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Bunuel
Is \(pq > 0\)?

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

(2) \(p < 0\)


Are You Up For the Challenge: 700 Level Questions

Let's start with easy-

#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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Bunuel
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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Bunuel
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) \(p < 0\)
Nothing about q
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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