The rim (circular boundary) of a hemispherical bowl is $88\text{cm}$. Assuming the bowl is completely filled, how many people can be served if the content is distributed into hemispherical glasses, where each glass has a top diameter of $7\text{cm}$? (Take $\pi =\frac{22}{7}$)

**Step 1: Find the radius of the hemispherical bowl ($R$)**
The boundary/rim of a hemispherical bowl is a circle. Therefore, its circumference is given by:

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

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

$$\frac{44}{7}\times R=88$$

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

The volume of this large hemispherical bowl ($V_1$) is:

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

**Step 2: Find the radius of the hemispherical glasses ($r$)**
The top diameter of the glass is $7\text{cm}$. Thus, its radius is:

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

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

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

**Step 3: Calculate the number of people served ($n$)**

$$n=\frac{\text{Volume of the bowl }(V_1)}{\text{Volume of each glass }(V_2)}$$

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

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

Thus, $64$ people can be served.