I think answer is B.
Rephrase the question. What is the sign of m or p? Since mv and pv are negative, either mv must be +- or -+ and pv must be +- or -+ to make a negative product and adhere to the restrictions in the stem.
1) m<p - Insufficient. doesn't help us to determine whether m or p is positive or negative
2) m<0 - Sufficient. Knowing that m is negative allows us to conclude that v is positive.
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Gmat prep 2 - Numbers properties
Source: Beat The GMAT — Data Sufficiency |
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VP_Tatiana
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In order for mv and pv to be less than 0, we have one of two cases:
1) v < 0 (and m>0, p>0)
or
2) m<0 and p<0 (and v>0)
The reason is that a negative number is the product of a positive number and a negative number.
Statement 1 tells us that m<p. This is not sufficient information to find out if the two numbers are negative or positive, though. So, statement 1 is not sufficient on its own.
Statement 2 tells us that m<0. So, we know we have "case 2", ie that v>0. So, statement 2 is sufficient and the answer is B.
1) v < 0 (and m>0, p>0)
or
2) m<0 and p<0 (and v>0)
The reason is that a negative number is the product of a positive number and a negative number.
Statement 1 tells us that m<p. This is not sufficient information to find out if the two numbers are negative or positive, though. So, statement 1 is not sufficient on its own.
Statement 2 tells us that m<0. So, we know we have "case 2", ie that v>0. So, statement 2 is sufficient and the answer is B.
Tatiana Becker | GMAT Instructor | Veritas Prep
Thanks for catching my mistake-
I now see how 1) is sufficient.
consider:
for mv<pv<0, m and p must be negative when m<p
Ex:
If m and p are positive then you can have 2*-1 < 3*-1 < 0 which violates the restriction that mp<pv
If m and p are negative then you can have -2*1 < -1*1 < 0 which satisfies all restrictions.
I now see how 1) is sufficient.
consider:
for mv<pv<0, m and p must be negative when m<p
Ex:
If m and p are positive then you can have 2*-1 < 3*-1 < 0 which violates the restriction that mp<pv
If m and p are negative then you can have -2*1 < -1*1 < 0 which satisfies all restrictions.
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cubicle_bound_misfit
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is it a right approach, someone please let me know.
mv<pv<0
(m-p)v<0
stmt 1 says m<p hence m-p<0 v>0
stmt 2 says m<0 given mv<0 hence v>0.
ANS D.
regards,
mv<pv<0
(m-p)v<0
stmt 1 says m<p hence m-p<0 v>0
stmt 2 says m<0 given mv<0 hence v>0.
ANS D.
regards,
Cubicle Bound Misfit
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Ian Stewart
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cubicle_bound_misfit wrote: mv<pv<0
(m-p)v<0
stmt 1 says m<p hence m-p<0 v>0
stmt 2 says m<0 given mv<0 hence v>0.
ANS D.
regards,
Yes, excellent approach- nicely done.cubicle_bound_misfit wrote:is it a right approach, someone please let me know.
