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In a race between A and B, both start simultaneously from the same poi

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In a race between A and B, both start simultaneously from the same poi  [#permalink]

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New post 06 Jan 2020, 07:03
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A
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C
D
E

Difficulty:

  45% (medium)

Question Stats:

54% (02:48) correct 46% (02:41) wrong based on 13 sessions

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In a race between A and B, both start simultaneously from the same point but A runs clockwise and B runs anticlockwise. They meet for the first time at a distance of 300 meters clockwise from the starting point and for the second time at a distance of 200 meters anticlockwise from the starting point. Find the ratio of speeds of A and B, if it is known that A has not completed one full round until the second meeting.

A. 3 : 2
B. 1 : 1
C. 3 : 5
D. 1 : 3
E. Cannot be determined

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Re: In a race between A and B, both start simultaneously from the same poi  [#permalink]

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New post 10 Jan 2020, 01:06
Bunuel wrote:
In a race between A and B, both start simultaneously from the same point but A runs clockwise and B runs anticlockwise. They meet for the first time at a distance of 300 meters clockwise from the starting point and for the second time at a distance of 200 meters anticlockwise from the starting point. Find the ratio of speeds of A and B, if it is known that A has not completed one full round until the second meeting.

A. 3 : 2
B. 1 : 1
C. 3 : 5
D. 1 : 3
E. Cannot be determined


Let the total distance of the track = d and the speeds of A & B = x & y respectively
To find, x : y

When they first met,
Distance traveled by A = 300 & that of B = (d - 300)

Time taken by both A & B to meet each other is same
--> \(\frac{x}{300}\) = \(\frac{y}{(d - 300)}\)
--> \(\frac{d - 300}{300}\) = \(\frac{y}{x}\)
--> \(\frac{d}{300}\) = \(\frac{y}{x} + 1\)
--> \(d\) = \((\frac{y}{x} + 1)300\) ....... (1)

When they met for the second time,
Distance traveled by A = d - 300 - 200 = d - 500 & that of B = 300 + 200 = 500

Time taken by both A & B to meet each other is same
--> \(\frac{x}{d - 500}\) = \(\frac{y}{500}\)
--> \(\frac{x}{y}\) = \(\frac{d - 500}{500}\)
--> \(\frac{d}{500 - 1}\) = \(\frac{x}{y}\)
--> \(\frac{d}{500}\) = \(\frac{x}{y} + 1\)
--> \(d\) = \((\frac{x}{y} + 1)500\) ....... (2)

Equation (1) & (2) and let x/y = a
--> \((\frac{1}{a} + 1)300\) = \((a + 1)500\)
--> \(\frac{3(a + 1)}{a} = 5(a + 1)\)
--> \((a + 1)(\frac{3}{a} - 5) = 0\)
--> \((\frac{3}{a} - 5) = 0\) or \(a + 1 = 0\)
--> a = 3/5 or a = -1 (-1 is not possible!)
--> x : y = 3 : 5

IMO Option C
GMAT Club Bot
Re: In a race between A and B, both start simultaneously from the same poi   [#permalink] 10 Jan 2020, 01:06
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