Difference between revisions of "2011 AMC 12A Problems/Problem 9"

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== Problem ==
 
== Problem ==
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At a twins and triplets convention, there were <math>9</math> sets of twins and <math>6</math> sets of triplets, all from different families. Each twin shook hands with all the twins except his/her siblings and with half the triplets. Each triplet shook hands with all the triplets except his/her siblings and with half the twins. How many handshakes took place?
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<math>
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\textbf{(A)}\ 324 \qquad
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\textbf{(B)}\ 441 \qquad
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\textbf{(C)}\ 630 \qquad
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\textbf{(D)}\ 648 \qquad
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\textbf{(E)}\ 882 </math>
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== Solution ==
 
== Solution ==
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There are <math>18</math> total twins and <math>18</math> total triplets. Each of the twins shakes hands with the <math>16</math> twins not in their family and <math>9</math> of the triplets, a total of <math>25</math> people. Each of the triplets shakes hands with the <math>15</math> triplets not in their family and <math>9</math> of the twins, for a total of <math>24</math> people. Dividing by two to accommodate the fact that each handshake was counted twice, we get a total of <math>\frac{1}{2}\times 18 \times (25+24) = 9 \times 49 = 441 \rightarrow \boxed{\textbf{B}}</math>
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==Video Solution==
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https://www.youtube.com/watch?v=6tlqpAcmbz4
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~Shreyas S
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== See also ==
 
== See also ==
 
{{AMC12 box|year=2011|num-b=8|num-a=10|ab=A}}
 
{{AMC12 box|year=2011|num-b=8|num-a=10|ab=A}}
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{{MAA Notice}}

Latest revision as of 23:47, 15 July 2020

Problem

At a twins and triplets convention, there were $9$ sets of twins and $6$ sets of triplets, all from different families. Each twin shook hands with all the twins except his/her siblings and with half the triplets. Each triplet shook hands with all the triplets except his/her siblings and with half the twins. How many handshakes took place?

$\textbf{(A)}\ 324 \qquad \textbf{(B)}\ 441 \qquad \textbf{(C)}\ 630 \qquad \textbf{(D)}\ 648 \qquad \textbf{(E)}\ 882$

Solution

There are $18$ total twins and $18$ total triplets. Each of the twins shakes hands with the $16$ twins not in their family and $9$ of the triplets, a total of $25$ people. Each of the triplets shakes hands with the $15$ triplets not in their family and $9$ of the twins, for a total of $24$ people. Dividing by two to accommodate the fact that each handshake was counted twice, we get a total of $\frac{1}{2}\times 18 \times (25+24) = 9 \times 49 = 441 \rightarrow \boxed{\textbf{B}}$

Video Solution

https://www.youtube.com/watch?v=6tlqpAcmbz4 ~Shreyas S

See also

2011 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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 AMC 12 Problems and Solutions

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