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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 --> \((x=-\frac{1}{2})<(x^3=-\frac{1}{8})<(x^2=\frac{1}{4})\).

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 \(1>x\).

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 OG Pls 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.

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Is x > 0 ?

(1) x < x^2 (2) x < x^3

If the range of the question covers that of the condition, the condition is sufficient There is one variable (x) in the original condition and 2 equations are given by the conditions, so there is high chance (D) will be the answer For condition 1, x<x^2, 0<x^2-x, 0<x(x-1)--> x<0, 1<x, and this is insufficient as this range is not included in the question, For condition 2, x<x^3, 0<x^3-x, 0<x(x-1)(x+1) --> -1<x<0, 1<x is also not sufficient as this is also not included in the question Looking at the conditions together, 1<x<0, 1<x is also insufficient, as the range of the question does not include that of the conditions Hence, the answer becomes (E).

For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E. _________________

(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!

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.

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. _________________

------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

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.

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) \(x<x^2\) --> \(x*(x-1)>0\) --> \(x<0\) or \(x>1\) --> ---------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.

(2) \(x<x^3\) --> \(x(x-1)(x+1)>0\) --> \(-1<x<0\) 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.

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) \(x<x^2\) --> \(x*(1-x)>0\) --> \(x<0\) or \(x>1\) --> ---------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.

(2) \(x<x^3\) --> \(x(x-1)(x+1)>0\) --> \(-1<x<0\) 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 \(x<x^2\) --> \(x*(1-x)>0\) --> \(x<0\) or \(x>1\).. From \(x*(1-x)>0\) can't we consider both x and (x-1) to be positive also ?

Similarly for \(x(x-1)(x+1)>0\) --> \(-1<x<0\) or \(x>1\), I seem to be missing your logic.

Could you please explain why you're considering only the negative factors while considering \(x<x^2\) --> \(x*(1-x)>0\) --> \(x<0\) or \(x>1\).. From \(x*(1-x)>0\) can't we consider both x and (x-1) to be positive also ?

Similarly for \(x(x-1)(x+1)>0\) --> \(-1<x<0\) or \(x>1\), I seem to be missing your logic.

Thanks

Actually, both cases were considered.

\(x<x^2\) --> \(x*(x-1)>0\) x>0 and x-1>0 (x>1) --> x>1; x<0 and x-1<0 (x<1) --> x<0.

Thus, \(x<x^2\) holds true for \(x<0\) and \(x>1\).

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 OG Pls help in getting my concept right.

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 non-negative (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 non-negative.

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: \(x^3<y^3\).

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