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Ashok and Brian are both walking east along the same path;

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Ashok and Brian are both walking east along the same path;

by AAPL » Thu Jul 19, 2018 10:17 am

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Ashok and Brian are both walking east along the same path; Ashok walks at a faster constant speed than does Brian. If Brian starts 30 miles east of Ashok and both begin walking at the same time, how many miles will Brian walk before Ashok catches up with him?

(1) Brian's walking speed is twice the difference between Ashok's walking speed and his own.
(2) If Ashok's walking speed were five times as great, it would be three times the sum of his and Brian's actual walking speeds.

The OA is D.

Statement 1 :
B= 2[A-B]-> 3B=2A,
So A=(3/2)B

We have both A and B's walking speed, so we can find the distance. Statement 1 is sufficient,

So eliminate : B, C and E.

Statement 2 :
A= 5A,
Then 5A= 3[A+B],

Here also we can find the distance since we have both A and B's walking speed. Statement 2 is also sufficient.

So the answer is D.

Has anyone another strategic approach to solving this DS question? Regards!
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by GMATGuruNY » Thu Jul 19, 2018 11:42 am

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AAPL wrote:Ashok and Brian are both walking east along the same path; Ashok walks at a faster constant speed than does Brian. If Brian starts 30 miles east of Ashok and both begin walking at the same time, how many miles will Brian walk before Ashok catches up with him?

(1) Brian's walking speed is twice the difference between Ashok's walking speed and his own.
(2) If Ashok's walking speed were five times as great, it would be three times the sum of his and Brian's actual walking speeds
Let A = A's rate and B = Brian's rate.

Ashok has to CATCH-UP by 30 miles.
The CATCH-UP rate is equal to the DIFFERENCE between the two rates.
If A = 3mph, while B = 2mph, then every hour A walks 1 more mile than B, with the result that every hour A catches up by 1 mile -- the DIFFERENCE between the two rates:
A-B = 3-2 = 1mph.

Statement 1: Brian's walking speed is twice the difference between Ashok's walking speed and his own.
Thus:
B = 2(A-B)
B = 2A - 2B
3B = 2A
A = (3/2)B.

Case 1: B = 10mph, A = (3/2)(10) = 15mph
Here, the catch-up rate = A-B = 15-10 = 5mph.
Time for A to catch up by 30 miles = (catch-up distance)/(catch-up rate) = 30/5 = 6 hours.
In 6 hours, the distance traveled by B at a rate of 10mph = r*t = 10*6 = 60 miles.

Case 2: B = 20mph, A = (3/2)(20) = 30mph
Here, the catch-up rate = A-B = 30-20 = 10mph.
Time for A to catch up by 30 miles = (catch-up distance)/(catch-up rate) = 30/10 = 3 hours.
In 3 hours, the distance traveled by B at a rate of 20mph = r*t = 20*3 = 60 miles.

Since B travels the SAME DISTANCE in each case, SUFFICIENT.

Statement 2: If Ashok's walking speed were five times as great, it would be three times the sum of his and Brian's actual walking speeds.
Thus:
5A = 3(A+B)
5A = 3A + 3B
2A = 3B
A = (3/2)B.
Same information as statement 1.
SUFFICIENT.

The correct answer is D.
Statement 1 :
B= 2[A-B]-> 3B=2A,
So A=(3/2)B

We have both A and B's walking speed, so we can find the distance. Statement 1 is sufficient,
Nice work!
But to clarify:
Each statement enables us to determine not the actual speeds but the RATIO of Ashok's speed to Brian's speed.
In my solution above, I show why this ratio is SUFFICIENT to determine the distance that has been traveled by Brian when he catches up to Ashok.
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