The sum of the radii of two circles is 140 cm

Let the radius of two circles be $r_{1} \mathrm{~cm}$ and $r_{2} \mathrm{~cm}$ respectively. Then their circumferences are $C_{1}=2 \pi r_{1} \mathrm{~cm}$ and $C_{2}=2 \pi r_{2} \mathrm{~cm}$ respectively and their areas are $A_{1}=\pi r_{1}^{2} \mathrm{~cm}^{2}$ and $A_{2}=\pi r_{2}^{2} \mathrm{~cm}^{2}$ respectively.

It is given that the sum of the radii of two circles is 140 cm and difference of their circumferences is 88 cm. So,

$r_{1}+r_{2}=140 \mathrm{~cm} \ldots \ldots(\mathrm{A})$

$C_{1}-C_{2}=88 \mathrm{~cm}$

$2 \pi r_{1}-2 \pi r_{2}=88 \mathrm{~cm}$

$2 \pi\left(r_{1}-r_{2}\right)=88 \mathrm{~cm}$

$r_{1}-r_{2}=\frac{88}{2 \pi} \mathrm{cm}$

$r_{1}-r_{2}=\frac{88}{2 \times \frac{22}{7}} \mathrm{~cm}$

$r_{1}-r_{2}=\frac{88 \times 7}{44} \mathrm{~cm}$

$r_{1}-r_{2}=14 \mathrm{~cm}$..........(B)

Now, solving (A) and (B)

$r_{1}=77 \mathrm{~cm}$

$r_{2}=63 \mathrm{~cm}$

Thus diameters of circles are,

$2 r_{1}=154 \mathrm{~cm}$

$2 r_{2}=126 \mathrm{~cm}$