arithmetic mean and geometric mean of two numbers

Let x,y be the numbers.

AM = (x +y)/2 = 25 so x + y = 50 and y = 50 - x (1)

GM = sqrt(x y) = 20 so x y = 400 (2)

Substituting (1) into (2) gives x*(50-x) = 400 or 50x - x^2 = 400

Solving the quadratic x = 10 or x = 40

Smallest x = 10

If the A.M. of two numbers be $A$ and their G.M. be $G$ then the smaller number is $A-\sqrt {A^2-G^2}$ . Here $A=25$ and $G=20$ . Hence the smaller number is $25-\sqrt {25^2-20^2}=\boxed {10}$ .