2015/16 AHSMC Part 1 Problem 5

Problem 5 of 16

Each of Alan, Bailey, Clara and Diane has a number of candies. Compared with the average of the number of candies each person has, Alan has $6$ more than the average, Bailey has $2$ more than the average, Clara has $10$ fewer than the average and Diane has $k$ candies more than the average. Determine $k$ .

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Solution

Let $m$ be the average in question. Then the four people have a total of $4m$ candies. Alan, Bailey, and Clara have $m+6$ , $m+2$ , $m-10$ candies, which totals $3m-2$ candies. Therefore, Diane has $4m - (3m-2) = m+2$ and thus has $2$ more candies than the average.

An alternate approach: since the total of the differences from the average must be zero, Diane should have just $10-6-2=2$ candies more than the average.

The answer is (B).

University of Alberta

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