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# A closed aluminum rectangular box has inner dimensions x centimeters

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A closed aluminum rectangular box has inner dimensions x centimeters [#permalink]

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21 Oct 2012, 11:07
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Question Stats:

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A closed aluminum rectangular box has inner dimensions x centimeters by y centimeters by z centimeters. Each of the six sides of the box is 1 centimeter thick. Calculate the volume of the aluminium, in cubic centimeters?

A. $$xyz + 8$$

B. $$2(xy + xz + yz + 4)$$

C. $$2(xy + xz + yz) – xyz$$

D. $$2(xy + xz + yz + x + y + z + 4)$$

E. $$2(xy + xz + yz + 2x + 2y + 2z + 4)$$

[Reveal] Spoiler:
Are we really concerned with the extra 1cm thickness in calculating the volume? The thickness does not add to spaces to
be filled; so, how come it is affecting the volume?
[Reveal] Spoiler: OA

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Last edited by Bunuel on 06 Oct 2017, 01:16, edited 3 times in total.
Renamed the topic.

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Re: A closed aluminum rectangular box has inner dimensions x centimeters [#permalink]

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21 Oct 2012, 11:27
gmatbull wrote:
A closed aluminum rectangular box has inner dimensions x centimeters by y centimeters by z centimeters.
Each of the six sides of the box is 1 centimeter thick. Calculate the volume of the plastic, in cubic centimeters?

A. xyz + 8
B. 2(xy + xz + yz + 4)
C. 2(xy + xz + yz) – xyz
D. 2(xy + xz + yz + x + y + z + 4)
E. 2(xy + xz + yz + 2x + 2y + 2z + 4)

Are we really concerned with the extra 1cm thickness in calculating the volume? The thickness does not add to spaces to
be filled; so, how come it is affecting the volume?

What does plastic do? questions looks incomplete.

The most logical meaning we can draw out of it is that you are going to cover the alumunium box with the plastic. In such case you need to add +1 cm because inner dimensions are given and cover would be calculated based on outer dimensions.
Second case, if you are going to fill in the plastic - in that case you dont need to care about +1 cm. and inner volume should suffice.
Hope it helps.
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Re: A closed aluminum rectangular box has inner dimensions x centimeters [#permalink]

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22 Oct 2012, 07:58
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gmatbull wrote:
A closed aluminum rectangular box has inner dimensions x centimeters by y centimeters by z centimeters. Each of the six sides of the box is 1 centimeter thick. Calculate the volume of the aluminium, in cubic centimeters?

A. xyz + 8
B. 2(xy + xz + yz + 4)
C. 2(xy + xz + yz) – xyz
D. 2(xy + xz + yz + x + y + z + 4)
E. 2(xy + xz + yz + 2x + 2y + 2z + 4)

Are we really concerned with the extra 1cm thickness in calculating the volume? The thickness does not add to spaces to
be filled; so, how come it is affecting the volume?

Inner volume of the box is $$xyz$$ cubic centimeters.

Now, since each of the six sides of the box is 1 centimeter thick, then outer dimensions are $$x+2$$ by $$y+2$$ by $$z+2$$ centimeters. Therefore, the volume of the box with aluminium is $$(x+2)(y+2)(z+2)$$ cubic centimeters.

The volume of the aluminium is the difference of these two: $$(x+2)(y+2)(z+2)-xyz=2(xy + xz + yz + 2x + 2y + 2z + 4)$$.

OR:

Plug numbers: say $$x=y=z=1$$, then the inner volume is 1 cubic centimeters and the volume of the whole box is $$(1+2)(1+2)(1+2)=27$$ cubic centimeters. The volume of the aluminium is 27-1=26 cubic centimeters.

Now, plug $$x=y=z=1$$ and see which options yields 26: only answer choice E fits.

P.S. For plug-in method it might happen that for some particular numbers more than one option may give "correct" answer. In this case just pick some other numbers and check again these "correct" options only.
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Re: A closed aluminum rectangular box has inner dimensions x centimeters [#permalink]

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23 Oct 2012, 19:45
Bunuel wrote:
$$(x+2)(y+2)(z+2)-xyz=2(xy + xz + yz + 2x + 2y + 2z + 4)$$.

The question itself is very clear and not hard but it took me ages to multiple three brackets without a mistake. Is there a trick?
Thanks
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Re: A closed aluminum rectangular box has inner dimensions x centimeters [#permalink]

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23 Oct 2012, 21:16
fukirua wrote:
Bunuel wrote:
$$(x+2)(y+2)(z+2)-xyz=2(xy + xz + yz + 2x + 2y + 2z + 4)$$.

