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A driver completed the first 20 miles of a 40 miles trip at

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A driver completed the first 20 miles of a 40 miles trip at [#permalink]

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A driver completed the first 20 miles of a 40 miles trip at an average speed of 50 miles per hour. At what average speed must the driver complete the remaining 20 miles to achieve an average speed of 60 miles per hour for the entire 40-miles trip? ( Assume that the driver did not make any stops during the 40-miles trip)

A. 65
B. 68
C. 70
D. 75
E. 80

[Reveal] Spoiler:
Why not 70?
50 mph + x mph / 2 = 60 mph, so x = 70 since the first 20 miles ans the other 20 miles are the same distnace.
What's wrong with my reasoning? :cry:
[Reveal] Spoiler: OA

Last edited by Bunuel on 21 Oct 2013, 01:31, edited 2 times in total.
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Re: OG Quant PS 142 [#permalink]

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eybrj2 wrote:
A driver completed the first 20 miles od a 40 miles trip at an average speed of 50 miles per hour. At what average speed must the driver complete the remaining 20 miles to achieve an average speed of 60 miles per hour for the entire 40-miles trip? ( Assume that the driver did not make any stops during the 40-miles trip)

a) 65

b) 68

c) 70

d) 75

e) 80

Why not 70?
50 mph + x mph / 2 = 60 mph, so x = 70 since the first 20 miles ans the other 20 miles are the same distnace.
What's wrong with my reasoning? :cry:


LET X=20 MILES

x/50+x/y=2x/60
=> 1/y=1/30-1/50=1/75
=>y=75

HENCE D.

P.S.: You are doing direct average/ weighted average of speed, thats wrong. you need to check that the time it takes to cover the two individual 20 miles trip should be equal to the total time its takes to cover 40 miles with average speed 60 mph.

Hope this helps...!!
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Re: OG Quant PS 142 [#permalink]

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eybrj2 wrote:
A driver completed the first 20 miles od a 40 miles trip at an average speed of 50 miles per hour. At what average speed must the driver complete the remaining 20 miles to achieve an average speed of 60 miles per hour for the entire 40-miles trip? ( Assume that the driver did not make any stops during the 40-miles trip)

a) 65

b) 68

c) 70

d) 75

e) 80

Why not 70?
50 mph + x mph / 2 = 60 mph, so x = 70 since the first 20 miles ans the other 20 miles are the same distnace.
What's wrong with my reasoning? :cry:


avg speed = total distance/total time

t1 = 20/50h = 0.4h
t2 = 20/x h

60 = 40/(0.4 + 20/x)

x= 75
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Re: A driver completed the first 20 miles od a 40 miles trip at [#permalink]

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eybrj2 wrote:
A driver completed the first 20 miles od a 40 miles trip at an average speed of 50 miles per hour. At what average speed must the driver complete the remaining 20 miles to achieve an average speed of 60 miles per hour for the entire 40-miles trip? ( Assume that the driver did not make any stops during the 40-miles trip)

A. 65
B. 68
C. 70
D. 75
E. 80

Why not 70?
50 mph + x mph / 2 = 60 mph, so x = 70 since the first 20 miles ans the other 20 miles are the same distnace.
What's wrong with my reasoning? :cry:




Since the distance is same in both stretches of the journey,therefore average speed is Harmonic mean of the speed

Average speed = 2uv/(u + v)

Average speed = 60 ,
U or V = 50

60 = 2*50*v/(50 + v)
3 = 5v/(50 + v)
150 + 3v = 5v
2v=150
v = 75

Hence D

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Re: A driver completed the first 20 miles of a 40 miles trip at [#permalink]

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New post 09 Nov 2013, 08:31
I know how to solve this problem, but what happens if the distances traveled are different? In other words, in this problem we were told he covered one half of the distance @50 mph but what if he covered 1/3rd of the distance at 50 mph?

