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# Car X left Town T traveling at an average speed of 40 miles per hour.

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Re: Car X left Town T traveling at an average speed of 40 miles per hour. [#permalink]
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GMATinsight wrote:
Bunuel wrote:
Car X left Town T traveling at an average speed of 40 miles per hour. Car Y left Town T 18 minutes after car X left Town T, and car Y traveled at an average speed of 54 miles per hour. When car Y had traveled for z minutes, car Y had traveled 23 miles more that car X had from the time that car X left Town T. What is the value of z?

For Y to have travelled 23 miles more than X, Extra DIstance to be travelled by Y iver X during it's journey = 12+23 = 35 miles

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GMATinsight Thanks for this! I had a quick doubt..

When you say Y has to travel 12+23 = 35 miles, why did we not consider the distance that even X travels? A bit confused here.. I got via approximation as below

X travels 12 miles in 18 minutes.

Y starts and both travel at 14 miles (relative speed)

Now from this time, they both meet after (6/7)th of an hour... [Given d=12 and s=14 and hence t=12/14 which is 6/7)

This is the T1 say, and now to calculate remaining time where Y makes that gap of 23 miles, we continue the relative journey.

Over next 1 hour (60 minutes), distance of 14 is created and another 9 is created as follows

60 minutes -- 14 miles
X minutes -- 9 miles

X minutes = (9*60)/14 = 38 odd minutes.

Now we have 38+60+(6/7 of an hour)

98 + 51 minutes which is 149 minutes and hence 150
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Car X left Town T traveling at an average speed of 40 miles per hour. [#permalink]
I solved this question as following:

If car X is 18 minutes ahead of car Y when the latter starts moving, that means that car Y got 12 miles to catch up: 2/3 mpmin x 18 min= 12 miles (40mph=2/3mpmin).

Using the concept of relative speed, it will take car Y : t= 12miles/14mph= 6/7 hours to close the gap and meet car X.

Now car Y is 23 miles ahead of car X.

Car Y is 14 mph faster than car X, or in other words, every hour that goes by car Y gets 14 miles ahead of car X.

Therefore (again using the concept of relative speed), It will take car Y; t=23miles/14mph to create a gap of 23 miles between itself and car X.

Hence the total time it took car Y to catch up to car X and advance it by 23 miles is: (6/7+23/14) hours= (35/14) hours= (35x60/7) minutes= 150 minutes.

RahulJain293 I beleive the method followed by GMATinsight is similar to your and my method, he just added the 12 miles to the 23 miles from the start and applied the concept of relative speed at one go (12/14+23/14 is the same thing as 35/14).
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Car X left Town T traveling at an average speed of 40 miles per hour. [#permalink]
Car y left 18 minutes after car x, this distance covered within period

d = 40 x 3/10

= 12 miles.

Car y then travelled 23 miles ahead.

Total distance covered

= 12 + 23

= 35

Effective speed

54 - 40

= 14

Time that will happen

35/14 x 60

= 150 minutes.

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Car X left Town T traveling at an average speed of 40 miles per hour. [#permalink]
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