Question 300978
Let {{{n}}} = number of nickels
Let {{{d}}} = number of dimes
Let {{{q}}} = number of quarters
given:
{{{5n + 10d + 25q = 450}}} (in cents)
{{{d = 2n}}}
{{{q < d}}}
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By substitution:
{{{5n + 10*(2n) + 25q = 450}}}
{{{25n + 25q = 450}}}
{{{n + q = 18}}}
{{{d/2 + q = 18}}}
I know that {{{q<18}}}, since quarters must add up to less than $4.50
If {{{q<d}}} then, suppose {{{q = 11}}}, {{{d}}} could be {{{10}}}
That's $2.75 plus $1.00 = $3.75. If there are twice as many dimes as
nickels, {{{n = 5}}}, totaling $.25
$3.75 + $.25 = $4.00
My next guess is {{{q = 13}}},{{{d = 12}}}, {{{n = 6}}}
You can go on from here, since you get the idea