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Re: A driver completed the first 20 miles of a 40 miles trip at [#permalink]

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New post 31 Dec 2013, 08:27
eybrj2 wrote:
A driver completed the first 20 miles of a 40 miles trip at an average speed of 50 miles per hour. At what average speed must the driver complete the remaining 20 miles to achieve an average speed of 60 miles per hour for the entire 40-miles trip? ( Assume that the driver did not make any stops during the 40-miles trip)

A. 65
B. 68
C. 70
D. 75
E. 80

[Reveal] Spoiler:
Why not 70?
50 mph + x mph / 2 = 60 mph, so x = 70 since the first 20 miles ans the other 20 miles are the same distnace.
What's wrong with my reasoning? :cry:


Is there any short approach to solve problems such as this one? Is it possible to use harmonic mean for instance?

Let us know please

Cheers!
J :)

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A driver completed the first 20 miles of a 40-mile trip at an av [#permalink]

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sem wrote:
A driver completed the first 20 miles of a 40-mile trip at an average speed of 50 miles per hour. At what average speed must the driver complete the remaining 20 miles to achieve an average speed of 60 miles per hour for the entire 40-mile trip? (Assume that the driver did not make any stops during the 40-mile trip.)

A. 65 mph
B. 68 mph
C. 70 mph
D. 75 mph
E. 80 mph


Basically, the first 20 miles took 24 minutes, so the second 20 miles need to take 16 minutes in order for the average to be 60miles/h..


We need to pick an option from A-E (which we call X), that in the denominator makes 120/(16*x) = 1. Note that the value needs to be divided by 10 before it is multiplied by 16.. The only value that works is D (16 * 7.5 = 120), and thus D is the answer.


Don't even ask me how I came to solve it with this convoluted mumbo jumbo but my brain worked on full gear and at the time that I did this it made perfect sense, even though Im not that good at explaining the whole process in hindsight.

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Re: A driver completed the first 20 miles of a 40 miles trip at [#permalink]

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You can use the "normal" distance/rate approach.

First, divide the trip:
For the whole trip he has to take 40 minutes (2/3 h) because he is driving at an average speed of 60m/h.
So first 20 miles at 50 m/h means that he takes 24 min (2/5 h) for half the distance.
This means that he has to take 16 minutes = 16/60 = 4 / 15 h for the last 20 miles. This gives us the equation:

distance = rate(x) * time
20 = x * 4/15
20*15/4 = x
300/4 = x
75 = x

Answer D.

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Re: OG Quant PS 142 [#permalink]

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New post 21 Jan 2014, 16:09
harshavmrg wrote:
Quote:
avg speed = total distance/total time

t1 = 20/50h = 0.4h
t2 = 20/x h

60 = 40/(0.4 + 20/x)

x= 75


can you show your steps for solving for X

\(60=\frac{40}{(0.4+20/x)}\)

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Re: A driver completed the first 20 miles of a 40-mile trip at a [#permalink]

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New post 21 Jan 2014, 16:16
aeglorre wrote:
sem wrote:
A driver completed the first 20 miles of a 40-mile trip at an average speed of 50 miles per hour. At what average speed must the driver complete the remaining 20 miles to achieve an average speed of 60 miles per hour for the entire 40-mile trip? (Assume that the driver did not make any stops during the 40-mile trip.)

A. 65 mph
B. 68 mph
C. 70 mph
D. 75 mph
E. 80 mph


Basically, the first 20 miles took 24 minutes, so the second 20 miles need to take 16 minutes in order for the average to be 60miles/h..


We need to pick an option from A-E (which we call X), that in the denominator makes 120/(16*x) = 1. Note that the value needs to be divided by 10 before it is multiplied by 16.. The only value that works is D (16 * 7.5 = 120), and thus D is the answer.


Don't even ask me how I came to solve it with this convoluted mumbo jumbo but my brain worked on full gear and at the time that I did this it made perfect sense, even though Im not that good at explaining the whole process in hindsight.


why are you setting your problem to 40 miles in 40 mins? Where did 120 come from?

If you cant explain ANYTHING why write an explanation?

