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We are given that, in a two digit number the one’s digit is double the digit in ten’s place. The sum of the digits is $9$ . Let us assume that the ten’s digit is x then the one’s digit will be $2{\text{x}}$.

Then according to the question, sum of digits= $9$

$ \Rightarrow {\text{x + 2x = 9}}$

On solving we get,

$ \Rightarrow 3{\text{x = 9}} \Rightarrow {\text{x = }}\dfrac{9}{3} = 3$

Now we know the ten’s digit =x= $3$ then we can find the one’s digit.

One’s digit= $2{\text{x}}$=$2 \times 3 = 6$

Now we can express the two digit number as-

$ \Rightarrow $ Number =ten’s digit ×$10$ + one’s digit

Here we multiplied the ten’s digit by $10$ because it is in tenth’s place in the number.

On putting the value of the digit we get,

$ \Rightarrow $ Number=$\left( {3 \times 10} \right) + 6$

On solving we get,

$ \Rightarrow $ Number=$30 + 6 = 36$

Number= ($100$ ×hundred’s digit)+ (ten’s digit ×$10$) + one’s digit

See here we multiplied the hundred’s digit with $100$ because it is a place value of hundred’s digit.