When you have a problem that requires you to add coins together. Is it better to use fractions?

For example, you have 14 one dollar bills, 24 quarters, 12 dimes, 78 nickles, and 20 pennies.

So I know that I can just take: $$\begin{align*} 14 \times 1 &= \$14\\ 24 \times .25 &= \$6\\ 12 \times .10 &= \$1.20\\ 78 \times .05 &= \$3.90\\ 20 \times .01 &= \$0.20 \end{align*}$$

Easy stuff, but I'm trying to figure out the best way to add these up. Assuming I have only 15-30 seconds to get this addition done without calculator. I figured the best way is to use fractions.

ps. Can someone tell me how to write real fractions in here instead of using slash? I just want to write them to look better in the future, I tried to look for a way in FAQ but didn't see anything on formatting.


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    $\begingroup$ To write a fraction, you should use LaTeX, enclosed in appropriate dollar signs. To get $\frac{1}{2}$, for example, type $\frac{1}{2}$. $\endgroup$ – Arturo Magidin May 23 '11 at 3:43
  • $\begingroup$ Try using LaTeX: \frac{numerator}{denominator} $\endgroup$ – F M May 23 '11 at 3:43
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    $\begingroup$ Is the usual algorithm for addition with carry of numbers in decimal form not very efficient? That seems quick to me. $\endgroup$ – Jonas Meyer May 23 '11 at 3:47
  • $\begingroup$ For multiplying by .05, it may help to think of .05 as $\frac{1}{2}\cdot\frac{1}{10}$, so you can just divide by 2 and then shift the decimal point (and of course $.25=\frac{1}{4}$, or even $\frac{1}{2}\cdot\frac{1}{2}$, so you can divide by 4 or divide by 2 twice to multiply by .25). $\endgroup$ – Jonas Meyer May 23 '11 at 3:57
  • $\begingroup$ The MathJax tutorial/reference is here. $\endgroup$ – OJFord Jun 4 '14 at 1:01

I would try to "simplify" the problem a little. For instance, you know 4 quarters is a dollar, so 24 / 4 is 6 dollars. 10 dimes to a dollar, so you have 1 dollar and twenty cents. 20 nickels to a dollar, so you have 3 dollars (60 nickels), and 18 nickels left over.

So now we're at 14 + 6 + 1 + 3 = 24 dollars (and don't forget the remaining 2 dimes, 18 nickels, and 20 pennies).

Simplifying the problem like this might make it easier - at least, that's how I would do it in my head.


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