OG 13 79

[PGMAT]

After driving to a riverfront parking lot, Bob plans to run south along the river, turn around, and return to the parking lot, running north along the same path. After running 3.25 miles south, he decides to run for only 50 minutes more. If Bob runs at a constant rate of 8 minutes per mile, how many miles farther south can he run and still be able to return to the parking lot in 50 minutes?

A. 1.5
B. 2.25
C. 3.0
D. 3.25
E. 4.75

OA: A

Can some one explain in detail how to tackle this problem? Thanks.

[GMATGuruNY]

After driving to a riverfront parking lot, Bob plans to run south along the river, turn around, and return to the parking lot, running north along the same path. After running 3.25 miles south, he decides to run for only 50 minutes more. If Bob runs at a constant rate of 8 minutes per mile, how many miles farther south can he run and still be able to return to the parking lot in 50 minutes?

A. 1.5
B. 2.25
C. 3.0
D. 3.25
E. 4.75

OA: A

Can some one explain in detail how to tackle this problem? Thanks.

8 minutes/1 mile = 50 minutes/x miles
x = 50/8 = 25/4 = 6.25.
Thus, in 50 minutes, Bob can run 6.25 miles.

Since Bob has already run 3.25 miles, the total distance there and back = 3.25 + 6.25 = 9.5 miles.
Distance southward = 9.5/2 = 4.75 miles.
Since Bob has already run 3.25 miles southward, the remaining distance that can be run southward = 4.75-3.25 = 1.5.

The correct answer is A.

[PGMAT]

8 minutes/1 mile = 50 minutes/x miles
x = 50/8 = 25/4 = 6.25.
Thus, in 50 minutes, Bob can run 6.25 miles.

Since Bob has already run 3.25 miles, the total distance there and back = 3.25 + 6.25 = 9.5 miles.
Distance southward = 9.5/2 = 4.75 miles.
Since Bob has already run 3.25 miles southward, the remaining distance that can be run southward = 4.75-3.25 = 1.5.

The correct answer is A.

Mitch, great explanation. Thank you.

[pbrmoney]

Can someone please address why this problem doesn't require the Distance = Rate x Time formula?

By that, I mean that the first step in the OG explanation is to multiply the Distance (3.25 miles) by the rate (8 mph) which yields the time it took Bob to run the first leg.

I understand how to do the problem according to the explanations above, but I don't understand why or even how we do this problem without employing D=RT. When I see these distance problems, my first inclination is to write out D=RT.

If I did this problem using D=RT, I would end up with 3.25/8, which does not yield 26 minutes. If I tried to approach this problem with D=RT on the test, I'd be in trouble, which is disconcerting.

[Whitney Garner]

Can someone please address why this problem doesn't require the Distance = Rate x Time formula?

By that, I mean that the first step in the OG explanation is to multiply the Distance (3.25 miles) by the rate (8 mph) which yields the time it took Bob to run the first leg.

I understand how to do the problem according to the explanations above, but I don't understand why or even how we do this problem without employing D=RT. When I see these distance problems, my first inclination is to write out D=RT.

If I did this problem using D=RT, I would end up with 3.25/8, which does not yield 26 minutes. If I tried to approach this problem with D=RT on the test, I'd be in trouble, which is disconcerting.

Hi pbrmoney!

This is a very tricky problem because of the UNITS - check out the rate we are given...

If Bob runs at a constant rate of 8 MINUTES per MILE.

As you read problems, ALWAYS pay attention to the units. This should have sent up a red flag - WAIT a minute!! We usually see rates as Distance OVER Time (miles per hour, feet per second, etc.). But this gives us Time OVER Distance. It is the opposite.

So if I want to go from a rate that is Time/Distance - I need to "cancel" that distance in the denominator, so rather than DIVIDE by distance, I would have to Multiply!

So (3.25 miles) x (8 min/mile) = (3.25x8) x (miles x min/miles) = 26 minutes.

Hope this helps clear up that bit of the confusion!

Whit