# Moderate Time, Speed & Distance Solved QuestionAptitude Discussion

 Q. $P$ and $Q$ walk from $A$ to $B$, a distance of 27 km at 5 km/hr and 7 km/hr respectively. $Q$ reaches $B$ and immediately turns back meeting $P$ at $T$. What is the distance from $A$ to $T$?
 ✖ A. 25 km ✔ B. 22.5 km ✖ C. 24 km ✖ D. 20 km

Solution:
Option(B) is correct

Let the distance be $a$ km from $A$.

So total distance travelled by $P = a$ at a speed of 5 km/hr.

Total distance travelled by $Q=27+(27−a)=(54−a)$ at a speed of 7 km/hr.

Total time taken by $P$

$=\dfrac{a}{5}$

and that by $Q$

$=\dfrac{(54âˆ’a)}{7}$

Since they have met at the same time, they would have travelled for the same time.

Hence,

$\dfrac{a}{5}=\dfrac{54-a}{7}$

$⇒ a$ = 22.5 km

## (3) Comment(s)

Abhijeet
()

distance from A to B is 27

assume distance from A to T as x

time taken for Q to travel from A to B to T is same as P for A to T

so 54-x/7 = x/5

x = 22.5

()

$7a = 5(54-a)$

$7a = 270-5a$

$2a=270$

$a= 135$

How come $a$ is 22.5 km?

Ziah
()

Zoha, in your 3rd step you have made a mistake. It will be $7a+5a=270$

Which is $12a=270$ and NOT $2a=270$.

Thus giving the final answer as, $a=\textbf{22.5 kms}$