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Is x > 0 ?
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21 Sep 2010, 08:14
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Is x > 0 ? (1) x < x^2 (2) x < x^3 Can some one explain how do we eliminate choice "C" for this DS question using plug in values thanks
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Re: DS Algebra
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21 Sep 2010, 08:35
(1) x < x^2 so x can be less than 1 and greater than 1 ! insufficient! (2) x < x^3 so x is greater than 1 so it can be negative or positive insufficent!
(1) and (2) only said than x is greater than 1 so x can be positive or negative.... insufficient!
ANS:. E



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Re: DS Algebra
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21 Sep 2010, 08:36
rxs0005 wrote: Is x > 0
x < x^2
x < x^3
Can some one explain how do we eliminate choice "C" for this DS question using plug in values
thanks On DS questions when plugging numbers, goal is to prove that the statement is not sufficient. So we should try to get a YES answer with one chosen number(s) and a NO with another. Question: is \(x>0\)? When considering statements together we can see that YES answer is easy to get: \(x=2>0\), so answer YES > \((x=2)<(x^2=4)<(x^3=8)\); Now, for NO answer we should pick negative \(x\) > ANY negative \(x\) would be less than \(x^2\) as \(x^2\) would be positive, so satisfying (1) is easy. Next, in order negative \(x\) to be less than \(x^3\), it shoul be a fraction in the range (1,0) > \(x=\frac{1}{2}<0\), so answer NO > \((x=\frac{1}{2})<(x^3=\frac{1}{8})<(x^2=\frac{1}{4})\). We have two different answers, not sufficient. Answer: E.
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Re: DS Algebra
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21 Sep 2010, 09:22
Bunuel thanks for your patience with your explanation ! cheers R
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Re: DS Algebra
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21 Sep 2010, 13:33
Can't I write first equation x < x2 as 1 < x.
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Re: DS Algebra
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21 Sep 2010, 13:45
onedayill wrote: Can't I write first equation
x < x2 as 1 < x. I'm assuming you are trying to divide both sides by x. The problem with an inequality is that you don't know if x is positive or negative. If it is negative, you need to change the sign to greater than in this case.



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Re: DS Algebra
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21 Sep 2010, 14:32
rxs0005 wrote: Is x > 0
x < x^2
x < x^3
Can some one explain how do we eliminate choice "C" for this DS question using plug in values
thanks Can I deduce as following : 1) \(x < x^2\) holds true for x > 1 and x < 0 Thus, not sufficient 2) \(x < x^3\) holds true for x > 0 and 0 < x < 1 Not sufficient No unique value is determined when combining 1 & 2. Thus, E. Please correct me if I am incorrect in the above logic.



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Re: DS Algebra
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21 Sep 2010, 15:16
Orange08 wrote: rxs0005 wrote: Is x > 0
x < x^2
x < x^3
Can some one explain how do we eliminate choice "C" for this DS question using plug in values
thanks Can I deduce as following : 1) \(x < x^2\) holds true for x > 1 and x < 0 Thus, not sufficient 2) \(x < x^3\) holds true for x > 0 and 0 < x < 1 Not sufficient No unique value is determined when combining 1 & 2. Thus, E. Please correct me if I am incorrect in the above logic. You are correct. Statement 1: Positive fractions will never make this statement true. The square of a positive fraction is always smaller than the original fraction itself. Statement 2: Only positive numbers greater than 1 make this statement true. Therefore, by combining them, there is only one possibility to make both true: Numebrs greater than 1, which allows us to answer the original question.
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Re: DS Algebra
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21 Sep 2010, 23:43



