The sum of the digits of two digit number is 10. 10 less than the number is 16 more than twice the number reversing the digits of the number. find the number?
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let one digit be x
therefore the second digit will be 10-x
and hence the number will be 10x+10-x
9x+10
given:9x+10-10=16+2(10(10-x)+x)
9x=16+2(100-10x+x)
9x=16+200-18x
27x=216
x=72/9
x=8
and the old no=9×8+10
=82
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therefore the second digit will be 10-x
and hence the number will be 10x+10-x
9x+10
given:9x+10-10=16+2(10(10-x)+x)
9x=16+2(100-10x+x)
9x=16+200-18x
27x=216
x=72/9
x=8
and the old no=9×8+10
=82
mark me as brainlist
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