Question

A person starts from a point A and travels 3 km eastwards to B and then turns left and travels thrice that distance to reach C. He again turns left and travels five times the distance he covered between A and B and reaches his destination D. The shortest distance between the starting point and the destination is:

• A. 12 km
• B. 15 km
• C. 16 km
• D. 18 km

Answer 1

Answer: (B)

15 km

The movement of the person are as shown in fig.
Clearly AB = 3 km
$BC=3AB=(3\times 3)KM=9KM$
$CD=5AB=(5\times 3)KM=15KM$
Draw AE $\perp$ CD.
Then CE = AB = 3 km and
AE = BC = 9Km
DE = (CD-CE)=(15-3)Km = 12 Km.
In $\Delta$ AED, $A{D}^2=A{E}^2+D{E}^2$
= $AD=\left(\sqrt{9^2+{\left(12\right)}^2}\right)km=15km$
$\therefore$ Required distance = AD = 15 Km.