confused13 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.
Last edited by
GMATGuruNY on Mon May 04, 2015 9:47 am, edited 1 time in total.
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