Question 1176731


let {{{10}}} cents coins be {{{d}}} and {{{25}}} cents coins {{{q}}}

if she has twenty coins altogether, we have

{{{d+q=20}}}
{{{d=20-q}}}....eq.1

{{{1d}}}=${{{0.10}}}
{{{1q}}}=${{{0.25}}}

if Ann has ${{{3.20}}}, then

{{{0.10d+0.25q=3.20}}}.....substitute {{{d}}} from eq.1

{{{0.10(20-q)+0.25q=3.20}}}

{{{2-0.10q+0.25q=3.20}}}

{{{0.15q=3.20-2}}}

{{{0.15q=1.20}}}...both sides multiply by {{{100}}} to get rid of decimals

{{{15q=120}}}

{{{q=120/15}}}

{{{q=8}}}

go to eq.1

{{{d=20-q}}}....eq.1, substitute {{{q}}}
{{{d=20-8}}}
{{{d=12}}}

so Ann has {{{12}}} of {{{10}}} cents coins and {{{8}}} of {{{25}}} cents coins