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Car Trip Problem

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by kakz » Mon Feb 27, 2012 10:19 am
A car drives part of a trip at 48 kilometers per hour and the remainder of the same trip at 64 kilometers per hour. What fraction of the total distance covered during the trip must be driven at 48 kilometers per hour so that the average speed over the entire trip is 60 kilometers per hour?

Answer is [spoiler]1/5[/spoiler]
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by Mike@Magoosh » Mon Feb 27, 2012 10:40 am
Hi, there. I'm happy to help with this. :)

Let the distance be D. Let x be the variable we seek, the fraction of the trip the car drives at 48 mph. That means, the driver drives (1-x) of the trip at 64 mph.

First leg:
Distance = xD
Speed = 48
Time = xD/48

Second leg:
Distance = (1-x)D
Speed = 64
Time = (1-x)D/64

Average speed = [total distance]/[total time] = D/[xD/48 + (1-x)D/64]
= D/[4xD/192 + 3(1-x)D/192] = D/[D*(4x + 3(1-x))/192]
= 192/[3 + x]

Set this equal to the average of 60.

60 = 192/[3 + x]

3 + x = 192/60 = 3.2

x = 0.2 = 1/5

Does that make sense? Please let me know if I can answer any further questions.

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by LalaB » Mon Feb 27, 2012 10:51 am
average speed=total distance/ total time

D/((x/48)+(d-x)/64)=60

64*48*d/(16x+48d)=60

1/5D=x
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by LalaB » Mon Feb 27, 2012 11:29 am
hm, have a very very strange method )) and will appreciate if any expert corrects me
i seem to try to invent a bike again :)

ok, I used a weighted average method

S2/S1=(64-60/60-48)=1/3 or S1/S2=3/1

so, it means that t1/t2=1/3 or t1=1/4*Ttotal .lets assume that Ttotal =1

then the 1st distance is D1=s1*t1=48*1/4=12, 2nd distance D2=s2*t2=64*3/4=48


so D1/(D1+D2)=12/60=1/5
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by GMATGuruNY » Mon Feb 27, 2012 12:50 pm
kakz wrote:A car drives part of a trip at 48 kilometers per hour and the remainder of the same trip at 64 kilometers per hour. What fraction of the total distance covered during the trip must be driven at 48 kilometers per hour so that the average speed over the entire trip is 60 kilometers per hour?

Answer is [spoiler]1/5[/spoiler]
We can use alligation, which dictates the following:

The PROPORTION OF THE TOTAL TIME traveled at each speed is equal to the POSITIVE DIFFERENCE BETWEEN THE OTHER TWO SPEEDS.

The proportion of the total time traveled at 48 kph = 64-60 = 4.
The proportion of the total time traveled at 64 kph = 60-48 = 12.
Time traveled at 48 kph : time traveled at 64 kph = 4:12 = 1:3.

Thus, of every 4 hours of travel time, 1 hour will be traveled at 48 kph, 3 hours will be traveled at 64 kph.
Thus, if the car travels for 4 hours:
Distance traveled in 1 hour at 48 kph = 1*48 = 48 kilometers.
Distance traveled in 4 hours at 60 kph = 4*60 = 240 kilometers.
48/240 = 1/5.
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