After driving to a riverfront parking lot, Bob plans to run

[swerve]

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

The OA is A

Source: Official Guide

[deloitte247]

Bob runs at 8 minutes per mile so in 50 minutes he would have covered $$\frac{50\cdot1}{8}=6.25miles$$
Total distance covered by Bob = 3.25 + 6.25 = 9.50 miles
Distance south = 1/2 of total distance covered
= 1/2 * 9.5 = 4.75 miles
Out of 4.75 miles he has already covered 3.25 to the south.
The remaining distance to cover = 4.75 - 3.25 = 1.50 miles
Answer = option A

[Scott@TargetTestPrep]

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

The OA is A

Source: Official Guide

We are given that Bob plans to run south along the river, turn around, and return to where he started.

We know that his run south (from the parking lot) and his run north (back to the parking lot) are equal in distance. We will use this information later in the solution.

We are also given that Bob's rate is 8 minutes per mile, or, in other words, (since Rate = Distance/Time) his rate is 1 mile per 8 minutes or 1/8.

We are told that Bob has already run 3.25 miles south, and he wants to run for 50 minutes more. Thus, we calculate how far Bob will go in the remaining 50 minutes.

Distance = Rate x Time

Distance = 1/8 x 50

Distance = 50/8 = 25/4 = 6.25 miles

Thus, we know that Bob's total running distance will be 6.25 + 3.25 = 9.5 miles. Because we know the distance is the same both ways, we know that each leg of his trip is 9.5/2 = 4.75 miles. Since Bob has ALREADY RUN 3.25 miles south, he can run 4.75 - 3.25 = 1.5 miles more. At that point, he will have to turn around and head back north to the parking lot.

Answer: A

[Brent@GMATPrepNow]

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

The OA is A

Source: Official Guide

Let's sketch the situation:


If we let x = the extra distance Bob jogs south, then.....

... he must travel the same x miles north PLUS the additional 3.25 miles.

So the total distance Bob travels (AFTER jogging 3.25 miles south) = x + x + 3.25 miles
= 2x + 3.25 miles.

Since Bob's jogging speed is 8 minutes per mile, we can say that he travels 1 mile in 8 minutes.
So his speed = distance/time = 1/8 miles per minute.
Finally, Bob wants his remaining travel time to be 50 minutes.

Remaining travel time = (remaining distance)/(jogging speed)
Plug in our values to get: 50 = (2x + 3.25)/(1/8)
Simplify: 50 = (2x + 3.25)(8)
Expand: 50 = 16x + 26
Subtract 26 from both sides: 24 = 16x
Solve: x = 24/16 = 3/2 = 1.5

So, Bob can run an additional 1.5 miles.

Answer: A