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Director
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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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07 Jun 2018, 00:02
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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
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07 Jun 2018, 00:55
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
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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 trueThere 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 falseWe 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
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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. 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
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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.



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



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Re: If (x^2+8)yz < 0, wz > 0, and xyz < 0, the which of the following mus
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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: xyzw +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
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25 Sep 2019, 16:26






