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by BlackDog » Thu Aug 15, 2013 8:37 am
Jim takes a seconds to swim c meters at a constant rate from point P to point Q in a pool. Roger, who is faster than Jim, can swim the same distance in b seconds at a constant rate. If Jim leaves point P the same time that Roger leaves point Q, how many fewer meters will Jim have swum than Roger when the two swimmers pass each other?


A) c(b-a)/ a+b
B) c(a-b)/a+b
C) c(a+b)/a-b
D) ab(a-b)/a+b
E) ab(b-a)/a+b


The solution I am looking at tells me that I have to solve out for 't' It seems to me that for similar problems (for example: two trains leave point a and b and meet somewhere in between) I do not have to solve out for t. I am told that t= ab/a+b but I cannot figure out why that is necessary.
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by GMATGuruNY » Thu Aug 15, 2013 9:09 am
BlackDog wrote:Jim takes a seconds to swim c meters at a constant rate from point P to point Q in a pool. Roger, who is faster than Jim, can swim the same distance in b seconds at a constant rate. If Jim leaves point P the same time that Roger leaves point Q, how many fewer meters will Jim have swum than Roger when the two swimmers pass each other?


A) c(b-a)/ a+b
B) c(a-b)/a+b
C) c(a+b)/a-b
D) ab(a-b)/a+b
E) ab(b-a)/a+b
Since there are variables in the answer choices, we can plug in values.

Let c = 30 meters, a = 3 seconds, and b = 2 seconds.

Jim's rate = d/t = 30/3 = 10 meters per second.
Roger's rate = d/t = 30/2 = 15 meters per second.

Jim's rate : Roger's rate = 10:15 = 2:3 = 12:18.
Implication: for every 12 meters that Jim swims, Roger swims 18 meters.
Thus, when the two travel the 30 meters between them, Roger's distance - Jim's distance = 18-12 = 6 meters. This is our target.

Now we plug c=30, a=3 and b=2 into the answers to see which yields our target of 6.
Only B works:
c(a-b)/(a+b) = 30(3-2)/(3+2) = 6.

The correct answer is B.
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by BlackDog » Thu Aug 15, 2013 9:55 am
Interesting way to solve. Thanks!

I ask about solving for time because there are other seemingly similar problems that do not require this (for example, many problems involving passing trains seem to require we solve for any number of variables but the time they travel is always the same. Any insight?

Regards,

Zach
GMATGuruNY wrote:
BlackDog wrote:Jim takes a seconds to swim c meters at a constant rate from point P to point Q in a pool. Roger, who is faster than Jim, can swim the same distance in b seconds at a constant rate. If Jim leaves point P the same time that Roger leaves point Q, how many fewer meters will Jim have swum than Roger when the two swimmers pass each other?


A) c(b-a)/ a+b
B) c(a-b)/a+b
C) c(a+b)/a-b
D) ab(a-b)/a+b
E) ab(b-a)/a+b
Since there are variables in the answer choices, we can plug in values.

Let c = 30 meters, a = 3 seconds, and b = 2 seconds.

Jim's rate = d/t = 30/3 = 10 meters per second.
Roger's rate = d/t = 30/2 = 15 meters per second.

Jim's rate : Roger's rate = 10:15 = 2:3 = 12:18.
Implication: for every 12 meters that Jim swims, Roger swims 18 meters.
Thus, when the two travel the 30 meters between them, Roger's distance - Jim's distance = 18-12 = 6 meters. This is our target.

Now we plug c=30, a=3 and b=2 into the answers to see which yields our target of 6.
Only B works:
c(a-b)/(a+b) = 30(3-2)/(3+2) = 6.

The correct answer is B.
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by GMATGuruNY » Thu Aug 15, 2013 10:15 am
BlackDog wrote:Interesting way to solve. Thanks!

I ask about solving for time because there are other seemingly similar problems that do not require this (for example, many problems involving passing trains seem to require we solve for any number of variables but the time they travel is always the same. Any insight?

Regards,

Zach
I suspect that the solution uses the time to determine the distance traveled by each swimmer.

Jim:
Since Jim swims c meters in a seconds, Jim's rate = c/a.
Thus, in the time for the two swimmers to meet -- ab/(a+b) seconds -- the distance traveled by Jim = r*t = (c/a)(ab/(a+b) = bc/(a+b).

Roger:
Since Roger swims c meters in b seconds, Rogers's rate = c/b.
Thus, in the time for the two swimmers to meet -- ab/(a+b) seconds -- the distance traveled by Roger = r*t = (c/b)(ab/(a+b) = ac/(a+b).

Roger's distance - Jim's distance = ac/(a+b) - bc/(a+b) = (ac-bc)/(a+b) = c(a-b)/(a+b).
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by spark » Tue Mar 15, 2016 10:39 am
This problem contains an error. As written, the problem does not specify that the distance from P to Q is c meters. Intuitively, the total distance from P to Q should matter, and in fact it does.

The wording of the problem could be fixed by changing the first sentence to: "Swimming at a constant rate, Jim takes a seconds to cover the c meters from point P to point Q in a pool." If the problem were worded this way, the official solution and the other solutions posted here would be correct.

The number picking approach used in the official solution conveniently assumes that the distance from P to Q is c meters, so that's why this approach works. If any other distance were selected, the official answer would not work.

As the problem is currently worded, the correct answer should be D(a - b) / (a + b), where D is the distance from P to Q.

See the attached for a detailed solution.Image
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by Jeff@TargetTestPrep » Mon Dec 11, 2017 4:24 pm
BlackDog wrote:Jim takes a seconds to swim c meters at a constant rate from point P to point Q in a pool. Roger, who is faster than Jim, can swim the same distance in b seconds at a constant rate. If Jim leaves point P the same time that Roger leaves point Q, how many fewer meters will Jim have swum than Roger when the two swimmers pass each other?


A) c(b-a)/ a+b
B) c(a-b)/a+b
C) c(a+b)/a-b
D) ab(a-b)/a+b
E) ab(b-a)/a+b
We are given that Jim takes a seconds to swim c meters. Since rate = distance/time, Jim's rate is c/a. We are also given that Roger can swim c meters in b seconds. Roger's rate = c/b.

Since Jim leaves point P at the same time Roger leaves point Q, we can let t = the time it takes them to pass each other, and we can use the following formula:

distance of Jim + distance of Roger = total distance

(c/a)t + (c/b)t = c

Let's multiply both sides of the equation by ab:

bct + act = abc

Now divide both sides of the equation by c and solve for t:

bt + at = ab

t(b + a) = ab

t = ab/(a +b)

In t seconds (when they pass each other), Roger has swum (c/b)[ab/(a +b)] = ac/(a + b) meters and Jim has swum (c/a)[ab/(a +b)] = bc/(a + b) meters. Therefore, the difference between the distances swum is:

ac/(a + b) - bc/(a + b) = (ac - bc)/(a + b) = c(a - b)/(a + b)

Answer: B

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