I dont know how to solve this...I think its D

(x^2yz-8x^2z+3yz-24z\) is being compared against 0
the first step is to recognise that the expression is not equal to 0
next is to recognise that it can be factorized
The factored form of the expression will be equal to
(x^2z + 3z)(y-8) compared to 0
since we are not given any value for y, we should divide both sides by (y-8) leaving only the expressions that can be compared since 0 divided by any expression / number will always be zero
this would be (x^2z + 3z) and 0
given that we know that the values of x and z are both positive, it automatically means that A would always be greater
Therefore the answer is A

Great explanation.

The answer is A NOT D

Ask if you do need further explanations

Regards

but why cant we factorise it

Of course you can

$z(x^2y - 8x^2+3y-24)$

$zx^2(y-8) + (y-8)$

At this point basically what we looking for is

$zx^2+1$

$z(x^2+1)$

$x>9$

$z>2$

$2(9^2+1) >0$

But we still don't know what y is. Since x>9 and z>2, we can pick x=10 and z=3
Plugging those in gives us 3(10^2)(y-8)+(y+8) = 300(y-8)+(y+8).
y could still be any value.
If y=1, we get 300(1-8)+(1+8) = -2091
If y=10 we get 300(10-8)+(10+8) = 618

It could still be both greater than or less than 0

if we take (y-8) common then the answer will depend on the value of y.

can u please elaborate how u moved to line 2 of the factorization ?? (in bold) how did u get this??

Shouldnt the second step be--- zx^2(y-8) + 3z(y-8)

Yes. Good catch. Totally true.

$x^2yz-8x^2z+3yz-24z$

$x^2x (y-8) + yz-8z$ notice here how we can simplify by 3

$x^2 (y-8)+z(y-8)$ drop out $(y-8)$

$x^2+z$

At this point, we do know that $x>9$ and $z>2$

$A>B$

So whenever the question is comparing any value to "0" in these quantitative comparison questions, start thinking about factoring because that might be a way out. As it turns out, if you DONT factor, you'll get the same answer I got when I originally tried this question (D, which isn't right). You won't see the answer unless you factor, but again, if you see "0" in a question quant comparison question like this, go ahead and just set the other side equal to 0 and start factoring away.

Even if you don't recognize the full factoring of the equation, you can still factor out a "z" from each of the components.

z * (x^2y - 8x^2 + 3y - 24)

That second part of the equation should be a lot easier to factor with some trial and error

z * (x^2 +3) ( y-8)

Great, so now you know that if you're hypothetically setting this all equal to 0, you can divide both sides by y-8.

Now you only have positive values on the left hand side because of the given values of x and z, so A must be greater than 0.

A is the answer.

This is a tricky question, but with some factoring we can make sense of it.

$x^2yz-8x^2z+3yz-24z$

First, notice that $x^2yz-8x^2z$ has common factors of $x^2$ and $z$, so lets factor those out.

$x^2z(y-8)+3yz-24z$

Now notice that $3yz-24z$ has common factors of $3$ and $z$, so let's factor those out:

$x^2z(y-8)+3z(y-8)$

From here, we can factor out a factor $(y-8)$:

$(y-8)(x^2z+3z)$

Then from here we can factor out a factor of $z$ from $(x^2z+3z)$:

$(y-8)(z)(x^2+3)$

Since $y > 9$, $y - 8 > 0$, so the $(y-8)$ portion of that equation is positive.
Since $z > 2$, we know that the $(z)$ portion of that equation is positive.
Finally, since $x^2$ is always positive or 0, and $3$ is positive, $(x^2+3)$ is positive.

So we have a positive*positive*positive in Quantity A, and 0 in Quantity B.

Therefore, Quantity A is greater.

x^2yz-8x^2z+3yz-24z

z(x^2y-8x^2+3y-24)
z(x^2y+3y-8x^2-24)
z[y(x^2+3)-8(x^2+3)]
z(y-8)(x^2+3)

since y>9, z>2

z>0, y-8>0, x^2+3>0

Therefore

z(y-8)(x^2+3) > 0

Final Answer: A

Hi,

I have a quick question on the scope of gre quant as if we assume that x can be imaginary number, answer may be D.

Definition:
x^2 = x * x
-i ^2 = -1

Case1:
ex, x= 0 , y = 10 , z =3
x^2yz−8x^2z+3yz−24z = 0-0 + 3*10*3 - 24*3 = 18 > 0

Case2:
( ex, x= 100i , y = 10 , z = 3 )
x^2 = -10000 and the equation in the question will be less than 0.

x^2yz−8x^2z+3yz−24z = -10000*3*10 -(8*-10000*3) + 3*10*3 - 24*3 = -10000(30-24) + 90 - 72 = -59982< 0

Is it fair and safe that imaginary numbers are out of scope from gre guant test ?
I picked A as answer but was not so sure.

Best,
Gocha

The GRE specifically deal with real numbers!!

Hi Carcass,

Thank you for clarification, understood. It helps me a lot where to focus.

Best,
Gocha

Look at here https://gre.myprepclub.com/forum/ets-gre-m ... tml#p38263

Regards