Question

sum of the digits of two digit number is 9 when we interchange the digits it is found that the resulting new number is greater than the original number by 27 what is the two-digit number

Answer 1

$\bf{\large{\underline{\underline{Answer}}}}$

Original number = 36

$\bf{\large{\underline{\underline{Solution:-}}}}$

$\text{Let unit digit be x}$

∴ $\text{Digit at tens place be 9-x}$

$\text{Original number = 10(9-x) + x}$

$\text{\bf{\:\:\:\:\:\:\:\: \:\: \:\:\:\:\: \:\:\:\:\: \:\:\:\:\: \: = 90 - 10x + x}}$

$\text{\bf{\:\:\:\:\:\:\: \:\: \:\:\:\:\: \:\:\:\:\:\:\:\:\:\: \:\:\: = 90 - 9x}}$

$\text{New number = 10(x) + 9-x}$

$\text{\bf{ \:\:\:\:\:\: \:\:\:\:\:\:\: \: \:\:\:\:\:\: = 10x + 9-x}}$

$\text{\bf{\:\:\:\:\:\: \:\: \: \:\: \: \: \:\:\:\:\: = 9x + 9}}$

Now lets solve the equation:

90 — 9x + 27 = 9x + 9

90 + 27 — 9 = 9x + 9x

108 = 18x

x = $\frac{108}{18}$= 6

Therefore, original number

= 90 — 9x

= 90 — 9×6

= 90— 54

= 36

$\huge{\boxed{Original \:number \:= \:36}}$

Answer 2

Answer:

$\huge\green{\boxed{\sf Answer}}$

Given

The sum of the two digits = 9

On interchanging the digits, the resulting new number is greater than the original number by 27.

Let us assume the digit of units place = x

Then the digit of tens place will be = (9 – x)

Thus the two-digit number is 10(9 – x) + x

Let us reverse the digit

the number becomes 10x + (9 – x)

As per the given condition

10x + (9 – x) = 10(9 – x) + x + 27

⇒ 9x + 9 = 90 – 10x + x + 27

⇒ 9x + 9 = 117 – 9x

On rearranging the terms we get,

⇒ 18x = 108

⇒ x = 6

So the digit in units place = 6

Digit in tens place is

⇒ 9 – x

⇒ 9 – 6

⇒ 3

Hence the number is 36