The sum of the radii of two circles is 7 cm, and the difference of their circumferences is 8 cm.

Solution:

Let the radii of the two circles be r1 cm and r2 cm.
Now,
Sum of the radii of the two circles = 7 cm

$r_{1+} r_{2}=7 \quad \ldots(i)$

Difference of the circumferences of the two circles = 88 cm

$\Rightarrow 2 \pi \mathrm{r}_{1}-2 \pi \mathrm{r}_{2}=8$

$\Rightarrow 2 \pi\left(\mathrm{r}_{1}-\mathrm{r}_{2}\right)=8$

$\Rightarrow\left(\mathrm{r}_{1}-\mathrm{r}_{2}\right)=\frac{8}{2 \pi}$

$\Rightarrow \mathrm{r}_{1}-\mathrm{r}_{2}=\frac{8}{2 \times \frac{22}{7}}$

$\Rightarrow \mathrm{r}_{1-} \mathrm{r}_{2}=\frac{8 \times 7}{44}$

$r_{1}-r_{2}=\frac{56}{44}$

$r_{1}-r_{2}=\frac{14}{11} \quad \cdots(i i)$

Adding (i) and (ii), we get:

$2 r_{1}=\frac{91}{11}$

$r_{1}=\frac{91}{22}$

$\therefore$ Circumference of the first circle $=2 \pi r_{1}$

$=2 \times \frac{22}{7} \times \frac{91}{22}$

$=26 \mathrm{~cm}$

Also,

$r_{1}-r_{2}=\frac{14}{11}$

$\frac{91}{22}-r_{2}=\frac{14}{11}$

$\frac{91}{22}-\frac{14}{11}=r_{2}$

$r_{2}=\frac{63}{22}$

$\therefore$ Circumference of the second circle $=2 \pi \mathrm{r}_{2}$

$=2 \times \frac{22}{7} \times \frac{63}{22}$

$=18 \mathrm{~cm}$

Therefore, circumferences of the first and second circles are 18 cm and 26 cm, respectively.