Question

The sum of the digits of a two digit number is 9.If the number obtained by reversing the order of digits is 27 more than the original number,find the original number.​

Answer 1

Answer:

36

Step-by-step explanation:

let the digit at one's place be x

and the digit at tenth's place be y

=> x + y = 9 --- (i)

=> 10y + x = 10x + y - 27

   =>  9y - 9x = -27

   =>  y - x = -3 --- (ii)

adding (i) and (ii)

2y = 6

y = 3

therefore, x  = 6

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Answer 2

$\mathbb{ANSWER}$

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Let digit at tens place be: $x$

Then, digit at ones place would be $(9-x)$ (Since number of digits is $9$

So number would be: $10x+(9-x)=9x+9$

$\rule{300}{1}$

On reversing the order of the digit, new number would be :

$10(9-x)+x=90-10x+x=90-x$

$\rule{300}{1}$

According To question, we have :

$\rightarrow$ $(90-9x)=(9x+9)+27$

$\rightarrow$ $90-9x=9x+36$

$\rightarrow$ $-9x-9x=36-90$

$\rightarrow$ $-18x=-54$

$\rightarrow$ $x=$$\dfrac{-54}{-18}$$=3$

$\rule{300}{1}$

so, number would be, $(9*3+9)=27+9=36$

$\rule{300}{1}$

Hence, The number of two digit is :

$\rightarrow$ $=36$

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