Question

please help me.

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

Answer 1

Ans

explanation:

Given :-

The sum of digits f a two digit number is 9.

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

To Find :-

The Number

Solution :-

Let the unit digit and tens digits of the number be x and y

Number = 10y + x

Number after reversing the digits = 10x + y

According to the question,

⇒ x + y = 9 ... (i)

⇒ 9(10y + x) = 2(10x + y)

⇒ 88y - 11x = 0

⇒ -x + 8y =0 ... (ii)

Adding equation (i) and (ii), we get

⇒ 9y = 9

⇒ y = 1 ... (iii)

Putting the value in equation (i), we get

⇒ x = 8

Hence, the number is 10y + x = 10 × 1 + 8 = 18..

Hope this help you

Answer 2

Answer :

$\large{\star\:\:\boxed{\bf{The\:Original\:Number\:is\:18\:.}}\:\:\star}$

Explanation :

Given :–

• Sum of a two digit number is 9 .

• 9 times the Original Number is twice the number obtained by reversing the order of the digits.

To Find :–

• The Original Number.

Solution :–

Let the Tens digit of the Original Number be x and the Ones digit be y .

☆ According to the First Condition :-

⇒ x + y = 9  ----------(1)

☆ According to the Second Condition :-

⇒ 9(10x + y) = 2(10y + x)

⇒ 90x + 9y = 20y + 2x

⇒ 90x - 2x + 9y -20y = 0

⇒ 88x - 11y = 0

⇒ 11(8x - y) = 0

⇒ 8x - y = 0/11

⇒ 8x - y = 0  ------------(2)

Adding Equation(1) and Equation(2) :-

⇒ (x + y) + (8x - y) = 9 + 0

⇒ x + 8x + y - y = 9

⇒ 9x = 9

⇒ x = 9/9

⇒ x = 1

Putting this value of 'x' in Equation(1) :-

⇒ 1 + y = 9

⇒ y = 9 - 1

⇒ y = 8

Now we have,

• Tens Digit = 1

• Ones Digit = 8

The number will be = 10(1) + (8) = 10 + 8 = 18

∴ The Number will be 18 .