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Is the number of seconds required to travel d1 feet at r1

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Is the number of seconds required to travel d1 feet at r1

by AAPL » Sat Jul 21, 2018 6:55 am

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B

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E

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Is the number of seconds required to travel d1 feet at r1 feet per second greater than the number of seconds required to travel d2 feet at r2 feet per second?

(1) d1 is 30 greater than d2.
(2) r1 is 30 greater than r2.

The OA is E.

Can not be answered. Reason.
If D2=30 and r2=30 then d2/r2=1
and also d1/r1=1
But if d2=40, r2=30 then d2/r2=4/3
and d1/r1=7/6

Has anyone another strategic approach to solving this DS question? Regards!
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Source: — Data Sufficiency |

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by Jay@ManhattanReview » Mon Jul 23, 2018 12:11 am

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00:00

Your Answer

A

B

C

D

E

Global Stats

AAPL wrote:Is the number of seconds required to travel d1 feet at r1 feet per second greater than the number of seconds required to travel d2 feet at r2 feet per second?

(1) d1 is 30 greater than d2.
(2) r1 is 30 greater than r2.

The OA is E.

Can not be answered. Reason.
If D2=30 and r2=30 then d2/r2=1
and also d1/r1=1
But if d2=40, r2=30 then d2/r2=4/3
and d1/r1=7/6

Has anyone another strategic approach to solving this DS question? Regards!
Given:

1. Number of seconds required to travel d1 feet at r1 feet per second = d1/r1
2. Number of seconds required to travel d2 feet at r2 feet per second = d2/r2

To find out: Is d1/r1 > d2/r2?

Let's take each statement one by one.

(1) d1 is 30 greater than d2.

No information about r1 and r2. Insufficient.

(2) r1 is 30 greater than r2.

No information about d1 and d2. Insufficient.

(1) and (2) together

d1 = d2 + 30 and r1 = r2 + 30

=> d1/r1 = (d2 + 30) / (r2 + 30)

We cannot compare (d2 + 30) / (r2 + 30) and d2/r2 since on one hand the numerator of (d2 + 30) / (r2 + 30) is greater by 30 than that of d2/r2, making it greater than d2/r2; on the other hand the denominator of (d2 + 30) / (r2 + 30) is smaller by 30 than that of d2/r2, making it smaller than d2/r2. So the net effect cannot be deduced. Insufficient.

The correct answer: E

Hope this helps!

-Jay
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