The sum of the squares of two positive integers is 100 and the difference of their squares is 28. The sum of the numbers is ?

Answer: Option C

Let the positive integers be a and b where a > b
According to the question,
$$\begin{array}{rl}& a^2+b^2=\mathrm{100.....}(\text{i})\\ & a^2-b^2=\mathrm{28.....}(\text{ii})\end{array}$$
By adding (i) and (ii), we get :
$$\begin{array}{rl}& \therefore a^2+b^2+a^2-b^2=100+28\\ & \Rightarrow 2a^2=128\\ & \Rightarrow a^2=\frac{128}{2}\\ & \Rightarrow a=\sqrt{64}\\ & \Rightarrow a=8\end{array}$$
From equation (i)
$$\begin{array}{rl}& \Rightarrow 8^2+b^2=100\\ & \Rightarrow b^2=100-64\\ & \Rightarrow b=\sqrt{36}\\ & \Rightarrow b=6\\ & \therefore a+b=8+6\\ & =14\end{array}$$