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

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answer is B.

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Detailed Solution

It is given that in a two-digit number, the sum of two-digit number is 9. Nine times the number is twice the number obtained by reversing the order of digits of the number.
Let, the ten’s place digit be

x

, and the one’s place digit be

y

.
Then, the number is

10 x + y

.
According to the first condition, the sum of two-digit number is 9.
So,

x + y = 9

According to the second condition, nine times the number is twice the number obtained by reversing the order of digits of the number.

⇒ 9 (10 x + y) = 2 (10 y + x) ⇒ 90 x + 9 y = 20 y + 2 x ⇒ 88 x = 11 y

Multiplying the equation

x + y = 9

by 11 on both the sides.
So, we get,

11 x + 11 y = 99

Adding the equation

88 x − 11 y = 0

, and

11 x + 11 y = 99

.
So,

\Rightarrow

88x+11x-11y+11y=0+99

\Rightarrow

99x=99

\Rightarrow

x=1
Substituting x=1 in the equation

x + y = 9

.
So, we get,

⇒ x + y = 9 ⇒ 1 + y = 9 ⇒ y = 8

Substituting x=1 and y=8 in the number

10 x + y

.
Number

= 10 x + y

= 101 + 8 = 10 + 8 = 18

The two-digit number is 18.
Hence, the correct option is 2.

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