Aptitude Discussion

Q. |
$A$, $B$ and $C$ start simultaneously from $X$ to $Y$. $A$ reaches $Y$, turns back and meet $B$ at a distance of 11 km from $Y$. $B$ reached $Y$, turns back and meet $C$ at a distance of 9 km from $Y$. If the ratio of the speeds of $A$ and $C$ is $3:2$, what is the distance between $X$ and $Y$? |

✔ A. |
99 m |

✖ B. |
100 m |

✖ C. |
120 m |

✖ D. |
142 m |

**Solution:**

Option(**A**) is correct

Let the distance between $X$ and $Y$ be $d$ km

In the first instance distance travelled by $A =d+11$

In the first instance distance travelled by $B =d−11$

The time taken by both is same

\(\Rightarrow \dfrac{d+11}{A}=\dfrac{d-11}{B}\)

\(\Rightarrow \dfrac{A}{B}=\dfrac{d+11}{d-11}\)-------(i)

In the second instance, distance travelled by $B =d+9$

While distance travelled by $C =d−9$

\(\Rightarrow \dfrac{B}{C}=\dfrac{d+9}{d-9}\)------(ii)

From (i) and (ii)

\(\dfrac{A}{C}=\dfrac{A}{B}\times \dfrac{B}{C}=\dfrac{3}{2}\)

Thus, By solving we get, $d= 1$ or **99**