Time, Speed & Distance
Aptitude

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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

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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 $Q$

\(=\dfrac{(54−a)}{7}\)

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



Zoha Amjad
 ()

$7a = 5(54-a)$

$7a = 270-5a$

$2a=270$

$a= 135$

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, $a=\textbf{22.5 kms}$