Question

The sum of digits of the two digit no is 9 .if the digits are interchanged,the no obtained exceeds the original no by 27 find the number

Answer 1

Let the digit in unit place be y and tens digit be x
Original no =10x+y
A.T.Q
X+y=9 (equation no 1)
If the digits are interchanged
10y+x=10x+y+27
10y+x-10x-y=27
9y-9x=27(equation no 2)
Multiplying eq no 1 by 9 and eq no 2 by 1
9x+9y=81 (equation no 3)
-9x+9y=27 ( equation no 4)
Adding 3 and 4 we will get
18y = 108
Y = 108/18
Y=6
Substituting the value of y in any of the equation to get the value of x
I will substitute y in eq 1
X+6=9
X=3
( hope it will help u.... Plz mark as brainliest

Answer 2

Let us assume, the x is the tenth place and y is the unit place digits of the two-digit number

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

Given:

x + y = 9 --------------1

Also given:

10y + x = 10x + y + 27
9y - 9x = 27
y - x = 3 --------------2

Adding equation 1 and equation 2

2y = 12
y = 6

Therefore, x = 9 - y = 9 - 6 = 3

The number = 10x + y = 10 * 3 + 6 = 36