Question

0. The sum of the digits of a two-digit number is 9. The number obtained by interchanging the digits exceeds the given number by 27. Then the given number is *

Answer 1

Solution:-

Let the number at 10th place = x

And one the place no. = y

The number obtained = 10x + y

The sum of two digit number = x + y = 9 ......(i)eq

The number obtained by interchanging the digits exceeds the number is 27 :- 10y + x

Eq:-

10y + x - 10x - y = 27

9y - 9x = 27

9(y - x ) = 27

y - x = 3

x -y = -3 ..............( ii ) eq

Now Adding (i) and (ii) eq

x + y + x - y = 9 + - 3

2x = 6

x = 3

Now put the value of x on (i) eq

x + y = 9

3 + y = 9

y = 9 - 3

y = 6

x = 3 and y = 6

Now put the value of 10x + y , we get

10 × 3 + 6 => 36

Thus the number is 36

Answer 2

Answer :

$\boxed{\bf{THE\:REQUIRED\:NUMBER\:IS\:35\:.}}$

Explanation :

Given :–

• Sum of a two digit number is 9 .

• The number obtained by interchanging the digits exceeds the original number by 27.

To Find :–

• THE ORIGINAL NUMBER .

Solution :–

Let the First (tens) digit of the Original Number be x and the Second (ones) digit be y .

→ According to the First Condition :-

⇒ x + y = 9   ------------(1)

→ According to the Second Condition :-

⇒ 10y + x - (10x + y) = 27

⇒ 10y + x -10x - y = 27

⇒ 10y - y + x - 10x = 27

⇒ 9y - 9x = 27

⇒ 9(y - x) = 27

⇒ y - x = 27 ÷ 9

⇒ y - x = 3

⇒ - (x - y) = 3

⇒ x - y = (-3)  -----------(2)

Adding Equation(1) and Equation(2) :-

⇒ x + y + (x - y) = 9 + (-3)

⇒ x + x + y - y = 9 - 3

⇒ 2x = 6

⇒ x = 6 ÷ 2

⇒ x = 3

Now , putting the value of 'x' in Equation(1) :-

⇒ 3 + y = 9

⇒ y = 9 - 3

⇒ y = 6

So now we have both the digits :

→  Tens Digit => 3

→  Ones Digit => 6

Now the number will be in the form of :

→ 10(x) + (y)

→ 10(3) + (6)

→ 30 + 6

→ 36

∴ The required number is 36 .