Fish
Farm I
Posted March
5, 2001 as a Math Forum EPoW
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Introduction: In this problem,
students try to place male and female
fish in different ponds, in combinations
that satisfy certain restrictions
on the ratios of male fish to female
fish.
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Where’s the Math: Depending
on the student’s approach to
this problem, many types of math
can be incorporated to solve this
problem. The concept of ratio is
used throughout, and students can
solve the problem by experimenting
with the number of fish so that
the ratios fit the constraints.
Algebra can also be used, by coming
up with simultaneous equations to
represent the given conditions.
Standards: Number
& operations
Role of Components: AgentSheets
is used to display the simulation.
Pie Chart is used to display ratios
of fish in each pond with pie graphs.
A button
panel allows the simulation
to be run, stopped, and reset. ImageView
allows the fish icons to be displayed.
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Try the applet!
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Sample submitted solutions:
From: Mary, age 25+
School: North Carolina State University,
Raleigh, North Carolina
1. How many male fish and female fish
does each triplet get in their pond? Describe
the work you did to find the solution. (Sample
questions you can answer: Into which pond
did you put fish first? How many fish of
each kind went into that pond? Why? What
was your next step? How were you sure a
pond had the correct ratio?)
I began by putting 1 male and 1 female
in angel's pond. Then I went straight down
to Molly's pond and put in 3 males and 1
female. Then I went straight down again
and put 1 male and 2 females in Gar's pond.
Then I repeated the above process bringing
me to a grand total of 2 males and 2 females
in Angel's pond, 6 males and 2 females in
Molly's pond, and 2 males and 4 females
in Gar's pond. This left me with 8 fish
left to place in the ponds divided among
3 males and 5 females. I chose to place
2 more males and 4 more females in Gar's
pond and then 1 more male and 1 more female
in Angel's pond. Angel's ratio had to be
1:1 which means that for every 1 male present
there had to be 1 female present, or in
other words, the number of males had to
be equal to the number of females. My solution
gave her 3males:3females for a total of
6 fish. Molly's ratio had to be 3males:1female.
This ratio can be written as a fraction
of 3/1. My solution gave her 6males:2females
for a total of 8 fish. This ratio can be
written as a fraction of 6/2. If I were
to factor a 2 out of the top and bottom
of my 6/2, it would become 2/2 * 3/1. The
2/2 = 1 and so my fraction of 6/2 can be
simplified to 3/1 because 2/2*3/1=1*3/1
and 1*3/1= 3/1. These fractions, 6/2 and
3/1 are equivalent fractions. This shows
that the two ratios of 3males:1female is
equivalent to my solution of 6males:2females.
Gar's ratio had to be 1male:2females which
can be written as a fraction as 1/2. My
solution gave him 4males:8females for a
total of 12 fish. This ratio can be written
as a fraction of 4/8. If I were to factor
a 4 out of the top and bottom of my 4/8,
it would become 4/4 * 1/2. The 4/4 = 1 and
so my fraction of 4/8 can be simplified
to 1/2 because 4/4*1/2=1*1/2 and 1*1/2=
1/2. This shows that the two ratios of 1male:2females
is equivalent to my solution of 4males:8females.
You can add the totals of fish in each pond
to get 6+8+12=26 total fish
2. Given the 13 males and 13 females,
what are ALL the possible amounts of male
and female fish that would satisfy the ratio
of 1 male to 2 female fish in Gar's pond?
Explain why these different amounts are
equivalent to the ratio 1:2.
The possible amounts of fish in Gar's pond
could have been 1male:2females, 2males:4females,
3males:6females, 4males:8females, 5males:10females,
6males:12females. This is without regard
to the ratios set for the other two triplets!
Each of these ratios when written as a fraction
of males over females is equivalent to 1/2.
A ratio can be written as a fraction and
simplified. To simplify each fraction of
1/2, 2/4, 3/6, 4/8, 5/10, 6/12, you must
factor out their greatest common factor.
For example, 5/10=5/5*1/2. Then you can
simplify and replace the 5/5 with its equivalent
value of 1. That gives us 5/10=1*1/2. Since
1 is the multiplicative identity, then 5/10=1/2.
The same process can be followed for each
of the proposed fractions or ratios in this
problem. They are also equivalent if you
write them as females over males to 2/1.
Bonus: Explain why all possible answers
in #2 result in the same pie graph for Gar's
pond.
The pie chart remains the same each time
because it is comparing the number of males
and females to the whole pie or total number
of fishes in Gar's pond. If Gar has a ratio
of 1male:2females then the male fish will
from a pie slice of 1/3 which associates
1 male fish to 3 total fish (1male and 2females)
and the female fish will form a pie slice
of 2/3 which associates 2 female fish to
3 total fish (1male and 2females). If we
raise our ratio to 2 males:4 females then
the resulting part to whole comparison remains
the same. The number of males will be 2/6
or 1/3 and the number of females will be
4/6 or 2/3. This will continue for each
of the possible ratios listed in #2. The
reason why the pie slices never change size
is because of the equivalent fractions that
I discussed above. Lets return to the example
above of 5males:10females. This translates
to a fraction in our pie graph of 5/15 which
represents 5 male fish with the 15 total
fish (5male and 10female). The fraction
of 5/15 can be simplified to 1/3 by using
the same process that I outlined above by
factoring out the greatest common factor.
