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# Is pq > 0? (1) (2q)^p = 1 (2) p < 0

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Re: Is pq > 0? (1) (2q)^p = 1 (2) p < 0 [#permalink]
Bunuel wrote:
Is $$pq > 0$$?

(1) $$(2q)^p = 1$$

(2) $$p < 0$$

Are You Up For the Challenge: 700 Level Questions

#2. p is <0 but q can be anything </>/= zero. Insufficient
#1. p can be 0 and at the same time p can be 1 so Insufficient.

1+2.

p has to be a -ve value say -1,-2,-1.5 etc.
Inside the braces q can be +ve as well as -ve. Put q= 1/2 and (-1/2) to check. So overall #1+2 is also Insufficient.

E
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Re: Is pq > 0? (1) (2q)^p = 1 (2) p < 0 [#permalink]
Bunuel wrote:
Is $$pq > 0$$?

(1) $$(2q)^p = 1$$

(2) $$p < 0$$

Are You Up For the Challenge: 700 Level Questions

1) (2q)^p=1
P could be 0 ---- pq =0 OR q could be 0.5 and p can be anything . so Insufficient..

(2) p<0 ... q can be 0.5 or -0.5 .. accordingly value can change .. So insufficient

Combining (1&2)

p<0≠0: q=0.5…p=-2…2(0.5)^(-2)=1…pq=-1
p<0≠0: q=-0.5…p=-2…2(-0.5)^(-2)=(-1)^2=1/(-1)^2=1…pq=1

still insufficient
Hence
Ans (E)
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Re: Is pq > 0? (1) (2q)^p = 1 (2) p < 0 [#permalink]
1
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Bunuel wrote:
Is $$pq > 0$$?

(1) $$(2q)^p = 1$$

(2) $$p < 0$$

Are You Up For the Challenge: 700 Level Questions

pq>0 basically means both p and q have same sign.

(1) $$(2q)^p = 1$$
If p=0, q could be anything...so pq=0, that is answer is NO
If $$p=1$$, and $$q=\frac{1}{2}$$.....$$pq=2*\frac{1}{2}=1>0$$, that is answer is YES
Insuff

(2) $$p < 0$$
Insuff

Combined
If $$p=-1,$$and $$q=\frac{1}{2}....(2*\frac{1}{2})^{-1}=1 and p*q=-1*\frac{1}{2}=-\frac{1}{2}$$.......answer is NO
If $$p=-2$$, and $$q=-\frac{1}{2}....(2*-\frac{1}{2})^{-2}=(-1)^{-2}=\frac{1}{(-1)^2}=1$$ and $$pq=-2*-\frac{1}{2}=1>0$$...answer is Yes
Insuff
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Re: Is pq > 0? (1) (2q)^p = 1 (2) p < 0 [#permalink]
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Re: Is pq > 0? (1) (2q)^p = 1 (2) p < 0 [#permalink]
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