Jim takes a seconds to swim c meters at a constant rate

[AAPL]

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 than 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 OA is B.

I solved this PS question as follows,

The speed of Jim is c/a meter per second. The speed of Roger is c/b meter per second.
Let us suppose when they both meet or cross each other Jim has covered x distance and Roger has covered c-x distance. Since the time taken is same, as they both started together we can have the equation:

x/(c/a) = (c-x)/(c/b)

ax/c = b(c-x)/c or ax = b(c-x)
Solving for x, x = cb/(a+b)
Jim travelled, x = cb/(a+b)
Roger travelled, c-x = ca/(a+b)
Finally, the correct answer is, ca/(a+b) - cb/(a+b) = c(a-b)/(a+b). Option B.

Has anyone another strategic approach to solve this PS question? Regards!

[swerve]

Jim rate = c/a

Roger rate = c/b

b<a

Combined rate = c(a+b)/ab

Time to complete course = c/[c(a+b)/ab] = ab/(a+b)

Roger Distance - Jim Distance = [ab/(a+b)][c/b)-(c/a)] = c(a-b)/(a+b). Option B.

Regards!

[GMATGuruNY]

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 than 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)

Let c=6 meters, a=2 seconds, and b=1 second.
Since Jim takes 2 seconds to travel the 6-meter distance, Jim's rate = d/t = 6/2 = 3 meters per second.
Since Roger takes 1 second to travel the 6-meter distance, Roger's rate = d/t = 6/1 = 6 meters per second.

When Jim and Roger travel toward each other, they WORK TOGETHER to cover the 6 meters between them, so we ADD THEIR RATES:
3+6 = 9 meters per second.
Of every 9 meters traveled by Jim and Roger working together, 6 meters are traveled by Roger.
Implication:
Roger travels 6/9 of the 6-meter distance:
(6/9)(6)= 4 meters.
Since Roger travels 4 meters, the remainder of the 6-meter distance is traveled by Jim:
6-4 = 2 meters.

Difference between Roger's distance and Jim's distance = 4-2 = 2 meters.
The value in blue is the target.
Now plug c=6, a=2 and b=1 into the answer choices to see which yields the target value.
Only B works:
[c(a-b)]/(a+b) = [6(2-1)]/(2+1) = 2.

The correct answer is B.

[Jeff@TargetTestPrep]

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 than 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 meters per second. We are also given that Roger can swim c meters in b seconds. Roger's rate = c/b meters per second.

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