Coins making up a certain sum A cash register contains only dimes and quarters. There are $65$ coins equaling $\$12.80$ in the register. How many dimes and how many quarters are in the register?
 A: We first give a conventional "algebra" solution. Measure everything in cents. So the total in the cash register is $1280$.
Let $d$ be the number of dimes, and let $q$ be the number of quarters.
Then 
$$d+q=65.$$
The value, in cents, of the dimes is $10d$. The value of the quarters is $25q$. But the combined value of all the coins is $1280$. This gives us the equation
$$10d+25q=1280.$$
(If we measured everything in dollars, the equation would be $0.10d+0.25q=12.80$, not really very different.)
There are various ways to solve the system of two equations
$$d+q=65,\qquad 10d+25q=1280.$$
One way is to note that $d=65-q$.  Substitute in the second equation. We get
$$10(65-q)+25q=1280.$$
Expand. We get $650-10q+25q=1280$, and then $15q=630$, and then $q=630/15=42$.  It follows that $d=65-42=23$.
Or else multiply both sides of the equation $d+q=65$ by $10$. We get $10d+10q=650$. Then $(10d+25q)-(10d+10q)=1280-650$. Simplify. We have eliminated the variable $d$, and get $15q=630$, and therefore $q=630/15$.
Another way: Let us guess what the number of quarters is.  I will guess there are $0$ quarters.  Let's check whether my guess is right. If there are $0$ quarters, there are $65$ dimes.  So the total in the till is $650$, way too low.  Bad guess!
Obviously my guessed amount of quarters is too small.  So let's replace, one by one, dimes with quarters. Such a replacement does not change the total number of coins.  But with each replacement we gain $15$ cents.  We need to gain $1280-650$ cents, that is, $630$ cents.  The number of trades we need to make is therefore $630/15$, that is, $42$. So there are $42$ quarters, and therefore $23$ dimes.
Comment: Of course my guess that there are $0$ quarters was not a serious guess.  The point is that making a single "guess," and thinking about it, is enough to yield a complete fully mathematical solution.  We don't keep guessing: analysis of the error in the guess leads in one step to the answer.   
The above technique is very old. It was taught, under a name that translates to "too much and not enough," in China some $2000$ years ago. Later it was a standard method in India, the Islamic world, and Europe.  The method was taught routinely in the schools, in Europe and elsewhere, until some time in the nineteenth century.  
A: Iet's denote dimes as a and quarters as b
$a+b=65$
$0.1a+0.25b=12.80$
$b=65-a \Rightarrow 0.1a+0.25(65-a)=12.80 \Rightarrow a=23 \Rightarrow b=42 $
