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If xy > 0 and yz < 0, then which of the following must be negative?

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If xy > 0 and yz < 0, then which of the following must be negative?  [#permalink]

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New post 23 Jan 2019, 01:55
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A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

55% (01:29) correct 45% (01:36) wrong based on 41 sessions

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Re: If xy > 0 and yz < 0, then which of the following must be negative?  [#permalink]

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New post 23 Jan 2019, 03:59
Bunuel wrote:
If \(xy > 0\) and \(yz < 0\), then which of the following must be negative?

A \(xyz\)

B \(xy^2z\)

C \(x^2y^2z\)

D \(x^2y^2z^2\)

E \(\frac{xy}{z}\)


IMO B

If \(xy > 0\) and \(yz < 0\)

there can be 2 cases

x can be +ive, y can be +ive & y can be +ive, z can be -ive

x can be -ive, y can be -ive & y can be -ive, z can be +ive

Now just go through the options one by one

A \(xyz\), -ive or +ive, depending on the mentioned cases

B \(xy^2z\), always -ive
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Re: If xy > 0 and yz < 0, then which of the following must be negative?  [#permalink]

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New post 23 Jan 2019, 05:07
Bunuel wrote:
If \(xy > 0\) and \(yz < 0\), then which of the following must be negative?


A \(xyz\)

B \(xy^2z\)

C \(x^2y^2z\)

D \(x^2y^2z^2\)

E \(\frac{xy}{z}\)



xy>0 and yz<0.

multiply these 2 .

\(xy^2z<0.\)

one is positive and another one is negative. ultimate result must be negative.

B is the correct answer.
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Re: If xy > 0 and yz < 0, then which of the following must be negative?  [#permalink]

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New post 23 Jan 2019, 05:29
Bunuel wrote:
If \(xy > 0\) and \(yz < 0\), then which of the following must be negative?


A \(xyz\)

B \(xy^2z\)

C \(x^2y^2z\)

D \(x^2y^2z^2\)

E \(\frac{xy}{z}\)



test the cases
xy >0
when both x & y are either + or -
so in that case yz<0 when either of values are of opposite sign

we can say that
option B stands out

IMO B
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Re: If xy > 0 and yz < 0, then which of the following must be negative?  [#permalink]

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New post 27 Jan 2019, 18:59
Bunuel wrote:
If \(xy > 0\) and \(yz < 0\), then which of the following must be negative?


A \(xyz\)

B \(xy^2z\)

C \(x^2y^2z\)

D \(x^2y^2z^2\)

E \(\frac{xy}{z}\)


Looking at the first inequality we see that either x and y are both positive or they are both negative.

Combine that with the second inequality. If y is positive, z is negative, and when y is negative, z is positive.

Thus, our scenarios are:

x = pos, y = pos, z = neg

Or

x = neg, y = neg, z = pos

We see that x and z always have opposite signs, and thus, (x)(y^2)(z) is always negative.

Alternate solution:

We can use the following fact: If a > 0 and b < 0, then ab < 0. Therefore, we can multiply the two inequalities to obtain:

x(y^2)z < 0

Answer: B
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Re: If xy > 0 and yz < 0, then which of the following must be negative?   [#permalink] 27 Jan 2019, 18:59
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If xy > 0 and yz < 0, then which of the following must be negative?

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