Question 1042758
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In a collection of dimes and quarters, the number of dimes is eight less than twice the number of quarters. 
The value of the collection is $8.20. Find the number of dimes and quarters.
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Let me solve it mentally, without using equations.


<pre>
Add 8 dimes to the collection (mentally).
Then the total would be $ 8.20 + $0.80 = $9.00.

Then $9.00 consists of some number of dimes (d) and twice the number of quarters.

So, we have 3d coins, which we can re-group into d groups consisting of two quarters and one dime each.

d groups of coins worth 2*50+10 = 60 cents each group.

Obviously, the number of groups is {{{900/60}}} = 15.

Hence, originally we had 2*15 = 30 quarters and 15-8 = 7 dimes.

<U>Check</U>. 30*25 + 70 = 750 + 70 = 820 cents.

<U>Answer</U>.  30 quarters and 7 dimes.
</pre>

Solved.


For 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.