Question

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.​

Answer 1

S O L U T I O N :

Let the tens digit place be x & ones digit place be y respectively.

$\boxed{\bf{Original \:number = 10x + y}}$

$\boxed{\bf{Reversed \:number = 10y + x}}$

A/q

$\mapsto\tt{x+y = 9............(1)}$

&

$\mapsto\tt{9(10x + y) = 2(10y+x)}$

$\mapsto\tt{90x+ 9y = 20y+2x}$

$\mapsto\tt{90x - 2x = 20y-9y}$

$\mapsto\tt{\cancel{88}x = \cancel{11}y}$

$\mapsto\tt{8x = y...............(2)}$

∴ Putting the value of y in equation (1),we get;

$\mapsto\tt{x + 8x = 9}$

$\mapsto\tt{9x = 9}$

$\mapsto\tt{x = \cancel{9/9}}$

$\mapsto\bf{x=1}$

∴ Putting the value of x in equation (2),we get;

$\mapsto\tt{y = 8(1)}$

$\mapsto\bf{y = 8}$

Thus,

The number = 10x + y

The number = 10(1) + 8

The number = 10 + 8

The number = 18

Answer 2

Answer:

Given :-

• The sum of the digits of a two-digit number is 9.
• Nine times this number will be twice of that number obtained by reversing the order of the digits.

To Find :-

• What is the number.

Solution :-

Sum of the digits of the two number is 9

Let, the digits of unit place be x

And, the digits at tens place be (9 - x)

Hence, two numbers will be,

⇒ 10(9 - x) + 10

⇒ 90 - 10x + 10

⇒ 90 - 9x

Numbers will be reversing the digits,

⇒ 10x + 9 - x

⇒ 9x + 9

According to the question,

⇒ 9(90 - 9x) = 2(9x + 9)

⇒ 810 - 81x = 18x + 18

⇒ 810 - 18 = 18x + 81x

⇒ 792 = 99x

⇒ x = $\sf\dfrac{\cancel{792}}{\cancel{99}}$

➥ x = 8

Now, the required number is,

⇒ 90 - 9x

⇒ 90 - 9(8)

⇒ 90 - 72

➠ 18

$\therefore$ The number is 18.