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SajjadAhmad
A ball is dropped 192 inches above level ground, and after the third bounce, it rises to a height of 24 inches. If the height to which the ball rises after each bounce is always the same fraction of the height reached on its previous bounce, what is this fraction?

(A) \(\frac{1}{8}\)

(B) \(\frac{1}{4}\)

(C) \(\frac{1}{3}\)

(D) \(\frac{1}{2}\)

(E) \(\frac{2}{3}\)

Since the height reduces by same fraction at every bounce, this is a geometric progression. Say the rate at which the reduction happens every time is r.
If the first term is 192, the fourth term (after 3 bounces) is 24.

\(t_n = t_1 * r^{n-1}\)

\(24 = 192 * r^3\)
r = 1/2

Answer (D)
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