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Can you help me explain how to simplify statement 1. Is the key to test values between 0 and 1?
Thanks
Is \(k>3\)?
(1) \((k-3)(k-2)(k-1)>0\)
The product of 3 numbers is positive if all three are positive (+++) OR two of them are negative and the third one is positive (+--).
Note that: out of 3 numbers \(k-3\) is the least one and \(k-1\) is the biggest one.
\((+)(+)(+)\) is when even the least one is positive so when \(k-3>0\) --> \(k>3\); \((+)(-)(-)\) is when the biggest one is positive (\(k-1>0\) --> \(k>1\)) and the next one (hence the leas one too) negative (\(k-2<0\) --> \(k<2\)), so when \(1<k<2\);
So \((k-3)(k-2)(k-1)>0\) means that: \(k>3\) or \(1<k<2\) --> \(k\) may or may not be more than 3. Not sufficient.
(2) \(k>1\). Clearly insufficient.
(1)+(2) Intersection of the ranges from (1) and (2) is the range we had in (1) \(k>3\) or \(1<k<2\), so \(k\) may or may not be more than 3. Not sufficient.
Re: tricky DS inequalities problem from an old GMAT paper exam [#permalink]
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08 Oct 2010, 23:27
ranjanav22 wrote:
If (k-3)(k-2)(k-1)>0, doesnt that mean, K>3 or K>2 or K>1? Why is it K>3 or 1>k>2? Please explain. Thank you.
When you have inequalities like this one A*B*C>0 Either all three terms are greater than 0 (k>3) Or exactly one term is greater than 0, and other two are less than 0 (1<k<2)
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Re: tricky DS inequalities problem from an old GMAT paper exam [#permalink]
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15 Oct 2010, 11:33
shrouded1 wrote:
ranjanav22 wrote:
If (k-3)(k-2)(k-1)>0, doesnt that mean, K>3 or K>2 or K>1? Why is it K>3 or 1>k>2? Please explain. Thank you.
When you have inequalities like this one A*B*C>0 Either all three terms are greater than 0 (k>3) Or exactly one term is greater than 0, and other two are less than 0 (1<k<2)
Is there a reason for choosing K>3 ? Can it be K<3,K<2 and K>1?
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62% (00:53) correct