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Re: A driver completed the first 20 miles of a 40-mile trip at a [#permalink]

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TroyfontaineMacon wrote:
aeglorre wrote:
sem wrote:
A driver completed the first 20 miles of a 40-mile trip at an average speed of 50 miles per hour. At what average speed must the driver complete the remaining 20 miles to achieve an average speed of 60 miles per hour for the entire 40-mile trip? (Assume that the driver did not make any stops during the 40-mile trip.)

A. 65 mph
B. 68 mph
C. 70 mph
D. 75 mph
E. 80 mph


Basically, the first 20 miles took 24 minutes, so the second 20 miles need to take 16 minutes in order for the average to be 60miles/h..


We need to pick an option from A-E (which we call X), that in the denominator makes 120/(16*x) = 1. Note that the value needs to be divided by 10 before it is multiplied by 16.. The only value that works is D (16 * 7.5 = 120), and thus D is the answer.


Don't even ask me how I came to solve it with this convoluted mumbo jumbo but my brain worked on full gear and at the time that I did this it made perfect sense, even though Im not that good at explaining the whole process in hindsight.


why are you setting your problem to 40 miles in 40 mins? Where did 120 come from?



It's an instinctive method you often use when you learn to play with numbers in your head.
You want the average speed to be 60 miles/hr i.e. you need to cover 60 miles in 60 mins which means you must cover 40 miles in 40 mins.
The first 20 miles were covered at an average speed of 50 mph i.e. time taken (in mins) to cover the first 20 miles is 20/50 * 60 = 24 mins
You need to cover 40 miles in total 40 mins and you have already taken 24 mins during the first 20 miles. This means, you need to speed up now and cover the rest of the 20 miles in the leftover 16 mins.
What will be your speed in mph if you cover 20 miles in 16/60 hrs?
Speed = 20/(16/60) = 75 mph
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Re: A driver completed the first 20 miles of a 40 miles trip at [#permalink]

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A driver completed the first 20 miles of a 40 miles trip at an average speed of 50 miles per hour. At what average speed must the driver complete the remaining 20 miles to achieve an average speed of 60 miles per hour for the entire 40-miles trip? ( Assume that the driver did not make any stops during the 40-miles trip)

A. 65
B. 68
C. 70
D. 75
E. 80


Let \(x\) be the average speed during remaining 20 miles

Total Trip Time = Time to cover first 20 miles + time to cover remaining 20 miles
Since total trip distance is twice that of first part of the trip, we may write above equation as

\(\frac{1}{50} + \frac{1}{x}= \frac{2}{60}\)

Or,\(\frac{1}{x} =\frac{2}{60} - \frac{1}{50}\)
Or, \(x = \frac{300}{4}=75\) miles/hr

Answer: (D)

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Re: A driver completed the first 20 miles of a 40-mile trip at a [#permalink]

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New post 26 Feb 2014, 17:14
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TroyfontaineMacon wrote:
aeglorre wrote:
sem wrote:
A driver completed the first 20 miles of a 40-mile trip at an average speed of 50 miles per hour. At what average speed must the driver complete the remaining 20 miles to achieve an average speed of 60 miles per hour for the entire 40-mile trip? (Assume that the driver did not make any stops during the 40-mile trip.)

A. 65 mph
B. 68 mph
C. 70 mph
D. 75 mph
E. 80 mph


Basically, the first 20 miles took 24 minutes, so the second 20 miles need to take 16 minutes in order for the average to be 60miles/h..


We need to pick an option from A-E (which we call X), that in the denominator makes 120/(16*x) = 1. Note that the value needs to be divided by 10 before it is multiplied by 16.. The only value that works is D (16 * 7.5 = 120), and thus D is the answer.


Don't even ask me how I came to solve it with this convoluted mumbo jumbo but my brain worked on full gear and at the time that I did this it made perfect sense, even though Im not that good at explaining the whole process in hindsight.


why are you setting your problem to 40 miles in 40 mins? Where did 120 come from?

If you cant explain ANYTHING why write an explanation?


