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Percent, speed distance time

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Percent, speed distance time

by josh80 » Wed Dec 11, 2013 5:14 pm
During a 40 mile trip, Marla traveled at an average speed of x mph for the first y miles of the trip and at an average speed of 1.25x mph 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 traveled at an average speed of x mph the entire trip?

1) x = 48

2) y = 20
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Source: — Data Sufficiency |
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by Patrick_GMATFix » Wed Dec 11, 2013 10:56 pm
This is QID 1351 in the GMATFix Solutions Engine. Follow the link for a more thorough solution.

Basically, we want to know what % of time did Marla gain by speeding up from x to 1.25x mph. Understand that the value of x is irrelevant. What's important is that she sped up by 25%.

Notice that if the whole trip had been done at 25% faster speed (5/4 of regular speed), we would know that her time would have been 20% less (4/5 of regular time) because time and speeds are inversely proportional (doubling speed cuts time in half).

The later Marla switched to her higher speed, the less time she saved. So to answer the question, we basically need to know at what point during the trip did Marla switch to the higher speed?

Rephrase: What is y?

Statement 1 is Not Sufficient
Statement 2 is Sufficient

Answer: B
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by GMATGuruNY » Mon Dec 16, 2013 5:03 am
josh80 wrote:During a 40 mile trip, Marla traveled at an average speed of x mph for the first y miles of the trip and at an average speed of 1.25x mph 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 traveled at an average speed of x mph the entire trip?

1) x = 48

2) y = 20
Statement 1: x = 48, implying that the rate for the last 40-y miles = (5/4) * 48 = 60 miles per hour.
Time for the entire trip at x=48 miles per hour:
d/r = 40/48 = 5/6 hour.

Case 1: y=20
Time for the first 20 miles = d/r = 20/48 = 5/12 hour.
Time for the last 20 miles = d/r = 20/60 = 4/12 hour.
Total time = 5/12 + 4/12 = 9/12 hour.
Resulting ratio:
(shorter time)/(longer time) = (9/12) / (5/6) = 9/10.

Case 2: y=10
Time for the first 10 miles = d/r = 10/48 = 5/24 hour.
Time for the last 30 miles = d/r = 30/60 = 1/2 hour.
Total time = 5/24 + 1/2 = 17/24 hours.
Resulting ratio:
(shorter time)/(longer time) = (17/24) / (5/6) = 17/20.

Since different ratios are possible, INSUFFICIENT.

Statement 2: y = 20
Case 1 also satisfies statement 2.

Case 3: x=4 miles per hour, 1.25x = 5 miles per hour
Time for the first 20 miles = d/r = 20/4 = 5 hours.
Time for the last 20 miles = d/r = 20/5 = 4 hours.
Total time = 5+4 = 9 hours.
Time for the entire trip at x=4 miles per hour:
d/r = 40/4 = 10 hours.
Resulting ratio:
(shorter time)/(longer time) = 9/10.

Case 1 and Case 3 illustrate that -- if y=20 -- then the resulting ratio will always be the SAME:
(shorter time)/(longer time) = 9/10.
SUFFICIENT.

The correct answer is B.
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