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Is \(x^5 > x^4\)? --> reduce by x^4: is \(x > 1\)?

(1) x^3 > −x --> \(x(x^2 + 1) > 0\). Since \(x^2 + 1\) is always positive, then the second multiple, x, must also be positive. So, this statement implies that x > 0. Not sufficient.

(2) 1/x < x --> multiply by x^2 (it has to be positive, so we can safely do that): \(x < x^3\) --> \(x(x^2 -1) > 0\) --> \((x + 1)x(x - 1) > 0\) --> \(-1 < x < 0\) or \(x > 1\). Not sufficient.

(1)+(2) Intersection of ranges from (1) and (2) is \(x > 1\). Sufficient.

Bunuel, can you please explain this :"x < x^3 --> x(x^2 -1) > 0 --> (x + 1)x(x - 1) > 0 --> -1 < x < 0 or x > 1. Not sufficient."

" i understand that " (x + 1)x(x - 1) > 0" and then the three roots are = 1,-1,0. Since the inequality is ">", x is " outside these values. But how did we come to "-1 < x < 0 or x > 1."

I came to : x(x+1)(x-1)>0 either x>0 or (x+1)(x-1)>0

and then im lost. Please Help
_________________

'I read somewhere... how important it is in life not necessarily to be strong, but to feel strong... to measure yourself at least once.'

Bunuel, can you please explain this :"x < x^3 --> x(x^2 -1) > 0 --> (x + 1)x(x - 1) > 0 --> -1 < x < 0 or x > 1. Not sufficient."

" i understand that " (x + 1)x(x - 1) > 0" and then the three roots are = 1,-1,0. Since the inequality is ">", x is " outside these values. But how did we come to "-1 < x < 0 or x > 1."

I came to : x(x+1)(x-1)>0 either x>0 or (x+1)(x-1)>0

and then im lost. Please Help

(x + 1)x(x - 1) > 0 --> the "roots", in ascending order, are -1, 0 and 1, this gives us 4 ranges:

x < -1; -1 < x < 0; 0 < x < 1; x > 1.

Next, test some extreme value for x: if x is some large enough number, say 10, then all three multiples will be positive which gives the positive result for the whole expression, so when x>1, the expression is positive. Now the trick: as in the 4th range expression is positive, then in the 3rd it'll be negative, in the 2nd it'll be positive and finally in the 1st it'll be negative: - + - + . So, the ranges when the expression is positive are: -1 < x < 0 and x > 1.

It can also quickly be solved thinking about ranges.

Is \(

x^5

>

x^4

\)?

If you think about it, the question is implicitly asking whether a value is greater than one. Indeed every other option (numbers smaller than -1; numbers between -1 and 0; numbers between 0 and 1) if plugged into the expression results in a false statement.

1) picking numbers you can figure out that:

I. the inequality is satisfied when our value is greater than zero. ( pick -1/3. (-1/3)^3 is never gonna be greater than [-(-1/3)]. Same reasoning holds for numbers <= than -1) II. the inequality is satisfied for both proper fractions and values greater than 1.

NS

2) plugging in numbers you can figure out that the inequality is satisfied for values between both -1 and 0 and greater than 1.

NS

1+2) First statement rules out any value smaller than zero. second statement rules out any proper fraction. Our overlapping result is going to be a value grater than one, making us able to claim an answer.

C.

Hope it helps gmat6nplus1.
_________________

learn the rules of the game, then play better than anyone else.

Is \(x^5 > x^4\)? --> reduce by x^4: is \(x > 1\)?

(1) x^3 > −x --> \(x(x^2 + 1) > 0\). Since \(x^2 + 1\) is always positive, then the second multiple, x, must also be positive. So, this statement implies that x > 0. Not sufficient.

(2) 1/x < x --> multiply by x^2 (it has to be positive, so we can safely do that): \(x < x^3\) --> \(x(x^2 -1) > 0\) --> \((x + 1)x(x - 1) > 0\) --> \(-1 < x < 0\) or \(x > 1\). Not sufficient.

(1)+(2) Intersection of ranges from (1) and (2) is \(x > 1\). Sufficient.

Answer: C.

