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22 Sep 2020, 05:15
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Difficulty:
95%
(hard)
Question Stats:
25% (01:56) correct
75%
(02:14)
wrong
based on 76
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A clock sounds to indicate the time of 3 o'clock. At what time between 3 and 4 will the hour and minute needles overlap?
A. 3 hours, 16 minutes.
B. 3 hours, 16 minutes and 20 seconds.
C. 3 hours, 16 minutes and 21 seconds.
D. 3 hours, 16 minutes and 22 seconds.
E. 3 hours, 16 minutes and 23 seconds.
A. 3 hours, 16 minutes.
B. 3 hours, 16 minutes and 20 seconds.
C. 3 hours, 16 minutes and 21 seconds.
D. 3 hours, 16 minutes and 22 seconds.
E. 3 hours, 16 minutes and 23 seconds.
22 Sep 2020, 05:34
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A clock is actually an instrument which uses the concept of Relative Speed in the same direction
A clock can be thought of as a circle who's total angles is \(360^o\)
The Minute hand moves \(360^o\) in 60 minutes. Therefore it moves \(6^o\) per minute
The Hour Hand moves 5 minutes in 1 hour = 5 * 6 = \(30^o\) in 1 hour = \(\frac{30}{60} = \frac{1}{2}^o\) per minute
Therefore in 1 hour, the minute hand gains \(6 - \frac{1}{2} = 5.5^o\) over the hour hand. This is our Relative Speed.
We know that at approximately 3:15 or a little ahead, the 2 hands will meet.
That means from 3 o clock, the minute hand has to move 15 minutes or 15 * 6 = \(90^o\) to meet the hour hand
Therefore \(Time = \frac{Distance}{Relative \space Speed} = \frac{90}{5.5} = \frac{180}{11} = 16 \space \frac{4}{11} \space minutes\)
\(\frac{4}{11} \space minutes = \frac{4}{11} \space * \space 60 = 22 \space seconds\)
Therefore the 2 hands are together at 3 hours 16 minutes and 22 seconds
Option D
Arun Kumar
A clock can be thought of as a circle who's total angles is \(360^o\)
The Minute hand moves \(360^o\) in 60 minutes. Therefore it moves \(6^o\) per minute
The Hour Hand moves 5 minutes in 1 hour = 5 * 6 = \(30^o\) in 1 hour = \(\frac{30}{60} = \frac{1}{2}^o\) per minute
Therefore in 1 hour, the minute hand gains \(6 - \frac{1}{2} = 5.5^o\) over the hour hand. This is our Relative Speed.
We know that at approximately 3:15 or a little ahead, the 2 hands will meet.
That means from 3 o clock, the minute hand has to move 15 minutes or 15 * 6 = \(90^o\) to meet the hour hand
Therefore \(Time = \frac{Distance}{Relative \space Speed} = \frac{90}{5.5} = \frac{180}{11} = 16 \space \frac{4}{11} \space minutes\)
\(\frac{4}{11} \space minutes = \frac{4}{11} \space * \space 60 = 22 \space seconds\)
Therefore the 2 hands are together at 3 hours 16 minutes and 22 seconds
Option D
Arun Kumar
General Discussion
Archit3110
Major Poster
Updated on: 22 Sep 2020, 05:38
Originally posted by Archit3110 on 22 Sep 2020, 05:24.
Last edited by Archit3110 on 22 Sep 2020, 05:38, edited 1 time in total.
Last edited by Archit3110 on 22 Sep 2020, 05:38, edited 1 time in total.
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Bunuel
hour and minute hand overlap means angle is 0
11/2* min- 30* hour
option D ; is where we get 0*
.3 hours, 16 minutes and 22 seconds.
11/2 * ( 16+22/60)-3*30 = 0
