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XAT 2011 — QA & DI Question 16

2 Variable EquationsEasy
Passage / Data

Answer the following question based on the information given below.

A man standing on a boat south of a light house observes his shadow to be 24 meters long, as measured at the sea level. On sailing 300 meters eastwards, he finds his shadow as 30 meters long, measured in a similar manner. The height of the man is 6 meters above sea level.

The height of the light house above the sea level is:

Answer & solution

  • A

    90 meters

  • B

    94 meters

  • C

    96 meters

  • D

    100 meters

  • l06 meters

Solution

​​​​​​​

Let LM denote the light house of height h above the sea level.

Let KN denote the man and MN denote the south direction.

NS is the shadow of the man.

Then, KN = 6, NS = 24

Also, ∠ KNS = 90° and ∠ LMS = 90°.

By similarity of ΔLMS and ΔKNS,

LMMS = 624 = 14​​​​​​​

∴ If LM = h, MS = 4h and MN = 4h – 24

The boat moves from N to P along the east.

∴ NP = 300

The man’s new position is AP.

∴ AP = 6, PB = 30

Δ APB ~ Δ LMB

∴LMMB = 630 = 15 

∴ MP = 5h – 30

But MN2 + 3002 = MP2

∴ 16(h – 6)2 + 3002 = 25(h – 6)2

∴ 3002 = 25(h – 6)2 – 16(h – 6)2

∴ 90000 = 9(h – 6)2

∴ h = 106

Hence, option (e).

XAT 2011 QA & DI Q16: The height of the light house above the sea level is: — Solution | TheCATExam