Jim takes a seconds to swim S 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?  Choices:
A) c(b-a)/ a+b  B) c(a-b)/a+b  C) S(a-b)/(a+b)  D) ab(a-b)/a+b  E) ab(b-a)/a+b
Both JimRoger are travelling at constant speedin opposite direction:   So, speed of Jim = S/a speed of Roger = S/b   Let say Jim travelled distance x from P where it met Roger, it means that Roger travelled (S-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/S....................... (1)   Also, time taken by Roger to travel (c-x) = (S-x)b/S.....................(2)   Time taken by both JimRoger is same, so (1) = (2)   xa/S = (S-x)b/S,   Solving further, x = bS/(a+b).................... (3)   We require to find how many fewer meters will Jim have swum i.e   additional distance travelled by Roger = (S - x) - x   = S-2x   Substituting value of x from (3)solving the equation further, we get Answer = S(a-b)/a+b   Answer = (S(a-b)/(a+b))
<result>C</result>
Jim takes a seconds to swim S 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?