Solving Word Problems With Equations Of One Degree And One Unknown Worksheet With Answers Page 5
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20MONEY (COIN) PROBLEMS
EXAMPLE
1. The value of a bag of coins consisting of nickels, dimes and quarters is
$1.90. If there were half as many quarters as nickels, and three more nickels
than dimes, how many coins of each kind are in the bag?
Read the problem carefully.
!
Recognize that you are dealing with a coin problem. Note: Not every problem
!
involving dollars and cents is a money (coin) problem. This type always deals
with the number of coins, bills, or tickets. And the value of each type.
The chart for coin problems looks like this
!
# Of Coins
Value Of One Coin
Amount Of Money
Dimes
0.10
Nickels
0.05
Quarters
0.25
Go back and reread the problem. You are asked to find how many of
!
each kind of coin are there. Since you know the least about the number
of dimes, you let the number of dimes be X. The number of nickels is
three more than number of dimes, so it is represented by “X+3.” The
number of quarters is half as many as the number of nickels, so in terms
of X is ½(X+3)-- the grouping symbol is essential.
!
Fill in the chart.
# Of Coins
Value Of One Coin
Amount Of Money
X
Dimes
0.10
0.10 X
Nickels
X + 3
0.05
0.05(X+3)
Quarters
1/2(X + 3)
0.25
1/2(0.25)(X+3)
The “Amount Of Money Column” column is filled in using your common sense.
!
If you have five dimes, each worth ten cents, you have 5*10 or 50. What we’re
saying is that the “Amount Of Money” equals the “# of coins” times “the value of
one coin”. You multiply left-to-right in the rows of the chart.
After completion of the chart we return to the problem and look for clues to set
!
the equation
Since the sum of money is $1.90. Your equation is:
!
0.10X + 0.05(X + 3) + 1/2(0.25)(X+3) = 1.90
Solving the equation:
!
0.10X + 0.05X +0.15 + 0.125X + 0.375 = 1.90
0.275X + 0.525 = 1.90
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