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Need smart way to scan answer of this problem

by Mo2men » Thu Nov 10, 2016 2:42 am
Guy drives 60 miles to attend a meeting. Halfway through, he increases his speed so that his average speed on the second half is 16 miles per hour faster than the average speed on the first half. His average speed for the entire trip is 30 miles per hour. Guy drives on average how many miles per hour during the first half of the way?

A. 12
B. 14
C. 16
D. 24
E. 40

Source: Economist

I have solved this question using 2 way methods either by solving quadratic equation or plug in answers.

Is there any way to scan answer quickly?
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by GMATGuruNY » Thu Nov 10, 2016 4:24 am
Mo2men wrote:Guy drives 60 miles to attend a meeting. Halfway through, he increases his speed so that his average speed on the second half is 16 miles per hour faster than the average speed on the first half. His average speed for the entire trip is 30 miles per hour. Guy drives on average how many miles per hour during the first half of the way?

A. 12
B. 14
C. 16
D. 24
E. 40
Generally, when the SAME DISTANCE is traveled at TWO DIFFERENT SPEEDS that are relatively close in value, the average speed for the entire trip will be just a bit closer to the SLOWER SPEED than to the FASTER SPEED.
Only D -- which implies a slower speed of 24 mph and a faster speed of 40 mph -- satisfies the constraint that the average speed of 30 mph is just a bit closer to the slower speed than to the faster speed.

The correct answer is D.
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by DavidG@VeritasPrep » Thu Nov 10, 2016 12:53 pm
Mo2men wrote:Guy drives 60 miles to attend a meeting. Halfway through, he increases his speed so that his average speed on the second half is 16 miles per hour faster than the average speed on the first half. His average speed for the entire trip is 30 miles per hour. Guy drives on average how many miles per hour during the first half of the way?

A. 12
B. 14
C. 16
D. 24
E. 40

Source: Economist

I have solved this question using 2 way methods either by solving quadratic equation or plug in answers.

Is there any way to scan answer quickly?
If you test C, the slower speed would be 16, making the faster speed 32. If the average speed for the whole trip is 30, this would imply that he'd spent virtually all of his traveling at the faster speed. That makes no sense. If the distances are the same, you'll spend more time at the slower speed. So 16 is way too low. Kill A, B, and C.

E makes no sense: how could the slower speed be greater than the average speed?

That leaves you with D
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by Matt@VeritasPrep » Fri Nov 11, 2016 2:54 pm
One speed has to be greater than 30, one speed has to be less than 30. Since we want the first (slower) speed, our answer is less than 30. Eliminate A, B, E.

Now consider C. 16 and 32 (the rates) and 60 miles (the distance) don't play well together: they give ugly fractions. By the lazy testwriter principle, this is impossible, so we'll try answer D instead.

First half distance = 30
First half speed = 24
First half time = 30/24 = 5/4

Second half distance = 30
Second half speed = 40
Second half time = 3/4

and hey, 5/4 + 3/4 = 2, which is the right time, so we're set!
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