Ashok and Brian

This topic has expert replies
Legendary Member
Posts: 510
Joined: Thu Aug 07, 2014 2:24 am
Thanked: 3 times
Followed by:5 members

Ashok and Brian

by j_shreyans » Mon May 04, 2015 9:44 am
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.

User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pm
Location: New York, NY
Thanked: 13060 times
Followed by:1906 members
GMAT Score:790

by GMATGuruNY » Mon May 04, 2015 9:48 am
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.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3