Question

the sum of the digits of a two digits number is 9.the number obtained by reserving the digits is 9 less than the original number find the number​

Answer 1

Answer:

let the digit at the ones place be y

and that of tens place be x

then the number= 10x+y

its reverse no. = 10y+x

A/Q x + y= 9 ---------------------------(1)

10y+x = 10x+y - 9

9y - 9x = -9

y - x = -1

x - y = 1 -------------(2)

from 1 and 2 by adding

x = 5 and y = 4

thus the original number is 10x+y

= 10 x 5 + 4

= 54

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Answer 2

Answer:

Let the tenth place of the digit be = x

Let the ones place of the digit be = y

Number formed will be (10x + y)

Given that,

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

Now reversing the digit,

ones place = x

tens place = y

Number formed will be (10y +x)

Given that,

(10y +x) =  (10x + y) - 9

10y +x = 10x + y - 9

9y = 9x  - 9

y = x - 1

putting the value of y from above equation (i)

9 - x = x - 1

10/2 = x

x = 5

putting this value in equation (i)

5 + y = 9

y = 4

The number that we want was (10x + y) i.e. 10 x 5 + 4 => 54