Sorting Pennies, A Wagering Activity for Algebra 1
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David Drymiller Marie Sklodowska Curie Metro H.S.
4959 S. Archer Ave.
Chicago IL 60632
To abstract the idea of a line.
To introduce other methods of solving simultaneous equation besides
To develop problem solving skills.
A scale able to read (to the nearest tenth of a gram) up to 400 grams
at least 1,000 pennies
a plastic cup to hold pennies on the scale
numbered plastic ziplock bags 1 through 8
16 index cards, 2 per ziplock bag
The following instructions are directed to the instructor:
Place the 1,000 pennies in a large pile on the front desk along with
the scale, the plastic cup and the ziplock bags. Divide the class into 8
groups. From each group have one member come up and take a handful of pennies.
Weigh and record the amount 'W'. Place the weighed pennies in the
corresponding ziplock bag and give the group member the instructions to
separate the pennies into two groups: 1981 and before 'B', 1983 and after 'A'.
They are to record the number of each kind of penny on one of the index cards
provided and keep it from your view. On the other index card, they are to write
the total number of pennies in the bag 'T'. Place this second index card and
all pennies in the plastic bag and return to front desk. As the bags are
returned record the number of pennies and calculate the predicted number of
pennies for 1982 and before for each group. When all the pennies have been
returned and the calculations finished, announce that through the miracle of
algebra you know how many of each type they have sorted with a margin of error
of plus or minus two pennies. (I find a small wager, if possible, with each
group enlivens the activity). Collect your winnings and explain to the class
why you won.
Pennies for 1981 and before weigh 3.1 grams
Pennies for 1983 and after weigh 2.5 grams
(Note: avoid 1982 pennies because the mint produced some of each.)
Number Weight Total Weight
1981 and before B 3.1 3.1B
1983 and after A 2.5 2.5A
before and after T W
Which gives us the equations: 3.1B + 2.5A = W
B + A = T
The problem becomes just finding the intersection of two lines.
Solving by Substitution: A = T - B
3.1B + 2.5( T - B ) = W
3.1B + 2.5T - 2.5B = W
0.6B + 2.5T = W
- 2.5T = - 2.5T
0.6B = W - 2.5T
B = (W - 2.5T) / 0.6
I use this activity after I have taught the graphing of two lines and
finding their intersection but before other methods of finding the solution
of systems of equations. This is an introductory lesson that abstracts the
intersection of two lines to find a solution.