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Both runners and walkers are participating in a marathon on

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Both runners and walkers are participating in a marathon on

by alanforde800Maximus » Wed Oct 19, 2016 8:14 pm
Both runners and walkers are participating in a marathon on exactly the same course. The walkers start at 6:45 AM, and the runners begin at 8:00 AM. Bill is entered as a walker and is going to walk at a constant rate of 4 miles per hour. The fastest runner is going to be running at a constant rate of 9 miles per hour. How far will Bill have walked before he is passed by the fastest runner?

a) 4 miles
b) 5 miles
c) 8 miles
d) 9 miles
e) 10.5 miles

Please assist with above problem.
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by GMATGuruNY » Thu Oct 20, 2016 2:40 am
alanforde800Maximus wrote:Both runners and walkers are participating in a marathon on exactly the same course. The walkers start at 6:45 AM, and the runners begin at 8:00 AM. Bill is entered as a walker and is going to walk at a constant rate of 4 miles per hour. The fastest runner is going to be running at a constant rate of 9 miles per hour. How far will Bill have walked before he is passed by the fastest runner?

a) 4 miles
b) 5 miles
c) 8 miles
d) 9 miles
e) 10.5 miles
Since Bill's rate is 4 miles per hour, the distance traveled by Bill between 6:45am and 8am = rt = (4)(5/4) = 5 miles.

Time for the runner to catch up = (distance behind)/(catch-up rate).

The CATCH-UP rate is the DIFFERENCE between the runner's rate and Bill's rate:
9-4 = 5 miles per hour.
Here is the reasoning:
Every hour the runner travels 9 miles, while Bill travels 4 miles.
Result:
Every hour the runner travels 5 MORE MILES THAN than Bill, allowing the runner to CATCH UP by 5 miles every hour.

Time for the runner to catch up = (distance behind)/(catch-up rate) = 5/5 = 1 hour.
Additional distance traveled by Bill in this hour = rt = 4*1 = 4 miles.
Thus:
Total distance traveled by Bill = 5+4 = 9 miles.

The correct answer is D.
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by fiza gupta » Thu Oct 20, 2016 5:02 am
distance travelled by bill from 6:45 to 8 = 1hour 15 mints : 5/4 hours
d = s*t = 4*5/4 = 5

let t be the time when runner started and crossed Bill
let d be the additional distance travelled by Bill(after 5miles) when runner crossed him or in t time
Runner will travel 5+d distance in t time
distance travelled by Bill in t time : 4*t = (d) (i)
distance travelled by runner: (9)*t= (5 + d) (ii)
divide i and ii
d = 4

total distance travelled by bill = 5+1 = 9

SO D
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by [email protected] » Thu Oct 20, 2016 9:38 am
Hi alanforde800Maximus,

This is an example of a 'chase down' question - the first entity gets out ahead and the second entity chases the first down.

Based on the information in the prompt, the walkers have a 5/4 hour head start on the runners. Since the walkers average 4 miles/hour, the 'head start' of a walker will be...

D = (R)(T)
D = (4 miles/hour)(5/4 hours)
D = 5 miles

So the walker is 5 miles ahead when the runner gets started. Since both walker and runner are now moving, the walker will continue to move at 4 miles/hour wile the runner will move at 9 miles/hour. Thus, the runner will 'catch up' 5 miles every hour on the walker. Since the walker is currently 5 miles ahead of the runner, the runner will need 1 hour to catch up to the walker. In that 1 hour, the runner will travel 9 miles.

Final Answer: D

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Last edited by [email protected] on Thu Oct 20, 2016 6:05 pm, edited 1 time in total.
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by Mo2men » Thu Oct 20, 2016 11:52 am
[email protected] wrote:Hi alanforde800Maximus,

This is an example of a 'chase down' question - the first entity gets out ahead and the second entity chases the first down.

Based on the information in the prompt, the walkers have a 11/4 hour head start on the runners. Since the walkers average 4 miles/hour, the 'head start' of a walker will be...

D = (R)(T)
D = (4 miles/hour)(5/4 hours)
D = 5 miles

So the walker is 5 miles ahead when the runner gets started. Since both walker and runner are now moving, the walker will continue to move at 4 miles/hour wile the runner will move at 9 miles/hour. Thus, the runner will 'catch up' 5 miles every hour on the walker. Since the walker is currently 5 miles ahead of the runner, the runner will need 1 hour to catch up to the walker. In that 1 hour, the runner will travel 9 miles.

Final Answer: D

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Hi Rich,

Just for clarity, can you make 11/4 to be 5/4 or 1 1/4? In the beginning, I thought you had a mistake.

Thanks always for your help.
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by Jeff@TargetTestPrep » Thu Oct 20, 2016 3:59 pm
alanforde800Maximus wrote:Both runners and walkers are participating in a marathon on exactly the same course. The walkers start at 6:45 AM, and the runners begin at 8:00 AM. Bill is entered as a walker and is going to walk at a constant rate of 4 miles per hour. The fastest runner is going to be running at a constant rate of 9 miles per hour. How far will Bill have walked before he is passed by the fastest runner?

a) 4 miles
b) 5 miles
c) 8 miles
d) 9 miles
e) 10.5 miles
We can classify this problem as a "catch-up" rate problem, in which we use the formula:

distance of Bill = distance of fastest runner

We are given that Bill starts at 6:45 AM and walks at a constant rate of 4 mph and that the fastest runner starts at 8 AM and runs at a constant rate of 9 mph.

Since Bill started an hour and 15 minutes before the fastest runner, we can let Bill's time = t + 5/4 hours and the fastest runner's time = t. Since distance = rate x time, we can calculate each person's distance in terms of t.

Bill's distance = 4 x (t + 5/4) = 4t + 5

fastest runner's distance = 9t

We can equate the two distances and determine t.

4t + 5 = 9t

5 = 5t

t = 1 hour

Now we can determine how far Bill had walked by the time he was passed by the fastest runner.

Bill's distance = 4(1) + 5 = 9 miles.

Answer: D

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by [email protected] » Thu Oct 20, 2016 6:06 pm
Mo2men wrote:
Hi Rich,

Just for clarity, can you make 11/4 to be 5/4 or 1 1/4? In the beginning, I thought you had a mistake.

Thanks always for your help.
Hi Mo2Men,

Done.

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by crackverbal » Thu Oct 20, 2016 8:50 pm
Hi,

I will just add my 20 cents for this question.

Remember whenever it's a Speed and distance question, always backsolving and POE comes in handy.

Also picturing will help you understand better.

So here the scenario is,

Bill had head start of 1 hour 15 mins. Since his speed as a walker is 4 miles/hour.

He would have covered the Distance = 1.25 * 4 = 5 miles.

So right now he is 5 miles ahead of the fastest runner.

We know that bill definitely has walked 5 miles, so eliminate answer choices A(4) and B(5).

Now if we look at the other answer choices, D(9) is the easiest one to analyze.

Because to cover the 9 miles, bill needs another 1 hour (Because he has already covered 5 miles and his speed is 4 miles/hour).

Fastest runner speed is 9 miles/ hour, So he will cover 9 miles in hour. So he passes the bill. So bingo � .

Eliminate answer choice E(because they meet itself 9 miles).

You can stop here answer is D.

Just for the explanation purpose, answer choice C is not possible because to reach 8 miles bill requires another 45 mins (3/4th of an hour).

In the 45 mins - fastest runner can cover only 7.75 miles (because his speed 9 miles/hour).

So C could not be the answer.

Hope this helps
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