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Strategy 7: Change the Representation
Johnny's Orchards [Problem #3316]
In the early 1800's, Johnny Appleseed traveled throughout the Midwestern states planting apple trees and seeds.
On one of his trips he planted orchards for three neighboring farmers, Mr. Keillor, Mr. Bunsen, and Mr. Krebsbach. To be fair, he planted the
same number of seeds for each farmer.
He did Mr. Keillor's orchard first, creating a rectangle by planting the same number of seeds in each row.
Mr. Bunsen's orchard was also a rectangle, but it had 1 more row and 9 fewer seeds per row than Mr. Keillor's.
Finally, for Mr. Krebsbach's rectangular orchard Johnny increased the number of rows by 5 and decreased the number of seeds in a row by 15 compared
with Mr. Bunsen's.
How many total seeds did Johnny plant on the three farms?
Comments and Sample Solutions
This problem gives you a chance to explore situations where there are lots of unknown quantities and lots of relationships among them. In this problem, you
did have enough information to find specific values for every unknown quantity to make all of the relationships true, but it wasn't easy to see that at first.
There were two successful types of strategies for solving this problem. The first was to build up a mathematical model out of all of the relationships, until
you finally had a model that related a single variable to a number (or two linear equations in two variables). The other approach was to algebraically define
down as many relationships as possible, and then use a table or systematic guess and check to find a value that made every given relationship true.
No one who used algebra combined with guess and check had a fully explained solution, but Harry from Rosemont School of the Holy Child came close
enough that I think you can follow his methods. The steps he didn't explain were that when he "put all the equations together" he set (x + 1)(y - 9) = (x +
6)(y - 24) and simplified them until he had an expression for y in terms of x. Then, he generated a table of (x, y) values where y = 3x + 27. That gave him a
very narrow set of possibilities for the number of rows and number of seeds per row in Mr. Keillor's orchard. He checked those possibilities until he found a
set of dimensions that made each farmer have an equal number of seeds. I thought it was slick how he used algebra to narrow down the possible guesses
quite a bit. Guessing and checking for this problem, as many submitters noted, could have taken forever, since there were so many unknown quantities! In a
way, Harry's solution is like what you had to do to solve Math Club Mystery -- use algebra to narrow the possibilities as far as you can, and then use other
kinds of reasoning to take it from there.
The fact that Harry was able to use a table to find a single value for the number of seeds, and the number of rows and seeds per row in Mr. Keillor's
orchard, suggests that there is a way to use algebra to tackle this problem. The three other highlighted solutions show three different algebraic approaches.
There were so many ways you could think about combining and simplifying the relationships!
Student 1320g from Waterford Elementary School used the fact that the number of seeds in Mr. Keillor's orchard = the number of seeds in Mr. Bunsen's
orchard to find a way to express the number of seeds per row in Mr. Keillor's orchard in terms of the number of rows in Mr. Keillor's orchard. Then, Student
1320g used the fact that the number of seeds in orchard Mr. Bunsen's orchard = the number of seeds in Mr. Krebsbach's orchard to express the
relationship between seeds per row and rows another way. Having two linear relationships between the two variables meant the problem could be solved, in
this case by substitution.
Brian from Rancho San Joaquin Middle School used a method that was similar to Student 1320g's. What I liked about Brian's solution was that he started
by noticing every quantitiy in the problem he could think of, and then listing as many relationships as he could. And then, at the end, he used those
relationships again to check his work. Checking to see if your answer makes all the given relationships true is a really important step of problem-solving that
too many people ignore.
Hayne, also from Rancho San Joaquin Middle School, used the relationships a little differently, in a solution I thought was particularly unique. Rather than
use different variables for rows, seeds per row, and total number of seeds, Hayne used two variables, seeds (s) and rows (r). Then seeds per row, rather
than being a unique variable, is s/r. Hayne defined two variables, and used two relationships to solve the problem. The first was the relationship between Mr.
Keillor's orchard and Mr. Bunsen's orchard, and the second was between Mr. Keillor's orchard and Mr. Krebsbach's orchard.
