Problem 14 · 2025 Math Kangaroo
Medium
Arithmetic & Operations
total-then-dividesum-constraint
Sanja has two bowls of numbered balls. The left bowl holds seven balls numbered 1, 2, 6, 7, 10, 11 and 12, with arithmetic mean 7.0. The right bowl holds five balls numbered 3, 4, 5, 8 and 9, with arithmetic mean 5.8. Sanja wants to increase the arithmetic mean of both bowls. Which ball must she move from the left bowl to the right bowl to do this?
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Answer: A — 6
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Hint 1 of 2
Moving a ball must push BOTH averages up — think about what value keeps each average rising.
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Hint 2 of 2
The moved ball must be below the left average (so the left mean rises) yet above the right average (so the right mean rises).
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Approach: bound the ball value from both averages
- Left sum is 7×7 = 49, right sum is 5.8×5 = 29.
- To raise the left mean the removed ball must be below 7; to raise the right mean it must be above 5.8.
- The only left-bowl ball between 5.8 and 7 is 6, which is (A).
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