Question

A two digit number is such that the product of the digit is 20 if 9 is added to the number the digits interchange there places find the number

Answer 1

Let the digit at tens place be x and ones place be y.

So the no. will be 10x+y

CONDITION-1
                       xy = 20
                         y = 20/x
CONDITION-2
                       10x+y+9=10y+x
                            9x-9y=-9
                               x-y=-1
                               x-20/x=-1
                               x^2+x-20=0
                               x^2+5x-4x-20=0
                               x(x+5) -4(x+5)=0
                               (x+5)(x-4)=0
                                x=4 and can't be 5 because its negative
   So,                        y=5
  Hence the no. is 10(4)+5=45

Answer 2

Let the digit in the tens place be x and once place be y

Original number = 10y + x
Reverse number = 10x+ y

Product of digits = 20
xy = 20    (1)

original number + 9 = reverse number
10y + x + 9 = 10x + y
9x - 9y = 9
x - y = 1   (2)

(x + y)² = (x-y)² + 4xy
(x + y)² = (1)² + 4 × 20   (from 1 and 2)
(x + y)² = 1+ 80 = 81
x + y = 9  (3)

solving 2 and 3

x - y = 1
x +y = 9/ 2x = 10

x = 10/2 = 5
when x = 5
5 + y = 9
y = 9 - 5 = 4

original number 10 y + x = 10 × 4 + 5
40 + 5 = 45

HOPE THIS HELPS!!