# Difficult Time, Speed & Distance Solved QuestionAptitude 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 km ✖ B. 100 km ✖ C. 120 km ✖ D. 142 km

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

Atul
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