I think there were two places people got stuck solving this problem. The first was trying to work with three equations at once, rather than comparing one
pair at a time. Several students wrote equations like xy = (y+1)(x-9) = (y+5)(x-15). It's ok to simplify each expression that makes up that equation, to write xy
= xy+x-9y-9 = xy+5x-15y-75. But if you start adding or subtracting from "both sides" you really need to do the same thing to all three expressions in the
equation. So even if xy = xy+x-9y-9 = xy+5x-15y-75 is true, xy = x-9y-9 = 5x-15y-75 won't be. Now that we know x = 36 and y = 3, you can check. (36)(3)
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The Math Forum @ Drexel University
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= (36)(3) + 36-9(3)-9 = (36)(3) + 5(36)-15(3)-75. That's true, all three expressions equal 108. But if you subtract xy from two of the expressions, you're left
with xy = x-9y-9 = 5x-15y-75 or 108 = 36 - 9(3) - 9 = 5(36) - 15(3) - 75, which is no longer true. As soon as you start working with equations that aren't
actually equal, you won't get the correct answer for x or y.
The other place people got stuck was not being able to find ways to write variables in terms of other variables in order to have an equation in one variable,
or two equations in two variables. When I started to solve this problem, I had the same issue. I had so many algebraic equations and was doing so much
simplifying that I kept making algebraic mistakes and I wasn't getting any closer to having a single equation or two to solve. I thought you might be
interested in seeing what I did then. I chose to use the Change the Representation strategy. Rather than use equations, what if I drew pictures of the
orchards?
That is Mr. Keillor's orchard. It has y rows and x seeds per row. The total area represents the total number of seeds.
That is Mr. Bunsen's orchard. It has y + 1 rows and x - 9 seeds per row. The total pink area represents the total number of seeds. What I noticed is that the
number of seeds in the white striped part has to be the same as the number of seeds in the pink striped part. In other words, the number of seeds in the
extra row has to be equal to the number of seeds lost by shortening the rows by nine seeds each. That let me write an equation: area of white striped part
= area of pink striped part. 9*y = 1*(x - 9).
Last is Mr. Krebsbach's orchard. Again, the white striped part has the same area as the purple striped part. So I could write another equation: 24*y = 6*(x 24).
I used those two equations with two variables, 9*y = 1*(x - 9) and 24*y = 6*(x - 24), to solve the problem. I used the substition method, but I could have
used elimination.
The two main ideas I want you to take away from this commentary are:
1. If you can't solve a problem completely using algebra, try using algebra to narrow down the possiblities as much as you can and then use logical
reasoning or a table to find all the possible solutions.
2. If you can't see how to write algebraic expressions, try changing the representations. How can you show the relationships visually?
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The Math Forum @ Drexel University
11-04-28 10:13 PM
Armed with those two ideas, you can solve a lot of the problems we pose in the Algebra Problem of the Week and, even better, problems you face in your
daily life.
-Max, for the Algebra Problem of the Week
From: Harry B , age 14 , Rosemont School of the Holy Child
Johnny planted a total of 324 seeds
First I made an equations for all the people yards and the amount of seed in each. I knew they all equaled each other.
K=xy
he was first so he got the basic
B=(x+1)(y-9)
you added one because he had one more row, and subtracter
nine because there was nine less per row
Kr=(x+6)(y-24)
you added 5 per row and subtracted 15 to Mr Bunsens yard
I next put all the equations together and was able to solve it.
y=3x+27
Next I made a table for the answer and plugged the numbers into the original equations. The only numbers that worked for all three was 3,36
3*36=108 per garden * 3 gardens=
324
From: student 1 , age 14 , Waterford Elementary School
Johnny planted a total of 324 seeds in all.
WHAT'S THE PROBLEM: The problem is that I need to find out how many seeds Johnny Appleseed planted if each orchard has the same number of seeds,
each garden is rectangular, Mr. Bunsen's (B) orchard has one more row and nine fewer seeds then Mr. Keillor's (A), and Mr. Kresbach (C) has five more
rows and 15 fewer seeds then Mr. Bunsen's rectangular orchard.
MAKE A PLAN: First I will find a formula for each person's number of seeds in each row and how many many rows, then set the numbers equal to each
other and solve the problem.
CARRY OUT THE PLAN: A's formula would be X·Y if X is the number of seeds in each row, and Y equals the number of rows.
So XY= the total number of seeds.
If X·Y= the number of seeds in A's Orchard, then:
(Y+1)·(X-9)=XY for B's orchard. Then:
(Y+8)·(X-24)=XY for C's orchard.
To set the numbers equal to each other, I would have to simplify the formula's first.
For B: (Y·X)+Y·(-9)+(1X)+(1·[-9])=XY (distributive property)
YX
XY
are cancelled out, so:
-9Y+X-9=0, since XY cancelled out.
