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A ball dropped from 36 m above the ground rebounds to 1/3rd of the hei [#permalink]
­1) The ball's first drop is from a height of 36m.
2) The ball now rebounds to 1/3 of the above height (1/3 of 36m = 12m)
3) The ball will again fall from this point (12m)
4) The ball will again rebound to a height of 1/3 x 12m.
5) And so on.

- Distance travelled in 1st fall = 36m
- Distance traveled between 1st fall and 2nd fall (first rebound plus subsequent fall) = (1/3 x 36) + (1/3 x 36)m
- And so on.

So, Overall distance covered = 36 + (2)(36)(1/3) + (2)(36)(1/3^2) + (2)(36)(1/3^3) + .....
= [(2)(36) + (2)(36)(1/3) + (2)(36)(1/3^2) + (2)(36)(1/3^3) + .....] - 36 (Adding an subtracting 36 to create a proper Geometric Progression)
= [(2)(36)/(1 - 1/3)] - 36 (Sum of infinite gp with first term a (2x36) and common ratio r (1/3), when |r| < 1, = a/(1-r))
= 108 - 36 = 72

Choice B (72) is the answer.

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A ball dropped from 36 m above the ground rebounds to 1/3rd of the hei [#permalink]
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