Aptitude Discussion

Q. |
Laila drives to the station each day to pick up her husband Majnu, who usually arrives on a train at 6 o’clock. Last Monday, Majnu finished work earlier, caught an earlier train and arrived at the station at 5 ‘o clock. He started to walk home and eventually met Laila who drove him the rest of the way, getting home 20 minutes earlier than usual. On Tuesday, he again finished early and found himself at the station at 5:30. Again he began to walk home, again he met Laila on the way, and she drove him home the rest of the way, Assume constant speed throughout with no wasted time for waiting, backing of the car etc. How earlier than the usual time were they home on Tuesday? |

✖ A. |
6 min |

✖ B. |
8 min |

✔ C. |
10 min |

✖ D. |
12 min |

**Solution:**

Option(**C**) is correct

Every day, Laila leaves from $A$ (the time of her departure is same everyday) to reach point $D$ (the station) exactly at 6 ‘O clock where he finds Majnu waiting for him.

**On Monday:**

$A$-------------------$C$-------------------------$D$

Laila leaves from $A$ at her usual time but Majnu reaches the station ($D$) at 5’O clock (i.e. an hour earlier than his normal time ) and he starts walking towards home (i.e. towards point $A$).

Laila meets him on the way at $C$ and from there both of them head towards $A$.

A total of 20 minutes are saved this way. Where have they saved these 20 minutes’ time?

These 20 minutes are saved just because of the fact that today Laila did not have to travel the distance ($C→D)+(D→C) $

So we can deduce that Laila must be taking a total of 20 minutes time in a to-and-fro travel between $C$ and $D$ daily.

As Laila (and for that matter Majnu also) have a constant speed throughout the journey, we can also deduce that for

Laila, the time taken in going from $C$ to $D$ is the same as the time taken in going from $D$ to $C$.

So we can say, had Majnu been at D at 6’O clock on Monday as well, then Laila must have travelled the $C$ to $D$ distance and she must have passed point $C$ at exactly 10 minutes before 6 ‘ O clock i.e. at 5 : 50 PM.

This only means that on Monday when they meet at point $C$, the time of their meeting was exactly 5:50 pm.

Now what should this tell us about their speeds?

We know that Laila takes 10 minute’s time to move from $C$ to $D$.

And now we also know that Majnu had been walking till 5:50 PM (he reaches the station and starts walking till he reaches the point $C$).

So Majnu takes exactly 50 minutes in traveling the same distance. If $VL$ and $VM$ be the speeds of Laila and Majnu respectively, then we must have:

**On Tuesday:**

$A$--------------------$E$--------------------$D$

Majnu reaches $D$ at 5:30 PM and starts walking towards $A$.

Let’s assume that he walks for $x$ minutes and reaches the point $E$, where he meets Laila.

So that Laila has saved time in moving from $E$ to $D$ and from $D$ to $E$.

Today, Majnu reaches at $D$ at $5:30+x$ PM.

From Laila’s point, we can say, Laila must have reached the point $E$ at exactly $30–x$ minutes before 6:00 PM.

For example if $5:30+x$ had been something like 5:45 PM, then she must have reached 15 minutes before 6:00 PM.

In other words, today both of them have saved a total of $2\times (30–x)$ minutes in reaching back to home (point $A$).

**To find the value of x:**

Majnu Moved from $D$ to $E$ in $x$ minutes, and Laila (would have moved) from $E$ to $D$ in $30–x$ minutes. Now,

\(\dfrac{x}{30-x}=\dfrac{V_L}{V_M}=5\)

Which gives $x = 25$ minutes.

So, today they have saved a total of $2×(30–25)$ = **10 minutes**

Hence (C) is the correct option.