Question 1098751
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        There are different methods to solve the problem. 
        I will show you 3 (three) basic methods in this post.


<U>1- Algebra solution, &nbsp;&nbsp;2 equations approach</U>


<pre>
Let d be the number of dimes and q be the number of quarters.

Then you have this system of two equations

  d +   q =   77,        (1)    (counting dimes)
10d + 25q = 1280.        (2)    (counting cents)


To solve it, multiply eq(1) by 10. You will have

10d + 10q =  770,        (1')
10d + 25q = 1280.        (2')


Next subtract (1') from (2'). You will get

25q - 10q = 1280 - 770,

15q = 510  ====>  q = {{{510/15}}} = 34.


The parking meter contains 34 quarters and 77-34 = 43 dimes.


<U>Check</U>.   43*10 + 34*25 = 33*10 + 34*25 = 1280.    !  Correct !


<U>Answer</U>.  34 quarters and 43 dimes.
</pre>


<U>2- Algebra solution, &nbsp;&nbsp;1 equation approach</U>


<pre>
Let  q be the number of quarters.

Then the number of dimes is (77-q). 

The quarters contribute 25q cents toward the total.
The dimes contribut 10*(77-q) toward the total.

The total is  25q + 10*(77-q).

From the other side, it is 1280 cents, according to the condition.

It gives you an equation

25q + 10*(77-q) = 1280.     


Simplify and solve for q:

25q + 770 - 10q = 1280 

15q = 1280 - 770  

15q = 510  ====>  q = {{{510/15}}} = 34.


The parking meter contains 34 quarters and 77-34 = 43 dimes.


You got the same answer.
</pre>


<U>2- Logical analysis (MENTAL solution without using equations)</U>


<pre>
Imagine for a minute that all coins in the parking meter are DIMES.

Then you would have 10*77 = 770 cents in the parking meter.

But in reality the money amount there is 1280 cents, which is 1280-770 = 510 cents more than 770 cents.

Why we have this difference ?  - But of course, because we counted 25-cent quarters as 10-cent coins.

Then it is clear, that the number of quarters is  {{{510/(25-10)}}} = {{{510/15}}} = 30 + 4 = 34.


And you get the same answer.
</pre>


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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.


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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.