Manipal Entrance Test 2013 Math Paper

Let PQ = 4m be the height of pole and, AB= 1.6m be height of man. Let the end of shadow is ft and it is at a distance of l from A when the man is at a distance x from PQ at some instantSince, ΔPQR and ΔABR are similar,
we have,

PQ

AB

=

PR

AR

⇒

4

1.6

=

x+1

1

⇒ 2x=31
⇒

2dx

dt

=3.

dl

dt

{given‌

dx

dt

=30m∕min}
⇒

dl

dt

=

2

3

×30m∕min =20m∕min