Coin problems are also common algebra applications. Usually the total number of coins is given as well as the total dollar value. The question is normally ‘‘How many of each coin is there?’’
Let x represent the number of one specific coin and put the number of other coins in terms of x. The steps involved are:
1. Let
x represent the number of a specific coin;
2. Write
the number of other coins in terms of x (this skill was developed in the
‘‘age’’ problems);
3. Multiply
the value of the coin by its number; this gives the total amount of money
represented by each coin;
4. Add
all of the terms obtained in Step 3 and set equal to the total money value;
5. Solve
for x;
6. Answer
the question. Don’t forget this step! It is easy to feel like you are done when
you have solved for x, but sometimes the answer to the question requires one
more step.
Examples
As in all
word problems, units of measure must be consistent. In the following problems,
this means that all money will need to be in terms of dollars or in terms of
cents. In the examples that follow, dollars will be used.
Terri has
$13.45 in dimes and quarters. If there are 70 coins in all, how many of each
coin does she have?
Let x represent the number of dimes. Because
the number of dimes and quarters is 70, 70 – x represents the number of
quarters. Terri has x dimes, so she has $0.10x in dimes. She has 70 – x quarters,
so she has $0.25(70 – x) in quarters. These two amounts must sum to $13.45.
Terri has 27
dimes and 70 – x = 70 – 27 = 43 quarters.
Bobbie has
$1.54 in quarters, dimes, nickels, and pennies. He has twice as many dimes as
quarters and three times as many nickels as dimes. The number of pennies is the
same as the number of dimes. How many of each coin does he have?
Nickels are being compared to dimes, and
dimes are being compared to quarters, so we will let x represent the number of
quarters. Bobbie has twice as many dimes as quarters, so 2x is the number of
dimes he has. He has three times as many nickels as dimes, namely triple 2x: 3(2x)
= 6x. He has the same number of pennies as dimes, so he has 2x pennies.
How much of the total $1.54 does Bobbie have
in each coin? He has x quarters, each worth $0.25, so he has a total of 0.25x (dollars)
in quarters. He has 2x dimes, each worth $0.10; this gives him 0.10(2x) = 0.20x
(dollars) in dimes. Bobbie has 6x nickels, each worth $0.05. The total amount
of money in nickels, then, is 0.05(6x) = 0.30x (dollars).
Finally, he
has 2x pennies, each worth $0.01. The pennies count as 0:01(2x) = 0.02x (dollars).
The total amount of money is $1.54, so,
Bobbie has 2
quarters; 2x = 2(2) = 4 dimes; 6x = 6(2) = 12 nickels; and 2x = 2(2) = 4
pennies.
Practice
1. A
vending machine has $19.75 in dimes and quarters. There are 100 coins in all.
How many dimes and quarters are in the machine?
2. Ann
has $2.25 in coins. She has the same number of quarters as dimes. She has half
as many nickels as quarters. How many of each coin does she have?
3. Sue
has twice as many quarters as nickels and half as many dimes as nickels. If she
has a total of $4.80, how many of each coin does she have?
Solutions
1. Let
x represent the number of dimes. Then 100 – x is the number of quarters. There
is 0.10x dollars in dimes and 0.25(100 – x) dollars in quarters.
There
are 35 dimes and 100 – x = 100 – 35 = 65 quarters.
2. Let
x represent the number of quarters. There are as many dimes as quarters, so x also
represents the number of dimes. There are half as many nickels as dimes, so ½ x
(or 0.50x, as a decimal number) is the number of nickels.
There
are 6 quarters, 6 dimes, 0.50x = 0.50(6) = 3 nickels.
3. As
both the number of quarters and dimes are being compared to the number of
nickels, let x represent the number of nickels. Then 2x represents the number
of quarters and ½ x (or 0.50x) is the number of dimes.
There
are 8 nickels, 0.50x = 0.50(8) = 4 dimes, and 2x = 2(8) = 16 quarters.
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