2015 AMC 8 Problems/Problem 23

Revision as of 16:41, 25 November 2015 by Matongxu (talk | contribs) (Solution presented on the problems page, modified to fit standards)

Tom has twelve slips of paper which he wants to put into five cups labeled $A$, $B$, $C$, $D$, $E$. He wants the sum of the numbers on the slips in each cup to be an integer. Furthermore, he wants the five integers to be consecutive and increasing from $A$ to $E$. The numbers on the papers are $2, 2, 2, 2.5, 2.5, 3, 3, 3, 3, 3.5, 4,$ and $4.5$. If a slip with $2$ goes into cup $E$ and a slip with $3$ goes into cup $B$, then the slip with $3.5$ must go into what cup?

$\textbf{(A) } A \qquad \textbf{(B) } B \qquad \textbf{(C) } C \qquad \textbf{(D) } D \qquad \textbf{(E) } E$

Solution

The numbers have a sum of $6+5+12+4+8=35$, which averages to $7$, which means $A, B, C, D, E$ will have values $5, 6, 7, 8, 9$, respectively. Now it's process of elimination: Cup $A$ will have a sum of $5$, so putting a $3.5$ slip in the cup will leave $5-3.5=1.5$; However, all of our slips are bigger than $1.5$, so this is impossible. Cup $B$ has a sum of $6$, but we are told that it already has a $3$ slip, leaving $6-3=3$, which is too small for the $3.5$ slip. Cup $C$ is a little bit trickier, but still manageable. It must have a value of $7$, so adding the $3.5$ slip leaves room for $7-3.5=3.5$. This looks good at first, as we do have slips smaller than that, but upon closer inspection, we see that no slip fits exactly, and the smallest sum of 2 slips is $2+2=4$, which is too big, so this case is also impossible. Cup $E$ has a sum of $9$, but we are told it already has a $2$ slip, so we are left with $9-2=7$, which is identical to the Cup C case, and thus also impossible. With all other choices removed, we are left with the answer: Cup $\boxed{D}$

See Also

2015 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AJHSME/AMC 8 Problems and Solutions

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