### 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

### 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.

A. 22
B. 25
C. 23
D. 24

A. 5
B. 6
C. 4
D. 3

A. -45
B. -55
C. -50
D. 0