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

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

EXPLANATION :-

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

Number = 10y + x

•Number after reversing the digits = 10x + y

A.T.Q

⇒ 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.

Answer 2

$\bold\red{GIVEN}$

The sum of two digits number is 9..

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

$\bold\red{TO \ FIND}$

• The number

$\bold\red{SOLUTION}$

↬Suppose the unit and tense digit of the number be x and y

↬Number=10y+x

↬Number after reserving the digits=10x+y

↬x+y=9 [step-1]

↬9(10y+x)=2(10x+y)

↬88y-11x=0

↬-x+8y=0 [step-2]

Add step 1+step 2

↬9y=9

↬y=1 [step-3]

Putting the value in equation 1 we will get:

↬ x=8

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