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24 Feb 2019, 00:15
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
64% (01:30) correct
36%
(01:39)
wrong
based on 219
sessions
History
Date
Time
Result
Not Attempted Yet
siddharthkapoor
GMAT Club Reviews PM Intern
Joined: 10 Apr 2018
Last visit: 12 Sep 2025
Posts: 529
Given Kudos: 522
Location: India
Schools: ISB'22 (D)
GMAT 1: 680 Q48 V34
GPA: 3.3
Products:
Updated on: 10 Oct 2019, 02:37
Originally posted by siddharthkapoor on 24 Feb 2019, 00:43.
Last edited by siddharthkapoor on 10 Oct 2019, 02:37, edited 6 times in total.
Last edited by siddharthkapoor on 10 Oct 2019, 02:37, edited 6 times in total.
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Objective: To find if x>3
From statement 1) we know that x>0, but from this we can't say whether or not x>3. So option A and D can be eliminated.
From statement 2) we get:
x^3-9x+4>2^2
x^3-9x+4>4
=> x^3-9x>0
=> x(x+3)(x-3)>0, so from this we can't say whether x>3 or not as x>3 and -3<x<0 are both valid solutions, so option B and D can be eliminated.
Combining statement 1) and 2), we get:
x>0 and x>3 or -3<x<0
Thus, we can say that x>3, so option C is the correct option.
From statement 1) we know that x>0, but from this we can't say whether or not x>3. So option A and D can be eliminated.
From statement 2) we get:
x^3-9x+4>2^2
x^3-9x+4>4
=> x^3-9x>0
=> x(x+3)(x-3)>0, so from this we can't say whether x>3 or not as x>3 and -3<x<0 are both valid solutions, so option B and D can be eliminated.
Combining statement 1) and 2), we get:
x>0 and x>3 or -3<x<0
Thus, we can say that x>3, so option C is the correct option.
Updated on: 24 Feb 2019, 00:47
Originally posted by kumarankit01 on 24 Feb 2019, 00:46.
Last edited by kumarankit01 on 24 Feb 2019, 00:47, edited 1 time in total.
Last edited by kumarankit01 on 24 Feb 2019, 00:47, edited 1 time in total.
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Answer is C .
1. x> 0 no sufficent
2. if we square both side ,
we get x^3-9x+4 > 4
=> x^3-9x>4
x(x^2 -9)>0
x(x-3)(x+3) >0
Not sufficent .
If we combine both,
we get x>0 , it means , x-3 should be greater than 0 or x >3
So answer is C . Both are sufficient.
Hit Kudos now ??
Posted from my mobile device
1. x> 0 no sufficent
2. if we square both side ,
we get x^3-9x+4 > 4
=> x^3-9x>4
x(x^2 -9)>0
x(x-3)(x+3) >0
Not sufficent .
If we combine both,
we get x>0 , it means , x-3 should be greater than 0 or x >3
So answer is C . Both are sufficient.
Hit Kudos now ??
Posted from my mobile device
