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?
Coins making up a certain sum
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0Please use more descriptive titles. General subjects like "algebra" are covered by the tags; the title should be more specific to the question where possible. – 2011-09-22
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1Let the number of dimes be $x$ and the number of quarters $y$. There are two constraints: the total number of coins, and the total value. Write these down, and solve the system. If you get stuck, ask for another pointer – 2011-09-22
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2Say the number of dimes is $d$ and the number of quarters is $q$. (1) What does the fact there are 65 coins total tell you about $d$ and $q$ together? (2) If each dime contributes 0.10 to the total, and each quarter contributes 0.25 to the total, how can we represent the total value in terms of the number of dimes and quarters? (3) When we set this total value equal to 12.80, what equations are we left with and how should we solve them? – 2011-09-22
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0is there a way to do it with only 1 variable? we haven't learned to use 2 yet. – 2011-09-22
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1Ok, if you insist on using just 1 variable, here's a way. Suppose the number of dimes is $d$. Then what is the number of quarters in terms of $d$? Now, as @anon says, each dime contributes $0.10$ to the total amount, and each dollar contributes $0.25$. Can you represent the total amount in terms of the number of dimes and quarters? But you are given that the amount is $12.80$. So, ... – 2011-09-22
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0could I use d +.15 to represent quarters? i'm sorry i really have no clue what i'm doing... – 2011-09-22
2 Answers
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.
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0+1, especially for the too-much-and-not-enough method. It's really more convenient to do mentally. Wrt to the last paragraph, do you have any references about it or its history, so I can read more? – 2011-10-30
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0@ShreevatsaR: I had a good collection once, for a History of Mathematics course. Can't find it. There is analysis in most books about the history of Chinese mathematics. More recently it was called the Rule of False Position, or a Latin equivalent. There is lots of literature. – 2011-10-30
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0Ah ok, thanks... I'll look it up. I remember seeing "false position" in a few places, but I hadn't followed up to see what it means. :-) – 2011-10-31
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 $