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Re: tricky DS inequalities problem from an old GMAT paper exam [#permalink]
08 Oct 2010, 18:32

Expert's post

tonebeeze wrote:

Is k greater than 3?

(1) (k - 3)(k- 2)(k - 1) > 0

(2) k > 1

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]
08 Oct 2010, 22:27

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) _________________

Re: tricky DS inequalities problem from an old GMAT paper exam [#permalink]
15 Oct 2010, 10:33

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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