D. -> 16 $\text"4″"/11$ past
Answer: (d)
At 3 O' clock, the minute hand is 15 min. Spaces apart from the hour hand. To be coincident, it must gain 15 min. spaces. 55 min. are gained in 60 min. 15 min. are gained in (60/55 × 15)min. = 16 4/11 min.
Therefore, the hands are coincident at 16 4/11 min. past 3.
Answer: (d)
At 3 O' clock, the minute hand is 15 min. Spaces apart from the hour hand. To be coincident, it must gain 15 min. spaces. 55 min. are gained in 60 min. 15 min. are gained in (60/55 × 15)min. = 16 4/11 min.
Therefore, the hands are coincident at 16 4/11 min. past 3.
Answer: Option B. -> 6
Answer: (b)
Number of rotations = $72/12$ = 6
Answer: (b)
Number of rotations = $72/12$ = 6
Answer: Option C. -> 4557.67 cm
Answer: (c)
Number of rounds completed by the minute hand in 3 days 5 hrs
= (3 × 24 + 5) = 77.
Number of rounds completed by the hour hand in 3 days 5 hrs
= $(3 × 2 + 5/12) = 6 5/12$
Therefore, Difference between the distance traversed = $[77 × (2 × 22/7 × 10) - 6 5/12 × (2 × 22/7 × 7)]$ cm
= (4840 - 282.33) cm = 4557.67 cm.
Answer: (c)
Number of rounds completed by the minute hand in 3 days 5 hrs
= (3 × 24 + 5) = 77.
Number of rounds completed by the hour hand in 3 days 5 hrs
= $(3 × 2 + 5/12) = 6 5/12$
Therefore, Difference between the distance traversed = $[77 × (2 × 22/7 × 10) - 6 5/12 × (2 × 22/7 × 7)]$ cm
= (4840 - 282.33) cm = 4557.67 cm.
Answer: Option B. -> 36 $5/11$ min.
Answer: (b)
55 min. spaces are covered in $(60/55 × 60)$ min. = 65 $5/11$ min.
Loss in 64 min. = $(65 5/11 - 64)$ = $16/11$ min.
loss in 24 hrs = $(16/11 × 1/64 × 24 × 60)$ min. = 32 $8/11$ min.
Answer: (b)
55 min. spaces are covered in $(60/55 × 60)$ min. = 65 $5/11$ min.
Loss in 64 min. = $(65 5/11 - 64)$ = $16/11$ min.
loss in 24 hrs = $(16/11 × 1/64 × 24 × 60)$ min. = 32 $8/11$ min.
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