Refer diagram below:



Setting up the time equation:

Total time = Time required in first quarter + Time required in the remaining journey

\(\frac{d}{60} = \frac{d}{4*90} + \frac{3d}{4*s}\)

\(s = \frac{90*3}{5} = 54\)

Answer = C

Using the RTD chart, with A being the first quarter, B the rest of the trip and All the combined "trip".

_______R_____T______D
A.........90........3..........270
B.........54.......15.........810
All........60.......18........1080

Let me explain how we fill in the chart:
1) Starting with what we know, we add 90 and 60
2) Picking an easy number for the total distance (6*9=54), so I chose 540. Multiply by 2 to get the whole trip, and we get 1080. Add 1080 for all-distance.
3) 1/4 of the whole distance happened at a Rate of 90. So, 1080/4=270, add 270 under A-D.
4) 2080 - 270 = 810, for the rest of the trip. Add 810 under B-D.
5) Claculate the individual Times for A and All, using T=D/R. Add the results, 3 and 18, under A-T and All-T.
6) 18-3=15, this is the remaining Time for B. Add 15 under B-T.
7) Finally, 15R=810 -->R=810/15 --> R= 54 ANS C

We have an average rate problem in which we can use the following formula:

Avg speed = (distance 1 + distance 2)/(time 1 + time 2)

If we let d = total distance of the trip, then the first quarter of the trip, or (1/4)d = d/4, was traveled at 90 mph. Thus, the time was (d/4)/90 = d/360.

We can let the rate for the remaining part of the trip = r, and thus the time for the remaining part of the trip, or (3/4)d = 3d/4, is (3d/4)/r = 3d/(4r). Let’s use all of this information in the average rate equation:

60 = d/(d/360 + 3d/(4r))

60 = 1/(1/360 + 3/(4r))

60(1/360 + 3/(4r)) = 1

1/6 + 45/r = 1

Let’s multiply the above equation by 6r:

r + 270 = 6r

270 = 5r

r = 54

Answer: C
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Hi All,

This question can be solved by TESTing VALUES.

The prompt tells us that average speed of the train over the course of the entire trip is 60 miles/hr and that it travels the first quarter of the distance at a speed of 90 mi/hr. We're asked for the average speed of the train for the remaining three-quarters of the trip.

Let's choose 90 miles for the FIRST QUARTER of the distance. We can then immediately calculate two things...
1) Since the train was traveling 90 miles/hour, the first quarter of the trip took 1 hour to complete.
2) Since (1/4)(Total Distance) = 90 miles, then the FULL TRIP = 4(90) = 360 miles.

The total trip is 360 miles; with an average speed of 60 miles/hr, the FULL TRIP would take... (X)(60 mph) = 360 miles.... X = 6 hours to complete.

The first hour of the trip covered 90 miles, so the remaining 360 - 90 = 270 miles of the trip are covered in the remaining 5 hours...

Thus, the average speed for the remainder of the trip is... (270 miles)/(5 hours) = 54 miles/hour

Final Answer:

GMAT assassins aren't born, they're made,
Rich
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1/4. (90).........................Avg. 60................................3/4 (x)
So I wrote this (x-60)/(60-90)= 1/3
(x-60)/-30= 1/3
I am getting x= 50.

So for remaining trip the speed is 50.
Now please tell me where I did it wrong.

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