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PareshGmat
A train travels from city A to city B. The average speed of the train is 60 miles/hr and it travels the first quarter of the trip at a speed of 90 mi/hr. What is the speed of the train in the remaining trip?

A. 30
B. 45
C. 54
D. 72
E. 90

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

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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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PareshGmat
A train travels from city A to city B. The average speed of the train is 60 miles/hr and it travels the first quarter of the trip at a speed of 90 mi/hr. What is the speed of the train in the remaining trip?

A. 30
B. 45
C. 54
D. 72
E. 90

total time of the trip ; d/60
1/4 of trip time ; d/ 4 * 90
d/60 = d/360 + 3d/ 4*s
solve for s
54
option C
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gvij2017
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.
KarishmaB, I fell into the same trap. Can we not use weighted averages here?
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gvij2017
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.
KarishmaB, I fell into the same trap. Can we not use weighted averages here?

No, we cannot. Check out this blog post: https://anaprep.com/arithmetic-weights- ... d-average/
When averaging speed, the weights are "time taken", not distance.
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Archit3110
PareshGmat
A train travels from city A to city B. The average speed of the train is 60 miles/hr and it travels the first quarter of the trip at a speed of 90 mi/hr. What is the speed of the train in the remaining trip?

A. 30
B. 45
C. 54
D. 72
E. 90

total time of the trip ; d/60
1/4 of trip time ; d/ 4 * 90
d/60 = d/360 + 3d/ 4*s
solve for s
54
option C


Could someone please explain how we solve for s if we have two variables (s and d)?
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Archit3110
PareshGmat
A train travels from city A to city B. The average speed of the train is 60 miles/hr and it travels the first quarter of the trip at a speed of 90 mi/hr. What is the speed of the train in the remaining trip?

A. 30
B. 45
C. 54
D. 72
E. 90

total time of the trip ; d/60
1/4 of trip time ; d/ 4 * 90
d/60 = d/360 + 3d/ 4*s
solve for s
54
option C


Could someone please explain how we solve for s if we have two variables (s and d)?

d gets reduced there, so you'll be left with only one variable, s:

1/60 = 1/360 + 3/(4s)
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Bunuel

Thank you for the qick reply. What do you mean with it gets reduced? How does it? Sorry for asking again!! Making an extra post cause i can not quote the post anymore.
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Bunuel

Thank you for the qick reply. What do you mean with it gets reduced? How does it? Sorry for asking again!! Making an extra post cause i can not quote the post anymore.

"Gets reduced" means we can divide the equation d/60 = d/360 + 3d/(4s) by d to eliminate it.
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Bunuel
HaliGaliGmat
Bunuel

Thank you for the qick reply. What do you mean with it gets reduced? How does it? Sorry for asking again!! Making an extra post cause i can not quote the post anymore.

"Gets reduced" means we can divide the equation d/60 = d/360 + 3d/(4s) by d to eliminate it.
Thanks for the explanation! it helps!
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PareshGmat
A train travels from city A to city B. The average speed of the train is 60 miles/hr and it travels the first quarter of the trip at a speed of 90 mi/hr. What is the speed of the train in the remaining trip?

A. 30
B. 45
C. 54
D. 72
E. 90

total time of the trip ; d/60
1/4 of trip time ; d/ 4 * 90
d/60 = d/360 + 3d/ 4*s
solve for s
54
option C


Could someone please explain how we solve for s if we have two variables (s and d)?
Simply assume a figure, a value, like 360, Quarter of 360 is 90, 90 ms @ 90mph, we get 1 hour, and the remaining 270ms s mph equates to 6 hours, (since we have assumed d = 360 and the total speed given is 60 mph, 360/60 = 6 hours is the total journey time ). 1+270/s = 6, s = 54.
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Take a value for Distance for the whole trip. I chose 180

Distance for 1st quarter = D/4 = 45
Distance for remaining quarter = 180 - 45 = 135

Time for the whole trip =180/60 = 3
Time for 1st quarter =45/90 =0.5
Time for remaining quarter =45/90 =3 - 0.5 = 2.5

Speed for the remaining quarter = 135/2.5 = 54 miles/hour
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