The first time the ball hits the ground, it has fallen from a height of 125 cm. So it will bounce to a height of $\frac{3}{5}$ of 125 cm, which is 125$\div$5$\times$3 = 25$\times$3 = 75 cm.

The second time the ball hits the ground, it has fallen from a height of 75 cm. So it will bounce to a height of $\frac{3}{5}$ of 75 cm, which is 75$\div$5$\times$3 = 15$\times$3 = 45 cm.

The third time the ball hits the ground, it has fallen from a height of 45 cm. So it will bounce to a height of $\frac{3}{5}$ of 45 cm, which is 45$\div$5$\times$3 = 9$\times$3 = 27 cm.

125 = 5$\times$5$\times$5, and so $\frac{3}{5}$ of 125 is 3$\times$5$\times$5 (sinceĀ $\frac{1}{5}$ of 125 is 5$\times$5). So after hitting the ground once, the ball will bounce 3$\times$5$\times$5 cm.

Similarly, $\frac{3}{5}$ of 3$\times$5$\times$5 is 3$\times$3$\times$5, so after hitting the ground twice it will bounce 3$\times$3$\times$5 cm.

And $\frac{3}{5}$ of 3$\times$3$\times$5 is 3$\times$3$\times$3, so after hitting the ground three times it will bounce 3$\times$3$\times$3 = 27 cm.

Each time the ball hits the ground, it bounces to $\frac{3}{5}$ of the height from which it fell. So after each bounce, its height is multiplied by $\frac{3}{5}$.

So after 3 bounces, its height will be multiplied by $\left(\frac{3}{5}\right)^3=\frac{27}{125}$.

So after 3 bounces, its height will be $125\times\frac{27}{125}=27$ cm.

You can find more short problems, arranged by curriculum topic, in our short problems collection.