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Re: DS Algebra
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21 Sep 2010, 23:46
jallenmorris wrote: Orange08 wrote: rxs0005 wrote: Is x > 0
x < x^2
x < x^3
Can some one explain how do we eliminate choice "C" for this DS question using plug in values
thanks Can I deduce as following : 1) \(x < x^2\) holds true for x > 1 and x < 0 Thus, not sufficient 2) \(x < x^3\) holds true for x > 0 and 0 < x < 1Not sufficient No unique value is determined when combining 1 & 2. Thus, E. Please correct me if I am incorrect in the above logic. You are correct. Statement 1: Positive fractions will never make this statement true. The square of a positive fraction is always smaller than the original fraction itself. Statement 2: Only positive numbers greater than 1 make this statement true. Therefore, by combining them, there is only one possibility to make both true: Numebrs greater than 1, which allows us to answer the original question. No your ranges for (2) are not correct, it should be: \(1<x<0\) or \(x>1\). Negative fractions from the range (1,0) also make statement (2) true. Below is complete solution: Is \(x>0\)?(1) \(x<x^2\) > \(x*(x1)>0\) > \(x<0\) or \(x>1\) > 01. Not sufficient as if \(x\) is in the green zone answer is YES but if \(x\) is in the red zone answer is NO. (2) \(x<x^3\) > \(x(x1)(x+1)>0\) > \(1<x<0\) or \(x>1\) > (1)(0)(1). Not sufficient as if \(x\) is in the green zone answer is YES but if \(x\) is in the red zone answer is NO. (1)+(2) Intersection of the ranges from (1) and (2) is: (1)(0)(1)Again not sufficient as if \(x\) is in the green zone answer is YES but if \(x\) is in the red zone answer is NO. Answer: E. Hope it helps.
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Re: DS Algebra
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22 Sep 2010, 06:26
You're right. I hadn't thought of negative fractions. 1/2 ^3 = 1/8, which is greater than 1/2. Thanks.
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Re: DS Algebra
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16 Oct 2010, 08:39
+1 for E...took more than 3 mins though..



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Re: DS Algebra
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01 Nov 2012, 13:49
Bunuel wrote:
You are correct.
Statement 1: Positive fractions will never make this statement true. The square of a positive fraction is always smaller than the original fraction itself.
Statement 2: Only positive numbers greater than 1 make this statement true. Therefore, by combining them, there is only one possibility to make both true: Numebrs greater than 1, which allows us to answer the original question.
No your ranges for (2) are not correct, it should be: \(1<x<0\) or \(x>1\). Negative fractions from the range (1,0) also make statement (2) true. Below is complete solution: Is \(x>0\)?(1) \(x<x^2\) > \(x*(1x)>0\) > \(x<0\) or \(x>1\) > 01. Not sufficient as if \(x\) is in the green zone answer is YES but if \(x\) is in the red zone answer is NO. (2) \(x<x^3\) > \(x(x1)(x+1)>0\) > \(1<x<0\) or \(x>1\) > (1)(0)(1). Not sufficient as if \(x\) is in the green zone answer is YES but if \(x\) is in the red zone answer is NO. (1)+(2) Intersection of the ranges from (1) and (2) is: (1)(0)(1)Again not sufficient as if \(x\) is in the green zone answer is YES but if \(x\) is in the red zone answer is NO. Answer: E. Hope it helps.[/quote] Hi Bunuel, Could you please explain why you're considering only the negative factors while considering \(x<x^2\) > \(x*(1x)>0\) > \(x<0\) or \(x>1\).. From \(x*(1x)>0\) can't we consider both x and (x1) to be positive also ? Similarly for \(x(x1)(x+1)>0\) > \(1<x<0\) or \(x>1\), I seem to be missing your logic. Thanks



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Re: DS Algebra
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01 Nov 2012, 14:49
Bunuel wrote: onedayill wrote: Can't I write first equation
x < x2 as 1 < x. Never multiply (or reduce) an inequality by variable (or by an expression with variable) if you don't know the sign of it or are not certain that variable (or expression with variable) doesn't equal to zero.So you can not divide both parts of inequality \(x<x^2\) by \(x\) as you don't know the sign of this unknown: if \(x>0\) you should write \(1<x\) BUT if \(x<0\) you should write \(1>x\). Hope it helps. Hi Bunnuel With ref to your above post i would like to contradictory expalnation from DS section of OG 13 question no 52 TO make things easier i ll post the question If X and Y are positive , Is X<10<Y? 1. X<Y and XY = 100 2. X^2 <100< Y^2 In the explanation part 2nd stem is directly reduced to X<10<Y Isnt your explanation contradictory to above expalantion of OGPls help in getting my concept right.



