A family has a total of 80 coins, consisting of dimes and quarters. If the total value of the coins is $15.50, how many dimes and quarters are there?
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Let's denote the number of dimes as $d$ and the number of quarters as $q$. We know that $d+q=80$, and that the value of the dimes is $0.10d$ and the value of the quarters is $0.25q$. Additionally, we can express the total value of the coins as $0.10d+0.25q$. We can now set up a system of equations to represent the given information:
$\begin{cases} d+q=80 \\ 0.10d+0.25q=15.50 \end{cases}$
We can multiply the first equation by 0.10, in order to eliminate the $d$ term:
$\begin{cases} 0.10d+0.10q=8 \\ 0.10d+0.25q=15.50 \end{cases}$
By subtracting the first equation from the second equation, we obtain $0.15q=7.5$. Dividing both sides by 0.15, we find $q=50$. Substituting this value into the first equation, we get $d+50=80$, and hence $d=80-50=30$.
Thus, there are $30$ dimes and $50$ quarters.
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