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

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Joined: 19 Feb 2018
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x<y<z, where x,y & z are [#permalink]  19 Feb 2018, 01:05
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Question Stats:

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]  19 Feb 2018, 02:30
Expert's post

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

Regards
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GRE 1: Q167 V156
WE: Business Development (Energy and Utilities)
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Re: x<y<z, where x,y & z are [#permalink]  19 Feb 2018, 11:53
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Expert's post
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]  19 Feb 2018, 21:08
1
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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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This is my response to the question and may be incorrect. Feel free to rectify any mistakes

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