Difference between revisions of "2003 AMC 8 Problems/Problem 4"
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==Solution== | ==Solution== | ||
| + | ===Solution 1=== | ||
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. | ||
| − | + | ===Solution 2=== | |
| + | 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. | ||
| − | + | ~AllezW | |
| + | |||
| + | ==See Also== | ||
| + | {{AMC8 box|year=2003|num-b=3|num-a=5}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 16:15, 14 February 2024
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
Solution 1
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.
Solution 2
Setting up an equation, we have 
children and 
. Solving for the variables, we get, 
tricycles.
~AllezW
See Also
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
