Question 1121214
.
<U>1. One-equation setup and solution</U>


<pre>
Let x be the number of dimes (10-cents coins).

Then the number of quarters (25-cents coins) is 30-x.


The "value" equation is


    10x + 25(30-x) = 480   cents.     (1)


Simplify and solve for x 


    10x + 750 - 25x = 480

    750 - 480 = 25x - 10x

    270 = 15x

    x = 270/15 = 18.


<U>Answer</U>.  18 dimes and (30-18) = 12 quarters.


<U>Check</U>.   18*10 + 12*25 = 480 cents.     ! Correct !
</pre>

Solved.



<U>2. Two equations setup and solution</U>


<pre>
     x +   y =  30          (2)    (counting coins)
    5x + 25y = 480          (3)    (counting cents)


From equation (2), express y = 30-x  and then substitute it into equation (4). You will get 


    5x + 25*(30-x) = 480    (4)


Equation (4) is identical to equation (1) of the 1-st solution.


So, in this way you get the same answer.
</pre>

Solved.



<U>3. Mental/logic solution without using equations</U>


<pre>
Assume for a minute that all 30 coins are 10-cents dimes - then the total would be 300 cents, making the shortage of 480-300 = 180 cents.


So, we need to replace some dimes by quarters to compensate the difference.


At each replacement, we diminish the difference by 25-10 = 15 cents, so {{{180/15}}} = 12 replacements are required.


Thus, the collection must have 12 quarters, and the rest 30-12 = 18 are dimes.
</pre>


Solved (3 times by 3 different methods).



--------------


Congratulations ! &nbsp;&nbsp;You are now familiar with 3 basic methods for solving typical coin problems.


I suggest that algebraic methods will be your basic methods for such problems,

and the logical analysis method will allow you to solve the problems MENTALLY without using equations.

I will be happy if it will make your horizon wider.


--------------


To see other similar solved coin problems, look in 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/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/Solving-coin-problems-mentally-by-grouping-without-using-equations.lesson>Solving coin problems mentally by grouping without using equations</A>

in this site.


To see how the logical method works for other similar problems, look into the lessons

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/misc/Problem-on-two-wheels-and-three-wheels-bicycles-in-a-sale.lesson>Problem on two-wheel and three-wheel bicycles</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/misc/Problem-on-animals-at-a-farm.lesson>Problem on animals at a farm</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/misc/Problem-on-tablets-in-containers.lesson>Problem on pills in containers</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/misc/What-type-of-problems-are-these.lesson>What type of problems are these?</A> 

in this site.