The digit at the tens place of a two digit number is three times the digit at the ones place .if the sum of the number and the number formed by reversing its digit is 88 ,find the number ?
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Let one's place digit is x. tens place digit is 10-x
The value of two digit number =10(10-x)+x
=100-9x
If the digits are reversed then the number =9x+10
According to the problem
2(9x+10)-1=100-9x
18x+20-1=100-9x
18x+19=100-9x
18x+9x=100-19
27x=81
X=81/27
X=3
There fore the two digit number =73
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