Question

7. The sum of the digits of a two-digit number is 5. If the digits are reversed, the number is reduced by 27. Find the number.

Answer 1

Let tens digit number be x and ones digit number be y.

So, original number = 10x + y

The sum of the digits of a two-digit number is 5.

Tens digit number + Ones digit number = 5

⇒ x + y = 5

⇒ x = 5 - y............(1)

If the digits are reversed, the number is reduced by 27.

Interchanged number = 10y + x

According to question,

⇒ 10y + x = 10x + y - 27

⇒ 10y - y + x - 10x = -27

⇒ 9y - 9x = -27

⇒ y - x = -3

⇒ x - y = 3

⇒ x = 3 + y...............(2)

On comparing both the equations we get,

⇒ 5 - y = 3 + y

⇒ y + y = 5 - 3

⇒ 2y = 2

⇒ y = 1

Substitute value of y in (1)

⇒ x = 5 - 1

⇒ x = 4

Therefore,

Original number = 10(4) + 1 = 41

Answer 2

$\huge \red{SOLUTION}$

Let ten's place of the digit be x and unit's place be y.

According to first condition

x+y=5...(1)

Original number=10x+y

Reversed number=10y+x

According to second condition

10x+y=10y+x+27

9x-9y=27

9(x-y)=27

x-y=27/9

x-y=3...(2)

Adding equations (1) and (2)

x+y=5

+

x-y=3

--------------

2x=8

x=8/2

x=4

Substituting x=4 in equation (1)

4+y=5

y=5-4

y=1

Number=10x+y

=10(4)+1

=41

The number is 41

Hope it helps you:)