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Is kr<0?
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17 May 2017, 01:31
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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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Re: Is kr<0?
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17 May 2017, 09:49
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.



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Re: Is kr<0?
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17 May 2017, 11:31
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 = 32 = 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



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Re: Is kr<0?
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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 \(k^2+r^2+2kr<k^2+r^2+2kr\), and if you get rid of \(k^2+r^2\) from both sides, you get 2kr<2kr, then kr<kr, which becomes kr<0. The answer is B. Answer: B
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Re: Is kr<0?
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09 Aug 2018, 10:26
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