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If xyz ≠ 0, xy > 0, and x2 + xy + xz < 0, then which of the following

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If xyz ≠ 0, xy > 0, and x2 + xy + xz < 0, then which of the following  [#permalink]

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New post 25 Jan 2019, 08:37
1
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A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

57% (02:11) correct 43% (02:29) wrong based on 48 sessions

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If xyz ≠ 0, xy > 0, and x^2 + xy + xz < 0, then which of the following must be true?

I. x(y + z) < 0

II. x + y + z < 0

III. If x < 0, then z > 0.

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

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Re: If xyz ≠ 0, xy > 0, and x2 + xy + xz < 0, then which of the following  [#permalink]

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New post 25 Jan 2019, 09:16
CAMANISHPARMAR wrote:
If xyz ≠ 0, xy > 0, and x^2 + xy + xz < 0, then which of the following must be true?

I. x(y + z) < 0

II. x + y + z < 0

III. If x < 0, then z > 0.

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


IMO C

xy > 0, and x^2 + xy + xz < 0

xy > 0, and x (x + y + z) < 0

Some cases before marking the options

Case 1, x = -ive and y = -ive & For x (x+y+z) < 0, z > 0 and z > x + y

Case 2, x = + ive and y = + ive & For x (x+y+z) < 0, z > -(x + y)

III. If x < 0, then z > 0
This will be definitely true, case 1 holds good.

I. x(y + z) < 0
True when it is Case 1, True when it is Case 2

II. x + y + z < 0
For case 1, this is not true

For case 2, this is true
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Re: If xyz ≠ 0, xy > 0, and x2 + xy + xz < 0, then which of the following  [#permalink]

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New post 26 Jan 2019, 14:18
CAMANISHPARMAR wrote:
If xyz ≠ 0, xy > 0, and x^2 + xy + xz < 0, then which of the following must be true?

I. x(y + z) < 0

II. x + y + z < 0

III. If x < 0, then z > 0.

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


Fun problem! The key with these is to do as much work as you can using logic and simplification, before you start testing cases. That's especially true on this problem, because there are three variables. If there are more than two variables in a problem, I usually try to avoid testing cases as much as possible - it just takes too long to consider all of the possibilities.

The first thing I noticed is that x and y have the same sign.

I also noticed that the inequality, \(x^2 + xy + xz < 0\), can be simplified like this: \(x(x+y+z) < 0\). This means that either x is negative, or x+y+z is negative, but not both.

Now, go look at options I, II, and III.

Option II stood out to me the most, because I already know something about x+y+z from my reasoning above. It doesn't HAVE to be negative, since the alternative is that x could just be negative. Then, if y and z were big enough, x+y+z would be positive. Eliminate B, D, and E, because they include option II.

This would be a good point to guess! Or, we could keep going if we have enough time.

Notice that the difference between (A) and (C) is that (A) doesn't include III, while (C) does. They both include I! So, at this point, we know that I has to be true without testing it. We only have to worry about III. That's a good reason to use the process of elimination!

III says that if x < 0, z > 0. Well, if x < 0, from the above reasoning, we know that x + y + z > 0. We also know that x and y have the same sign. So, here's what we have:

If x is negative, y is also negative, but x + y + z is positive.

This should only happen if z is positive! You can't add three negative numbers together and get a positive result. So, z definitely has to be positive. III is true.

Eliminate (A) and choose (C).
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Re: If xyz ≠ 0, xy > 0, and x2 + xy + xz < 0, then which of the following   [#permalink] 26 Jan 2019, 14:18
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