Rate problem!

[bobdylan]

Train Y leaves NY at 1:00 am and travels east at an average speed of x mph. If train Z leaves NY at 2:00 am and travels east, at what average rate of speed will train Z have to travel in order to catch train Y at exactly 5:30 am?
a. 5/6 x
b. 9/8 x
c. 6/5 x
d. 9/7 x
3/2 x

[dimochka]

Let Dy = Distance train Y traveled, and Dz = Distance train Z traveled
When the trains catch up, they will have traveled the same distance, so Dy = Dz

We know that Distance = rate * time, and we know that train Y travels for 4.5 hours (from 1am till 5:30am) whereas train Z travels 3.5 hours. We need to solve for the speed of the 2nd train (Sz) using this information:

x * 4.5 = Sz * 3.5
[spoiler]Sz = 4.5x/3.5 = 9x/7[/spoiler]

[Ashujain]

Train Y leaves NY at 1:00 am and travels east at an average speed of x mph. If train Z leaves NY at 2:00 am and travels east, at what average rate of speed will train Z have to travel in order to catch train Y at exactly 5:30 am?
a. 5/6 x
b. 9/8 x
c. 6/5 x
d. 9/7 x
3/2 x

Distance traveled by train Y = 4.5 * x miles
Therefore, speed of train z = 4.5x/3.5 = 9x/7
Hence, D

[raunekk]

Yup answer is D

Distance traveled by both the trains would be the same.

x * 4.5 = 3.5 * a (Let a = avg. speed of Z)

a = 9/7x

Hope this helps!

[GMATGuruNY]

Train Y leaves NY at 1:00 am and travels east at an average speed of x mph. If train Z leaves NY at 2:00 am and travels east, at what average rate of speed will train Z have to travel in order to catch train Y at exactly 5:30 am?
a. 5/6 x
b. 9/8 x
c. 6/5 x
d. 9/7 x
3/2 x

Time and rate are RECIPROCALS.
Y takes 4.5 hours to travel the same distance that Z travels in 3.5 hours.
Since the ratio of the TIMES = (4.5)/(3.5) = 9/7, the ratio of the RATES = 7/9.

Thus, if Y's rate = x = 7 miles per hour, then Z's rate = 9 miles per hour. This is our target.

Now we plug x=7 into the answers to see which yields our target of 9.
Only answer choice D works:
(9/7)x = (9/7)7 = 9.

The correct answer is D.