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# The hands of a strange clock move such that they would meet twice as

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Manager
Joined: 18 Jul 2019
Posts: 54
The hands of a strange clock move such that they would meet twice as  [#permalink]

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25 Nov 2019, 05:58
6
00:00

Difficulty:

65% (hard)

Question Stats:

33% (02:01) correct 67% (02:05) wrong based on 12 sessions

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The hands of a strange clock move such that they would meet twice as frequently if they run in opposite directions than if they run in the same direction. How many times would the faster hand meet the slower hand in the time that the slower hand completes 20 rotations, given that the hands run in opposite directions?

a) 60
b) 40
c) 80
d) 120
e) 150
VP
Joined: 19 Oct 2018
Posts: 1174
Location: India
Re: The hands of a strange clock move such that they would meet twice as  [#permalink]

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25 Nov 2019, 11:08
2
1
Speed of faster hand= v

speed of slower hand= u

$$2*\frac{360}{v+u}=\frac{360}{v-u}$$

v=3u

Time taken for the first meeting=$$\frac{ 360}{v+u}$$=$$\frac{90}{u}$$

Distance travelled by slower hand till they meet first time= $$\frac{90}{u}*u$$=90

Total number of meeting while slower hand completes first rotation= $$\frac{360}{90}$$=4

Total number of meetings= 20*4=80

CaptainLevi wrote:
The hands of a strange clock move such that they would meet twice as frequently if they run in opposite directions than if they run in the same direction. How many times would the faster hand meet the slower hand in the time that the slower hand completes 20 rotations, given that the hands run in opposite directions?

a) 60
b) 40
c) 80
d) 120
e) 150
Re: The hands of a strange clock move such that they would meet twice as   [#permalink] 25 Nov 2019, 11:08
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