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Expert replies
by teejaycrown » Fri Nov 02, 2012 10:10 pm
During a 40 mile trip, Marla traveled at an average speed of x miles/hour for the first y miles of the trip at an average speed of 1.25x miles/hr for the last 40-y miles of the trip. The time that Marla took to travel the 40 miles was what % of the time it would have taken her if she had travelled at an average speed of x miles for the enitre trip

1. x = 48
2. y =20
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Source: — Problem Solving |

by Anurag@Gurome » Fri Nov 02, 2012 10:14 pm
teejaycrown wrote:During a 40 mile trip, Marla traveled at an average speed of x miles/hour for the first y miles of the trip at an average speed of 1.25x miles/hr for the last 40-y miles of the trip. The time that Marla took to travel the 40 miles was what % of the time it would have taken her if she had travelled at an average speed of x miles for the enitre trip

1. x = 48
2. y =20
Time would have taken if speed was x mph, T = 40/x hour
Time taken in current situation, t = [(y/x) + (40 - y)/(5x/4)] hour
= [(y/x) + 4*(40 - y)/(5x)] hour
= [(5y + 160 - 4y)/(5x)] hour
= [(y + 160)/(5x)] hour

Required percentage = 100*(t/T) = 100*[(y + 160)/(5x)]/(40/x )
= 100*[(y + 160)/200] = (y + 160)/2

Thus the percentage depends upon y not on x.

Hence statement 1 is NOT sufficient whereas statement 2 is SUFFICIENT.

The correct answer is B.
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by GMATGuruNY » Sat Nov 03, 2012 1:18 am
teejaycrown wrote:During a 40 mile trip, Marla traveled at an average speed of x miles/hour for the first y miles of the trip at an average speed of 1.25x miles/hr for the last 40-y miles of the trip. The time that Marla took to travel the 40 miles was what % of the time it would have taken her if she had travelled at an average speed of x miles for the enitre trip

1. x = 48
2. y =20
Time and rate are RECIPROCALS.
x = the REGULAR speed.
1.25x = the FASTER speed.
1.25x = (125/100)x = (5/4)x.
5/4 the rate implies 4/5 the time.
Thus:
The time needed at the FASTER speed is equal to 4/5 the time needed at the REGULAR speed.

The question stems asks for the following ratio:
(time needed when some of the distance is traveled at the faster speed):(time needed if the entire distance is traveled at the regular speed).
Since the question is asking only for a ratio, the actual speed and the actual distance are not needed.
To calculate the requested ratio, all we need is the answer to the following question:

What FRACTION of the distance was traveled at the faster speed?

To illustrate:
Let the total distance = 40, the regular speed = 4mph, and the faster speed = 5mph.
Time needed to travel the entire 40 miles at the regular speed = 40/4 = 10 hours.
Time needed to travel half the distance at 4mph and half the distance at 5mph = (20/4 + 20/5) = 9 hours.
Faster time:regular time = 9:10.

Let the total distance = 80, the regular speed = 20mph, and the faster speed = 25mph.
Time needed to travel the entire 80 miles at the regular speed = 80/20 = 4 hours.
Time needed to travel half the distance at 20mph and half the distance at 25mph = (40/20 + 40/25) = 18/5 hours.
Faster time:regular time = (18/5):4 = (18/5):(20/5) = 18:20 = 9:10.

Notice that changing the actual distance and the actual rates does not affect the ratio.
In each case, since half the distance is traveled at the regular speed and half the distance is traveled at the faster speed, the ratio of the times is the same:
9:10.

Onto the statements at hand:

Statement 1: x=48
The rate is irrelevant: regardless of the rate, the time needed at the faster speed is 4/5 the time needed at the regular speed.
INSUFFICIENT.

Statement 2: y=20
Thus, half the distance was traveled at the faster speed.
SUFFICIENT.

The correct answer is B.
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