Aptitude 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**

**Abhijeet**

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**Zoha Amjad**

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$7a = 5(54-a)$

$7a = 270-5a$

$2a=270$

$a= 135$

How come $a$ is 22.5 km?

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, $a=\textbf{22.5 kms}$

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