What two-digit number is less than the sum of the square of its digits by 11 and exceeds their doubled product by 5?
(E) None of the Above
Let the digits be x and y. The number would be 10x + y.
We are given that 2xy + 5 = 10x +y = x^2 y^2 -11
Thus 2xy +5 = x^2 + y^2 - 11
x^2 + y^2 -2xy = 16
(x-y)^2 = 16
(x-y) = 4 or -4
Substituting the values of (x-y) in the equation 2xy +5 = 10x + y
x comes out to be 1 or 9... thus the two numbers can be 15 or 95
Thus the answer is A
Please +1 KUDOS if my post helped you in any way