Since 5/15=5/5*1/3 then 5/15=1*1/3 and then
1/15=1/3. This shows that a pie graph ratio
of 5males:10females yields 1/3 of the graph
because 5/15=1/3. The same process can be
used when discussing the pie graph interpretation
of female fish. The ratio is 5males:10females
which yields 10/15 and compares the 10 female
fish to the 15 total fish (5males and 10females).
This fraction will also yield 2/3 for the
female fish because 10/15=5/5*2/3 which
gives 10/15=1*2/3 which yields 10/15=2/3.
No matter what fraction is chosen from #2,
the resulting areas on the pie chart will
always be 1/3 for the males and 2/3 for
the females because of the equivalent fractions.
From: Katie,
age 13
School: Taipei American School, Taipei,
Taiwan
1. How many male fish and female fish does
each triplet get in their pond? Describe
the work you did to find the solution. (Sample
questions you can answer: Into which pond
did you put fish first? How many fish of
each kind went into that pond? Why? What
was your next step? How were you sure a
pond had the correct ratio?)
Angel had 8 male fish and 8 female fish
in her pond.
Molly had 3 male fish and 1 female fish
in her pond.
Gar had 2 male fish and 4 female fish in
her pond.
I first put one male and one female in Angel's
pond, 3 male and 1 female in Molly's pond,
and 2 males and 4 females in Gar's pond.
I thought I could put the rest into Angel's
pond, but I noticed that there was unequal
amounts of males and females. So to make
it equal, I put one more male fish and 2
more female fish in in Gar's pond, that
would still be the same as 1:2. That left
me with the same amount of male fish and
female fish, so they could all go into Angel's
pond (that would still equal 1:1). Since
I did everything slowly, I made sure that
my amounts of fish were equal to the ratios.
All I did was get the total amounts and
then reduce them, and the reduced number
should've equaled the ratio. I knew I got
everything right when the bricks turned
green.
2. Given the 13 males and 13 females,
what are ALL the possible amounts of male
and female fish that would satisfy the ratio
of 1 male to 2 female fish in Gar's pond?
Explain why these different amounts are
equivalent to the ratio 1:2.
All of the possible amounts of male and
female fish that would satisfy the ratio
of 1 male to 2 female fish are:
1:2
2:4
3:6
4:8
5:10
6:12
These work because when they are all reduced
to lowest terms, they equal 1:2. To reduce
something to its lowest terms is to divide
both sides of the ratio by the same number.
Then to make sure it's in lowest terms,
keep on dividing the sides by a number until
the different sides of the ratio can't be
divided anymore. Another way to check is
divide the first half of the fraction by
the second half you'd get .5, and if you
did this for every single ratio, then you
compared them, you will see that they are
all the same. If they're not the same, then
you know that they are not equivalent to
the ratio. This tells me that the ratios
are equivalent because when the above ratios
are all reduced, they equal the same as
1:2.
Bonus: Explain why all possible answers
in #2 result in the same pie graph for Gar's
pond.
All the ratios I got in question #2 result
in the same pie graph for Gar's pond because
they are the same thing. If you looked at
(for example) 1:2 by ratios, you would notice
that that the first number is one third
of the total (the 3, the total, is gotten
from the first and second number added together).
Also, you would notice that every single
other number in problem #2 also have the
first number 1/3 of the total of the ratio.
If you look at the pie graph for Gar's pond,
then you would also see that 1/3 of it is
one color, and 2/3 is the other color, which
is the same as the ratios above in problem
#2.
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Reflections:
Students used several different approaches
in thinking about how to distribute the
fish into the three ponds.
1. Many students starting by giving each
pond the minimum number of fish necessary
to meet the ratio (i.e., 1 male and 1 female
in Angel's pond, 3 males and 1 female in
Molly's pond, and 1 male and 2 females in
Gar's pond). Then they did another "round"
of giving fish to each pond in the same
ratios. After this there were 5 females
and 3 males left, and students then had
to think about how to distribute the remaining
fish. Many decided to concentrate on how
to put fish into Gar's or Molly's pond so
that an equal number of male and female
fish would be left to put into Angel's pond.
2. Some students realized up front that
they needed to put fish in Molly and Gar's
ponds so that the sum of the number of males
in Molly's and Gar's ponds would be equal
to the sum of the number of females in Molly's
and Gar's ponds. Then there would be an
equal number of female and male fish left
in the tank, and those could all be put
in Angel's pond to satisfy the 1:1 ratio.
3. A few students approached this problem
using algebra, and set up equations that
would model relations like those described
in (2) above.
Many students found correct solutions to
both question 1 and question 2, but gave
limited justifications for why the ratios
are equivalent. Some students made statements
about ratios "reducing" to the
same thing. In order to earn credit for
a solution and justification, a student
needed to give a reason why ratios are equivalent.
For example, 4:8 is the same as 1:2 because
in each case, the number of males is exactly
half the number of females, or in each ratio
there are twice as many females as males.
Several students expressed the ratio as
a fraction (1:2 as 1/2). Although a ratio
can be expressed in the form of a fraction,
this representation led some students to
reason that the pie graph for a 1:2 relation
would be half yellow and half brown: when
they tried to use a pie graph to represent
1:2, they tended to give it two equal parts
instead of the correct three equal parts.
Perhaps when people wrote the ratio as 1/2,
they got confused about the number of equal
parts because they were thinking of "one
half" instead of counting the total
number of parts, which in this case was
3 (1 and 2, rather than 1 divided by 2).
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