If (x^2+8)yz 0, and xyz

Question

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

EgmatQuantExpert · Accepted Answer

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

pushpitkc · Answer

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.

GyMrAT · Answer

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!!

KSBGC · Answer

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.

rakeshmishra5690 · Answer

(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

TheNightKing · Answer

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.!

mimie20 · Answer

(x^2+8) is always >0, so y*z<0 -> y or z has to be negative III. This statement is needed to fulfill the equation (see above) II. We know that y or z has to be negative Two possible combinations that w*y<0: -w*y<0 then -w*-z>0 OR w*-y<0 then w*z>0 -> y and z have in both cases the "opposite signs" -> y*z<0 -> is needed to fulfill the equation I. This statement is wrong because x cannot be negative. x has to be positive as z or y has to be negative. Possible combinations: x*-y*z<0 OR x*y*-z<0 Solution: III. and II. are right -> D

Fdambro294 · Answer

Tough question. Have to start by making deductions with the easiest inequality and work your way through possible cases.

WZ > 0

W and Z must have the same sign (1)

(X^2 + 8) YZ < 0

(X’2 + 8) as a factor will always be positive. Thus in order for the expression to be negative —— YZ < 0

Y and Z must have opposite signs (2)

XYZ < 0

We know from (2) that YZ must be negative and have opposite signs.

Thus: X must be positive (3)

Case 1: W = +pos.

Then this means: From (1) —- Z = +

And this means: From (2) ——- Y = (-)neg.

And finally: from (3) ——— X = +

Or

Case 2: W = (-)neg.

This means from (1): Z = (-)neg.

And this means from (2): Y = (+)

And from (3): X = (+)

I. X < 0

False, X is positive as shown above.

II. WY < 0

Either in case 1 or case 2, W and Y always have opposite signs. must be TRUE.

III. YZ < 0

Either in case 1 or case 2, Y and Z always have opposite signs. Must be TRUE.

II AND III Only

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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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