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30 May 2026, 09:39
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At a community center, Nora and five other volunteers packed snack bags for an event. If altogether they packed 43 snack bags, and no two volunteers packed the same number of snack bags, did Nora pack more than 6 snack bags?
(1) Each volunteer packed at least 4 snack bags, and the greatest number of snack bags packed by a volunteer was greater than 12.
(2) Nora packed fewer snack bags than the median number of snack bags packed by the six volunteers.
(1) Each volunteer packed at least 4 snack bags, and the greatest number of snack bags packed by a volunteer was greater than 12.
(2) Nora packed fewer snack bags than the median number of snack bags packed by the six volunteers.
30 May 2026, 09:39
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Official Solution:
At a community center, Nora and five other volunteers packed snack bags for an event. If altogether they packed 43 snack bags, and no two volunteers packed the same number of snack bags, did Nora pack more than 6 snack bags?
(1) Each volunteer packed at least 4 snack bags, and the greatest number of snack bags packed by a volunteer was greater than 12.
The smallest possible total of 6 distinct integers, each at least 4, is:
\(4 + 5 + 6 + 7 + 8 + 9 = 39\)
Since the total is 43, there are only 4 “extra” bags beyond this minimum. Statement (1) also says the greatest count is greater than 12, so it must be at least 13. The only way to reach 13 with only 4 extra bags is to raise 9 to 13, which forces the full set of counts to be \(\{4, 5, 6, 7, 8, 13\}\). But we still do not know which of these counts is Nora’s, so Nora could be 6, giving a No answer, or 7, giving a Yes answer. Not sufficient.
(2) Nora packed fewer snack bags than the median number of snack bags packed by the six volunteers.
Many different distributions are possible, so Nora could have packed more than 6 in some cases and 6 or fewer in others. Not sufficient.
(1)+(2) From (1), the counts must be \(\{4, 5, 6, 7, 8, 13\}\), so the median is \(\frac{6 + 7}{2} = 6.5\). If Nora packed fewer than the median, Nora must be 4, 5, or 6, so she did not pack more than 6. Sufficient.
Answer: C
At a community center, Nora and five other volunteers packed snack bags for an event. If altogether they packed 43 snack bags, and no two volunteers packed the same number of snack bags, did Nora pack more than 6 snack bags?
(1) Each volunteer packed at least 4 snack bags, and the greatest number of snack bags packed by a volunteer was greater than 12.
The smallest possible total of 6 distinct integers, each at least 4, is:
\(4 + 5 + 6 + 7 + 8 + 9 = 39\)
Since the total is 43, there are only 4 “extra” bags beyond this minimum. Statement (1) also says the greatest count is greater than 12, so it must be at least 13. The only way to reach 13 with only 4 extra bags is to raise 9 to 13, which forces the full set of counts to be \(\{4, 5, 6, 7, 8, 13\}\). But we still do not know which of these counts is Nora’s, so Nora could be 6, giving a No answer, or 7, giving a Yes answer. Not sufficient.
(2) Nora packed fewer snack bags than the median number of snack bags packed by the six volunteers.
Many different distributions are possible, so Nora could have packed more than 6 in some cases and 6 or fewer in others. Not sufficient.
(1)+(2) From (1), the counts must be \(\{4, 5, 6, 7, 8, 13\}\), so the median is \(\frac{6 + 7}{2} = 6.5\). If Nora packed fewer than the median, Nora must be 4, 5, or 6, so she did not pack more than 6. Sufficient.
Answer: C