+9Y +9 +9Y
+9
-9Y and -9 cancels out, leaving:
X=9Y+9
For C: using distributive property, (YX)+(-24Y)+(6X)+(-144)=0
+144 +144
-144 cancels out, leaving:
-24Y+6X or -24Y+6(9Y+9)=144
6·9Y=54Y
6·9=54 so,
-24Y+54Y+54=144
-54
-54
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The Math Forum @ Drexel University
11-04-28 10:13 PM
54 cancels out leaving,
-24Y+54Y=90
or
30Y=90
making Y=3
so if Y is three, i can find X
if X=9Y+9 and Y=3
X=9(3)+9= 36.
There are three rows in A's orchard, and 36 seeds in each row, so 3·36=108 seeds in each orchard since all three orchards have the same number of seeds
planted, and if I need to find out how many seeds Johnny planted in all, 108·3=324 seeds.
LOOK BACK: Johnny planted 324 seeds.
I would tell another student to use a formula for this problem because it is much easier to figure out the problem.
I was stuck when i thought X=9Y-9, so the answer did not come out even, but i checked my math again and realized it would be X=9y+9.
This problem was difficult because there were so many formulas and equations I had to use to find the answer.
From: Brian B , age 14 , Rancho San Joaquin Middle School
The total number of seeds used for all 3 farmers combined is 324 seeds. Mr. Keillor's farm had 3 rows, 36 seeds per row, Mr. Bunsen's farm had 4 rows,
27 seeds per row, and Mr. Krebsbach's farm had 9 rows, 12 seeds per row.
A = number of rows in Mr. Keillor's farm
x = number of seeds per row in Mr. Keillor's farm
B = number of rows in Mr. Bunsen's farm = A + 1
y = number of seeds per row in Mr. Bunsen's farm = x - 9
C = number or rows in Mr. Krebsbach's farm = B +... 5 = (A + 1) + 5 = A + 6
z = number of seeds per row in Mr. Krebsbach's farm = y - 15 = (x - 9) - 15 = x - 24
T = total number of seeds that Johnny planted on the three farms
Equations:
Eqn1: Ax = By = Cz; he planted the same number of seeds for each farmer.
Eqn2: T = Ax + By + Cz
Ax = By
Ax = (A + 1)(x - 9)
Ax = Ax - 9A + x - 9
x = 9A + 9; Eqn3
Ax = Cz
Ax = (A + 6)(x - 24)
Ax = Ax - 24A + 6x - 144
6x = 24A + 144
x = 4A + 24; Eqn4
Equate Eqn3 with Eqn4, then solve for A:
9A + 9 = 4A + 24
9A - 4A = 24 - 9
5A = 15
A=3
Substitute to Eqn4, then solve for x:
x = 4A + 24
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The Math Forum @ Drexel University
11-04-28 10:13 PM
x = 4(3) + 24
x = 12 + 24
x = 36
Note: substituting the value of A to Eqn3 should yield the same x = 36.
Now,
T = Ax + By + Cz
Since Ax = By = Cz, T = 3Ax = 3(3)(36) = 324
Answer:
Johnny planted 324 seeds on the three farms
Check:
Ax = (3)(36) = 108
By = (A + 1)(x - 9) = (4)(27) = 108
Cz = (A + 6)(x - 24) = (9)(12) = 108
Ax + By + Cz = 108 + 108 + 108 = 324 Yes!
From: Hayne B , age 14 , Rancho San Joaquin Middle School
Johnny planted total of 324 seeds by planting 108 seeds for each orchard.
To start with, I made a variable to solve the problem.
r = number of rows in Keilor's
s = number of seeds in each orchard
Then, seeds per row in Keilor's orchard becomes s/r.
Let's solve Bunsen's orchard by these information.
He had one more row and 9 fewer seeds per row than Keillor's.
Then this becomes s = (r+1)(s/r-9)
s= s -9r +s/r -9
-s/r=-9r-9
s/r=9r+9
s=9r2 +9r
Now we can make a equation for Krebsbach's orchard.
The number of rows increased by 5 and number of seeds in a row decreased by than Mr. Bunsens.
s = (r+1+5)(s/r-9-15)
s = (r+6)(s/r-24)
s = s-24r+6s/r-144
24r = 6s/r-144
4r = s/r -24
4r 2 = s-24r
s =4r 2 + 24r
Since I have equations for each orchards, I'll combine those equations.
<Equation 1 and 2>
s = 9r 2 +9r = 4r 2 + 24r
9r 2 +9r = 4r 2 + 24r
5r 2 =15r
r 2 =3r
r=3
I'll put r=3 into s = 9r 2 +9r.
s = 9(32 )+9(3)
s = 81 + 27
s = 108
Johnny planted 108 seeds per orchard.
108 * 3= 324
The number of total seeds planted is 324.
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