Question 1051682
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Joe has 2 more quarters than nickels and 5 times as many nickels as dimes.  Altogether he has $6.90. How many dimes does Joe have?
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<pre>
Let "d" be the number of dimes.
Then the number of nickels is 5d,
and the number of quarters is (5d+2).

The dimes contribute 10d cents to the total.
The nickels contribute 5*(5d) = 25d cents to the total.
The quarters contribute 25*(5d+2) = 125d + 50 cents to the total.

Therefore, the "value" equation is

10d + 25d + 125d + 50 = 690   (cents).

Simplify and solve for d:

160d + 50 = 690,

160d = 690-40,

160d = 640  --->  d = {{{640/160}}} = 4.

<U>Answer</U>. The number of dimes is 4.  The number of nickels is 4*5 = 20.  The number of quarters is 5*4+2 = 22.

<U>Check</U>.  5*20 + 10*4 + 22*25 = 690.  Correct !
</pre>

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&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/coins/Coin-problems.lesson>Coin problems</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/coins/More-Coin-problems.lesson>More Coin problems</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/coins/Solving-coin-problem-without-equations.lesson>Solving coin problems without using equations</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/coins/Kevin-and-Randy-Muise-have-a-jar.lesson>Kevin and Randy Muise have a jar containing coins</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/coins/Typical-coin-problems-from-the-archive.lesson>Typical coin problems from the archive</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/coins/More-complicated-coin-problems.lesson>More complicated coin problems</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/coins/OVERVIEW-of-lesson-on-coin-word-problems.lesson>OVERVIEW of lessons on coin word problems</A>

in this site.


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