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 Post subject: GMAT Speed / Distance / Time
PostPosted: Sat Jun 19, 2010 1:53 am 
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Joined: Sun May 30, 2010 3:15 am
Posts: 424
A similar question is here:
http://800score.com/forum/viewtopic.php?t=139
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Image

In the figure above, car A and car B travel around a circular park with a radius of 20 miles. Both cars leave from the same point, the START location shown in the figure. Car A travels counter-clockwise at 60 miles/hour and car B travels clockwise at 40 miles/hour. Car B leaves 20 minutes after car A. Approximately how many minutes does it take for the cars to meet after car A starts?

A. 63
B. 75
C. 83
D. 126
E. 188

(C) To visualize this question, think about a circle as just a line segment joined at its two ends. Let's first cut the circle at the starting point and make it in to a straight line. Notice that cars A and B are at opposite ends of the line traveling towards each other.

A [ ______________________ ] B

We start by solving for the circumference of the park:
C = 2πr.

Plugging 20 into the equation, we get:
C = 2 × 3.14 × 20 = 125.6 miles.

To calculate how long it takes the cars to meet, use the variable T as the time, in hours, that it takes for the cars to meet after car A starts.

When the cars meet, car A will have traveled 60T miles (distance = rate × time) and car B will have traveled 40 × (T – (1/3)) miles (since it left 20 minutes later, it will have been traveling 1/3 of an hour less). Furthermore, the distance both cars will travel combined is 125.6 miles. So we have the equation:

Total Distance = Distance A travels + Distance B travels
125.6 = 60T + 40(T – (1/3))
125.6 = 100T – (40/3).

Then we have approximately: 139 = 100T. So, T = 1.39 hours.This is equivalent to 1.39 × 60 = 83 minutes.

The correct answer is choice (C).
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The answer should be 2 × π × 20 = 60 × T + 60 × 1/3 + 40 × T
T= (2 × π – 1) / 5 Then T should be a little bit bigger than one hour.


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 Post subject: Re: math (test 1, question 19): speed/distance
PostPosted: Sat Jun 19, 2010 1:57 am 
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Joined: Sun May 30, 2010 2:23 am
Posts: 498
In your calculations you've denoted by T the time that car B has traveled. Therefore you need to add 20 minutes to get the time that car A has traveled and you'll get the same answer.


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 Post subject: Re: math (test 1, question 19): speed/distance
PostPosted: Mon Feb 21, 2011 6:24 pm 
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Joined: Sun May 30, 2010 3:15 am
Posts: 424
Why is it assumed that both cars complete the full circumference? Isn't it possible that they meet somewhere along the track?


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 Post subject: Re: math (test 1, question 19): speed/distance
PostPosted: Mon Feb 21, 2011 6:43 pm 
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Joined: Sun May 30, 2010 2:23 am
Posts: 498
They do meet somewhere along the track. We do NOT assume that EACH car completes the circumference.

Since they must meet somewhere on the circumference, each car travels from the starting point till the meeting point. Clearly, these distances make up the circumference:
Image

(The path of car A is marked in blue and the path of car B is marked in red).


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