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**If a rubber ball consistently bounces back $\Large\frac{2}{3}$ of the height from which it is dropped, what fraction of its original height will the ball bounce after being dropped and bounced four times without being stopped?**

A. $\Large\frac{16}{81}$
B. $\Large\frac{16}{27}$
C. $\Large\frac{4}{9}$
D. $\Large\frac{37}{81}$
**Answer: Option A**

## Show Answer

Solution(By Apex Team)

Each time the ball is dropped and it bounces back, it reaches $\Large\frac{2}{3}$ of the height it was dropped from.
After the first bounce, the ball will reach $\Large\frac{2}{3}$ of the height from which it was dropped – let us call it the original height. After the second bounce, the ball will reach $\Large\frac{2}{3}$ of the height it would have reached after the first bounce.
So, at the end of the second bounce, the ball would have reached $\Large\frac{2}{3}$ × $\Large\frac{2}{3}$
of the original height = $\frac{4}{9}$th of the original height.
After the third bounce, the ball will reach $\Large\frac{2}{3}$ of the height it would have reached after the second bounce.
So, at the end of the third bounce, the ball would have reached $\Large\frac{2}{3} \times \frac{2}{3} \times \frac{2}{3}=\frac{8}{27}$th of the original height.
After the fourth and last bounce, the ball will reach $\Large\frac{2}{3}$ of the height it would have reached after the third bounce.
So, at the end of the last bounce, the ball would have reached $\Large\frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} \times \frac{2}{3}$ of the original height =$\Large\frac{16}{81}$ of the original height.

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