OG12 Question 122, OG13 Question 130.

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OG12 Question 122, OG13 Question 130.

by zorroprateek » Sun Oct 07, 2012 11:32 am
What is the volume of a certain rectangular solid?
(1) Two adjacent faces of the solid have areas 15
and 24, respectively.
(2) Each of two opposite faces of the solid has
area 40.

Answer of the above question is C.

I know the first statement alone is not sufficient but statement two alone should be sufficient as EACH face area 40, that is l*b=40, b*h=40 and h*l=40.

But the explanation in OG for statement 2 is :-

If the edge lengths of the rectangular solid
are 5, 8, and x, where x is a positive real
number, then the rectangular solid will have
a pair of opposite faces of area 40, namely
the two faces that are 5 by 8. However, the
volume is (5)(8)(x), which will vary as x
varies; NOT sufficient.

i dont understand the above explanation as with edges 5,8 and x only two opposite faces will have area 40 and not EACH of the two opposite face, isn't it?
Source: — Data Sufficiency |

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by Anurag@Gurome » Sun Oct 07, 2012 6:50 pm
zorroprateek wrote:What is the volume of a certain rectangular solid?
(1) Two adjacent faces of the solid have areas 15
and 24, respectively.
(2) Each of two opposite faces of the solid has
area 40.

Answer of the above question is C.

I know the first statement alone is not sufficient but statement two alone should be sufficient as EACH face area 40, that is l*b=40, b*h=40 and h*l=40.

But the explanation in OG for statement 2 is :-

If the edge lengths of the rectangular solid
are 5, 8, and x, where x is a positive real
number, then the rectangular solid will have
a pair of opposite faces of area 40, namely
the two faces that are 5 by 8. However, the
volume is (5)(8)(x), which will vary as x
varies; NOT sufficient.

i dont understand the above explanation as with edges 5,8 and x only two opposite faces will have area 40 and not EACH of the two opposite face, isn't it?
In a rectangular solid, there are 6 faces let's say A, B, C, D, E and F from among which we can have 3 pairs such that two faces of each pair will have the same area and will be opposite. For example A and D will be opposite and have the same area.
B and E will be opposite and have the same area.
C and F will be opposite and have the same area.
What the second statement is saying that out of these 3 pair one pair will be of area 40.

Solution:
Obviously each statement alone is not sufficient to answer the question.
Let us check after combining both the statements.
Let the sides of the solid be a, b and c.
So ab =15, bc = 24 and ac = 40.

What we need is abc.
Now (abc)^2 = 15*24*40 = 14400
Or abc is 120.

The correct answer is C.
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by Whitney Garner » Mon Oct 08, 2012 7:16 pm
zorroprateek wrote:i dont understand the above explanation as with edges 5,8 and x only two opposite faces will have area 40 and not EACH of the two opposite face, isn't it?
Hi zorroprateek!

I agree that the explanation in the OG is (as typical of the OGs) really confusing!! So rather than get all hung up on it, let's simplify!

So here is a nice simple little box diagram. I have shaded in 3 sides so that I can clearly talk about them, and just assume that the sides opposite those sides have the same color (i.e. the front AND back sides are green, the top and bottom are red and the 2 sides are blue).

Image

Okay, what is the volume? Well, I would need to know the dimensions of this thing.

Statement (1): "Two adjacent faces of the solid have areas 15 and 24, respectively."
Okay, adjacent means next to each other, so that could be like the red and the green, or the green and the blue, or the blue and the red. So what if I make it the following:

red area = 15
green area = 24.

They share one dimension, that long line across the top. So we could say that that could be the common integer factor of the 2 areas (3), so the red would be 3x5 and the green could be 3x8 (so that means the missing side would be 5x8), but that isn't the only common factor. What if the line they share is only 1 in length? Then red would be 1x15 and the green would be 1x24, making the blue area 15x24. This gives a VERY different volume for the box! And that doesn't even scratch the surface! We actually have NO CLUE that the sides are integers so there are an infinite number of possible shapes for the red and green sides!

Statement (2): "Each of two opposite faces of the solid has area 40."
Well, this only tells us about 2 opposite sides, but it doesn't say WHICH sides. Just that one pair has areas of 40 each. So maybe it is the red sides, or the green sides, or the blue sides. So we don't know which sides it is. AND, even if we did know, we don't know the dimensions of the other 2 so we're sunk for the same reasons as statement (1)

Statement (1+2): "Two adjacent faces of the solid have areas 15 and 24, respectively, and Each of two opposite faces of the solid has area 40."

Well, then it doesn't really matter which colors we assign, but we know the surface areas of the 3 sides, 15, 24 and 40. Now, that doesn't mean we have a fixed volume though. We need a new diagram (or some labels).
Image

Okay, so lets go with the labels from earlier:
red area = LW = 15
green area = LH = 24
blue area = WH = 40

What is the volume? LWH. You have a system of 3 equations with 3 variables, you can solve (and since it is DS I'm not going to!)

Hope this helps!
:)
Whit
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Math is a lot like love - a simple idea that can easily get complicated :heart-eyes:

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by zorroprateek » Tue Oct 09, 2012 11:14 am
Thanks Anurag and Whitney :)