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

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

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.