Problem 17 · 2014 Math Kangaroo
Stretch
Logic & Word Problems
sum-constraint
Six boys live together in an apartment that has two bathrooms. Each morning from 7:00 they use both bathrooms before breakfast, each boy being alone in one of the two bathrooms for 8, 10, 12, 17, 21, and 22 minutes respectively. What is the earliest time that all six boys can have breakfast together?
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Answer: B — 7:46
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Hint 1 of 2
The two bathrooms run in parallel, so the finish time is the larger of the two bathrooms' total minutes.
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Hint 2 of 2
Split the six times into two groups to make the bigger group's total as small as possible.
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Approach: balance the two parallel bathroom totals
- Total bathroom time is 8+10+12+17+21+22 = 90 minutes, shared between two bathrooms running at the same time.
- Everyone is done when the busier bathroom finishes, so split the times to minimise the larger total.
- The best balance is {22,12,10} = 44 and {21,17,8} = 46; no split reaches 45–45, so the larger total is 46 minutes.
- Starting at 7:00, the earliest common breakfast time is 7:46.
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