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