Question

The sum of the digit of two number is 9 if the number obtained by reversing the order of digit is 27 more than the original number find the original number

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

The sum of the digit of two number is 9.

Let assume that

Digit at ones place be x

Digit at tens place be 9 - x

Thus,

Original number = 10 × (9 - x) + x × 1 = 90 - 10x + x = 90 - 9x

Reverse number = 10 × x + 1 × (9 - x) = 10x + 9 - x = 9x + 9

According to statement,

The number obtained by reversing the order of digit is 27 more than the original number.

$\rm \: 9x + 9 = 90 - 9x + 27$

$\rm \: 9x + 9x = 117 - 9$

$\rm \: 18x = 108$

$\rm\implies \:x = 6$

So,

Original number = 90 - 9x = 90 - 54 = 36

Verification

Original number = 36

So, Sum of digits = 3 + 6 = 9

Reverse number = 63

So, 63 = 36 + 27

This implies, the number obtained by reversing the order of digit is 27 more than the original number.

Hence, Verified

Answer 2

Answer:

36

Step-by-step explanation:

Given that the sum of the digit of two number is 9 and the number obtained by reversing the order of digit is 27 more than the original number.

We need to find out the original number.

Let's say the ten's digit number is x and one's digit number is y.

• Original number = 10x + y
• Reversed number = 10y + x

As the sum of those numbers (x & y) is 9. So, we can write it like:

→ x + y = 9

→ x = 9 - y ------------ (eq 1)

As per second condition,

→ 10y + x - 27 = 10x + y

→ 10y - y + x - 10x = 27

→ 9y - 9x = 27

Take 9 as common,

→ y - x = 3

Substitute value of x in above equation

→ y - (9 - y) = 3

→ y - 9 + y = 3

→ 2y = 12

→ y = 6

Substitute value of y = 6 in (eq 1)

→ x = 9 - 6

→ x = 3

Original number = 10x + y = 10(3) + 6 = 36

Reversed number = 10y + x = 10(6) + 3 = 63

Therefore, the original number is 36.