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

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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, 06:21
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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. 10
B. 20
C. 25
D. 30
E. 37.5
[Reveal] Spoiler: OA

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

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27 Mar 2013, 06:51
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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.

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

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27 Mar 2013, 20:08
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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.
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Re: A car traveled 75% of the way from town A to town B at an [#permalink]

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20 Nov 2013, 02:35
Total distance = 100 miles (easier to work with %)
75% of the distance = 75 miles
25% of the distance = 25 miles

1st part of the trip → 75/50 = 1.5
2nd part of the trip → 25/S = t
Total trip → (75+25)/40 = 1.5 + t » 100/40 = 1.5 + t » 2.5 = 1.5 + t » t = 1

Back to 2nd part of the trip formula: 25/S = 1 » S = 25

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

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26 May 2014, 00:39
I just solved it algebraically, and better believe me - Assuming a Value for the Distance is way faster.
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Re: A car traveled 75% of the way from town A to town B at [#permalink]

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26 May 2014, 12:05
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.

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

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26 May 2014, 19:36
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ronr34 wrote:
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.

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.

Ratio of 'Distance traveled' cannot act as the weight. See this post for a discussion of this concept: bill-travels-first-40-of-the-distance-to-his-destination-at-137000.html#p1172411

For this question, use the regular formula:

Avg Speed = Total Distance/Total Time $$= \frac{100}{75/50 + 25/S} = 40$$
This gives S = 25 mph
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews SVP Status: The Best Or Nothing Joined: 27 Dec 2012 Posts: 1858 Location: India Concentration: General Management, Technology WE: Information Technology (Computer Software) Followers: 47 Kudos [?]: 1935 [0], given: 193 Re: A car traveled 75% of the way from town A to town B at an [#permalink] ### Show Tags 27 May 2014, 02:16 Setting up the equation for time Total time consumed in the Journey = Addition of time of individual journeys Refer the diagram below Attachments spe.jpg [ 37.13 KiB | Viewed 6938 times ] _________________ Kindly press "+1 Kudos" to appreciate Intern Joined: 15 May 2014 Posts: 10 Followers: 0 Kudos [?]: 6 [0], given: 6 Re: A car traveled 75% of the way from town A to town B at an [#permalink] ### Show Tags 23 Jun 2014, 20:23 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! Math Expert Joined: 02 Sep 2009 Posts: 36590 Followers: 7091 Kudos [?]: 93344 [0], given: 10557 Re: A car traveled 75% of the way from town A to town B at an [#permalink] ### Show Tags 24 Jun 2014, 03:12 Expert's post 1 This post was BOOKMARKED ravih wrote: 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! Well, the question talks only about the percentages of the total distance, so whatever the actual distance is the answer must remain the same. _________________ Manager Joined: 24 Oct 2012 Posts: 65 WE: Information Technology (Computer Software) Followers: 0 Kudos [?]: 17 [0], given: 5 Re: A car traveled 75% of the way from town A to town B at an [#permalink] ### Show Tags 24 Jun 2014, 21:58 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? Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7125 Location: Pune, India Followers: 2137 Kudos [?]: 13677 [0], given: 222 Re: A car traveled 75% of the way from town A to town B at an [#permalink] ### Show Tags 24 Jun 2014, 22:44 GMatAspirerCA wrote: 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. 100 km at 50 kmph 100 km at 100 kmph Avg speed = 2*50*100/(100+50) = 66.7 kmph Avg Speed = 200/3 = 66.7 kmph _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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

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26 Jun 2014, 11:31
Thanks Karisma. I donot have many kudos to give away. but your reply meant alot to clarify concepts.
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Re: A car traveled 75% of the way from town A to town B at an   [#permalink] 22 Dec 2016, 10:39
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