Question

The sum of the digits of a two digit number is 13. If 27 is added to the number, the place of the digits are reversed. Then the required number is??

Answer 1

The number is = 10x+y

if reverse the digits, the number is: 10y+x

The sum of its digits is = x+y=5

I suggest this system of equations:

(10y+x)-(10x+y)=27 ⇒9y-9x=27

x+y=5

We can solve this system of equations by the equalization method

x=5-y

9y-9(5-y)=27

9y-45+9y=27

18y=27+45

18y=72

y=72/18

y=4

x=5-y

x=5-4=1

The number is: 10x+y=10(1)+4=14

Answer: the number is 14

Answer 2

Given:

Sum of the digits = 13

If 27 is added to the number, the digits are reversed

Find:

the required number

Solution:

Let the digit at ten's place be x and digit at one's place be y.

⟶ The number = 10x + y.

Sum of the digits = 13

⟹ x + y = 13

⟹ x = 13 - y -- equation (1)

Also given that,

If 27 is added to the number, the digits are reversed.

Number formed by reversing the digits = 10y + x.

According to the above condition,

⟹ 10x + y + 27 = 10y + x

⟹ 27 = 10y + x - 10x - y

⟹ 27 = 9y - 9x

Substitute the value of x from equation (1).

⟹ 27 = 9y - 9(13 - y)

⟹ 27 = 9y - 117 + 9y

⟹ 27 + 117 = 18y

⟹ 144 = 18y

⟹ 144/18 = y

⟹ 8 = y

Substitute the value of y in equation (1).

⟹ x = 13 - 8

⟹ x = 5

Hence,

• The number = 10(5) + 8 = 50 + 8 = 58.

∴ The required two digit number is 58.

I hope it will help you.

Regards.