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M14-13

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Re: M14-13  [#permalink]

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New post 08 Feb 2018, 01:29
gmatc2018 wrote:
Hello all,

how can we just rewrite (1-2x)(1+x) as (2x-1)(x+1) ? How did we turn around the signs for the first braket?

Best


\((1 - 2x)(1 + x) \lt 0\);

Multiply by -1 and flip the sign since we are multiplying by negative value: \(-(1-2x )(x + 1) \gt 0\);

Next, since -(1-2x) = 2x-1, then we'd get \((2x - 1)(x + 1) \gt 0\).

Hope it's clear.
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Re: M14-13  [#permalink]

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New post 11 May 2018, 01:53
Dear Team,
In my opinion the answer for this question should be D

Q) If x is an integer is |x|>1

1.(1-2x)(1+x)<0
2.(1-x)(1+2x)<0
from question stem we have to find out whether x>1 or x<-1
From Statement 1 if we don't multiply by -1 and take critical key points
we have x=-1 and x=1/2
____________________________________________________________
+ (-1) - 1/2 +
As the inequality has a less than sign so valid region should be -1<x<1/2 so the value is neither greater than 1 nor less than -1 so the answer to the question is no
Similarly from statement 2 x=1 and x=-1/2 and if we take critical key points on a number line

___________________________________
+ (-1/2) - 1 +
So the valid region for the inequality will be -1/2<x<1 so the value is neither greater than 1 nor less than -1 so the answer to the question will be no and hence according to me Answer should be Option D. Please go through my analysis and let me know where i faltered.

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Re: M14-13  [#permalink]

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New post 11 May 2018, 03:51
Shikh27 wrote:
Dear Team,
In my opinion the answer for this question should be D

Q) If x is an integer is |x|>1

1.(1-2x)(1+x)<0
2.(1-x)(1+2x)<0
from question stem we have to find out whether x>1 or x<-1
From Statement 1 if we don't multiply by -1 and take critical key points
we have x=-1 and x=1/2
____________________________________________________________
+ (-1) - 1/2 +
As the inequality has a less than sign so valid region should be -1<x<1/2 so the value is neither greater than 1 nor less than -1 so the answer to the question is no
Similarly from statement 2 x=1 and x=-1/2 and if we take critical key points on a number line

___________________________________
+ (-1/2) - 1 +
So the valid region for the inequality will be -1/2<x<1 so the value is neither greater than 1 nor less than -1 so the answer to the question will be no and hence according to me Answer should be Option D. Please go through my analysis and let me know where i faltered.

Regards


Hi Shikh27

if you are not multiplying the inequality by -1, then note that coefficient of \(x^2\) in the inequality \((1-2x)(1+x)<0\) will be -1. So when you are plotting the zero points on the number line and assigning the signs to the right of the first root, it should be -ve, matching the coefficient of \(x^2\). so the signs should be -, +, - and you need to choose the negative region, implying x>1/2 or x<-1

hence your highlighted part is incorrect, giving you wrong range.
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M14-13  [#permalink]

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New post Updated on: 18 Jun 2018, 05:51
I don't agree with the explanation. Since it's given x is an integer,
1. x can have values 0 and 1.
|x|>1 no sufficient
2. x can have values 0 and -1.
|x|>1 no sufficient.

We don't necessarily need to prove |x|>1 as yes even if we can prove that |x|>1 as no, the statement will be sufficient.

Originally posted by Anishajha on 18 Jun 2018, 05:10.
Last edited by Anishajha on 18 Jun 2018, 05:51, edited 1 time in total.
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Re: M14-13  [#permalink]

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New post 18 Jun 2018, 05:19
Anishajha wrote:
I don't agree with the explanation. Since it's given x is an integer,
1. x can have values 0 and 1.
|x|>1 no sufficient
2. x can have values 0 and -1.
|x|>1 no sufficient.


Not sure what you are trying to say there but when we combine the statement we get that x < -1 and x > 1, which given an YES answer to the question.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: M14-13  [#permalink]

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New post 16 Sep 2018, 07:52
I felt that the solution was over complicated with the sign switching and all. Perhaps my approach is flawed but below is how I did it

st 1 says the product of 2 terms is negative, so one of the terms is negative and the other is positive
i.e. either (1-2x) is -ve and (1+x) is +ve
=> x>1/2 (Note that statement is already insufficient)
OR (1+x) is -ve and (1-2x) is +ve
=> x is <-1

similarly statement 2
Either (1-x) is -ve and (1+2x) is positive
=>x >1
OR (1+2x) is -ve and (1-x) is +ve
=> x<-1/2 ---> Insufficient

Together, combining the ranges, either x>1 or x< -1 ----> in either case |x|>1

Let me know if I'm flawed pls
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Re: M14-13 &nbs [#permalink] 16 Sep 2018, 07:52

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