2015/16 AHSMC Part 1 Problem 16

Problem 16 of 16

Each of Alvin, Bob and Carmen spent five consecutive hours composing problems. Alvin started alone, and was later joined by Bob. Carmen joined in before Alvin stopped. When one person was working alone $4$ problems were composed per hour. When two people were working together, each only composed $3$ problems per hour. When all three were working, each composed only $2$ problems per hour. No coming or going occurs during the composition of any problem. At the end, $46$ problems were composed. How many were composed by Bob?

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Solution

The total work period may be divided into five intervals by comings and goings. The respective numbers of people working during these intervals are $1$ , $2$ , $3$ , $2$ and $1$ respectively. Note that the total length of any three consecutive intervals is $5$ hours. Hence the fourth interval has the same length as the first and the fifth interval has the same length as the second. During each of the second, third and fourth interval, the number of problems composed was $6$ per hour since $3+3=6=2+2+2$ . Hence $5 \times 6 = 30$ problems were composed when Bob was working. The number of problems composed when Alvin or Carmen was working alone was $46-30=16$ . It follows that the total length of these two intervals is equal to $16 \div 4 = 4$ hours. Hence the total length of the second and the fourth interval is also $4$ hours, so that the length of the third interval is $5-4 = 1$ hour. Thus the number of problems composed by Bob was $1 \times 2 + 4 \times 3 = 14$ . The answer is (A).

University of Alberta

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