If xy > 0 and yz < 0, then which of the following must be negative? : Problem Solving (PS)

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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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23 Jan 2019, 01:55

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35% (medium)

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55% (01:29) correct

45% (01:36) wrong

based on 41 sessions

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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 $\frac{xy}{z}$

Spoiler: OA

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

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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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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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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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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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