Aptitude 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

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