Thread: One unknown variable - how to explain to a third grader

Question:
One half of the birds at a pet store are yellow. Tara buys one of the yellow birds. Then one third of the birds at the store are yellow. How many yellow birds were at the pet store before Tara bought one? Explain how you know.


Answer: (Me trying, hope it is right)
Let t be the total number of birds.
Number of yellow birds, y = t/2
Now, Tara bought one.
Number of yellow birds now = y-1=(t/2)-1
which is also equals to t/3.
So, (t/2)-1 = t/3 -------equation 1


To solve this equation, bring all t's to one side of = sign.
(t/2)-(t/3)=1
(3t-2t)/6=1
3t-2t=1*6=6
t=6


If we had total 6 birds before, and half of it was yellow birds, we had 3 yellow birds before Tara bought one.


Now, my question is, how would I explain steps after equation 1 to a third grader who knows fractions but not learned subtracting fractions yet. Thanks in advance.

Third graders are not typically learning algebra. First I will show the algebraic solution, then I will explain the guessing and checking method (which is probably what you will wind up with).

Total number of birds = t
Number of yellow birds = y
t = 2y (there are twice as many birds as there are yellow birds)
t-1 = 3(y-1) (removing one yellow bird means the total number of birds also decreased by one, and there are three times as many birds remaining as the new number of yellow birds)

2y = 3(y-1)+1
2y = 3y-2
y = 2
t = 2y = 2(2) = 4.

To explain it, you could create a chart.

$$\begin{matrix} & | & \text{# birds} & \text{# yellow birds} & \text{relationship} \\ \hline \text{Starting off} & | & t & y & t = 2y \\ \text{Take one away} & | & t-1 & y-1 & (t-1) = 3(y-1)\end{matrix}$$

To compare it to your solution, your equation #1 is incorrect.

(t/2)-1 = (t-1)/3 because it is one third of the new number of birds, which is t-1.

But, since third graders are not typically learning algebra, it may be easier to use the answer to figure it out. Starting off, there are 4 birds, 2 yellow, so 2/4 = 1/2 are yellow. You take away 1 yellow bird, so there are 3 birds, 1 is yellow, so 1/3 are yellow. Don't worry about figuring out the general case (namely that this is the only possible solution). The fractions are probably more important right now.

A guessing and checking method may be more appropriate. Start with, suppose there was only 1 yellow bird. Then, if that is half the total number of birds, then there must be 2 birds to start. One yellow, one not yellow. If Tara buys the yellow bird, now there are no yellow birds left, so it cannot be that 1/3 of the remaining birds are yellow.

So, we guess the starting number of yellow birds is 2. That means there are 4 birds total (because half the original number of birds are yellow). Then, Tara buys one, leaving one yellow bird and 3 birds total (one yellow, two not yellow), so that leaves 1/3 of the remaining birds are yellow. This is correct.

Is this the only solution? Let's guess the starting number of yellow birds is 3 and see what happens:
Now, there are 6 birds total, after Tara buys one, there are 2 yellow and 3 non-yellow, so 2/5 of the birds are yellow (not 1/3)
Suppose there are 4 to start, 8 birds total, after Tara buys one, there are 3 yellow and 4 non-yellow, so 3/7 of the birds are yellow (not 1/3)
.
.
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Suppose there are n to start, 2n birds total. After Tara buys one, there are n-1 yellow and n non-yellow, so (n-1)/(2n-1) of the birds are yellow. Notice the pattern, but do not give it an explicit formula that will likely confuse the 3rd grader.

If the 3rd grader is already converting fractions to decimals, it may be beneficial to convert to decimals to show that as the initial number of yellow birds increases, the resulting fraction of the remaining birds being yellow also increases (getting closer and closer to the original fraction of 1/2).

You can elude to the fact that in the future, there is math that can figure it out without guessing and checking, but simply giving that math to the 3rd grader without giving a reason to learn it may make it too abstract. Using a problem where the solution is a fraction can help, as well, as how would they know which fraction to guess and check? This would give even more incentive for them to learn the techniques that will yield the answer without any guessing and checking, and serves as a good lead-in to algebraic expressions.

Thank you so much. I see my incorrect equation. Well explained.

Originally Posted by wanttolearn

Question:
One half of the birds at a pet store are yellow. Tara buys one of the yellow birds. Then one third of the birds at the store are yellow. How many yellow birds were at the pet store before Tara bought one? Explain how you know.


Answer: (Me trying, hope it is right)
Let t be the total number of birds.
Number of yellow birds, y = t/2
Now, Tara bought one.
Number of yellow birds now = y-1=(t/2)-1
which is also equals to t/3.
So, (t/2)-1 = t/3 -------equation 1


To solve this equation, bring all t's to one side of = sign.
(t/2)-(t/3)=1
(3t-2t)/6=1
3t-2t=1*6=6
t=6


If we had total 6 birds before, and half of it was yellow birds, we had 3 yellow birds before Tara bought one.


Now, my question is, how would I explain steps after equation 1 to a third grader who knows fractions but not learned subtracting fractions yet. Thanks in advance.

Do you have to use a variable??
I would think like this. There are 2 birds and 1 is yellow OR there are 4 birds and 2 are yellow OR there are 6 birds and 3 are yellow. Do this more time if needed (see in next step if you need more).

Now, remove the yellow bird from the above number (if you do not get 1/3 of the birds are yellow, then add more numbers to the last step)
Using the numbers above we get, 1 bird, 0 yellow OR 3 bird, 1 yellow ==> stop here because the yellow birds make up 1/3 of all the birds.