Difference between revisions of "2003 AMC 8 Problems/Problem 4"
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==Solution== | ==Solution== | ||
If all the children were riding bicycles, there would be <math> 2 \times 7=14 </math> wheels. Each tricycle adds an extra wheel and <math> 19-14=5 </math> extra wheels are needed, so there are <math> \boxed{\mathrm{(C)}\ 5} </math> tricycles. | If all the children were riding bicycles, there would be <math> 2 \times 7=14 </math> wheels. Each tricycle adds an extra wheel and <math> 19-14=5 </math> extra wheels are needed, so there are <math> \boxed{\mathrm{(C)}\ 5} </math> tricycles. | ||
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| + | Setting up an equation, we have <math>a+b=7</math> children and <math>3a+2b=19</math>. Solving for the variables, we get, <math>a=\boxed{\mathrm{(C)}\ 5} </math> triangles. | ||
Revision as of 21:50, 18 October 2011
Problem
A group of children riding on bicycles and tricycles rode past Billy Bob's house. Billy Bob counted 
children and 
wheels. How many tricycles were there?
Solution
If all the children were riding bicycles, there would be 
wheels. Each tricycle adds an extra wheel and 
extra wheels are needed, so there are 
tricycles.
or
Setting up an equation, we have 
children and 
. Solving for the variables, we get, 
triangles.
