Question

a number consist of two digit whose sum is 9 if 9 is subtracted from the number the digit interchange their places find the number

Answer 1

Let the digits at ones place be x
Then the digits at tens place=9-x (sum of two digits is 9)
Therefore the no.=10 (9-x)+x=90-10x+x=90-9x
Now,no. obtained by reversing the digits=10x+9-x=9x+9
it is given if 9 is subtracted from the no.,it's digits are interchanged.
no.-9=no. obtained by reversing the digits.
90-9x-9=9x+9
81-9x=9x+9
81-9=9x+9x
18x=72
x=72/18
x=4
digits at
ones place=x=4
tens place=9-x=9-4=5
the no. is 54....

Answer 2

Given : a number consist of two digit whose sum is 9 if 9 is subtracted from the number the digit interchange their places

To find :  the number

Solution:

Let say number is AB

A + B = 9

Value of AB  = 10A + B

on interchanging digits BA

Value of BA  = 10B + A

10 A + B - 9  = 10B + A

=> 9A - 9B = 9

=> A - B = 1

A + B = 9

A - B = 1

=> 2A = 10

=> A = 5

B = 4

54 is the number

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