I need help with a math problem for my son...

[cynatnite]

Here it is...

****

Find the dimensions of a box that will hold half as many cubes as a box that is 2 by 8 by 10.

Volume of original box: 160

Volume of new box: 80

Dimensions of new box: ________ (this is where I have trouble)

Explain how you found the dimensions of the new box.

****

This is the whole problem. I don't know how to help him and I can't find anything on the tubes comparable to this.

[RandomThoughts]

1. that is not much of a question.

Any answer where X times Y times Z = 80 would solve that equation.

Although why would you ask?


Or are you saying that conclusions can make for many views of the same problem.

Specifically that finding a correct answer, does not mean you find the only correct answer.





Side note, I am still due money for beer and travel.

[Ptah]

7. 3.1416 X 4.000 X 6.3662 = 80

[gratuitous]

12. Side note: No you're not

I told you and told you to turn in your receipts, but noooooo!

Keep looking for tickee.

[LiberalAndProud]

2. Divide any one of the dimension by 2.

1 x 8 x 10 = 80
2 x 4 x 10 = 80
2 x 8 x 5 = 80

[Recovered Repug]

17. That, I believe, is the correct answer.

There are a large number of potential dimensions that would work. I think the "how did you find" part was the important part.

[Chan790]

3. 2x8x5.

Alternately 1x8x10 or 2x4x10. To reduce the volume of a rectangular solid by 50%, reduce any 1 dimension of the rectangular solid by 50%.

I'm shocked I still remembered that, it's been close to 20 years since I took 6th grade math.

[cynatnite]

5. I forgot...

Thanks so much for the help:)

[LeftyFingerPop]

4. If you cut the middle dimension in half...

2x4x10 = 80

Cut the first dimension in half:

1x8x10 = 80

Cut the last dimension in half: 2x8x5 = 80

Cut any of the dimensions in half to get half of the volume.

[rurallib]

11. 4X4X5 just to be difficult

that's darn near a cube.

[Ptah]

6. What size cubes?

[RandomThoughts]

8. great question.

The length of the side of each cube would have to be able to be a product of the all three dimensions.

Theoretically by increasing size of cube you could solve the problem without changing the dimensions of the containing object.



So the answer would be 2 x 8 x 10 container, with cubes double size by volume, but still with length of sides that divide integrally into 2, 8, 10.

[Ptah]

15. Huh?

The problem we are given is:
Find the dimensions of a box that will hold half as many cubes as a box that is 2 by 8 by 10.

Your answer is:
So the answer would be 2 x 8 x 10 container, with cubes double size by volume, but still with length of sides that divide integrally into 2, 8, 10.

[Mr. Ected]

9. I'm having problems visualizing the actual cubes

Cube

http://www.clown-cars.com/wp-content/uploads/2010/02/0609_z+2008_nissan_cube+front.jpg

Half a Cube



Do the math. :D

[ElboRuum]

22. I'm waiting for the Honda Tesseract.

It should be ugly in four dimensions instead of the usual three.

[pgodbold]

10. My head exploded.

[zipplewrath]

13. classic case of focusing on the numbers

I'd bet you if I just gave you a box and said cut it in half, you'd have little problem "solving" that problem. We get caught up in the numbers and don't think about the physical problem. Once you think about how you'd actually cut a box in half, the numerical answer becomes obvious. Pick the long side and divide it by 2.

[Duer 157099]

16. Best answer

:thumbsup:

[zipplewrath]

18. Thanks, I see this alot

I get asked by parents occasionally to help their kids who are "struggling with math". I typically say "bull". I've never met a kid that was "struggling with math". All the ones I've ever worked with understood math just fine, they just occasionally get caught up in the notation. It makes no sense to them. I frequently get to a point where I put away the pencil, I put away the paper, and I grab some objects, maybe some poker chips, or some marbles, anything really. We start manipulating objects. I'll put out 9 chips and say "divide it into two equal piles". It takes them about 2 seconds to split it in two and then complain that they have "one left over". So I tell them to split into 3 piles. That works out fine. Mind you, most don't "calculate" a single thing. They do it just by "looking". People instinctively understand "three equal piles". They don't really even have to know how to "count" per se since they can just look at 3 equal piles and see that there are 3 in each one.

"Congratulations", I'll say, "you've just mastered long division". In that simple exercise we've divided on number by another and also introduced the concept of the "remainder". They rarely get it immediately. But now we are approaching it from a stand point of "you know how to divide, let's just learn to do it on paper, instead of with poker chips".

We all know how to "do math", it's the particular notation and the language of solving math problems that confuses alot of people. With my wife, I joke regularly that I just have to put it in terms of shopping and money. She'll ask me what 1/3 of 4 cups is. I'll just say it's a "70% sale". She "gets it" right away. The study of math isn't so much a study of what we don't know, as it is a perfection of what we already do know.

[libnnc]

25. I wish I'd had you as a math teacher in 5th and 6th grades.

"The study of math isn't so much a study of what we don't know, as it is a perfection of what we already do know."

What you just wrote made much more sense to me than anything I've heard in the past 25 years.

[pokerfan]

14. 1 x 1 x 80

(It's for mailing snakes.)

[Deep13]

19. 2x8x5 nt

[Orsino]

26. Yes. Just cut any one of the three dimensions in half, and the volume is halved.

So 1x8x10 would work, too, or 2x4x10.

[qb]

20. Just to be sure... the most reliable answer is 2 x 4 x 10.

Since the size of the cubes is not given, assume the largest possible that would fit in the original box: 2 x 2 x 2. In that case each side must be an even number to exactly fit half of the original number of cubes.

