 ###### Carl Horowitz

University of Michigan
Runs his own tutoring company

Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!

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# Applied Linear Equations: Collection Problem - Problem 1

Carl Horowitz ###### Carl Horowitz

University of Michigan
Runs his own tutoring company

Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!

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We're now going to look at a coin problem, dealing with amount of money and various coins. I have \$2.20 cents comprised of nickels, dimes and quarters. I have twice as many dimes as quarters, and the same number of nickels as dimes, how many of these do I have?

So, when dealing with any sort of word problems, I always end up making a diagram or chart or, some sort of way of piecing together all my information. So, we have in this case, three different denominations. We have nickels, dimes and quarters, and we know that in total, we have \$2.20. So, we now have the denominations worked out, we also need to work. We can figure out the number of each coin we have.

So we know that we have twice as many dimes as quarters and common mistake in this stuff is trying to figure out which you actually have more, which variable is x, which variable is twice x, 2x. So we know that we have twice as many dimes as quarters so that tells us we have more dimes, so that means we have to take our amount of quarters multiply it by 2 to find the number of dimes.

So quarters I know I have x of those, we have twice as many dimes so this is then going to be 2x. You can always check your work if I have three quarters, I would then have 6 times, twice as many dimes as quarters.

The problem also tells us that we have the same numbers of nickels as dimes. So if we have 2x dimes, we also then have 2x nickels. For now dealing with coin problems, you need to make sure you distinguish between the number of coins and the actual amount that these coins is worth.

So in this case say x is 1, that's telling me I have one quarter, but I also know that means I have 25 cents. If I have one quarter that means I would have two dimes that are worth 20 cents, so that distinguishing factor between the two coins versus 20 cents.

If you figure out how much you have, you just multiply the number you have by the denomination, so if I have x quarters, I just have .\$25x, 4 quarters would give me \$1, .25 times 4. The same idea for dimes except this time I know I have 2x dimes, so you take the amount of dimes times how much it's worth, add that to the number of quarters. The same idea for nickels, the number of nickels times the amount they're worth.

Now we have a equation to make sure we're adding these all up and that is going to be equal to our total amount, the amount of quarters plus the amount in dimes, plus the amount in nickels. Distribute this through 2 times .05 is just .1x, 2 times .1, .2x plus .25x is equal to 220. Combine like terms so .1 plus .2 is .3, plus .25 is .55x is equal to 220, divide by .55 end up with x is equal to 4. Again making sure we always answer what the question is asking for.

So here we found x is 4, x we said is the number of quarters, the question is actually asking for the number of each denomination, so we know that we have 4 quarters and we know we have twice as many dimes, so that means we have 8 dimes and the same number of dimes as nickels which would mean we have 8 nickels as well.

So taking our word problem, making this sort of diagram, these sort of thought, turning into an equation which we then solved.