2003 AMC 8 Problems/Problem 4: Difference between revisions

Latest revision as of 15:15, 14 February 2024

Problem

A group of children riding on bicycles and tricycles rode past Billy Bob's house. Billy Bob counted $7$ children and $19$ wheels. How many tricycles were there?

$\mathrm{(A)}\ 2 \qquad\mathrm{(B)}\ 4 \qquad\mathrm{(C)}\ 5 \qquad\mathrm{(D)}\ 6 \qquad\mathrm{(E)}\ 7$

Solution

Solution 1

If all the children were riding bicycles, there would be $2 \times 7=14$ wheels. Each tricycle adds an extra wheel and $19-14=5$ extra wheels are needed, so there are $\boxed{\mathrm{(C)}\ 5}$ tricycles.

Solution 2

Setting up an equation, we have $a+b=7$ children and $3a+2b=19$ . Solving for the variables, we get, $a=\boxed{\mathrm{(C)}\ 5}$ tricycles.

~AllezW

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

@@ Line 5: / Line 5: @@
 ==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.
-or
+===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.
-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}}

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