Jim takes a seconds to swim c meters at a constant rate from point P

Speed of Jim is c/a meter per second. 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 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)
Ans: ca/(a+b) - cb/(a+b) = c(a-b)/(a+b)

Pick Numbers Approach:
Pick a=15 (time taken by Jim)
b=10 (time taken by Roger)
c=30 (Total distance PQ)

Rate for Jim = 30/15 = 2
Rate for roger = 30/10 = 3
Mutual rate = 2+3 = 5
Total time taken by both to cross each other = 30/5 = 6

Roger's distance - Jim's distance= (Roger's rate * mutual time) - (Jim's rate * mutual time) = mutual time * (Roger's rate - Jim's rate) = 6 (3-2) = 6

Choice analysis with Plug-in the numbers a=15, b=10, c=30
A. c(b-a)/(a+b) = 30(-5)/25= -6 (this is negative)
B. c(a-b)/(a+b) = 30(5)/25= 6 (this is correct)!
C. c(a+b)/(a-b) = 30(25)/5 = 150 (does not match with 6)
D. ab(a-b)/(a+b) = 150(5)/25= 30 (does not match with 6)
E. ab(b-a)/(a+b) = 150(-5)/25= -30 (does not match with 6)

OK. Unit of distance is mt. Check the unit of options only A B and C satisfies.
b<a so option would be negative.
Between B and C now. Educated guess is B because we have relative speed in denominator. But let me explain why.....

The relative speed will be b+a for objects travelling in opposite direction. Let both of them meet in t seconds. Distance traveled is c.
t = c / (c/a+c/b) = ab/(a+b).
Distance traveled by jim whose speed is c/a is c/a * ab/ (a+b) = cb/(a+b)
Distance traveled by roger... c/b*ab/(a+b) = ca/(a+b)
Difference c*(a-b)/(a+b)

Both Jim & Roger are travelling at constant speed & in opposite direction:
So, speed of Jim = c/a & speed of Roger = c/b
Let say Jim travelled distance x from P where it met Roger, it means that Roger travelled (c-x) from point Q
[x would be less than (c-x) as Jim is travelling slow]
From above, time taken by Jim to travel x = xa/c....................... (1)
Also, time taken by Roger to travel (c-x) = (c-x)b/c.....................(2)
Time taken by both Jim & Roger is same, so (1) = (2)
xa/c = (c-x)b/c,
Solving further, x = bc/(a+b).................... (3)
We require to find how many fewer meters will Jim have swum i.e
additional distance travelled by Roger = (c - x) - x
= c-2x
Substituting value of x from (3) & solving the equation further, we get Answer = c(a-b)/a+b
Answer = (B)

We know from the question stem that a>b, as because for the same distance, Roger will take lesser time than Jim.Eliminate options A and E(as they will lead to negative answer).

Also, as the answer is representing distance, we can straightaway eliminate D, which is representing \(time^2\)

Out of the 2 options remaining , i.e. B and C, we know that the answer has to be less than c(The difference between the distance covered by Jim and Roger can-not be more than c) For option C, the expression (a+b)/(a-b) WILL always be greater than 1. Thus, by process of elimination, the answer is B.

B.

Please refer screenshot below:

How come Roger's distance be taken as c-x ? I dont see it being mentioned that distance between the two points is c.

Jim takes a seconds to swim c meters at a constant rate from point P to point Q in a pool.

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

let d=roger's distance at passing
c-d=jim's distance at passing
d/(c-d)=a/b
d=ac/(a+b)
d-(c-d)=2d-c
substituting, 2d-c=2*ac/(a+b)-c➡
c(a-b)/(a+b)
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

(1) Time taken for Jim and Roger to meet: c/(a+b)... (Since they are swimming towards each other their combined speeds will come into play). So Jim and Roger each swim for c/(a+b) secs before they meet.
(2) Jim swims a*[c/(a+b)] meters.
(3) Roger swims b*[c(a+b)] meters.
(4) Difference: bc/(a+b) - ac/(a+b)...> c(a-b)/(a+b)

Check out my video solution here:

https://www.youtube.com/watch?v=adcVyZLrPvY&ab_channel=DantheGMATMan

If you do not want to pick numbers, take ratios.

Over same distance, time taken by Jim:Rogers = a:b
Speed of Jim:Rogers = b:a (Inverse of time ratio since distance is constant)
Distance covered in same time by Jim:Rogers = b:a
So diff in distance covered on ratio scale = a - b (because a > b)
Total distance of ratio scale = a+b
Total actual distance = c
Multiplier = c/(a+b)

So diff in distance covered = (a-b)*c/(a+b)
Answer (B)

Check this post for a discussion on ratios:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/0 ... of-ratios/
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/0 ... os-in-tsd/

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