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

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Solution

The correct option is D
81

Let the number = 10x + y

x + y = 9……(i)

2 (10x + y) = 9 (10y + x)

i.e 20x + 2y = 90 y + 9 x

11x - 88y = 0

⇒x – 8y = 0….(ii)

Subtracting (ii) from (i) , we get 9y = 9 ⇒ y = 1

x + y = 9 ⇒ x + 1 = 9 ⇒ x = 8

⇒ the number is 10 (8) + 1 = 81

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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.

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

The correct option is D 81Let the number = 10x + y x + y = 9……(i) 2 (10x + y) = 9 (10y + x) i.e 20x + 2y = 90 y + 9 x 11x - 88y = 0 ⇒x – 8y = 0….(ii) Subtracting (ii) from (i) , we get 9y = 9 ⇒ y = 1 x + y = 9 ⇒ x + 1 = 9 ⇒ x = 8 ⇒ the number is 10 (8) + 1 = 81

The correct option is D
81

Let the number = 10x + y

x + y = 9……(i)

2 (10x + y) = 9 (10y + x)

i.e 20x + 2y = 90 y + 9 x

11x - 88y = 0

⇒x – 8y = 0….(ii)

Subtracting (ii) from (i) , we get 9y = 9 ⇒ y = 1

x + y = 9 ⇒ x + 1 = 9 ⇒ x = 8

⇒ the number is 10 (8) + 1 = 81