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(A) Let the average of the $10$ numbers be $y$.Then,the sum of the $10$ numbers is $10y$.When the digits of one number are interchanged,the new averag

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$4$

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(A) Let the average of the $10$ numbers be $y$.Then,the sum of the $10$ numbers is $10y$.When the digits of one number are interchanged,the new average becomes $y + 3.6$.Therefore,the new sum of the $10$ numbers is $10(y + 3.6) = 10y + 36$.The increase in the sum is $(10y + 36) - 10y = 36$.Let the two-digit number be $10a + b$,where $a$ is the tens digit and $b$ is the units digit.After interchanging the digits,the new number is $10b + a$.The difference between the new number and the original number is $(10b + a) - (10a + b) = 9b - 9a = 9(b - a)$.Since the increase in the total sum is $36$,we have $9(b - a) = 36$.Dividing both sides by $9$,we get $b - a = 4$.

$A$ student finds the average of $10$ two-digit numbers. If the digits of one of the numbers are interchanged,the average increases by $3.6$. The difference between the digits of that two-digit number is:

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