Question

(m) The sum of a two-digit number and the number obtained by reversing its digits
is 176. Find the number, if its tens place digit is greater than the units place
digit by 2.
​

Answer 1

The sum of a two-digit number and the number obtained by reversing its digits is 176. If its tens place digit is greater than the units place digit by 2 then the number is 97.

Let the digits at ones and tens place be x and y respectively

Then, the number obtained = 10y+x

Number obtained by reversing the digits = 10x+y

According to the question

(10y+x)+(10x+y)=176

\implies 11x+11y=176

\implies x+y=16 ...........(1)

Also given that

y=x+2

\implies x-y=-2 ............(2)

Adding equation (1) and (2)

2x=14

\implies x=7

Therefore,

y=16-7=9

Therefore, the number is

10\times 9+7

or, 97

Hope this answer is helpful.

PLZZ MARK AS BRAINLIEST

Answer 2

Answer:

$\boxed{\text{Number = 97}}$

Step-by-step explanation:

$\text{Let the number be 10x + y}\\\\\text{Hence, the digit at the ten's place is x and digit at one's is y}\\\\\text{Therefore, x - 2 = y - \boxed{1} [tens place digit is greater than the units}\\\text{place digit by 2]}\\\\\text{Also, reversed number = 10y + x and it is given that:}\\\\\text{10x + y + 10y + x = 176}\\\\11x + 11y = 176\\\\x + y = 16\\\\x + (x - 2) = 16 \text{ - From \boxed{1}}\\\\2x - 2 = 16\\\\2x = 18\\\\\boxed{x = 9}\\\\x - 2 = y \text{ - From \boxed{1}}$

$y = 9 - 2\\\\\boxed{y = 7}\\\\\text{Now, number is 10x + y = 10 \text{ x } 9 + 7 = 90 + 7}\\\\\boxed{\text{Number = 97}}$

Hope this helps! If it does, please mark it brainliest! Thanks :D