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Re: After driving to a riverfront parking lot, Bob plans to run [#permalink]
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How do we know that he ran half the distance south ? Thanks

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Re: After driving to a riverfront parking lot, Bob plans to run [#permalink]
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Walkabout wrote:
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

shahir16 wrote:
How do we know that he ran half the distance south ? Thanks


Hi Shair. The question states that Bob will run south along the river and then turn around to return to the parking lot. Since there are only two legs to the journey, namely south and back, this implies that Bob will run half the total distance south.

Hope that helps.
Jesse
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Re: After driving to a riverfront parking lot, Bob plans to run [#permalink]
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JesseWhelan wrote:
Walkabout wrote:
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

shahir16 wrote:
How do we know that he ran half the distance south ? Thanks


Hi Shair. The question states that Bob will run south along the river and then turn around to return to the parking lot. Since there are only two legs to the journey, namely south and back, this implies that Bob will run half the total distance south.

Hope that helps.
Jesse


Another method :Solve through equations.
Total remaining distance ,Dsouth + Dnorth = 50/8 = 6.25 miles
(1): 3.25 + Dsouth = Dnorth
(2): Dnorth + Dsouth = 6.25

Solve (1) and (2) i.e (1) + (2) => 3.25 + 2*Dsouth = 6.25 => Dsouth = 3/2= 1.5 miles
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Re: After driving to a riverfront parking lot, Bob plans to run [#permalink]
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another way to solve this:

Let x be the extra mile that bob can run and make it in 50 mins back to the parking.
So while returning he has to travel (x+3.25) miles
Total distance = 2x+3.25 miles
time equation = (2x+3.25)miles*8mins/miles = 50 mins
solving for x, we get it as 1.5 miles
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Re: After driving to a riverfront parking lot, Bob plans to run [#permalink]
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Bob has only 50 minutes to complete the whole run. With rate of 8 minutes/mile the distance he can cover is 50/8 mile.

We can picture the path as follow

Running south
<-------3.25 miles--------><---- d = ? ----->

Running north back to origin
<-------3.25 miles--------><------- d ------->

From this we can conclude if he want to go back to his first point the distance would be d + d + 3.25.

Because he only has 50/8 mile left, we can make equation as follows d+d+3.25 = 50/8; 2d+3.25 = 6.25 --> d=1.5
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Re: After driving to a riverfront parking lot, Bob plans to run [#permalink]
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1. He runs 3.25 mile down the south.
2. Then he runs 6.25miles in 50 min. (i.e. 50/8).
So he runs 3.0 miles in North direction. (6.25 total - 3.25 south).
Hence his one-way trip in North is 1.5miles. (3.0/2).
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After driving to a riverfront parking lot, Bob plans to run [#permalink]
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He runs 3.25 south.
Now we know he has only 6.25-3.25 to run (as 3.25 is already reserved for going back to the parking lot)
so we have 3 miles left, to get back to where he is now.... divide that by 2, and get 1.5 left in each direction

Originally posted by ronr34 on 03 Nov 2013, 22:53.
Last edited by ronr34 on 28 Aug 2014, 13:22, edited 1 time in total.
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After driving to a riverfront parking lot, Bob plans to run [#permalink]
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Walkabout wrote:
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


Solution:

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



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 had 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

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After driving to a riverfront parking lot, Bob plans to run [#permalink]
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Walkabout wrote:
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


1. Converting speed to miles / minute, speed =1/8 miles/minute
2. Miles run = 3.25
3. To cover 3.25 miles in the return he needs 3.25/(1/8) minutes = 26 min
4. In the remaining 24 minutes, down south is 12 min and up north is 12 minutes
5. So down south further, he can run 12*1/8 = 1.5 miles
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Re: After driving to a riverfront parking lot, Bob plans to run [#permalink]
ScottTargetTestPrep

Do you mind explaining this?
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Re: After driving to a riverfront parking lot, Bob plans to run [#permalink]
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ms1862 wrote:
ScottTargetTestPrep

Do you mind explaining this?


Solution:

I have actually posted a solution for this question (which can be found at https://gmatclub.com/forum/after-drivin ... l#p1680552), here’s an alternate solution:

Suppose that Bob can run x miles further and make it back in 50 minutes (in total). This means that he will run x miles and turn around to run x + 3.25 miles for a total of x + x + 3.25 = 2x + 3.25 miles in 50 minutes. We are given that Bob can run 1 mile in 8 minutes, so 2x + 3.25 miles will take him (2x + 3.25)*8 minutes. We want this quantity to equal 50, hence we can create the following equation:

(2x + 3.25)*8 = 50

2x + 3.25 = 50/8 = 6.25

2x = 6.25 - 3.25 = 3

x = 1.5

So, Bob can run 1.5 miles further if he is to run for 50 more minutes.

Answer: A
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Re: After driving to a riverfront parking lot, Bob plans to run [#permalink]
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Walkabout wrote:
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


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Re: After driving to a riverfront parking lot, Bob plans to run [#permalink]
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Distance = speed ×time. So why have they done distance = time ÷ speed in some solutions?
12mins ÷ 8mins/mile? Why is it done in this way can someone explain?

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Re: After driving to a riverfront parking lot, Bob plans to run [#permalink]
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Video solution from Quant Reasoning:
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1
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Re: After driving to a riverfront parking lot, Bob plans to run [#permalink]
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Walkabout wrote:
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


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
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