Question

Hemant and ajay start a two-length swimming race at the same moment but from opposite ends of the pool.  they swim in lane and at uniform speed, but hemant is faster than ajay.they first pass at a point 18.5 m  from the deep end and having completed one length, each one is allowed to rest on the edge for exactly 45  seconds.after setting off on the return length, the swimmers pass for the second time just 10.5 m from the  shallow end.how long is the pool?

Answer 1

Answer:

Ans is 45

Step-by-step explanation:

let the speed is a & b (b>a)

total distance =x

two equation will form

18.5/a=(x-18.5)/b (Deep end equation)

x=18.5(a+b)/a..................1

also,

x/b+45+10.5/b=x/a+45+(x-10.5)/a (swallow end equation)

x=10.5(a+b)/(2a-b).........2

equating 1 & 2

we get a/b=37/53

so, x=18.5*90/37=45