Difference between revisions of "2003 AMC 8 Problems/Problem 4"
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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> tricycles. | 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> tricycles. | ||
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| + | {{AMC8 box|year=2003|num-b=3|num-a=5}} | ||
Revision as of 08:50, 25 November 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, 
tricycles.
| 2003 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 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 | ||
