# Difficult Time, Speed & Distance Solved QuestionAptitude Discussion

 Q. In an industry, the raw materials and the finished goods are transported on the conveyor belt. There are two conveyor belt, one for carrying parts from $P$ to point $Q$ and another for carrying parts from $R$ to point $Q$. $P$, $Q$ and $R$ in that order are in a straight line. Sometimes, the belt serves to transport cart, which can themselves move with respect to the belts. The two belts move at a speed of 0.5 m/s and the cart move at a uniform speed of 2 m/s with respect to the belts. A cart goes from point $P$ to $R$ and back to $P$ taking a total of 64 s. Find the distance $PR$ in meters. Assume that the time taken by the cart to turn back at $R$ is negligible?
 ✖ A. 48 ✖ B. 54 ✔ C. 60 ✖ D. 64

Solution:
Option(C) is correct

Let the speed of the belts be '$a$' and that of the trolley is '$b$'

Let $PQ = x$ and $QR = y$

Time taken for cart to cover $PR$$=\dfrac{x}{a+b}+\dfrac{y}{b-a}$

Time taken for trolley to cover $RP$$=\dfrac{y}{a+b}+\dfrac{x}{b-a}$

Total time

$= (x+y)\left[\dfrac{1}{a+b}+\dfrac{1}{b-a}\right]=64$s

Thus

$x^2+y^2=\left(\dfrac{2^2-0.5^2}{2\times 2}\right)=60$