I always wonder if problems are written this ambiguously on purpose to make students think, or if the instructor is just sloppy.

[jmowreader]

24. It doesn't matter what size the cubes are

They're probably an inch on each side, but they could be anything. They could even be seven inches cubed, but if we call seven inches a "unit," then the original box is 2 units x 8 units x 10 units.

[petronius]

27. If you make the assumption that the cubes are 1 unit in size, then the

'halve any edge' approach works. But, since we're not told anything about the size of the cubes, it's safer to not assume anything about the cubes sizes relative to the box, we just know that some number of cubes fit in the box. Thus, if the cubes are 2x2x2, the solution with a 5-unit edge wouldn't work. The volume of the box would be halved, but half the number of cubes wouldn't fit - you'd have to slice up some of the cubes to fit them in.

If the cubes can be any size that fits in the original box, then the edges of the new box have to be multiples of the smallest original side - 2x4x10 and 2x2x20 are the only solutions...

[Chan790]

28. Considering that it's a math problem being given to an elementary school student.

We do in fact know the size and dimensions of the cubes, more precisely we know them to be irrelevant.

The dimensions of the box are given in "cubes". I know someone who writes this stuff...that's considered an inviolable rule of writing textbooks for that level so that we don't have these sorts of arguments in the classroom. "If no explicit measurement is given, assume all measurements to be expressed in standard and equal units.

If you needed to assign dimensions to the cube for you own sanity, assume them to be 1x1x1.

Sorry to pick on you for this but it's the same argument that's been going on for days and it simply does not matter.

[petronius]

29. The best possible answer is the one that requires the fewest assumptions,

and that's the one I gave. The problem as written appears to incorporate a critical thinking component, and it's never too early to encourage students to apply logic. Relying on assumptions and unwritten rules results in poorly prepared students, so I say you're mistaken - it matters very much (in both the specific and the general sense)...

[jmowreader]

30. The only thing we need to assume is the cubes are equal in size

Assume the cubes are all the same size, and because it says "cube" we're going to assume all the sizes are the same length. We can then say whatever the length is, that's one unit.

Once we get past that issue, we know that if box B has half the volume of box A, half as many cubes will fit in it--NO MATTER how big the cubes are!

So...what I would enter, were this my paper:

"Since we don't know how big the cubes are, whatever length they really are will be termed a 'unit.' The current box has a volume of 160 cubic units--it is 2 units high, 8 wide and 10 long. If you split the box exactly in half along any axis, you will reduce its capacity from 160 cubic units to 80 cubic units. Therefore...the box can be either 1 x 8 x 10, 2 x 4 x 10 or 2 x 8 x 5. All three do the exact same thing."

[petronius]

31. But you are also making the unsuppported and unnecessary assumption that

the cubes are all 1x1x1.

Consider it this way: The original box is 2 units x 8 units x 10 units, giving it a volume of 160 units³. If the cubes are 1x1x1 the box will hold 160 of them. However, what if the cubes are 2x2x2, or 0.4 x 0.4 x 0.4, or any other edge length that divides evenly into 2? Some number of cubes, call it X, will fill the box. But, we don't know how many cubes the original box holds, and we have no reason to assume the cubes have 1 unit edges.

Put another way, we can't define a unit as the length of a cube, because we are already told that the length of one side of the box is 2 units, and we don't know how many cubes span that length (but we do know that the length of a cube, however many units it is, divides evenly into 2, 8, and 10).

Since we know that the edge length of the cubes must divide evenly into 2 (the shortest edge of the original box), we can see that all the edges of the final box must be evenly divisible by 2. Solutions with box edge lengths of 1 or 5 won't work. With that criterion, there are only two possible final boxes that work.

Your assumption that the cubes are 1x1x1 may be correct, but we have no basis for making it. Therefore, solutions that don't require that assumption are superior (we only have to assume that the cubes are the same size, and that the original and final boxes are filled by them without gaps or overflow)...

[jmowreader]

32. We're going round in loops here

Think of this as an algebra equation--the box is 2u x 8u x 10u, and it doesn't matter if u = 1 inch or u = 37.25 feet. Bust that 160u^3 box in half along any axis and it's going to contain 80u^3. The math works the same either way.

[petronius]

33. The box is 2u x 8u x 10u, but we don't know that the cubes are 1u x 1u x1u

The question simply says "cubes" and a cube is just a 3D object - we have no reason to believe that the cubes contained in the box are cubes with 1 unit³ volume.

We can assume that, but why make the assumption when there are possible answers that don't require the assumption?

[Sentath]

21. zipplewrath gave the right way to do it, but I'm ..

Well, I'm just going to be a complete bastard here.

Given 2x8x10 the largest cube it will hold is 2x2x2 and the smallest is subatomic, so we won't bother with them. Starting with the 2 on a side cubes the original box will hold 20 cubes (in an array 1x4x5). Reducing the cubes to 1.99 (are they working with more than 2 decimal places in his class yet?) the given box will still only hold 20 of them, with a little air space. The new box could be as large as 3.97x5.96x11.93 (to maintain the size ratios of the old array cut in half along the even numbered axis) and still able to hold only 10 cubes (in an array 1x2x5)

edited to insert the word way in the title

[jmowreader]

23. Pretty easy: take any one dimension and divide by two

If a box 2 x 8 x 10 will hold 160 cubes (note we don't know what the unit of measure is, and here it doesn't really matter: it could be 2 x 8 x 10 miles and still work) then a box 1 x 8 x 10 or 2 x 4 x 10 or 2 x 8 x 5 will hold 80 cubes because it's half the size of the first box.