Question

a 2 digit is such that the product of its digits is 20.if 9 is added to the number the digits interchange their places find the number. with all the steps

Answer 1

Given :Let the x be th number at units place and y be the number at digits place
number will be =x+10y
Reverse number will be 10x+y
Product of digit is 20
Therefore,xy=20 -----{1}
original number + 9=reverse number
x+10y +9=10x +y
9x-9y=9
x-y=1-------{2}
using formula (x+y)²=(x-y)²+4xy
 (x+y)²=(1)²+4(20)   [ by using equation 1 and 2]
 (x+y)²=1+80=81
 (x+y)=9-----(3)
solve equation 3 and 2
x+y=9
x-y=1
_______
by adding, we get,
2x=10 
⇒x=5
y=9-5
y=4
so x and y are 5 and 4
so number is x +10y =5 +10x4=5+40=45

∴Required number is 45

Answer 2

Solution -

Let us assume x and y are the two digits of the two-digit number

Therefore, the two-digit number = 10x + y
and reversed number = 10y + x

given -

xy = 20 --------------1

and 10x + y + 9 = 10y + x ---------------2

Solve equation 2

10x + y + 9 = 10y +x

9y - 9x = 9

y - x = 1

y = 1 + x ---------------3

substitute equation 3 in equation 1

xy = 20

x (1 + x) = 20

x^2 + x - 20 = 0

Solve this quadratic equation....

x^2 + 5x - 4x - 20 = 0

x(x + 5) - 4 (x + 5) = 0

(x + 5) (x -4) = 0

x = -5 and x = 4

As required number is positive two digit number, consider x = 4

Therefore, xy = 20  y = 20/x  y = 20/4 = 5

therefore, the two digit number = 10x + y = 10*4 + y = 45.