Question

The sum of digits of a two digit number is 9 .Also 9 times the number is twice the number obtain by reversing the order of digits.find the number​

Answer 1

Answer:

Number = 18

Step-by-step explanation:

Let one's digit number be x and ten's digit number be y.

According to question, given conditions are..

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

We have to find the number. Means we have to find the value of x and y.

→ x + y = 9

→ x = 9 - y

Also,

→ 9(10y + x) = 2(10x + y)

→ 90y + 9x = 20x + 2y

→ 88y = 11x

Substitute value of x

→ 88y = 11(9 - y)

→ 88y = 99 - 11y

→ 99y = 99

→ y = 1

Substitute value if y in x

→ x = 9 - 1

→ x = 8

Therefore,

Number = 10y + x

→ 10 + 8 = 18

Answer 2

Question :--- The sum of digits of a two digit number is 9 .Also 9 times the number is twice the number obtain by reversing the order of digits.find the number ?

Solution :---

Let the original Two digit Number be = 10x+y .

It is given that sum of both digits is 9.

So,

→ x + y = 9 ----------------- Equation (1)

Now, it has been said that , 9 times the number is twice the number obtain by reversing the order of digits..

→ Reverse Number is = 10y+x .

So, According to Question now,

→ 9 * original number = 2 * Reverse Number

→ 9 *(10x+y) = 2 * (10y+x)

→ 90x + 9y = 20y + 2x

→ 90x - 2x = 20y - 9y

→ 88x = 11y

Dividing both sides by 11 now, we get,

→ y = 8x ------------------ Equation (2)

Putting value of Equation (2) in Equation (1) now we get,

→ x + 8x = 9

→ 9x = 9

Dividing both sides by 9,

→ x = 1 .

Putting this value in Equation (2) now we get,

→ y = 8x = 8*1 = 8

So, the original number will be = 10x+y = 10*1+8 = 10+8 = 18 ...

Hence, the Required Number is 18..