Jim takes a seconds to swim t meters at a constant rate from point M to point N in a pool.
Jim takes a seconds to swim t meters at a constant rate from point M to point N 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 M the same time that Roger leaves point N, how many fewer meters will Jim have swum than Roger when the two swimmers pass each other?
Answer/Solution
t(a-b)/a+b
Steps/Work
Both JimRoger are travelling at constant speedin opposite direction:
So, speed of Jim =t /a speed of Roger = t/b
Let say Jim travelled distance x from M where it met Roger, it means that Roger travelled (t-x) from point N
[x would be less than (t-x) as Jim is travelling slow]
From above, time taken by Jim to travel x = xa/t....................... (1)
Also, time taken by Roger to travel (t-x) = (t-x)b/t.....................(2)
Time taken by both JimRoger is same, so (1) = (2)
xa/t = (t-x)b/t,
Solving further, x = bt/(a+b).................... (3)
We require to find how many fewer meters will Jim have swum i.e
additional distance travelled by Roger = (t - x) - x
= t-2x
Substituting value of x from (3)solving the equation further, we get Answer = t(a-b)/a+b
Answer = (D)
So, speed of Jim =t /a speed of Roger = t/b
Let say Jim travelled distance x from M where it met Roger, it means that Roger travelled (t-x) from point N
[x would be less than (t-x) as Jim is travelling slow]
From above, time taken by Jim to travel x = xa/t....................... (1)
Also, time taken by Roger to travel (t-x) = (t-x)b/t.....................(2)
Time taken by both JimRoger is same, so (1) = (2)
xa/t = (t-x)b/t,
Solving further, x = bt/(a+b).................... (3)
We require to find how many fewer meters will Jim have swum i.e
additional distance travelled by Roger = (t - x) - x
= t-2x
Substituting value of x from (3)solving the equation further, we get Answer = t(a-b)/a+b
Answer = (D)