Question

the sum of two digit number and the number obtained by reversing its digit is 176 find the number if its ten's place digit is greater than the units place digit by 2​

Answer 1

Answer:97

Step-by-step explanation:

Let the unit's place be x. And ten's place will be (x+2).

Original number=10(x+2)+x

=10x+20+x

=11x+20

Reversed number=10(x)+x+2

=10x+x+2

=11x+2

11x+2+11x+20=176

22x+22=176

x+1=8

x=8-1

x=7

Therefore, original number =10(7+2)+7

=10*9+7

=90+7

=97

Answer 2

Answer :

• The two digit number is 97.

Given :

• The sum of two digit number and the number obtained by reversing its digit is 176.

To find :

• The number if its ten's place digit is greater than the units place digit by 2 = ?

Step-by-step explanation :

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

⟹ 11x+11y=176

⟹ x + y = 16 ...........(1)

Also given that

y = x + 2

⟹ x − y = − 2 ............(2)

Adding equation (1) and (2)

2x = 14

⟹ x = 7

Therefore,

y = 16 − 7 = 9

Hence , the two digit number is 97.