If the average rate is 60mph, then 40mi must be completed in \(\frac{2}{3}\) hr (or 40 minutes). If \(\frac{40miles}{(2/3hrs)}\) is 60mph, then the rest of the distance must be covered in 16minutes. What rate is \(\frac{20miles}{16minutes}\)?


rate = \(\frac{20mi}{(16/60)hr}\) = \(\frac{20mi}{(4/15)hr}\) =\(\frac{300mi}{4hr}\) = 75mph

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Re: A driver completed the first 20 miles of a 40 miles trip at [#permalink]

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New post 17 Jun 2015, 07:15
eybrj2 wrote:
A driver completed the first 20 miles of a 40 miles trip at an average speed of 50 miles per hour. At what average speed must the driver complete the remaining 20 miles to achieve an average speed of 60 miles per hour for the entire 40-miles trip? ( Assume that the driver did not make any stops during the 40-miles trip)

A. 65
B. 68
C. 70
D. 75
E. 80

[Reveal] Spoiler:
Why not 70?
50 mph + x mph / 2 = 60 mph, so x = 70 since the first 20 miles ans the other 20 miles are the same distnace.
What's wrong with my reasoning? :cry:


Can we use this formula:

60 = 20/50 + 20/x ?? If yes, what is the way to solve for x ?

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Re: A driver completed the first 20 miles of a 40 miles trip at [#permalink]

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New post 17 Jun 2015, 07:29
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LaxAvenger wrote:
eybrj2 wrote:
A driver completed the first 20 miles of a 40 miles trip at an average speed of 50 miles per hour. At what average speed must the driver complete the remaining 20 miles to achieve an average speed of 60 miles per hour for the entire 40-miles trip? ( Assume that the driver did not make any stops during the 40-miles trip)

A. 65
B. 68
C. 70
D. 75
E. 80

[Reveal] Spoiler:
Why not 70?
50 mph + x mph / 2 = 60 mph, so x = 70 since the first 20 miles ans the other 20 miles are the same distnace.
What's wrong with my reasoning? :cry:


Can we use this formula:

60 = 20/50 + 20/x ?? If yes, what is the way to solve for x ?



NO, you can't use this Principle. In fact the only principle for calculating Average Speed is

Average Speed = Total Distance / Total Time

Solving Equation for you:

Here, Average Speed = 60
Total Distance = 40
Total Time = Time taken in Travelling 1st 20 Miles + Time taken in Travelling 2nd 20 Miles = (20/50) + (20/x) [Because Time = Distance/Speed]

i.e. 60 = 40/ [20/50 + 20/x]

i.e. 60 = 40/ 20[1/50 + 1/x] = 2/ [(x+50)/(50x)] = 2*50x / [(x+50)] = 100x / [(x+50)]

i.e. 60*[(x+50)] = 100x

i.e. 60x + 30000 = 100x

i.e. 40x = 3000

i.e. x = 300/4 = 75 mph

I hope It clears your Doubt! :)
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Re: A driver completed the first 20 miles of a 40 miles trip at [#permalink]

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New post 17 Jun 2015, 08:30
GMATinsight wrote:
LaxAvenger wrote:
eybrj2 wrote:
A driver completed the first 20 miles of a 40 miles trip at an average speed of 50 miles per hour. At what average speed must the driver complete the remaining 20 miles to achieve an average speed of 60 miles per hour for the entire 40-miles trip? ( Assume that the driver did not make any stops during the 40-miles trip)

A. 65
B. 68
C. 70
D. 75
E. 80

[Reveal] Spoiler:
Why not 70?
50 mph + x mph / 2 = 60 mph, so x = 70 since the first 20 miles ans the other 20 miles are the same distnace.
What's wrong with my reasoning? :cry:


Can we use this formula:

60 = 20/50 + 20/x ?? If yes, what is the way to solve for x ?



NO, you can't use this Principle. In fact the only principle for calculating Average Speed is

Average Speed = Total Distance / Total Time

Solving Equation for you:

Here, Average Speed = 60
Total Distance = 40
Total Time = Time taken in Travelling 1st 20 Miles + Time taken in Travelling 2nd 20 Miles = (20/50) + (20/x) [Because Time = Distance/Speed]

i.e. 60 = 40/ [20/50 + 20/x]

i.e. 60 = 40/ 20[1/50 + 1/x] = 2/ [(x+50)/(50x)] = 2*50x / [(x+50)] = 100x / [(x+50)]

i.e. 60*[(x+50)] = 100x

i.e. 60x + 30000 = 100x

i.e. 40x = 3000

i.e. x = 300/4 = 75 mph

I hope It clears your Doubt! :)


Thank you so far!

