Question 320851
p+n+d=21
.01p+.05n+.1d=100
if there are any pennies they have to be in 5,10,15,or 20
if there are 20 penies then only room for 1 more coin which cant add to 1.00
15 pennies then there are 6 coins left but even at 6 dimes thats only 0.75
10 pennies makes 11 coins left n+d=11 and .05n+.1d=.90 is a possibility
5  pennies makes 16 coins left n+d=16 and .05n+.1d=.95 is a 2nd possibility
according to the problem it seems there has to be a number for each coin
<p>

{{{.05n+.1d=.90}}}  multiply by 100
{{{n+d=11}}} multiply by 5
.
.
{{{5n+10d=90}}}
{{{5n+5d=55}}}
{{{5d=35}}}
{{{d=7}}}
<p>
{{{p+n+d=21}}}
{{{p=10}}}
{{{n=7}}}
{{{17+d=21}}}
{{{d=4}}}
check:
{{{highlight(10+7+4=21)}}}
10 pennies, 7 nickels, 4 dimes
this wasnt that hard...lol that other tutor confused even me.
<p>
other solution:
{{{n+d=16}}} multiply by 10
{{{.05n+.1d=.95}}} multiply by 100
.
.
{{{10n+10d=160}}}
{{{5n+10d=95}}}
{{{5n=65}}}
{{{n=13}}}
.
{{{13+d=16}}}
{{{d=3}}}
.
{{{highlight(5+13+3=21)}}}
5 pennies, 13 nickels, 3 dimes