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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8033
GMAT 1: 760 Q51 V42 GPA: 3.82

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Difficulty:   45% (medium)

Question Stats: 63% (01:17) correct 37% (01:20) wrong based on 178 sessions

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Is kr<0?

1) $$k^2r^3<0$$
2) $$|k+r|<|k|+|r|$$

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Intern  B
Joined: 29 Jul 2013
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Location: India
Concentration: Human Resources, Strategy
Schools: MBS '21 (A)
WE: Information Technology (Computer Software)
Re: Is kr<0?  [#permalink]

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1
condition 1 : k^2r^3<0, basically is the same as telling r<0, ignoring squares as it's always positive. We still dont know if k is positive or not, therefore Not Sufficient

Condition 2 : |k+r|<|k|+|r|

there are 4 scenarios we need to consider of which only 2 conditions are there where this rule will work
k is pos and r is pos (rule doesnt work)
k is pos and r is neg (rule works)
k is neg and r is pos (rule works)
k is neg and r is neg (rule doesnt work)

since for the conditions that work, since either k or r is negative, the product will be negative, hence this condition is Sufficient.
Retired Moderator P
Joined: 22 Aug 2013
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Location: India
Re: Is kr<0?  [#permalink]

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We need to know two things for this question.

First, product of two numbers is < 0 (negative) only if one of them is positive and the other is negative.

Secondly, lets compare |a+b| with |a| + |b|
|a| means absolute value of a, or the distance of a from zero on the number line.

If both a & b are positive, then |a+b| = |a| + |b| (you can check by substituting positive values for a & b)

If both a & b are negative, then also |a+b| = |a| + |b| (you can check by substituting negative values for a & b)

If both a & b are zero, then also |a+b| = |a| + |b|

If ONE of a & b is zero (other can be positive or negative), still |a+b| = |a| + |b|

But if one of a & b is positive and the other is negative, then |a+b| < |a| + |b|. Eg, if a=3 and b=-2, then |a+b| = |3-2| = 1,
but |a|+|b| = |3| + |-2| = 3+2 = 5

Now lets look at the statements.

Statement 1. k^2 r^3 <0
k^2 cannot be negative, which means r^3 < 0 or r < 0.
But we don't know anything about k, so we cannot say whether product of k & r will be negative or positive. Insufficient.

Statement 2. |k+r| < |k| + |r|
This can only happen when one of k and r is positive and other is negative (as discussed above).
So product kr < 0. Sufficient.

Hence answer is B
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8033
GMAT 1: 760 Q51 V42 GPA: 3.82
Re: Is kr<0?  [#permalink]

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==> For con 1), you ignore the square, so t<0, and for con 2), you get kr<0, hence yes, it is sufficient. The reason is that from $$(|k+r|)^2<(|k|+|r|)^2$$, you get $$k^2+r^2+2kr<k^2+r^2+2|kr|$$, and if you get rid of $$k^2+r^2$$ from both sides, you get 2kr<2|kr|, then kr<|kr|, which becomes kr<0.

The answer is B.
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Re: Is kr<0?  [#permalink]

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_________________ Re: Is kr<0?   [#permalink] 09 Aug 2018, 10:26
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