In part one I asked you to use your higher order thinking skills to solve this problem:

You have two identical glasses, both filled to exactly the same level. One contains red dye, the other water. You take exactly one spoonful of red dye and put it in the water glass. Then you take one spoonful of the mixture from the water glass and return it to the red dye glass.

Question: Is there more red dye in the water glass than water in the red dye glass? Or is there more water in the red dye glass than red dye in the water glass? In other words, the percentage of foreign matter in each glass has changed. Has the percentage changed more in one of the glasses, or is the percentage change the same for both glasses?

I even gave you a hint:

Instead of water and red dye, think of red balls and white balls.

Assume that each glass starts out with 100 balls of a single color. Now remove a number of red balls from the red-ball glass and put them in the white-ball glass. Then return the same number of balls from the glass with the “mixture” and put them in the red-ball glass. Do this with different numbers of red and white balls.

Now it's time to end your suffering and give you the answer.

To solve the problem (without resorting to math) you need to understand the concept of conservation of number--a Piagetian concept. According to Piaget, conservation of number is

the understanding that the number of objects remains the same when they are rearranged spatially.

Also according to Piaget, you should have developed this concept naturally by the age of seven:

Piaget proposed that number conservation develops when the child reaches the stage of Concrete Operations at around 7 years of age. Around this time, children also develop an understanding of other forms of conservation (e.g., weight, mass). However, number conservation is often the first form of conservation to develop. Before the stage of Concrete Operations, children may believe that the number of objects can increase or decrease when they are moved around.

In Piaget's classic example he arranged objects in two rows, then spread out the objects, and asked the child if the number had changed. Apparently, children younger than about seven don't realize that the number of objects stays the same, i.e., their number is conserved.

No doubt, as an adult you understand the concept of conservation of number and could answer Piaget's problem with the rows of objects readily. But why couldn't you answer my red dye and water example? It's based on the same concept. Liquids are made up of a fixed number of molecules whose number is conserved. So why the difficulty?

What I was trying to do was demonstrate the difference between flexible and inflexible knowledge. You understand that number is conserved and can solve problems involving concrete objects with ease. However, you may have struggled with the unfamiliar red dye and water problem I presented, not realizing that liquids are composed of molecules. Even those of you that solved the problem by resorting to solving a math ratio problem may not have realized that the problem is readily solvable by knowing the number is conserved without resorting to math.

Hopefully, this demonstrates that most people do not have a flexible understanding of conservation of number that is readily generalizable to unfamiliar examples. Your knowledge of conservation of number is likely not flexible. It did not develop naturally like Piaget said it would and it probably was not taught well to you either. The concept remains inflexible.

Answering the problem doesn't require higher order thinking skills or critical thinking skills or any other fancy jargon educators like to use. All it requires is a flexible understanding and the application of a basic concept that any seven year old readily understands--conservation of number. If the basic concept/skill is well taught to the student and the student is given sufficient practice, the student will eventually develop a flexible understanding of the concept and will be able to apply the concept to solve tricky problems. You don't need a super high IQ to understand such things if you were taught them beforehand.

However, if this concept is not well taught, the student must rely on other knowledge he may have learned to solve the problem indirectly. Setting up a math ratio problem to solve the problem may be one way to solve the problem without having a flexible concept of conservation of number. I bet many of you math heads solved the problem this way, demonstrating your flexible understanding of the concept of ratio--another basic concept.

But what about higher order thinking skills. Do these even exist? Or are they just the byproduct of not receiving good instruction of basic skills?

NB: I got this problem from pp. 7-10 of chapter 4 of Engelmann's book.

## 8 comments:

Ok, maybe I just don't have flexible higher order thinking skills =), but I didn't find the liquid example intuitive or a good example of number conservation. If a spoonful was taken from each pure liquid and then added to the opposite glass, it would have been obvious each had the same amount and percentage of foreign matter, and a clear example that changing the "piles" didn't change the total amount.

For me the issue was the fact that the pure dye was added to the water and then the mix was added back to the dye. I was surprised when the ratios came out even. This had nothing to do with understanding that the total amount of dye and water was unchanged no matter how it was distributed, or understanding that each glass would end up with the same amount of liquid it started with (just a different mixture), but knowing that even with the manner of redistribution, there would be the same amount of foreign matter in each. That is more than simple number conservation, and not intuitive unless you've worked the math yourself or have otherwise learned it.

Hi 42.

That is more than simple number conservation, and not intuitive unless you've worked the math yourself or have otherwise learned it.I think you can solve the problem using simple number conservation. But, I agree its not intuitive and must be learned through some means, such as being taught.

Think of it this way. No matter how you rearrange the water and dye in the end you must have some amount of dye in one glass and that amount minus x in the other glass. If both glasses end up with the same amount of liquid at the end then the amount of water in the (total minus x) glass must always equal x and thus the ratios will always be the same.

Here's one example that's closer to Piaget's. Make two rows of differently colored clips. Now rearrange the clips in whatever manner you wish. Keep on rearranging the clips for a full minute. Now mak both rows equal to ten clips again using whatever methos suits you. Is the percentage of foreign clips in each row the same?

How about a problem that can't be worked out using math.

I start out with a container with of 1 liter of oil and another container with 1 liter of vinegar. beside me is a box full of different quantity measuring spoons. I randomly select ten spoons and without looking proceed to use each spoon in turn and take a quantity of liquid from container one and put it into container two. Then I take the next spoon and take some liquid from teo and put it into container one. I repeat this procedure until I run out of spoons. At the end I pour an amount of liquid from the more full container into the less full container so that I have equal amounts of liquid in each container. Is the percentage change of foreign liquid in each container the same?

42, I had the same problem.

I knew that it ended up with the same amount of liquid is each glass, but it seemed wrong that what I put back in the dye glass wasn't the opposite of what I took out. (but it did reflect the opposite of what I left in the other glass, I guess.)

If you move the dye to the water glass, then take an amount equal to that, and add it to the dye glass, the quantities will be the same, but the mix will not.

The dye glass will have:

dye amount - 1 teaspoon dye + one teaspoon of dye and water mixture

The water glass will have:

water amount + 1 teaspoon dye - 1 teaspoon of dye and water mixture

There will be fewer water molecules in the dye glass than dye molecules in the water glass.

Linda, you might want to work that out again. It is impossible to have a different mix in the glasses. Instead of liquid pretend that each glass has ten marbles in it. Now take three red marbles out of the one glass and put them in the other glass. Now take out any three marbles from the other glass and put them back in the first. No matter what marbles you pick the mix will always wind up the same.

The way I understand the problem is this. Because the two glasses end with the same volume as they started with, the net effect of the movements is that some dye molecules have been transferred from dye glass to water glass and have been replaced by an equal number of water molecules that have been moved in the opposite direction. The concentration of dye in water is therefore equal to that of water in dye. This works even if you know nothing about molecules and see a quantity of liquid as a single entity.

I was very surprised no-one has challenged this. The teaspoon of red dye added to the water is 100% red dye. The teaspoon of mixture then taken to add to the red dye is not 100% water. Therefore there is less water in the red dye than there is red dye in the water. So 100a, -10a, +10b is not equal to 100b, +10a, -7b+3a.

Your last step does not conserve matter.

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