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"Coin" Word Problems

There are twenty-six coins in total. Some of them are dollar coins; let "d" stand for the number of dollar coins. The rest of the coins are quarters; let "q" stand for the number of quarters. Then d + q = 26.

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 part to work, you'll have to convert to cents. That is, one dollar is one hundred cents. Since he has d dollars, then he has 100d 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, I'll set up a table:

	number of coins	cents per coin	total cents
dollars	d	100	100d
quarters	q	25	25q
total	26		1700

If you know how to solve systems of linear equations, you can see that you have the following system:

d + q = 26
100d + 25q = 1700

If you don't know about systems yet, don't worry: we're nearly done. Since we have two equations related to the same situation, I'll solve one of them and plug the result into the other one. For instance, starting with the two equations::

d + q = 26
100d + 25q = 1700

...I'll solve the first equation above to get:

q = 26 – d

(I could have solved the first equation for d instead of for q; the end result would have been the same, though the computations would have looked different in the middle. Choosing to solve for q in this case was purely a matter of taste.)

Then I'll plug this into the second equation in place of q to get:

100d + 25(26 – d) = 1700

Then I'll solve:

100d + 25(26 – d) = 1700
100d + 650 – 25d = 1700
75d + 650 = 1700
75d = 1050
d = 14

In other words, fourteen of the coins are dollar coins. Since the remainder of the twenty-six coins are quarters, there are 12 quarters.

Let's do another one.

• 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?

Let "n" stand for the number of nickels, "d" for the number of dimes, and "q" for the number of quarters. Then n + d + q = 33. Also, since the number of nickels is three times the number of quarters, then n = 3q; since the number of dimes is one-half the number of nickels, then d = 0.5n = 0.5(3q) = 1.5q. With these relationships, we can now re-write that first equation above:

n + d + q = 33
3q + 1.5q + q = 33
5.5q = 33
q = 6

Then there are six quarters, and you 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, you can use that information to check your answer. You could also have used it to solve the problem in a different way. (Yes, there is more than one way!)

Instead of dealing with the problem only in terms of how the numbers of coins related to each other, we could have dealt with how much the coins totalled to. The relationships would still have been the same (q, n = 3q, and  d = 1.5q), but we would have taken into account how much each coin is worth (in cents);

10d + 5n + 25q = 330
10(1.5q) + 5(3q) + 25q = 330
15q + 15q + 25q = 330
55q = 330
q = 6   Copyright © Elizabeth Stapel 2006-2008 All Rights Reserved

So you get the same result again.

• 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?

I'll set up some variables:

number of pennies:  p
number of nickels:  n
number of dimes:  d

But there are an equal number of each of these, so I really have:

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 problems 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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