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Airplane (Knewton Prep)

Expert replies
by RadiumBall » Sun Apr 24, 2011 9:23 am
How long did it take a certain airplane to fly from a remote airstrip to Nairobi, Kenya?

1.If a constant tailwind had increased the airplane's average speed by 20 percent, the trip would have taken 90 minutes.
2.The average speed of the airplane for the trip was 490 miles per hour.
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Source: — Data Sufficiency |

by manpsingh87 » Sun Apr 24, 2011 10:33 am
RadiumBall wrote:How long did it take a certain airplane to fly from a remote airstrip to Nairobi, Kenya?

1.If a constant tailwind had increased the airplane's average speed by 20 percent, the trip would have taken 90 minutes.
2.The average speed of the airplane for the trip was 490 miles per hour.
1) let d be the distance between remote strip and nairobi; average speed be v and time be t;

under normal conditions time taken would be d/v=t;
as average speed is increased by 20% therefore it becomes 1.2v; and time taken would have been 90;
d/1.2v=90; we know that d/v=t; t=1.2*90; t=108; hence 1 alone is sufficient to answer the question.

2) since we don't know the distance; we can't find the time taken by traveling at the average speed of 490; hence 2 alone is not sufficient to answer the question..!!!

Hence A
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by RadiumBall » Sun Apr 24, 2011 8:58 pm
Thanks, OA: A
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by GMATGuruNY » Mon Apr 25, 2011 4:03 am
RadiumBall wrote:How long did it take a certain airplane to fly from a remote airstrip to Nairobi, Kenya?

1.If a constant tailwind had increased the airplane's average speed by 20 percent, the trip would have taken 90 minutes.
2.The average speed of the airplane for the trip was 490 miles per hour.
No math is needed here.

Statement 1: If a constant tailwind had increased the airplane's average speed by 20 percent, the trip would have taken 90 minutes.
If we know the time needed when the speed is increased by x%, we can determine the time needed at the actual speed.
Sufficient.

If this were a PS question and we needed to determine the time, we could plug in a number for the distance.
Let d = 540 miles.
Rate for 90 minutes = d/t = 540/90 = 6 miles per minute.
This rate is 20% greater than the actual rate.
Thus, the actual rate is 5 miles per minute.
Time at 5 miles per minute = d/r = 540/5 = 108 minutes.

Statement 2.The average speed of the airplane for the trip was 490 miles per hour.
Without knowing the distance, we can't determine the time.
Insufficient.

The correct answer is A.
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