2003 AMC 8 Problems/Problem 4

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
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All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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2003 AMC 8 Problems/Problem 4

Contents

Problem

Solution

Solution 1

Solution 2

See Also