Saturday, May 31, 2014

Intermediate Algebra, Chapter 2, 2.4, Section 2.4, Problem 10

Nana Nantambu found some coins while looking under her sofa pillows. There were equal numbers of nickels and quarters and twice as many half dollars as quarters. If she found $\$2.60$ in all, how many of each denomination of coin did she find?

$
\begin{array}{|c|c|c|}
\hline
\text{Number of coins} & \rm{Denomination} & \rm{Value} \\
\hline
x & 0.05 & 0.05x \\
\hline
x & \phantom{blank} & \phantom{blank} \\
\hline
2x & 0.50 & \phantom{blank} \\
\hline
\end{array}
$


If we fill in the table, then we have


$
\begin{array}{|c|c|c|}
\hline
& \text{Number of coins} & \text{Denomination} & \text{Value} \\
\hline
\rm{Nickels}& x & 0.05 & 0.05x \\
\hline
\rm{Quarters}& x & 0.25 & 0.25x \\
\hline
\rm{Half dollar}& 2x & 0.50 & 0.50(2x) \\
\hline
\end{array}
$


In the last column of the table, we know that the total value is equal to the sum of each
values of the coin she found.


$
\begin{equation}
\begin{aligned}
0.05x + 0.25x + 0.50(2x) &= 2.60\\
\\
0.30x + x &= 2.60\\
\\
1.30x &= 2.60 \\
\\
x &= 2
\end{aligned}
\end{equation}
$


Then by substitution,
$2x = 2(2) = 4$
In other words, she found 2 nickels, 2 quarters and 4 half dollars under the pillows.

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