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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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New post 27 Mar 2013, 06:21
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A
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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
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Re: A car traveled 75% of the way from town A to town B at  [#permalink]

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New post 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.

Answer: 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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New post 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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New post 20 Nov 2013, 02:35
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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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New post 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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New post 26 May 2014, 12:05
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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.

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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New post 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.

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.

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

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New post 27 May 2014, 02:16
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Setting up the equation for time

Total time consumed in the Journey = Addition of time of individual journeys

Refer the diagram below
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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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New post 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!
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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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New post 24 Jun 2014, 03:12
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.
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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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New post 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?
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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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New post 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
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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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New post 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 percent of the way from town A to town B at an avera  [#permalink]

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New post 31 Oct 2017, 09:44
Bunuel wrote:
A car traveled 75 percent 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


let D = 100
average speed = total distance / (T1 + T2)
T1 =75/ 50 = 3/2 ----> time taken to cover first part of the journey
T2 = 25/S ----> time taken to cover second part of the journey

Substituting
40 = 100/(3/2 + 25/S)
S = 25
IMO C
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Re: A car traveled 75 percent of the way from town A to town B at an avera  [#permalink]

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New post 31 Oct 2017, 18:06
Let distance= 100

Time taken to cover first part of the journey= \(\frac{75}{50}\)

................................second part..................= \(\frac{25}{S}\)

Average Speed= \(\frac{Total Distance}{Total Time}\)

40= 100/(\(\frac{3}{2}\)+\(\frac{25}{S}\))

Solving

S= 25

Answer: 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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Re: A car traveled 75% of the way from town A to town B at an   [#permalink] 27 May 2020, 05:10

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

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