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Re: DS Algebra
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01 Nov 2012, 14:55
Archit143 wrote: Bunuel wrote: onedayill wrote: Can't I write first equation
x < x2 as 1 < x. Never multiply (or reduce) an inequality by variable (or by an expression with variable) if you don't know the sign of it or are not certain that variable (or expression with variable) doesn't equal to zero.So you can not divide both parts of inequality \(x<x^2\) by \(x\) as you don't know the sign of this unknown: if \(x>0\) you should write \(1<x\) BUT if \(x<0\) you should write \(1>x\). Hope it helps. Hi Bunnuel With ref to your above post i would like to contradictory expalnation from DS section of OG 13 question no 52 TO make things easier i ll post the question If X and Y are positive , Is X<10<Y? 1. X<Y and XY = 100 2. X^2 <100< Y^2 In the explanation part 2nd stem is directly reduced to X<10<Y Isnt your explanation contradictory to above expalantion of OGPls help in getting my concept right. First of all, they are not reducing, they are taking square root. Also, notice that the stem says that x and y are positive numbers. P.S. This question is discussed here: ifxandyarepositiveisx10y139873.html
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Re: Is x > 0 ?
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01 Nov 2012, 15:07
Thanks BUnnel really missed the statement.Sorry
But is it ok to take square root,considering signs are unknown in above scenario.



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Re: Is x > 0 ?
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01 Nov 2012, 15:12
Archit143 wrote: Thanks BUnnel really missed the statement.Sorry
But is it ok to take square root,considering signs are unknown in above scenario. If we were not told that x and y are positive numbers, then from x^2<100<y^2, we would have x<10<y (since \(\sqrt{x^2}=x\)). Generally: A. We can raise both parts of an inequality to an even power if we know that both parts of an inequality are nonnegative (the same for taking an even root of both sides of an inequality).For example: \(2<4\) > we can square both sides and write: \(2^2<4^2\); \(0\leq{x}<{y}\) > we can square both sides and write: \(x^2<y^2\); But if either of side is negative then raising to even power doesn't always work. For example: \(1>2\) if we square we'll get \(1>4\) which is not right. So if given that \(x>y\) then we can not square both sides and write \(x^2>y^2\) if we are not certain that both \(x\) and \(y\) are nonnegative. B. We can always raise both parts of an inequality to an odd power (the same for taking an odd root of both sides of an inequality).For example: \(2<1\) > we can raise both sides to third power and write: \(2^3=8<1=1^3\) or \(5<1\) > \(5^2=125<1=1^3\); \(x<y\) > we can raise both sides to third power and write: \(x^3<y^3\). Hope it helps.
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Re: Is x > 0 ?
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13 Jan 2013, 21:50
I don't think that we need plugin here. It has very easy algebraical solution "solvable" within 80 sec
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Re: Is x > 0 ?
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15 Jan 2013, 07:22
rxs0005 wrote: Is x > 0 ? (1) x < x^2 (2) x < x^3
1. x < x^2 Test x = 2: 2 < 4 OK! Test x = 1: 1 < 1 OK! Test x = 1/4: 1/4 < 1/16 OK! INSUFFICIENT! 2. x < x^3 Test x = 2: 2 < 8 OK! Test x = 1: 1 = 1 NOT OK! Test x = 1/4: 1/4 < 1/64 OK! INSUFFICIENT! Together: When x is negative fraction or x is greater than 2, both inequalities work! Answer: E
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