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x<y<z, where x,y & z are

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x<y<z, where x,y & z are [#permalink] New post 19 Feb 2018, 01:05
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0% (00:00) correct 100% (00:34) wrong based on 3 sessions
\(x<y<z\), where \(x,y & z\) are the lengths of the edges of a cuboid

Quantity A
Quantity B
Volume of the cuboid with edges having lengths (x + 10),y & z
Volume of the cuboid with edges having lengths x,y & (z+ 10)


A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.



How to solve this question?
[Reveal] Spoiler: OA

Last edited by Carcass on 19 Feb 2018, 12:55, edited 2 times in total.
Edited by Carcass
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Re: x<y<z, where x,y & z are [#permalink] New post 19 Feb 2018, 02:30
Expert's post
Please follow the rules of the board for posting a question.

A QCQ has to go under the right subforum such the one we are now, NOT in general quant forum.

Moreover, follow the right way to post a question on the board here https://greprepclub.com/forum/qq-how-to ... -2357.html

Thank you for your collaboration.

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Re: x<y<z, where x,y & z are [#permalink] New post 19 Feb 2018, 11:53
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Write out the volume for both Quanities:

Qty A: \((x+10)yz\) and Qty B: \((z+10)yx\)

or

Qty A: \(xyz + 10yz\) and Qty B: \(xyz + 10xy\)

Cancelling \(xyz\) from both qunatiteis as \(xyz\) is positive

Qty A: \(10yz\) and Qty B: \(10yx\)

Now given that \(x<y<z\) multiplying \(y\) we get \(xy<y^2<zy\) multiplying 10 we get \(10xy<10y^2<10zy\).

Hence Quanitity A is greater.
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Re: x<y<z, where x,y & z are [#permalink] New post 19 Feb 2018, 21:08
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volume of a cuboid of length = \(x\); breadth = \(y\); and height = \(z\) is given by \(x * y * z\)
we are asked to calculate volume on both of the two quantities so let us have

Qty A as \((x+10) * Y * Z\) and Qty B as \(X * Y * (Z+10)\)

Notice that the measure of the sides are positive hence cancelling Y from both quantities will not affect our comparison and keep the syntax of the equation intact.


Now,

Qty A = \((x+10) * Z\)
Qty B = \(x* (z+10)\)

At this point chose any value for \(X\) and \(Z\) such that \(Z>x\) and both \(X\) and \(Z\) are +ve

For eg. \(x =1\) and \(Z = 2\)

Qty A = \((1+10) * 2 = 22\)
Qty B = \(1 * (2+10) = 12\)

Qty A is bigger
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Re: x<y<z, where x,y & z are   [#permalink] 19 Feb 2018, 21:08
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