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

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New post 07 Jun 2018, 00:02
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Question Stats:

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

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New post 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.

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

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New post 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.

Answer D.

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

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

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New post 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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