<PRE><B>Question 131957</B>
Number of nickels: n
Number of dimes: d
Number of quarters: q


Given:
{{{n=d+2}}}, two more nickels than dimes,
{{{q=n+3}}}, three more quarters than nickels


Value of n nickels is {{{5n}}} cents
Value of d dimes is {{{10d}}} cents
Value of q quarters is {{{25q}}} cents


Total money is $3.35 which can also be expressed as 335 cents.


{{{5n + 10d + 25q=335}}}


Since {{{n=d+2}}} and {{{q=n+3}}}, by substitution we can say that {{{q = (d+2)+3=d+5}}}


Now take the value equation and substitute the two expressions for nickels and quarters in terms of dimes in place of the n and q variables:


{{{5(d+2) + 10d + 25(d+5)=335}}}


Distribute and collect terms:
{{{5d+10+10d+25d+125}}}
{{{40d+135=335}}}


Add -135 to both sides:
{{{40d=200}}}


Divide by 40:
{{{d=5}}}


So there are 5 dimes.  That means there are {{{5 + 2=7}}} nickels and {{{5+5=10}}} quarters.


Check the answer:
10 quarters is $2.50,
5 dimes is $.50,
7 nickels is $.35, and finally,
$2.50 plus $.50 plus $.35 = $3.35.


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