Cubes and box question

inko

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This question is from my son's school text book.

"How many 2cm³ cubes could fit into a box with length 8cm, width 4cm and height 10cm?"

Could anyone figure out which answer is correct?

A. 160
B. 126
C. 40

First my son's answer was "A". Later we figured out "B" was actually correct. But his school teacher keeps saying "C" is correct.
 
C is correct. To see this consider the bottom of the box first. The dimensions of the bottom are 8 cm by 4 cm. If we were to make a single layer of 2cm³ cubes across the bottom of the box how many would it take? Eight, right (2 cubes wide and 4 cubes long)? Now, since our box is 10 cm tall, how many of these layers can we stack on top of one another before leaving the confines of our box? 10/2 = 5. Thus we have 5 layers each containing 8 cubes each. 5*8= 40. Do you understand?
 
C is correct. To see this consider the bottom of the box first. The dimensions of the bottom are 8 cm by 4 cm. If we were to make a single layer of 2cm³ cubes across the bottom of the box how many would it take? Eight, right (2 cubes wide and 4 cubes long)? Now, since our box is 10 cm tall, how many of these layers can we stack on top of one another before leaving the confines of our box? 10/2 = 5. Thus we have 5 layers each containing 8 cubes each. 5*8= 40. Do you understand?

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Thanks for answering:grin: Your explanation is the same as the teacher.

My son thought 2cm³ cube's side length is not 2cm but ³√2 (cubic root of 2) cm.
³√2 cm x ³√2 cm x ³√2 cm makes 2cm³. Then his answer became different.

I wanted to know how people comprehend this question. Maybe my son's was wrong.
 
Last edited:
It is ambiguous. The notation is poor. (2cm)^3 =/= 2cm^3. In cases like this you have to go with the conventions of the author.
 
It is ambiguous. The notation is poor. (2cm)^3 =/= 2cm^3. In cases like this you have to go with the conventions of the author.

Thanks again:p Yes, the question is not very clear. If the cube's side length is described as 2cm, I would think the same way as yours. Then in this case cube's volume would be 8cm³?

This is a primary school's textbook. Suppose I should check with the author or the editor.


 
Hello, inko!

This question is from my son's school textbook.

"How many 2 cm³ cubes could fit into a box
. . with length 8 cm, width 4 cm and height 10 cm?"

Could anyone figure out which answer is correct?

. . (A) 160 . . (B) 126 . . (C) 40

First, my son's answer was "A".
Later we figured out "B" was actually correct.
But his school teacher keeps saying "C" is correct.

If that is the exact wording of the problem,
. . then your son is correct.

The box has a volume of: \(\displaystyle 8\times4\times10 \:=\:320 \text{ cm}^3.\)

If the problem has said "2-cm cubes",
. . that means each cube is 2 cm on an edge.
Each cube has a volume of \(\displaystyle 2^3 = 8\text{ cm}^3.\)

Then there can be: \(\displaystyle 320 \div 8 \:=\:40\text{ cubes in the box.}\)


But it said "2 cm3 cubes".
. . This means that each cube has a volume of 2 cm3.

Then there can be: \(\displaystyle 320 \div 2 \:=\:160\text{ cubes in the box.}\)

The answer is (A).


~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


Ask the teacher to consider the difference of these two shapes:

(1) A 4-inch square.
. . .This square is 4 inches on a side.
. . .Its area is: \(\displaystyle 4 \times 4 \,=\,16\text{ in}^2.\)

(2) A 4 in2 square.
. . .This square is 2 inches on a side.
. . .Its area is (of course) \(\displaystyle 4\text{ in}^2.\)
 
Clearer:
How many 2by2by2 boxes could fit into a 8by4by10 box?

...but that's not confusing enough :roll:

Thank you very much:D

That's how the teacher wanted to say. But it means the side length of the cube is 2cm. The the cube's volume will be (2x2x2)cm³. It is 8cm³, not 2cm³. This is the confusing point.

In this way, the answer will be (8x4x10)/(2x2x2)=40. The box's volume is divided by a cube's volume (assume cube's side length is 2cm).

The teacher wanted students to do with another way. Divide each side length of the box by cube's side length. Then multiply those three numbers.

8/2 x 4/2 x 10/2 = 4 x 2 x 5 = 40

For this question, "2" is a convenient number because 2 is a factor of 8, 4, and 10. But in the question, the cube is described with its volume "2cm³". Then some people say this "2" is "2cm". That's how I was confused.

