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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> triangles.
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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.

Revision as of 21:51, 18 October 2011

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

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.

or

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.

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