A coin-based word problem presents you with numbers and types of coins, and usually the total value of the coins. You are expected to create a model (that is, an equation) that relates this information. Solve the equation, and then back-solve (as necessary) to find the numbers of the other types of coins in play.
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The following is a typical example of a coin problem:
Your uncle walks in, jingling the coins in his pocket. He grins at you and tells you that you can have all the coins if you can figure out how many of each kind of coin he is carrying. You're not too interested until he tells you that he's been collecting those gold-tone one-dollar coins. The twenty-six coins in his pocket are all dollars and quarters, and they add up to seventeen dollars in value.
How many of each coin does he have?
To solve your uncle's puzzle, you need to use the total number of coins, the values of the two types of coins, and the total value of those coins.
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There are twenty-six coins in total. Some of them are quarter coins; let "q" stand for the number of quarters. The rest of the coins are dollar coins. Since there are 26 coins in total and q of them are quarter coins, then there are 26 − q coins left to be dollar coins. (This "how much is left over" construction is important. Make sure you understand what it is and how to use it.)
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If your uncle has only one quarter, then 25×1 = 25 cents comes from quarters. If he has two quarters, then 25×2 = 50 cents comes from quarters. Since he has q quarters, then 25×q = 25q cents comes from quarters.
For the dollar coins, we need first to convert their value to cents; one dollar is one hundred cents. Since he has 26 − q dollars, then he has 100(26 − q) cents from the dollar coins.
He has seventeen dollars in total, or 1700 cents, part of which is from quarters and part of which is from dollars. To help keep things straight, we can set up a table:
number of coins |
cents per coin |
total cents |
|
---|---|---|---|
quarters | q | 25 | 25q |
dollars | 26 − q | 100 | 100(26 − q) |
total | 26 | 1700 |
The total value comes from adding the value of the quarters and the value of the dollar coins. So we add the "total cents" expressions from the right-hand column above, and set this sum equal to the given total:
25q + 100(26 − q) = 1700
Then solve:
25q + 100(26 − q) = 1700
25q + 2600 − 100q = 1700
−75q + 2600 = 1700
−75q = −900
q = 12
In other words, 12 of the coins are quarters. Since the remainder of the twenty-six coins are dollar coins, then there are 26 − 12 = 14 dollar coins. I can check to make sure this works:
14×$1 + 12×$0.25 = $14 + $3 = $17
Since the answer works in the original exercise, it must be right.
(Remember: You can *always* check your answers to "solving" problems by plugging your answers back into the original contexts. If everything checks out, then you know that your answer is right.)
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I'll start by picking and defining a variable, and then I'll use translation to convert this exercise into mathematical expressions.
Nickels are defined in terms of quarters, and dimes are defined in terms of nickels, so I'll pick a variable to stand for the number of quarters, and then work from there:
number of quarters: q
There are three times as many nickels as quarters, so:
number of nickels: 3q
And there are one-half as many dimes as nickels, so:
There is a total of 33 coins, so I can add my coin expressions, set the sum equal to that number, and solve:
8q + 3q = 66
11q = 66
q = 6
Checking my variable definition, I verify that I chose q to stand for the number of quarters. Then there are six quarters, and I can work backwards to find the counts of the other coins. So my answer is:
6 quarters
9 dimes
18 nickels
"But," I hear you say, "we never used the fact that the coins add up to $3.30. Shouldn't we have?" Well, we can use that information to check our answer. But it turns out that this information was not actually necessary for finding the solution.
Since there is the same number of each type of coin, I can use one variable to stand for each. They mention pennies first, so I'll use the variable p:
number of pennies: p
number of nickels: p
number of dimes: p
The value of the coins is the number of cents for each coin times the number of that type of coin, so:
value of pennies: 1p
value of nickels: 5p
value of dimes: 10p
The total value is $1.44, so I'll add the above, set equal to 144 cents, and solve:
1p + 5p + 10p = 144
16p = 144
p = 9
I picked the variable p to stand for the number of pennies, but the original exercise specifies that the wallet contains the same number of each type of coin. Then my answer is:
9 coins of each type
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Remember that you can always check your answers to "solving" problems by plugging them back in to the original question. In this case, I would check whether nine pennies plus nine nickels plus nine dimes totalled $1.44:
$0.09 + 0.45 + 0.90
= $0.54 + 0.90 = $1.44
So the solution is correct.
When you have the time, this type of checking is a good idea on tests. That way, you can make sure your answer is correct before you hand in your work.
The take-aways for solving coin word problems are these:
And, as always, label everything!
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