<PRE><B>Question 742235</B>
# of Coins: N-nickels, D-dimes, Q-quarters

there are 241 coins - means {{{N+D+Q=241}}}

worth $26.20 - means {{{0.05N+0.10D+0.25Q=$26.20}}}, but if we multiply 20 we get

{{{N+2D+5Q=524}}}

There are 4 more nickels than dimes means {{{D+4=N}}}

So we have 3 equations and 3 unknowns:

{{{N+2D+5Q=524}}}
{{{N+D+Q=241}}}
{{{D+4=N}}}

So we begin...

5 times the second equation minus the first equation will eliminate Q:

{{{5(N+D+Q)-(N+2D+5Q)=5(241)-524}}}
{{{5N+5D+5Q-N-2D-5Q=681}}}
{{{4N+3D=681}}}

Since N=D+4 we can write {{{4N+3D=681}}} as

{{{4(D+4)+3D=681}}}

{{{7D+16=681}}}
{{{7D=665}}}
{{{D=95}}} (95 dimes)

This means that there D+4=95+4=99 nickels

Finally we can see that there must be 241-99-95=47 quarters

(so there are 99 nickels, 95 dimes and 47 quarters)</PRE>