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"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 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 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 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.

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

 

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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.


  • A collection of 33 coins, consisting of nickels, dimes, and quarters, has a value of $3.30. If there are three times as many nickels as quarters, and one-half as many dimes as nickels, how many coins of each kind are there?

    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
      number of nickels:
      3q
      number of dimes:
      (½)(3q) = (3/2)q

    There is a total of 33 coins, so:

      q + 3q + (3/2)q = 33
      4q + (3/2)q = 33

      8q + 3q = 66

      11q = 66

      q = 6

    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 2000-2011 All Rights Reserved

  • A wallet contains the same number of pennies, nickels, and dimes. The coins total $1.44. How many of each type of coin does the wallet contain?

    Since there is the same number of each type of coin, I can use one variable to stand for each:

      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

    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:

  • convert the relationships between the numbers of coins (if given, as in the second examle) into equations, and
  • convert the statements about the values of the coins (if given, as in the first example) into equations that state the values all in the same unit (for instance, in cents).

And, as always, label everything!

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Cite this article as:

Stapel, Elizabeth. "'Coin' Word Problems." Purplemath. Available from
    http://www.purplemath.com/modules/coinprob.htm. Accessed
 

 

 

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