Let d = number of dimes
Let n = number of nickels
Let p = number of pennies
As given, there are two more nickels than dimes and three times as many pennies as dimes:
[tex]d+2=n[/tex]
[tex]3d=p[/tex]
[tex]0.10d+0.05n+0.01p=0.64[/tex]
Plug in the value to find d
[tex]0.10d+0.05(d+2)+0.01(3d)=0.64[/tex]
[tex]0.10d+0.05d+0.1+0.03d=0.64[/tex]
[tex]0.18d=0.54[/tex]
[tex]d=3[/tex]
As p=3d
p=[tex]3\times3=9[/tex]
As n=d+2
[tex]n=3+2=5[/tex]
Hence, there are 5 nickels, 3 dimes and 9 pennies.
