Question

the sum of a two digit number and the number obtained by reversing the digits is 66. If the digits of the number differ by 2 find the number ​

Answer 1

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

• Sum of a two digit number and the number obtained by reversing the digits is 66.
• The digits of a number differ by 2.

$\large\bf\underline {To \: find:-}$

• The number.

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

$\frak{let}\begin{cases}\textrm{ten's place digit be x .}\\\textrm{unit's place digit be y .}\\\end{cases}$

Then,

the Number = 10x + y

The number obtained by reversing the digits is 10y + x

$\underbrace{ \star \rm \:According \: to \: question \: \: }$

the sum of a two digit number and the number obtained by reversing the digits is 66.

$\leadsto \rm \: (10x + y) + (10y + x) = 66 \\ \\ \leadsto \rm \:10x + x + 10y + y = 66 \\ \\ \leadsto \rm \:11x + 11y = 66 \\ \\ \leadsto \rm \:11(x + y) = 66 \\ \\ \leadsto \rm \:x + y = \cancel\frac{66}{11} \\ \\ \leadsto \rm \:x + y = 6............(i)$

It is given that the digits of the number differ by 2.

$\leadsto \rm \:x - y = 2............(ii)$

Solving eq.(i) and (ii) by elimination method.

$\rm \: x + y = 6 \\ \rm \: x - y = 2 \\ \: - \: \: \: + \: \: \: - \\ \rm \underline{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: } \\ \: \: \: \rm \: \: \: \: \: 2y = 4 \\ \bf\: \: \: \: \: \: \: \: \: \:y = 2$

putting value of y in eq. (i)

$\longmapsto\rm \: x + y = 6 \\ \longmapsto \rm \: x + 2 = 6 \\ \longmapsto \rm \: x = 6 -2 \\ \longmapsto \bf \: x = 4$

So

the two digit number = 10x + y

the two digit number = 10×4 + 2

the two digit number = 40 +2

the two digit number = 42

Answer 2

Answer:-

The number is 42.

Given:

• The sum of a two digit number and the number obtained by reversing the digits is 66.

• The digits of number differ by 2.

To find:

• The number.

Solution:

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

$\sf{\therefore}$ Original number=10x+y

$\sf{\therefore}$ Number obtained by reversing the digits=10y+x

According to the first condition.

(10x+y)+(10y+x)=66

$\sf{\therefore}$ 11x+11y=66

$\sf{\therefore}$ 11(x+y)=66

$\sf{\therefore{x+y=\frac{66}{11}}}$

$\sf{\therefore}$ x+y=6...(1)

According to the second condition

x-y=2...(2)

Add equations (1) and (2), we get

x+y=6

+

x-y=2

______

2x=8

$\sf{\therefore{x=\frac{8}{2}}}$

$\sf{\therefore}$ x=4

Substitute x=4 in equation (1), we get

4+y=6

$\sf{\therefore{y=6-4}}$

$\sf{\therefore}$ y=2

$\sf{\therefore}$ Original number=10(4)+2=42

The number is 42.