If the sum of two numbers is $490$  and the difference between their squares is $4900$ , then smaller number is

Let the two number are $x,y$

Given sum of the numbers $=490$

$x+y=490\to \left(1\right)$

And also given difference between their squares is $4900$

$x^2-y^2=4900$

But we know that $a^2-b^2=\left(a+b\right)\left(a-b\right)$

$\left(x+y\right)\left(x-y\right)=4900$

$490\left(x-y\right)=4900$

$x-y=\frac{4900}{490}$

$x-y=10\to \left(2\right)$

By solving $\left(1\right)\&\left(2\right)$

$\begin{array}{c}x+y=490\\ \underset{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}{x-y=10}\\ \underset{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}{2x=500}\end{array}$

$2x=500$

$x=\frac{500}{2}$

$x=250$

Put $x=250$  in equation $\left(1\right)$

$x+y=490$

$250+y=490$

$y=-490-250$

$y=240$

$\therefore$  the smaller number $\left(y\right)=240$