Question

a number consist of 2 digits whose sum is 11. the number formed by reversing the digit is 9 less than the original number. find the number?​

Answer 1

Answer:

Let the tens digit be and the units digit be y. Then the number is 10x+y.

Sum of the digits is x+y=11.

The number formed by reversing the digits is 10y+x.

Given data, (10x+y)−9=10y+x

⇒10x+y−10y−x=9

9x−9y=9

Dividing by 9 on both sides, x−y=1 ........ (2)

Equation (2) becomes x=1+y .......... (3)

Substituting x in (1) we get, 1+y+y=11

⇒2y+1=11

2y=11−1=10

∴y=

2

10

=5

Substituting y=5 in (3) we get, x=1+5=6

∴ The number is 10x+y=10(6)+5=65

Answer 2

Given :-

• A number consist of 2 digits whose sum is 11. the number formed by reversing the digit is 9 less than the original number

To find :-

• Required numbers

Solution :-

Let the ones digit be y then tens digit be x

• Original number = 10x + y

A number consist of 2 digits whose sum is 11.

• x + y = 11 ----(i)

The number formed by reversing the digit is 9 less than the original number

• Reversed number = 10y + x

→ 10x + y - 9 = 10y + x

→ 10x - x + y - 10y = 9

→ 9x - 9y = 9

→ 9(x - y) = 9

→ x - y = 1 -----(ii)

Add both the equations

→ x + y + x - y = 11 + 1

→ 2x = 12

→ x = 12/2

→ x = 6

Put the value of x in equation (ii)

→ x - y = 1

→ 6 - y = 1

→ y = 6 - 1

→ y = 5

Hence,

• Tens digit = x = 6

• Ones digit = y = 5

Therefore,

• Original number = 10x + y = 65

• Reversed number = 10y + x = 56