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# If (x^2+8)yz < 0, wz > 0, and xyz < 0, the which of the following mus

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Director
Joined: 12 Feb 2015
Posts: 941
If (x^2+8)yz < 0, wz > 0, and xyz < 0, the which of the following mus  [#permalink]

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07 Jun 2018, 00:02
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Question Stats:

60% (02:23) correct 40% (02:26) wrong based on 157 sessions

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If $$(x^2 + 8)yz< 0$$ , $$wz > 0$$, and $$xyz < 0$$, the which of the following must be true?
I. $$x < 0$$
II. $$wy < 0$$
III. $$yz < 0$$

A) II only
B) III only
C) I and III only
D) II and III only
E) I, II, and III

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Manish

"Only I can change my life. No one can do it for me"
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Joined: 04 Jan 2015
Posts: 3142
Re: If (x^2+8)yz < 0, wz > 0, and xyz < 0, the which of the following mus  [#permalink]

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07 Jun 2018, 00:55
2
1

Solution

Given:
• $$(x^2 + 8) yz < 0$$
• $$wz > 0$$
• $$xyz < 0$$

To find:
• Out of the given options, which one is always true

Approach and Working:
From the given statements, we can say:
• $$yz < 0$$ as $$x^2$$ is always positive, therefore, $$x^2 + 8$$ is also positive
o as $$yz < 0$$, either of y or z is less than 0
o therefore, y and z will have opposite signs

• $$wz > 0$$ means either both w and z are positive or both of them are negative
o if w and z are positive, then y is negative
o if w and z are negative, then y is positive
o therefore, w and y will have opposite signs

• $$xyz < 0$$ means x is positive, as we already know $$yz < 0$$

Now, if we consider the choices,
• $$x < 0$$
o cannot be true as we know x is positive
• $$wy < 0$$
o always true as we know w and y will have opposite signs
• $$yz < 0$$
o always true as we know y and z will have opposite signs

Hence, the correct answer is option D.

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If (x^2+8)yz < 0, wz > 0, and xyz < 0, the which of the following mus  [#permalink]

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07 Jun 2018, 01:00
CAMANISHPARMAR wrote:
If $$(x^2 + 8)yz< 0$$ , $$wz > 0$$, and $$xyz < 0$$, the which of the following must be true?
I. $$x < 0$$
II. $$wy < 0$$
III. $$yz < 0$$

A) II only
B) III only
C) I and III only
D) II and III only
E) I, II, and III

Given: $$(x^2 + 8)yz< 0$$ | $$wz > 0$$ | $$xyz < 0$$

To prove III is true
$$x^2 + 8$$ is positive irrespective of whether x is positive/negative
yz must always be negative for the expression $$(x^2 + 8)yz < 0$$ to be TRUE.

To prove II is true
There are 2 possibilities for $$yz$$ to be negative:
Either y or z is negative and the other variable is positive.
If z is negative, w must be also negative because $$wz > 0$$
However, if z is positive, w must also be positive.

The expression $$wy$$ will always be negative as
Case 1: IF y is negative, z is positive and w is also positive.
Case 2: IF y is positive, z is negative and w is also negative.

To prove I is false
We know that $$yz$$ is always negative(as we prove III is true)
Since we are given that $$xyz < 0$$, x is always positive.

Therefore, Option D(II and III only) must always be true.
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Director
Joined: 14 Dec 2017
Posts: 509
Location: India
If (x^2+8)yz < 0, wz > 0, and xyz < 0, the which of the following mus  [#permalink]

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07 Jun 2018, 11:00
CAMANISHPARMAR wrote:
If $$(x^2 + 8)yz< 0$$ , $$wz > 0$$, and $$xyz < 0$$, the which of the following must be true?
I. $$x < 0$$
II. $$wy < 0$$
III. $$yz < 0$$

A) II only
B) III only
C) I and III only
D) II and III only
E) I, II, and III

Given,

$$(x^2 + 8)yz< 0$$ , which means $$yz<0$$

$$wz > 0$$, which means i) $$w>0, z>0$$ or ii) $$w<0, z<0$$

$$xyz < 0$$, since we know, $$yz<0$$, hence $$x>0$$.
Therefore, i) $$y>0, z<0$$ or ii) $$y<0, z>0$$

Now since z is common in all three expressions, lets use cases based around sign of z.

i) if $$z>0$$, then $$w>0, y<0, x>0$$
ii) if $$z<0$$, then $$w<0, y>0, x>0$$

Lets check the given expressions:
I. $$x < 0$$ - False
II. $$wy < 0$$ - True, since $$w$$ & $$y$$ always have opposite signs
III. $$yz < 0$$ - True, since $$y$$ & $$z$$ always have opposite signs.

Thanks,
GyM

Ps - Thanks Manish, its a good question!!
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If (x^2+8)yz < 0, wz > 0, and xyz < 0, the which of the following mus  [#permalink]

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07 Jun 2018, 16:10
CAMANISHPARMAR wrote:
If $$(x^2 + 8)yz< 0$$ , $$wz > 0$$, and $$xyz < 0$$, the which of the following must be true?
I. $$x < 0$$
II. $$wy < 0$$
III. $$yz < 0$$

A) II only
B) III only
C) I and III only
D) II and III only
E) I, II, and III

This question needs part to part treatment.

part 1:
($$x^2$$+8)yz<0

From this part we understand that x^2 will always be positive. When we add the value of x^2 with 8 ultimate result will be positive. Thus, yz has to be negative as ($$x^2$$+8)yz<0.
So, iii is one of the options.

part 2:

xyz<0

we know yz <0 , thus x has to be positive. We can eliminate option containing i.

part 3:

wz>0
yz<0...........................we have figure it out earlier.

we can multiply them to prove ii option.

after multiply we will get: ($$z^2$$)wy<0..............................................wz>0 and yz<0 . one is positive and another is negative. ultimately, we will get negative answer.

now, $$z^2$$ is always positive. Thus, wy <0. This is a must.

So , the best answer is II and III.
Intern
Joined: 18 May 2016
Posts: 26
Re: If (x^2+8)yz < 0, wz > 0, and xyz < 0, the which of the following mus  [#permalink]

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07 Jun 2018, 17:18
(x^2 + 8)yz < 0; as x^2 + 8 >0 so yz<0
this means either y>0, z<0 or y<0, z>0

wz >0
case 1: w>0 and z>0
from above statement, if z>0 then y<0 hence wy<0

case 2: w<0 and z<0
from above statement, if z<0 then y>0 hence wy<0

xyz<0
as yz<0 so x>0

Ans D
Director
Joined: 18 Dec 2017
Posts: 729
Location: United States (KS)
Re: If (x^2+8)yz < 0, wz > 0, and xyz < 0, the which of the following mus  [#permalink]

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25 Sep 2019, 16:26
CAMANISHPARMAR wrote:
If $$(x^2 + 8)yz< 0$$ , $$wz > 0$$, and $$xyz < 0$$, the which of the following must be true?
I. $$x < 0$$
II. $$wy < 0$$
III. $$yz < 0$$

A) II only
B) III only
C) I and III only
D) II and III only
E) I, II, and III

I made a small table with columns x y z w and wrote the possibilities:
x-----y-----z-----w
+ve +ve -ve -ve
+ve +ve +ve +ve
-ve +ve -ve -ve ---- NOT POSSIBLE
-ve -ve +ve ----NOT POSSIBLE

As we can see II and III are always True.!
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Re: If (x^2+8)yz < 0, wz > 0, and xyz < 0, the which of the following mus   [#permalink] 25 Sep 2019, 16:26
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