Time, Speed and Distance Question

The distance for 2nd meeting point from one end has decreased from the distance of 1st meeting point from another end, it means that whoever has traveled 10.5m distance for 2nd meeting point is slower swimmer, and that is y. It would not have been true if X had traveled 10.5

Outcome- Y starting from Deep end, X starting from Shallow end
suppose length of the pool is L and first time they meet time elapsed is t, speed of X&Y is x&y respectively
then L/(x+y) = t ........ i)
y.t= 18.5...........ii)


Now for total travel time of X:
t = time taken from Shallow end(SE) to 1st Meeting point(1MP)
tx1= time taken from 1MP to Deep end(DE)= 18.5/x
45s = rest
tx2= time taken from DE to 2MP( 2nd meeting point)= (L-10.5)/x

total travel time of Y:
t = time taken from DE to 1MP
ty1 = time taken from 1MP to SE = (L-18.5)/y
45s= rest
ty2 = time taken from SE to 2MP = 10.5/y


Now to meeting occur at 2MP total travel time of both X&Y must be equal i.e
t+ tx1+ 45s + tx2 = t + ty1 + 45s + ty2
which implies
tx1 + tx2 = ty1 + ty2
on substituting the values
18.5/x + (L-10.5)/x = (L-18.5)/y + 10.5/y
on solving
x/y = (L+8)/(L-8).............iii)


Now from eq. ii) substituting the value of t in eq. i)

L/(x+y) = 18.5/y
L/18.5 = (x/y + 1)
now from iii) substituting the value of x/y in above eq.

L/18.5 = ((L+8)/(L-8) + 1)
L/18.5 = ( 2L/(L-8))
L(L-8) = 37.L
L= 8+ 37 =45(ANS)