Question

the sum of the digits of atwo digut number is 3 if 9 is substracted from the number it's digits are reversed.find the number

Answer 1

Let the tens place digit be a

And Unit place digit be b

According to first condition,

a + b = 3 ------(1)

According to second condition,

10a + b - 9 = 10b +a

=> 10a + b - 10b - a = 9

=> 9a - 9b = 9

=> a - b = 1 --------(2)

On adding equation 1 and 2, we get

2a = 4

=> a = 2

Now,

On Substituting the value of a in equation 1, we get

2 + b = 3

=> b = 1

Required number = 10 × 2 + 1

= 20 + 1

= 21

Answer 2

Here is your Answer!


Let the number be x & y.

x + y = 3..........(1)

(10x + y) - 9 = 10y + x

multiplied by 10 because of place values)


Now solve
10x + y - 9 = 10y + x
10x-x -10y+y = 9
9x -9y =9
x-y = 1

solve one & two

X+y =3
x-y = 1
__________
2x = 4

X = 2

2+y = 3
y = 3-2
y = 1

hence the digit number 21
Ans