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Bunuel
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If –5 < k < 5, is k > 2 ? (1) (k – 1)(k – 5)(k – 2) < 0 (2) k – 1 > 0 14 Mar 2017, 03:05
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C
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55% (hard)

Question Stats:

57% (01:49) correct 43% (01:51) wrong based on 115 sessions

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If –5 < k < 5, is k > 2 ?

(1) (k – 1)(k – 5)(k – 2) < 0
(2) k – 1 > 0
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C
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If –5 < k < 5, is k > 2 ? (1) (k – 1)(k – 5)(k – 2) < 0 (2) k – 1 > 0 14 Mar 2017, 07:40
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Bunuel
If –5 < k < 5, is k > 2 ?

(1) (k – 1)(k – 5)(k – 2) < 0
(2) k – 1 > 0
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(1) (k – 1)(k – 5)(k – 2) < 0
means either k<1 or 2<k<5
insuff

(2) k>1
clearly insuff
k can be 3 or 1.5
insuff

combined
k<1 or 2<k<5 from (1)
also k>1 (from 2)

thus 2<k<5 only left out
so k>2
suff

Ans C
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If –5 < k < 5, is k > 2 ? (1) (k – 1)(k – 5)(k – 2) < 0 (2) k – 1 > 0 14 Mar 2017, 07:51
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Ste 1: This implies that any one of (k-1)(k-5)(k-2) is negative or all three are negative .then only (k-1)(k-5)(k-2) <0
not sufficient.

Ste 2 : Implies K>1 Not suff

Combining both
(k-1) will be greater than 0

and if (k-2) is negative (k-5 ) will also be negative ,which is not possible. therefor k>2
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If –5 < k < 5, is k > 2 ? (1) (k – 1)(k – 5)(k – 2) < 0 (2) k – 1 > 0 15 Mar 2017, 04:42
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St 1: from the expression we can say that -5<k<1 and 2<k<5. for region 1 it is not and for region 2 it is. INSUFFICIENT
St 2: k-1>0 or k>1 not necessarily greater than 2. INSUFFICIENT

St 1 & St 2: since k>1, region 1 is ruled out. hence 2<k<5. therefore k>2. ANSWER

Option C
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If –5 < k < 5, is k > 2 ? (1) (k – 1)(k – 5)(k – 2) < 0 (2) k – 1 > 0 31 Mar 2017, 00:03
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Bunuel
If –5 < k < 5, is k > 2 ?

(1) (k – 1)(k – 5)(k – 2) < 0
(2) k – 1 > 0
Show more

Bunuel
If –5 < k < 5, is k > 2 ?

(1) (k – 1)(k – 5)(k – 2) < 0
(2) k – 1 > 0
Show more

Given: Valid values of k = -4 to + 4

Question : Is k > 2 ?

Statement : 1 From St 1 we can say that k can't be equal to 1, 2 and 5 (also 5 can't be the value as it is out of range)

If k = 4 answer to the question is YES.

But if k = -4 then answer to the question is NO. So Insufficient

Statement : 2 , k > 1 , as k can be 1.5 or 3 Insufficient

Combining 1 & 2 we can surely say that k > 2. So (C)
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If –5 < k < 5, is k > 2 ? (1) (k – 1)(k – 5)(k – 2) < 0 (2) k – 1 > 0 03 Nov 2018, 06:23
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I followed the method given in a PS question for finding the limits when <0 and >0,( graphical representation of above and below the x axis) but not able to figure out how we arrived at the limits in the statement one. can somebody please help me with the concept.
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If –5 < k < 5, is k > 2 ? (1) (k – 1)(k – 5)(k – 2) < 0 (2) k – 1 > 0 23 Dec 2020, 14:20
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Bunuel
If –5 < k < 5, is k > 2 ?

(1) (k – 1)(k – 5)(k – 2) < 0
(2) k – 1 > 0
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(1) With the help wavy curve method we get, \(k<1, 2<k<5\); Insufficient

(2) \(k – 1 > 0; k>1, k \ can \ be \ 1.5, 1.7, or \ 2.5\); Insufficient.

Considering both:
\(K\) is not less than \(1\), so \(2<k<5\); Sufficient.

The answer is \(C\)
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