Difference between revisions of "2011 AMC 12A Problems/Problem 9"
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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} 18(25+24) = 9 \times 49 = 441 \rightarrow \boxed{\textbf{B}}</math> | + | 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> |
== 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}} | ||
Revision as of 21:12, 11 February 2011
Problem
At a twins and triplets convention, there were 
sets of twins and 
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?
Solution
There are 
total twins and total triplets. Each of the twins shakes hands with the 
twins not in their family and of the triplets, a total of 
people. Each of the triplets shakes hands with the 
triplets not in their family and of the twins, for a total of 
people. Dividing by two to accommodate the fact that each handshake was counted twice, we get a total of 
See also
| 2011 AMC 12A (Problems • Answer Key • Resources) | |
| 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 | |
