Question

the sum of two digit number and the number obtained by reversing the digits is 21 find the number if its units place digit is greater than the tens place digit by 7​

Answer 1

                 

Sorry. The question is found as wrong because the sum of a two digit number and the number obtained by reversing the digits can never be 21, as it's not a multiple of 11.

Also, if the difference of the digits is 7, then the lowest possible sum of the two digit numbers shall be 99.  So 21 can't be.

So I'm changing the question as follows:

Question:

The sum of a two digit number and the number obtained by reversing the digits is 121. Find the number if its units place digit is greater than the tens place digit by 7. ​

Answer: 29

Step-by-step explanation:

Let the number be 10y + x.

So that  x - y = 7     →     (1)

The two digit number obtained by reversing the digits will be 10x + y.

(10y + x) + (10x + y) = 121

⇒ 10y + x + 10x + y = 121

⇒ 11x + 11y = 121

⇒ 11(x + y) = 121

⇒ x + y = 11     →    (2)

(2) + (1)

⇒ (x + y) + (x - y) = 11 + 7

⇒ 2x = 18

⇒ x = 9

(2) - (1)

⇒ (x + y) - (x - y) = 11 - 7

⇒ 2y = 4

⇒ y = 2

So the answer is,

10y + x

⇒ 10 × 2 + 9

⇒ 20 + 9

⇒ 29

Hope this helps.

Plz ask me if you've any doubts.

Thank you. :-))