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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Car Trip Problem
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Mike@Magoosh
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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.
Mike
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.
Mike
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LalaB
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average speed=total distance/ total time
D/((x/48)+(d-x)/64)=60
64*48*d/(16x+48d)=60
1/5D=x
D/((x/48)+(d-x)/64)=60
64*48*d/(16x+48d)=60
1/5D=x
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LalaB
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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
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
Happy are those who dream dreams and are ready to pay the price to make them come true.(c)
In order to succeed, your desire for success should be greater than your fear of failure.(c)
In order to succeed, your desire for success should be greater than your fear of failure.(c)
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GMATGuruNY
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We can use alligation, which dictates the following: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]
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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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
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