If I confuse you, too, very sorry;)
 

But it said "2 cm3 cubes".
. . This means that each cube has a volume of 2 cm3.

Then there can be: \(\displaystyle 320 \div 2 \:=\:160\text{ cubes in the box.}\)

The answer is (A).



If each cube were to have a volume of 2 cubic centimeters,
then the edge lengths would equal \(\displaystyle \sqrt[3]{2} \ cm.\)

If these cubes are to to be placed so that their respective faces are parallel
or perpendicular, as the case may be, to the respective orientations of the
dimensons of the box, without tilting any cubes in any directions,
then choice B) can be justified. This leaves spaces (some volume) in the box
not filled by the cubes.



How many whole number of edges of the cubes fit into 4 cm?

\(\displaystyle 4/\sqrt[3]{2} \ \) must be rounded down to 3.



How many whole number of edges of the cubes fit into 8 cm?

\(\displaystyle 8/\sqrt[3]{2} \ \) must be rounded down to 6.



How many whole number of edges of the cubes fit into 10 cm?

\(\displaystyle 10/\sqrt[3]{2} \ \) must be rounded down to 7.


Then (3)(6)(7) = 126
 
Hello, inko!


If that is the exact wording of the problem,
. . then your son is correct.

The box has a volume of: \(\displaystyle 8\times4\times10 \:=\:320 \text{ cm}^3.\)

If the problem has said "2-cm cubes",
. . that means each cube is 2 cm on an edge.
Each cube has a volume of \(\displaystyle 2^3 = 8\text{ cm}^3.\)

Then there can be: \(\displaystyle 320 \div 8 \:=\:40\text{ cubes in the box.}\)


But it said "2 cm3 cubes".
. . This means that each cube has a volume of 2 cm3.

Then there can be: \(\displaystyle 320 \div 2 \:=\:160\text{ cubes in the box.}\)

The answer is (A).


~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


Ask the teacher to consider the difference of these two shapes:

(1) A 4-inch square.
. . .This square is 4 inches on a side.
. . .Its area is: \(\displaystyle 4 \times 4 \,=\,16\text{ in}^2.\)

(2) A 4 in2 square.
. . .This square is 2 inches on a side.
. . .Its area is (of course) \(\displaystyle 4\text{ in}^2.\)


Hello Soroban (abacus-san?)


Thank you very much:p

Yes, my son’s first instinct was to divide box’s volume by a cube’s volume.

Then we were wondering if the cubes can fit in the box perfectly or not. In other words, all side length of the box can be multiples of the cube’s side length or not.

The teacher suggested dividing each side length of the box by cube's side length. Then multiply those three numbers. This is the case if the side length of the cube is 2cm.

8/2 x 4/2 x 10/2 = 4 x 2 x 5 = 40



But my son and I did not think the side length of the cube is 2cm. 2cm³ cube’s side length should be ³√2 cm. Or it is described as 2^(1/3).

Then we tried to divide box’s side length with this number.

8/2^(1/3) = 6.3496042078728
4/2^(1/3)= 3.1748021039364
10/2^(1/3)= 7.937005259841

We discovered that the cubes can not fit into the box perfectly. Maximum number of the cubes to fit into the box is 6x3x7=126.

That’s how the Answer B came out.

I tried to explain all this to the teacher. But his idea never changed. I was thinking to explain with the way you suggested, using 2D shapes instead of 3D shapes (area instead of volume). But it was too late. I stopped arguing with the teacher. That’s why I came here:(
 
If each cube were to have a volume of 2 cubic centimeters,
then the edge lengths would equal \(\displaystyle \sqrt[3]{2} \ cm.\)

If these cubes are to to be placed so that their respective faces are parallel
or perpendicular, as the case may be, to the respective orientations of the
dimensons of the box, without tilting any cubes in any directions,
then choice B) can be justified. This leaves spaces (some volume) in the box
not filled by the cubes.



How many whole number of edges of the cubes fit into 4 cm?

\(\displaystyle 4/\sqrt[3]{2} \ \) must be rounded down to 3.



How many whole number of edges of the cubes fit into 8 cm?

\(\displaystyle 8/\sqrt[3]{2} \ \) must be rounded down to 6.



How many whole number of edges of the cubes fit into 10 cm?

\(\displaystyle 10/\sqrt[3]{2} \ \) must be rounded down to 7.


Then (3)(6)(7) = 126


That's how we thought, too. I will show this to my son.
This is a very good site to see different people's view points:eek:
 
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