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Re: Is |x| < 5? [#permalink]
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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.
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Is |x| < 5? [#permalink]
Expert Reply
STAT 1:
x^2 < 16
=> -4 < x < 4
So, |x| < 4 so will be less than 5
So, SUFFICIENT

STAT 2:
x(x-5) < 0
two cases
case 1: x < 0 and x-5 > 0 => NO Solution
case 2: x> 0 and x-5 < 0 => 0 < x < 5
So, |x| < 5
So, SUFFICIENT

So, Answer will be D
Hope it helps!

Watch the following video to learn How to Solve Inequality Problems



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How to Solve: Inequalities (Basic)
How to Solve: Inequality Problems - Algebra and Sine Wave/Wavy Method
How to Solve: Absolute Value (Basics)
How to Solve: Absolute Value Problems
How to Solve: Absolute Value + Inequality Problems

Originally posted by BrushMyQuant on 18 Aug 2014, 09:28.
Last edited by BrushMyQuant on 17 Jul 2021, 09:39, edited 5 times in total.
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Re: Is |x| < 5? [#permalink]
Could someone help me understand why inequality x(x-5) < 0 gives us x>0 and not x<0?
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Re: Is |x| < 5? [#permalink]
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.

D.
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Re: Is |x| < 5? [#permalink]
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? [#permalink]
skysailor wrote:
Could someone help me understand why inequality x(x-5) < 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? [#permalink]
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 & (x-5)>0 2) x>0 & (x-5)<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(x-5)>0?
Case 1 (Both +): x>0, x-5>0
x>5

Case 2 (Both -): x<0,x-5<0
x<0
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Re: Is |x| < 5? [#permalink]
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 & (x-5)>0 2) x>0 & (x-5)<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(x-5)>0?
Case 1 (Both +): x>0, x-5>0
x>5

Case 2 (Both -): x<0,x-5<0
x<0



Hey dabaobao

You 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.html

Hope 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(x-5)>0, then the solution should be x>5 or x<0 (which is what I got). I believe you mixed up x(x-5)>0 with x(x-5)<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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