charan79 wrote:

There are two cites P and Q. A and B start from P and Q and move on their motorcycles towards Q and P receptively. After reaching their destination, they reverse their direction and move back to where they started from. First time A and B meet at a distance of 65 km from P and second time while they were returning to the point where they have started from, they meet at a distance of 55 km from Q.Assuming that A and B started at the same time and travelled at uniform speeds, find the distance between P and Q.

A) 130

B) 140

C) 150

D) can't be determined

OA:B

let the Average speed of A be \(v_a\) and Average speed of B be \(v_b\)

Let the distance between P and Q be \(x\)

\(t_1\) = time after start, A and B meet first time

\(t_2\) = time interval between when they first meet and when they meet 2nd time

\(t_1=\frac{65}{v_a}=\frac{x-65}{v_b}\)

\(\frac{v_a}{v_b} = \frac{65}{x-65}\) ...........(1)

\(t_2=\frac{x-65+55}{v_a}=\frac{65+x-55}{v_b}\)

\(\frac{v_a}{v_b} =\frac{x-10}{x+10}\) ........(2)

Equating 1 and 2 , we get

\(\frac{65}{x-65}=\frac{x-10}{x+10}\)

\(65(x+10)=(x-65)(x-10)\)

\(65x+650=x^2-10x-65x+650\)

\(x^2-140x=0\)

\(x(x-140)=0\)

Rejecting \(x=0\), we get \(x =140\)

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