Gmat prep 2 - Numbers properties

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Gmat prep 2 - Numbers properties

by mttorii » Tue Jul 22, 2008 9:47 am
HI,

Can someone help me with this ...

If mv<pv<0, is v>0?

(1) m<p
(2) m<0

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by Canman » Tue Jul 22, 2008 9:54 am
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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by VP_Tatiana » Tue Jul 22, 2008 9:57 am
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.
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by mttorii » Tue Jul 22, 2008 1:03 pm
HI,

The OA is D.

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by Canman » Tue Jul 22, 2008 1:22 pm
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.

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by cubicle_bound_misfit » Tue Jul 22, 2008 10:10 pm
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.

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by Ian Stewart » Tue Jul 22, 2008 11:59 pm
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.

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cubicle_bound_misfit wrote:is it a right approach, someone please let me know.
Yes, excellent approach- nicely done.