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Old 05-14-2008, 08:57 PM
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Irv's cycle rental shop, Irv rents all kinds of cycles: unicycles,tandem bikes, regular bikes, and even tricycles for little kids. He parks all the cycles in front of his shop with a helmet for each rider strapped to the cycles. THis morning Irv counted 57 helmets and 115 wheels parked in front of his store. He knows he has an equal number of unicycles and tandem bikes. He also knows that he has 32 regular bikes. How many unicycles, tandem bikes, and tricycles does Irv have? (a tandem bike has two helmets)
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Old 05-14-2008, 10:23 PM
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Hello, fishnprincipal!

Quote:
At Irv's cycle rental shop, he rents all kinds of cycles:
unicycles, tandem bikes, regular bikes, and even tricycles for little kids.
He has a helmet for each rider strapped to the cycles.
This morning Irv counted 57 helmets and 115 wheels parked in front of his store.
He knows he has an equal number of unicycles and tandem bikes.
He also knows that he has 32 regular bikes.
How many unicycles, tandem bikes, and tricycles does Irv have?


\begin{array}{c|cc}
& \text{Helmets} & \text{Wheels} \\ \hline
\text{Unicycles (U)} & 1 & 1 \\
\text{Tandems (T)} & 2 & 2 \\
\text{Bicycle (B)} & 1 & 2 \\
\text{Tricycles (R)} & 1 & 3 \\ \hline
\text{Totals:} & 57 & 115 \end{array}


We have: . U + 2T + B + R \:=\:57
. . . and: .U + 2T + 2B + 3R \:=\:115

We are told that: .U = T,\:B = 32

The equations becomes: .\begin{array}{cccccccc}
U + 2U + 32 + R &=&57 & \Rightarrow & 3U + R &=&25 & {\color{blue}[1]}\\
U + 2U + 64 + 3R &=&115 & \Rightarrow & 3U + 3R &=&51 & {\color{blue}[2]} \end{array}

Subtract [1] from [2]: .2R \:=\:26 \quad\Rightarrow\quad\boxed{ R \:=\:13}

Substitute into [1]: .3U + 13 \:=\:25 \quad\Rightarrow\quad\boxed{U \:=\:4} \quad\Rightarrow\quad\boxed{T \:=\:4}


Therefore: .4 unicycles, 4 tandems, and 13 tricycles.

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