A hemispherical bowl is $88\text{cm}$ round the brim. Assuming it to be full, how many persons may be served from it in hemispherical glasses, $7\text{cm}$ in diameter at the top? (Where $\pi =\frac{22}{7}$)

Let the radius of the hemispherical bowl be $R$.
The circumference (round the brim) of the bowl is given as:

$$2\pi R=88\text{cm}$$

$$2\times \frac{22}{7}\times R=88$$

$$\frac{44}{7}\times R=88⟹R=2\times 7=14\text{cm}$$

The volume of the hemispherical bowl ($V_1$) is:

$$V_1=\frac{2}{3}\pi R^3=\frac{2}{3}\pi (14{)}^3$$

For the small hemispherical glasses, the diameter is $7\text{cm}$, so the radius ($r$) is:

$$r=\frac{7}{2}\text{cm}$$

The volume of one hemispherical glass ($V_2$) is:

$$V_2=\frac{2}{3}\pi r^3=\frac{2}{3}\pi {\left(\frac{7}{2}\right)}^3$$

Let $n$ be the number of persons that can be served:

$$n=\frac{V_1}{V_2}=\frac{\frac{2}{3}\pi R^3}{\frac{2}{3}\pi r^3}={\left(\frac{R}{r}\right)}^3$$

$$n={\left(\frac{14}{\frac{7}{2}}\right)}^3={\left(\frac{14\times 2}{7}\right)}^3=(4{)}^3=64$$