
"Coin" Word Problems 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 goldtone onedollar coins. The twentysix 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 his puzzle, you need to use the total number of coins, the values of the two types of coines, and the total value of those coins. There are twentysix 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. 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:
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 righthand column above, and set this sum equal to the given total: 25q + 100(26 – q) = 1700 Then solve: 25q + 100(26 – q) = 1700 In other words, 12 of the coins are quarters. Since the remainder of the twentysix 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.
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 is a total of 33 coins, so: q +
3q + (3/2)q = 33 Then there are six quarters, and I can work backwards to figure out that there are 9 dimes and 18 nickels. "But," 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 this information was not actually necessary to the solution. Copyright © Elizabeth Stapel 20002011 All Rights Reserved
Since there is the same number of each type of coin, I can use one variable to stand for each: number of pennies: 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 The total value is $1.44, so I'll add the above, set equal to 144 cents, and solve: 1p + 5p + 10p = 144 There are nine of each type of coin in the wallet. 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. Make sure your answer is correct before you turn it in. The tricks to these exercises are two:
And, as always, label everything!


MATHHELP LESSONS
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