The sum of the digits of a two-digit number is 7. If we put the digits of the number in reverse order, the new number is 49 less than twice the original number. Find 40% of the number.
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Step-by-step explanation:
Here ,
let x and y be the two digits such that
10x+y is the number
according to the question
x+y= 7
x= 7-y (1)
and again putting the digit in reverse i.e if 14(1×10+4) is the number, it would be 41 (4×10+1)
10y+x= 2(10x+y)-49 (2)
putting the value of x in equation 2
10y+(7-y)=2[ 10(7-y)+y]-49
9y+7=2[70-10y+y]-49
9y+7= 140-18y-49
9y+18y= 91-7
27y= 84
9y= 28
please recheck your question
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