Two digit numbers is9 more than the number obtained when its digits are reversed .if the sum of its digits is9 ,find the number
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Suppose unit place digit be x
So ten's place digit is 9-x.
Now the two digit number is 10(9-x)+x
As per question, 10(9-x)+x+9= 10x+9-x
or, 90-10x+x+9=9x+9
or,99-9x=9x-9
or, 18x=90
or,x=90/18=5
Hence required two number=10(9-5)+5=40+5=45
So ten's place digit is 9-x.
Now the two digit number is 10(9-x)+x
As per question, 10(9-x)+x+9= 10x+9-x
or, 90-10x+x+9=9x+9
or,99-9x=9x-9
or, 18x=90
or,x=90/18=5
Hence required two number=10(9-5)+5=40+5=45
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