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14 Jan 2015, 02:29
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
52% (01:07) correct
48%
(00:54)
wrong
based on 172
sessions
History
Date
Time
Result
Not Attempted Yet
Is q positive?
(1) qp^2 is not negative
(2) q^2 is positive
Kudos for a correct solution.
(1) qp^2 is not negative
(2) q^2 is positive
Kudos for a correct solution.
14 Jan 2015, 11:20
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Bunuel
st.1
if q=1, p=0 then answer to the original question is yes
if q=0, p=1 then answer to the original question is no.
hence st.1 alone is not sufficient
st.2
for both positive and negative values of q , q^2 can be positive. hence st.2 alone is not sufficient
st.1 + st.2
we know that q=0 is not a possible solution.because if q=0, then q^2 will not be positive. but if q takes some negative value, then p can still take zero as a value and can satisfy the condition given in st.1. hence st.1 and st.2 together are not sufficient. as both positive and negative values of q are possible.
therefore answer must be E
14 Jan 2015, 11:22
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We can't really rewrite the information in the question, so we directly go into the statements:
(1) Since 0 is not negative, we don't know enough about q. q can be either 0, positive or negative, as the product of q and \(p^2\) will be 0 if p is 0.
(2) \(q^2\) is positive, that only tells us that q cannot be 0, but nothing more. It is also insufficient.
Both statements combined: From the second statement we know that q cannot be 0, but q can still be positive or negative, as either way, \(qp^2\) can be 0 and therefore fulfil the condition in statement 1.
Answer E is correct.
(1) Since 0 is not negative, we don't know enough about q. q can be either 0, positive or negative, as the product of q and \(p^2\) will be 0 if p is 0.
(2) \(q^2\) is positive, that only tells us that q cannot be 0, but nothing more. It is also insufficient.
Both statements combined: From the second statement we know that q cannot be 0, but q can still be positive or negative, as either way, \(qp^2\) can be 0 and therefore fulfil the condition in statement 1.
Answer E is correct.
