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# Is kr<0?

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

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17 May 2017, 01:31
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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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MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
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"Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Intern Joined: 29 Jul 2013 Posts: 8 Location: India Concentration: Human Resources, Strategy Schools: MBS '21 (A) WE: Information Technology (Computer Software) Re: Is kr<0? [#permalink] ### Show Tags 17 May 2017, 09:49 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 Joined: 22 Aug 2013 Posts: 1428 Location: India Re: Is kr<0? [#permalink] ### Show Tags 17 May 2017, 11:31 1 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 Joined: 16 Aug 2015 Posts: 8033 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: Is kr<0? [#permalink] ### Show Tags 19 May 2017, 01:50 ==> 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. Answer: B _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 1 month Online Course"
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09 Aug 2018, 10:26
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Re: Is kr<0?   [#permalink] 09 Aug 2018, 10:26
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