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Average Speed problem as mixture problem

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Average Speed problem as mixture problem

by Mo2men » Sun Apr 09, 2017 3:07 am
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

OA: C

The question above is could be solved through different ways but I tried to treated as mixture but it goes wrong.

Let first speed V1= 90
The average speed V = 60
second speed V2= X

d1......... V...................d2
90..30...60..60-x.........V2

d1/d2= 60-x/30

1/3 = 60-x/30...........V2=x= 50

Where did I go wrong?
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by GMATGuruNY » Sun Apr 09, 2017 3:28 am
Mo2men wrote: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
Let the time for the first quarter = 1 hour, implying that the 90 miles traveled in the first quarter is equal to 1/4 of the total distance.
Thus, the total distance = 4*90 = 360 miles.
Since the average speed for the whole trip is 60mph, the time to travel the total distance = d/r = 360/60 = 6 hours.

After 90 miles is traveled in the first hour:
Remaining distance = 360-90 = 270 miles.
Remaining time = 6-1 = 5 hours.
Average speed for the remainder of the trip = d/t = 270/5 = 54 miles per hour.

The correct answer is C.
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by Mo2men » Sun Apr 09, 2017 3:31 am
GMATGuruNY wrote:
Mo2men wrote: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
Let the time for the first quarter = 1 hour, implying that the 90 miles traveled in the first quarter is equal to 1/4 of the total distance.
Thus, the total distance = 4*90 = 360 miles.
Since the average speed for the whole trip is 60mph, the time to travel the total distance = d/r = 360/60 = 6 hours.

After 90 miles is traveled in the first hour:
Remaining distance = 360-90 = 270 miles.
Remaining time = 6-1 = 5 hours.
Average speed for the remainder of the trip = d/t = 270/5 = 54 miles per hour.

The correct answer is C.
Thanks Mitch,

But really my question, what is the problem with using Alligation method. It seems a mixture problem.

Can you comment on my work above?
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by GMATGuruNY » Sun Apr 09, 2017 4:52 am
Mo2men wrote:Thanks Mitch,

But really my question, what is the problem with using Alligation method. It seems a mixture problem.

Can you comment on my work above?
In rate problems, alligation must be performed in terms of the TIME spent at each speed.
Let S = the slower speed.
As shown in my solution above, the time ratio for the time spent at 90mph (1 hour) to the time spent at the slower speed (5 hours) = x:5x.

Step 1: Plot the 3 rates on a number line, with the two individual rates (90 and S) on the ends and the average speed for the whole trip (60) in the middle.
90------------60------------S

Step 2: Calculate the distances between the rates.
The distances between the 3 speeds is equal to the RECIPROCAL of the time ratio:
90-----5x-----60-----x-----S

The number line indicates that the distance between 90 and 60 is equal to 5x:
(90-60) = 5x
30 = 5x
x = 6.

Thus, S = 60-x = 60-6 = 54.

The correct answer is C.
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by [email protected] » Sun Apr 09, 2017 10:11 am
Hi Mo2men,

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: C

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by Scott@TargetTestPrep » Wed Dec 13, 2017 12:25 pm
Mo2men wrote: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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