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A motorcyclist goes from city X to Y, a distance of 192Km, at an avera

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A motorcyclist goes from city X to Y, a distance of 192Km, at an avera  [#permalink]

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New post 24 Jan 2020, 01:53
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A motorcyclist goes from city X to Y, a distance of 192Km, at an average speed of 32Kmph. Another man starts from city X by car, 2.5hrs after the motorcyclist and reaches city Y 0.5hr earlier. What is the ratio of the speed of the motorcyclist to that of the car?

A. 1:2
B. 1:3
C. 5:4
D. 2:1
E. 2:3

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Re: A motorcyclist goes from city X to Y, a distance of 192Km, at an avera  [#permalink]

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New post 24 Jan 2020, 02:10
Time taken by motorcyclist = 192/32 = 6 hours

Lets say the motorcyclist starts at 12pm and reaches at 6pm

This means that the car starts at 2:30pm and reaches at 5:30pm which is 3 hours

So the speed of the car is 192/3 = 64kmph

Ratio of speed of motorcyclist to car is 32/64 = 1/2

Answer is (A)

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Re: A motorcyclist goes from city X to Y, a distance of 192Km, at an avera  [#permalink]

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New post 24 Jan 2020, 02:22
The answers is A.
Time taken by motorcyclist= 192/32= 6hrs
Then, it is given that the car driver reaches 0.5hrs before motorcyclist, therefore 6-0.5=5.5hrs.

Then further, it starts 2.5hrs after the motorcyclist = 5.5-2.5= 3hrs.
It takes car 3hrs to complete the distance of 192.
Then its speed be 192/3=64kmph

Ratio would be 32/64=1:2.

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Re: A motorcyclist goes from city X to Y, a distance of 192Km, at an avera   [#permalink] 24 Jan 2020, 02:22
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