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

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Joined: 02 Jan 2017
Posts: 81
Location: Pakistan
Concentration: Finance, Technology
GMAT 1: 650 Q47 V34
GPA: 3.41
WE: Business Development (Accounting)
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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 10 Dec 2017, 04:27
WholeLottaLove wrote:
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?




It will not matter if it is half of 50 or 1/3rd of 50 miles. Whatever the distance that may be traveled is to be divided by the speed it was covered at. Why? Because we want to find out one chunk of the time till that distance and find the rest chunk using the journey's average speed ( 60 Miles hour).

As average speed = total distance travelled/ Total time
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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 01 Jan 2018, 00:42
Average Speed = Total Distance / Total Time

60 = 40 / (t1+t2)

20 = 50 * t1 , 20 = x * t2

solving we get x = 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 22 Mar 2018, 09:05
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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


The total distance is 40 miles, and we want the average speed to be 60 miles per hour.
Average speed = (total distance)/(total time)
So, we get: 60 = (40 miles)/(total time)
Solve equation to get: total time = 2/3 hours
So, the TIME for the ENTIRE 40-mile trip needs to be 2/3 hours.

The driver completed the first 20 miles of a 40-mile trip at an average speed of 50 miles per hour.
How much time was spent on this FIRST PART of the trip?
time = distance/speed
So, time = 20/50 = 2/5 hours

The ENTIRE trip needs to be 2/3 hours, and the FIRST PART of the trip took 2/5 hours

2/3 hours - 2/5 hours = 10/15 hours - 6/15 hours
= 4/15 hours
So, the SECOND PART of the trip needs to take 4/15 hours


The SECOND PART of the trip is 20 miles, and the time is 4/15 hours
Speed = distance/time
So, speed = 20/(4/15)
= (20)(15/4)
= 75 mph

Answer: D

Cheers,
Brent
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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 27 Mar 2018, 08:44
A good formula to memorize is the harmonic mean formula. When the distance travelled in each segment is constant, the added rate for each leg of the trip (depending on how it's provided in the question) will be equal to the average rate for the whole trip.

(D1/S1) + (D2/S2) = Dtot/Avg S

D1 = distance 1
D2 = distance 2
S1 = speed 1
S2 = speed 2
Dtot = distance total
AvgS = average speed

Now you can plug in the given values and solve for the unknown (S2).

(20/50) + (20/X) = 40/60
X = 75
Re: A driver completed the first 20 miles of a 40 miles trip at   [#permalink] 27 Mar 2018, 08:44

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