Speed Distance Time /Proportion

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.

Hi ,

How come 12:18?

Thanks guru !

I was able to eliminate options D and E based on Dimensional Analysis (D and E ruled out since we need the distance, whereas D and E are missing variable C)..Any trick to figure out the incorrect option among A,B,C using some diff. approach?

GMATGuruNY 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.

raj44 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?

1. c(b-a)/a+b
2. c(a-b)/a+b
3. c(a+b)/a-b
4. ab(a-b)/a+b
5. ab(b-a)/a+b

OA 2

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.