What about if we use this equation: (from first answer)

60 = 40/(0.4 + 20/x)

x= 75

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Re: A driver completed the first 20 miles of a 40 miles trip at [#permalink]

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New post 17 Jun 2015, 08:38
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LaxAvenger wrote:
GMATinsight wrote:
LaxAvenger wrote:
A driver completed the first 20 miles of a 40 miles trip at an average speed of 50 miles per hour. At what average speed must the driver complete the remaining 20 miles to achieve an average speed of 60 miles per hour for the entire 40-miles trip? ( Assume that the driver did not make any stops during the 40-miles trip)

A. 65
B. 68
C. 70
D. 75
E. 80

[Reveal] Spoiler:
Why not 70?
50 mph + x mph / 2 = 60 mph, so x = 70 since the first 20 miles ans the other 20 miles are the same distnace.
What's wrong with my reasoning? :cry:


Can we use this formula:

60 = 20/50 + 20/x ?? If yes, what is the way to solve for x ?



NO, you can't use this Principle. In fact the only principle for calculating Average Speed is

Average Speed = Total Distance / Total Time

Solving Equation for you:

Here, Average Speed = 60
Total Distance = 40
Total Time = Time taken in Travelling 1st 20 Miles + Time taken in Travelling 2nd 20 Miles = (20/50) + (20/x) [Because Time = Distance/Speed]

i.e. 60 = 40/ [20/50 + 20/x]

i.e. 60 = 40/ 20[1/50 + 1/x] = 2/ [(x+50)/(50x)] = 2*50x / [(x+50)] = 100x / [(x+50)]

i.e. 60*[(x+50)] = 100x

i.e. 60x + 30000 = 100x

i.e. 40x = 3000

i.e. x = 300/4 = 75 mph

I hope It clears your Doubt! :)


Thank you so far!

What about if we use this equation: (from first answer)

60 = 40/(0.4 + 20/x)

x= 75

?[/quote]


This equation 60 = 40/(0.4 + 20/x) is same as the equation I mentioned in my solution

i.e. 60 = 40/ [20/50 + 20/x]
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Re: A driver completed the first 20 miles of a 40 miles trip at [#permalink]

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New post 22 Dec 2015, 22:50
Hey everyone,

I approached the question this way, First 20 miles at an average speed of 50mph, T1=20/50. For second leg, T2=20/x. (x is the required speed). Therefore 20/50+20/x=40/60. (T1+T2=T/D). Solving for x, the answer is 75.

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Re: A driver completed the first 20 miles of a 40 miles trip at [#permalink]

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New post 25 Jan 2016, 10:11
Average Speed = Total Distance/Total Time ...... (i)

So,
T1 = 20/50 hrs.
T2 = 20/x hrs.

Total Time = T1+T2 and total distance = 40

Putting values in (i),

60 = 40/(20/50 + 20/x)

x= 75

(D) - correct ans

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Kudos [?]: 306 [0], given: 17

Re: A driver completed the first 20 miles of a 40 miles trip at [#permalink]

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New post 01 Jul 2017, 16:38
eybrj2 wrote:
A driver completed the first 20 miles of a 40 miles trip at an average speed of 50 miles per hour. At what average speed must the driver complete the remaining 20 miles to achieve an average speed of 60 miles per hour for the entire 40-miles trip? ( Assume that the driver did not make any stops during the 40-miles trip)

A. 65
B. 68
C. 70
D. 75
E. 80


let t=second leg time
40 miles/(2/5+t) hrs=60 mph
t=4/15 hrs
20 miles/(4/15 hrs)=75 mph

Kudos [?]: 306 [0], given: 17

Re: A driver completed the first 20 miles of a 40 miles trip at   [#permalink] 01 Jul 2017, 16:38

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