Question 1044926
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Steve has $2.05 in quarters and nickels. He has 13 coins altogether. How many coins of each kind does he have?
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<pre>
Let "n" be the number of nickels Steve has in his collection.
Then the number of quarters is 13 - n.

The nickels contribute 5n cents to the total.
The quarters contribute 25*(13-n).
Altogether they make 5n + 25*(13-n) cents.

It is equal to 205 cents, according to the condition.

It gives you an equation

5n + 25*(13-n) = 205.

Simplify and solve it for n:

5n + 325 - 25n = 205.

-20n = 205 - 325,

-20n = -120,

n = {{{(-120)/(-20)}}} = 6.

<U>Answer</U>. 6 nickels and 13-6 = 7 quarters.
</pre>

For many other solved coin problems see the lessons

&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>

in this site.