Question - $[\text { AIME } 1991]$ Suppose that $r$ is a real number for which $ \left\lfloor r+\frac{19}{100}\right\rfloor+\left\lfloor r+\frac{20}{100}\right\rfloor+\cdots+\left\lfloor r+\frac{91}{100}\right\rfloor=546 $ Find $\lfloor 100 r\rfloor$
Now author proved using some simple techniques that
$ \left\lfloor r+\frac{56}{100}\right\rfloor=7 \quad \text {and} \quad\left\lfloor r+\frac{57}{100}\right\rfloor=8 $
It follows that $7.43 \leq r<7.44$ and hence that $\lfloor 100 r\rfloor=743$
but i did not see how they found that $7.43 \leq r<7.44$
$$\left\lfloor r+\frac{56}{100}\right\rfloor=7 \implies r+\frac{56}{100} \lt 8 \implies r < 7.44$$
$$\left\lfloor r+\frac{57}{100}\right\rfloor=8 \implies r+\frac{57}{100} \ge 8 \implies r \ge 7.43$$