sum of thw digits of a two digit nymber is 11.the given number is less than the number obtained.by interchanging the digits by 9.Find the number
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let the digit at tens place be x and at one place be 9-x
number= 10x+(9-x)
9x+9
if digits are interchanged then
digit at tens place =9-x and at ones place=x
number= 10(9-x)+x
90-10x+x
90-9x
a/q
90-9x-(9x+9)
90-9x-9x-9=9
81-18x=9
81-9=18x
72=18x
x=4
9-x= 5
therefore , original number= 9x+9
9×4+9
=45.ans
number= 10x+(9-x)
9x+9
if digits are interchanged then
digit at tens place =9-x and at ones place=x
number= 10(9-x)+x
90-10x+x
90-9x
a/q
90-9x-(9x+9)
90-9x-9x-9=9
81-18x=9
81-9=18x
72=18x
x=4
9-x= 5
therefore , original number= 9x+9
9×4+9
=45.ans
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