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A car traveled 75% of the way from town A to town B at an [#permalink]

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27 Mar 2013, 07:21

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69% (02:45) correct
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A car traveled 75% of the way from town A to town B at an average speed of 50 miles per hour. The car travels at an average speed of S miles per hour for the remaining part of the trip. The average speed for the entire trip was 40 miles per hour. What is S ?

A car traveled 75% of the way from town A to town B at an average speed of 50 miles per hour. The car travels at an average speed of S miles per hour for the remaining part of the trip. The average speed for the entire trip was 40 miles per hour. What is S ?

A. 10 B. 20 C. 25 D. 30 E. 37.5

Say the entire distance is 200 miles. 75% of the distance = 150 miles. 25% of the distance = 50 miles.

Total time = 200/40 = 5 hours; Time spent to cover 150 miles = 150/50 = 3 hours.

Thus 50 miles was covered in 5-3=2 hours --> S = (speed) = (distance)/(time) = 50/2 = 25 miles per hour.

Re: A car traveled 75% of the way from town A to town B at an [#permalink]

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27 Mar 2013, 21:08

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Let the distance from A to B be 200 miles. 75% of 200 miles = 150 miles.So time taken to cover that = 150/50 hrs = 3 hrs. [time taken = distance covered / speed ] So time taken to cover remaining distance = 50/s hrs.

Total time taken for the trip = 200/40 = 5 hrs. so, 3 + 50/s = 5 =>50/s = 2 => s = 25

Hence,Option C will be the answer. ------------------------------------------------ Please press KUDOS if you like my post. _________________

A car traveled 75% of the way from town A to town B at an average speed of 50 miles per hour. The car travels at an average speed of S miles per hour for the remaining part of the trip. The average speed for the entire trip was 40 miles per hour. What is S ?

A. 10 B. 20 C. 25 D. 30 E. 37.5

Say the entire distance is 200 miles. 75% of the distance = 150 miles. 25% of the distance = 50 miles.

Total time = 200/40 = 5 hours; Time spent to cover 150 miles = 150/50 = 3 hours.

Thus 50 miles was covered in 5-3=2 hours --> S = (speed) = (distance)/(time) = 50/2 = 25 miles per hour.

Answer: C.

I tried to solve it a different way... using weighted averages but hit a wall... Can you see where my logic fails?

Since the first part of the trip was 3/4 and the last was 1/4, this is what I got in my diagram:

S-----40-----50

----1-----3-----

When doing the cross multiplication, I get (50-40)/(40-S) = 3/1 -> I get S=110/3.

Why is this failing?

The weight when calculating average speed is time, not distance. This means that when you write (50-40)/(40-S) = 3/1, you are assuming that the car traveled 75% of the TIME at speed S and 25% of the time at speed 50 mph.

Re: A car traveled 75% of the way from town A to town B at [#permalink]

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26 May 2014, 13:05

Bunuel wrote:

summer101 wrote:

A car traveled 75% of the way from town A to town B at an average speed of 50 miles per hour. The car travels at an average speed of S miles per hour for the remaining part of the trip. The average speed for the entire trip was 40 miles per hour. What is S ?

A. 10 B. 20 C. 25 D. 30 E. 37.5

Say the entire distance is 200 miles. 75% of the distance = 150 miles. 25% of the distance = 50 miles.

Total time = 200/40 = 5 hours; Time spent to cover 150 miles = 150/50 = 3 hours.

Thus 50 miles was covered in 5-3=2 hours --> S = (speed) = (distance)/(time) = 50/2 = 25 miles per hour.

Answer: C.

I tried to solve it a different way... using weighted averages but hit a wall... Can you see where my logic fails?

Since the first part of the trip was 3/4 and the last was 1/4, this is what I got in my diagram:

S-----40-----50

----1-----3-----

When doing the cross multiplication, I get (50-40)/(40-S) = 3/1 -> I get S=110/3.

Re: A car traveled 75% of the way from town A to town B at an [#permalink]

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23 Jun 2014, 21:23

Hey Bunuel,

How do we know when is it ok to plug in an assumed value? In this case, I was hesitant to plug in a value thinking that it may impact the calculation wrongly. Also, there simply isnt enough time to plug in 2 values and check if they each give the same answer!

How do we know when is it ok to plug in an assumed value? In this case, I was hesitant to plug in a value thinking that it may impact the calculation wrongly. Also, there simply isnt enough time to plug in 2 values and check if they each give the same answer!

Well, the question talks only about the percentages of the total distance, so whatever the actual distance is the answer must remain the same.
_________________

Re: A car traveled 75% of the way from town A to town B at an [#permalink]

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24 Jun 2014, 22:58

Some where i read that average speed = 2s1*s2/(s1+s2). Clearly this is not applicable in current scenario. Can some one recollect when does this average speed formula holds good?

Some where i read that average speed = 2s1*s2/(s1+s2). Clearly this is not applicable in current scenario. Can some one recollect when does this average speed formula holds good?

It applies when the distance traveled at the two speeds is the same.

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