Math, asked by Mirhumaira256976, 4 months ago

the sum of the digits of a two digits number is 9.Also, nine times this number is twice the number obtained by reversing the order of the digits. find the number. ​

Answers

Answered by Dhaarini22
1

Answer:

hy

Step-by-step 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.

Answered by Anonymous
2

Let the two digit number is 10x + y. S

um of the digits of to digit number is

x + y = 9….... (i)

Number obtained by reversing the order of the digits,

we get 10y + x.

Twitching the 10y + x, it is nine times of first number.

∴ 2(10y + x) = 9(10x + 4)

20y + 2x = 90x + 9y

2x – 90x + 20y – 9y = 0

-88x + 11y = 0

88x – 11y = 0

8x – y = 0 ………….. (ii)

By adding eqn. (i) to eqn. (ii),

(in attachment)

Substituting the value of ‘x’ in eqn. (i),

x + y = 9

1 + y = 9

y = 9 – 1

y = 8

∴ x = 1, y = 8

∴ Two digit number = 10x + y

= 10 × 1 + 8

= 10 + 8

=18.

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