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problem with speed *difficult*

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problem with speed *difficult*

by Taniuca » Mon Nov 01, 2010 5:10 pm
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?

a) 81
b) 90
c) 105
d) 125
e) 150
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by pesfunk » Mon Nov 01, 2010 5:40 pm
After z minutes, Car Y travelled 54*z miles
Till then, Car X had travelled for 18+z minutes

so distance travelled by X = 40 (18 + z )

As per question: 54 * z - 40 ( 18 + z ) = 23

or 54z - 720 - 40z = 23

or 14Z = 743

or z = 53

Where am I going wrong ? :(


Taniuca 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?

a) 81
b) 90
c) 105
d) 125
e) 150
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by pesfunk » Mon Nov 01, 2010 5:46 pm
OK. Missed that the speed was in hours


so 23 has to be multiplied by 60

correct equation => 54 * z - 40 ( 18 + z ) = 23*60

or Z = 150
pesfunk wrote:After z minutes, Car Y travelled 54*z miles
Till then, Car X had travelled for 18+z minutes

so distance travelled by X = 40 (18 + z )

As per question: 54 * z - 40 ( 18 + z ) = 23

or 54z - 720 - 40z = 23

or 14Z = 743

or z = 53

Where am I going wrong ? :(


Taniuca 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?

a) 81
b) 90
c) 105
d) 125
e) 150
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by Testluv » Mon Nov 01, 2010 5:51 pm
Hi,

This is a catch-up problem in which the slower moving object has a headstart on the faster moving one. Car x travels for 18 minutes before Car y moves at all. In this 18 minutes, Car x gains a headstart of 12 miles (since it's travelling at 40mph).

When two objects are travelling in the same direction, we can subtract their rates. That is, Car y will "catch-up" or close this 12 mile headstart at a rate of 54-40 = 14 mph. Once Car y catches up to car x, it will then exceed car x at this same rate. We know that Car y needs to catch up 12 miles, and then exceed Car x for another 23 miles. So, that's a total of 35 miles. The rate of catch-up/excess is 14. So, z or the time of all of this is 35/14 or 2.5 hours or 150 minutes.

Choose E.
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