The question itself is very clear and not hard but it took me ages to multiple three brackets without a mistake. Is there a trick?
Thanks

You have to put pen to paper as verbal calculation of the bracket will make issues complicated.
Try solving first two brackets first and then multiply with the third one.
Also, to avoid careless mistakes, always follow a pattern so that you don't miss anything.
for eg ( a + b) ( c + d) = a( c + d) + b ( c + d) always multiply with a pattern .
Hope this helps.
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Re: A closed aluminum rectangular box has inner dimensions x centimeters [#permalink]

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24 Oct 2012, 02:25
fukirua wrote:
Bunuel wrote:
$$(x+2)(y+2)(z+2)-xyz=2(xy + xz + yz + 2x + 2y + 2z + 4)$$.

The question itself is very clear and not hard but it took me ages to multiple three brackets without a mistake. Is there a trick?
Thanks

Start with multiplying the first two in the product of the three factors:

$$(x+2)(y+2)(z+2)=(xy+2x+2y+4)(z+2)$$

No need to fully carry out the other multiplication. You can see that there will be one term of $$xyz$$, which will cancel out in the final expression with $$-xyz$$.
In addition, you will have terms containing products of two factors - like $$xy, \,\,xz$$, and $$yz$$. All will have a coefficient of 2 in front.
Then the terms with $$x, \,\,y,$$ and $$z$$, all have a coefficient of 4 in front.
Answers A, B, and C can be eliminated. Between D and E, obviously E wins.
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Re: A closed aluminum rectangular box has inner dimensions x centimeters [#permalink]

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11 Dec 2013, 13:48
A closed aluminum rectangular box has inner dimensions x centimeters by y centimeters by z centimeters. Each of the six sides of the box is 1 centimeter thick. Calculate the volume of the aluminium, in cubic centimeters?

A. xyz + 8
B. 2(xy + xz + yz + 4)
C. 2(xy + xz + yz) – xyz
D. 2(xy + xz + yz + x + y + z + 4)
E. 2(xy + xz + yz + 2x + 2y + 2z + 4)

We are trying to find the volume of the aluminum, NOT the empty space inside the box.

In the INNER dimensions are x*y*z then the dimensions of the box including the one inch thick sides are going to be two CM greater as shown in the attached diagram (in a way, this is similar to a "walkway surrounding a rectangular garden" problem, except in 3 dimensions) Therefore, the length of the box (not just the interior dimensions) is (L+2) The width is (W+2) and the height is (H+2) --> (L+2)*(W+2)*(H+2) is the volume of the box if we consider the volume of the sides included. If we are to figure out the volume of JUST the sides we subtract from (L+2)*(W+2)*(H+2) the volume of the empty space (L*W*H)

(L+2)*(W+2)*(H+2) - (LWH)
(LW+2L+2W+4)*(H+2) - (LWH)
LWH+2LH+2WH+4H+2LW+4L+4W+8 - (LWH)
2LH+2WH+4H+2LW+4L+4W+8
2(LH+WH+LW+2H+2L+2W+4)

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Re: A closed aluminum rectangular box has inner dimensions x centimeters [#permalink]

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02 Sep 2017, 09:51
Bunuel wrote:
gmatbull wrote:
A closed aluminum rectangular box has inner dimensions x centimeters by y centimeters by z centimeters. Each of the six sides of the box is 1 centimeter thick. Calculate the volume of the aluminium, in cubic centimeters?

A. xyz + 8
B. 2(xy + xz + yz + 4)
C. 2(xy + xz + yz) – xyz
D. 2(xy + xz + yz + x + y + z + 4)
E. 2(xy + xz + yz + 2x + 2y + 2z + 4)

Are we really concerned with the extra 1cm thickness in calculating the volume? The thickness does not add to spaces to
be filled; so, how come it is affecting the volume?

Inner volume of the box is $$xyz$$ cubic centimeters.

Now, since each of the six sides of the box is 1 centimeter thick, then outer dimensions are $$x+2$$ by $$y+2$$ by $$z+2$$ centimeters. Therefore, the volume of the box with aluminium is $$(x+2)(y+2)(z+2)$$ cubic centimeters.

The volume of the aluminium is the difference of these two: $$(x+2)(y+2)(z+2)-xyz=2(xy + xz + yz + 2x + 2y + 2z + 4)$$.

OR:

Plug numbers: say $$x=y=z=1$$, then the inner volume is 1 cubic centimeters and the volume of the whole box is $$(1+2)(1+2)(1+2)=27$$ cubic centimeters. The volume of the aluminium is 27-1=26 cubic centimeters.

Now, plug $$x=y=z=1$$ and see which options yields 26: only answer choice E fits.

P.S. For plug-in method it might happen that for some particular numbers more than one option may give "correct" answer. In this case just pick some other numbers and check again these "correct" options only.

Hi - can you please help me understand how you know that based on the 1cm thickness, you should expand each side x/y/z by +2?

Thanks

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Re: A closed aluminum rectangular box has inner dimensions x centimeters   [#permalink] 02 Sep 2017, 09:51
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