Answer
$12$ nickels, $30$ dimes, $8$ quarters.
Work Step by Step
Step 1. Assume there are $x$ number of nickels, $y$ number of dimes, and $z$ number of quarters.
Step 2. Based on the conditions given, we can set up the following system of equations:
$\begin{cases} x+y+z=50\hspace3cm(total-number) \\0.05x+0.1y+0.25z=5.6\hspace1cm(total-value) \\0.1y=5\times0.05x \end{cases}$
Step 3. Simplify the equation by removing the decimals:
$\begin{cases} x+y+z=50\hspace2cm(total-number) \\5x+10y+25z=560\hspace1cm(total-value) \\10y=25x \end{cases}$
Step 4. Further simplify the equation by removing common factors and use the last equation for substitution:
$\begin{cases} x+2.5x+z=50\\x+2(2.5x)+5z=112\\y=2.5x \end{cases}$
or
$\begin{cases} 3.5x+z=50\\6x+5z=112\\y=2.5x \end{cases}$
Step 5. The first equation gives $z=50-3.5x$, use it in the second equation to get $6x+5(50-3.5x)=112$ which gives $x=12$. Find other numbers as $y=30$ and $z=8$
Step 6. Conclusion: there are $12$ nickels, $30$ dimes, and $8$ quarters.
