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Is x < 5?
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Updated on: 18 Aug 2014, 09:10
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Is x < 5? (1) x^2 < 16 (2) x(x  5) < 0 Source: 4gmat
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Originally posted by alphonsa on 18 Aug 2014, 09:01.
Last edited by Bunuel on 18 Aug 2014, 09:10, edited 1 time in total.
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Re: Is x < 5?
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18 Aug 2014, 09:12



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Is x < 5?
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Updated on: 10 Jun 2018, 00:26
STAT1 X^2 < 16 => 4 < x < 4 So, X < 4 so will be less than 5 So, SUFFICIENT STAT2 x(x5) < 0 two cases case 1: x < 0 and x5 > 0 => NO Solution case 2: x> 0 and x5 < 0 => 0 < x < 5 So, X < 5 So, SUFFICIENT So, Answer will be D Hope it helps!
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Originally posted by BrushMyQuant on 18 Aug 2014, 09:28.
Last edited by BrushMyQuant on 10 Jun 2018, 00:26, edited 1 time in total.



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Re: Is x < 5?
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02 May 2017, 13:39
Could someone help me understand why inequality x(x5) < 0 gives us x>0 and not x<0?



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Re: Is x < 5?
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02 May 2017, 15:31
agree with the above statement. 1. can be rewritten as x<4. 2. x is definitely <5, but can't be negative, otherwise, it wouldn't be true 2. therefore 0<x<5 we have for each statement a definite answer.
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Re: Is x < 5?
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01 Jun 2017, 23:40
Bunuel wrote: Is x < 5?
(1) x^2 < 16. Since both sides are nonnegative, then we can take the square root from: x < 4. Sufficient.
(2) x(x  5) < 0. The "roots" are 0 and 5 and "<" sign indicates that the solution lies between the roots, so 0 < x < 5. Sufficient.
Answer: D. hey, do we include negative terms also in first condition? in X^2 the value of x can be negative also. pls, explain. thank you.



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Re: Is x < 5?
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02 Jun 2017, 03:50
skysailor wrote: Could someone help me understand why inequality x(x5) < 0 gives us x>0 and not x<0? x (x  5) < 0; we have two cases ; x < 0 & x  5 > 0 or x > 5 ; You can try some values in this range and you will see that this inequality will not hold true. x > 0 & x < 5 ; or 0 < x <5 ; solutions are between 0 and 5.
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Re: Is x < 5?
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08 Jun 2018, 12:53
Bunuel wrote: Is x < 5?
(1) x^2 < 16. Since both sides are nonnegative, then we can take the square root from: x < 4. Sufficient.
(2) x(x  5) < 0. The "roots" are 0 and 5 and "<" sign indicates that the solution lies between the roots, so 0 < x < 5. Sufficient.
Answer: D. Hey abhimahna, how do we know the range by just looking at "x(x  5) < 0" without having to consider the 2 scenarios (i.e. 1) x<0 & (x5)>0 2) x>0 & (x5)<0) as Bunuel mentioned? Is there a trick for it and can it apply to other situations as the example mentioned below? Thanks! How about x(x5)>0? Case 1 (Both +): x>0, x5>0 x>5 Case 2 (Both ): x<0,x5<0 x<0
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Re: Is x < 5?
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08 Jun 2018, 13:16
dabaobao wrote: Bunuel wrote: Is x < 5?
(1) x^2 < 16. Since both sides are nonnegative, then we can take the square root from: x < 4. Sufficient.
(2) x(x  5) < 0. The "roots" are 0 and 5 and "<" sign indicates that the solution lies between the roots, so 0 < x < 5. Sufficient.
Answer: D. Hey abhimahna, how do we know the range by just looking at "x(x  5) < 0" without having to consider the 2 scenarios (i.e. 1) x<0 & (x5)>0 2) x>0 & (x5)<0) as Bunuel mentioned? Is there a trick for it and can it apply to other situations as the example mentioned below? Thanks! How about x(x5)>0? Case 1 (Both +): x>0, x5>0 x>5Case 2 (Both ): x<0,x5<0 x<0Hey dabaobaoYou have made a mistake in the highlighted portion If x>0 and x>5, then the solution will be x > 0 Similarly, if x < 0 and x < 5, the solution is x < 5. Therefore, the solution to this inequality must be 0 < x < 5 You might want to read the following link https://gmatclub.com/forum/inequalities ... 91482.htmlHope this helps you!
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Re: Is x < 5?
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15 Jun 2018, 08:27
pushpitkc wrote: dabaobao wrote: Bunuel wrote: Is x < 5?
(1) x^2 < 16. Since both sides are nonnegative, then we can take the square root from: x < 4. Sufficient.
(2) x(x  5) < 0. The "roots" are 0 and 5 and "<" sign indicates that the solution lies between the roots, so 0 < x < 5. Sufficient.
Answer: D. Hey abhimahna, how do we know the range by just looking at "x(x  5) < 0" without having to consider the 2 scenarios (i.e. 1) x<0 & (x5)>0 2) x>0 & (x5)<0) as Bunuel mentioned? Is there a trick for it and can it apply to other situations as the example mentioned below? Thanks! How about x(x5)>0? Case 1 (Both +): x>0, x5>0 x>5Case 2 (Both ): x<0,x5<0 x<0Hey dabaobaoYou have made a mistake in the highlighted portion If x>0 and x>5, then the solution will be x > 0 Similarly, if x < 0 and x < 5, the solution is x < 5. Therefore, the solution to this inequality must be 0 < x < 5You might want to read the following link https://gmatclub.com/forum/inequalities ... 91482.htmlHope this helps you! pushpitkc Thanks a lot bruh for that link. Remember studying that trick in high school but couldn't recall it. That trick makes it really easy to do such questions. Btw I don't think I made a mistake above. If we are looking at x(x5)>0, then the solution should be x>5 or x<0 (which is what I got). I believe you mixed up x(x5)>0 with x(x5)<0. Irrespective of that issue, I think we should always discard the less limiting inequality. Example 1: If x>0 and x>5, then solution is x>5 [x>0 is the less limiting inequality] Example 2: If x<0 and x<5, then solution is x<0 [x<5 is the less limiting inequality]
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