The sum of the digits of a two - digit number is 9. Also, nine times t

1.	The sum of the digits of a two - digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.
Answer» Let the one’s digit be ‘a’ and ten’s digit be ‘b’. therefore no is = 10b + a Reversed no = 10a + b Given, Sum of the digits of a two - digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. ⇒ a + b = 9 ----- (1) And 9(10b + a) = 2(10a + b) ⇒ 90b + 9a = 20a + 2b ⇒ 88b = 11a ⇒ a = 8b Substituting value of a in eq1 ⇒ 8b + b = 9 ⇒ 9b = 9 ⇒ b = 1 Thus, a = 8(1) = 9 Hence, no is 10(1) + 8 = 18

The sum of the digits of a two - digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

Answer»

Let the one’s digit be ‘a’ and ten’s digit be ‘b’.

therefore no is = 10b + a

Reversed no = 10a + b

Given,

Sum of the digits of a two - digit number is 9.

Also, nine times this number is twice the number obtained by reversing the order of the digits.

⇒ a + b = 9 ----- (1)

And 9(10b + a) = 2(10a + b)

⇒ 90b + 9a = 20a + 2b

⇒ 88b = 11a

⇒ a = 8b

Substituting value of a in eq1

⇒ 8b + b = 9

⇒ 9b = 9

⇒ b = 1

Thus,

a = 8(1) = 9

Hence, no is 10(1) + 8 = 18