Xavier Aptitude Test 2014 Paper

Question : 62 of 103

Marks: +1, -0

There are two windows on the wall of a building that need repairs. A ladder 30 m long is placed against a wall such that it just reaches the first window which is 26 m high. The foot of the Ladder is at point A After the first window is fixed, the foot of the ladder is pushed backwards to point B so that the ladder can reach the second window. The angle made by the ladder with the ground is reduced by half, as a result of pushing the ladder. The distance between points A and B is

< 9 m
≥ 9 m and < 9.5 m
≥ 9.5 m and <10 m
≥ 10 m and < 10.5 m
≥ 10.5 m

Validate

Solution:

(e)

x=\√{30^{2}-26^{2}}=\√{224} \≈ 15 \ {m}

∴\ \cos 2 \θ=\ {15}/{30}=\ {1}/{2}

\⇒2 \θ=60^{\°} ⇒θ=30^{°}

\cos 30^{°}= {y}/{30}⇒ {√{3}}/{2}=\ {y}/{30}

⇒ y=15 \√{3}{m}

\ {AB}=\ {y}-\ {x}=15 \√{3}-15=10.98 {m}

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