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Difference between revisions of "2015 AMC 8 Problems/Problem 23"

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

Latest revision as of 18:12, 20 January 2024

Problem

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 $A, B, C, D,$ and $E$ will have the values $5, 6, 7, 8$ and $9$, respectively. Now, it's the 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 have slips smaller than that, but upon closer inspection, we see that no slip fits exactly. And the smallest sum of two 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 Cup C and thus also impossible. With all other choices removed, we are left with the answer: Cup $\boxed{\textbf{(D)}~D}$.

Video Solution

https://youtu.be/SgEXCLfk_AU

~savannahsolver

See Also

2015 AMC 8 (Problems • Answer Key • Resources)
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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