Hi Bunuel

I was able to solve the ques by plugging in values but can you explain how did x < x^3 become x(x^2-1) > 0 how did you know x^3 would be a negative quantity and took it LHS and change the inequality sign? Please explain.
_________________

The given question can be re-written as as \(x^{5}-x^{4} >0\)----> \(x^{4}*(x-1)>0\) Now we know that \(x^{4}\geq{0}\) so we need to know whether x=0 or \(x\neq{0}\)

St 1 says \(x^{3}+x>0\) or \(x(x^2+1)>0\)...If this true we can safely say that \(x\neq{0}\) and \(x>0\) Now if x=2 the above expression \(x^{4}*(x-1)>0\) is true but if x=1/2 then \(x^{4}*(x-1)>0\) is false

Option A and D ruled out

St 2 says 1/x < x or \(\frac{(x^2-1)}{x} >0\)

Now this expression is true when Numerator and Denominator are of same sign..

Case 1: When N and D are positive so we have x>0 and\(x^{2}\) > 1 or |x|>1 or x>1 or x<-1 but we know x>0 so we have x>1..For x>1 the expression is always positive Case 2: When N and D are of negative sign so we have x<0 and\(x^{2}<1\) or |x|<1 or -1<x<1...but if we know x<0 so we have x in the range -1<x<0..

So if x=-1/2 the expression \(x^{4}*(x-1)>0\) is false Thus from St2 also we have 2 answers possible..

Combining both statements we get that x>1 and the expression \(x^{4}*(x-1)>0\) is true for all values x>1

Ans C
_________________

“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”

Is \(x^5 > x^4\)? --> reduce by x^4: is \(x > 1\)?

(1) x^3 > −x --> \(x(x^2 + 1) > 0\). Since \(x^2 + 1\) is always positive, then the second multiple, x, must also be positive. So, this statement implies that x > 0. Not sufficient.

(2) 1/x < x --> multiply by x^2 (it has to be positive, so we can safely do that): \(x < x^3\) --> \(x(x^2 -1) > 0\) --> \((x + 1)x(x - 1) > 0\) --> \(-1 < x < 0\) or \(x > 1\). Not sufficient.

(1)+(2) Intersection of ranges from (1) and (2) is \(x > 1\). Sufficient.

Answer: C.

Hi Bunuel

I was able to solve the ques by plugging in values but can you explain how did x < x^3 become x(x^2-1) > 0 how did you know x^3 would be a negative quantity and took it LHS and change the inequality sign? Please explain.

x < x^3 0 < x^3 - x 0 < x(x^2 - 1), which is the same as x(x^2 - 1) > 0 .
_________________

You are asked is \(x^5 > x^4\) ---> \(x^5-x^4>0\)---> \(x^4(x-1) > 0\) ----> as \(x^4 > 0\) for all x ---> the question basically asks us whether x>1 ?

Per statement 1,\(x^3 > −x\) ---> \(x^3 + x> 0\) ---> x(x^2+1)>0 ---> x>0 (as \(x^2+1 > 0\) for all x) but is x>1 ? no definite answer. Not sufficient.

Per statement 2, 1/x < x ---> 1/x - x<0 ---> (1-x^2)/x < 0 ---> (x^2-1)/x > 0 ---> (x+1)(x-1)/x > 0 ---> -1<x<0 or x>1 , again not sufficient to give you a definite answer.

Combining the 2 statements you get, x>1, thus giving a definite yes to the question asked.

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^5 > x^4?

(1) x^3 > −x (2) 1x < x

When it comes to an inequality, when a range of 1) square and 2) que includes a range of con, the con is sufficient. These two facts are important. When you modify the original condition and the question, it becomes x^5-x^4>0?, x^4(x-1)>0?, x-1>0?, x>1?.(As x^4 is a positive number, the direction of the sign is not changed when the both equations are divided.) There is 1 variable(x), which should match with the number of equations. So you need 1 equation. For 1) 1 equation, for 2) 1 equation, which is likely to make D the answer. For 1), x^2+1>0 is always valid from x^3+x>0, x(x^2+1)>0, which makes x>0. Since the range of que doesn’t contain the range of con, it is not sufficient. For 2), -1<x<0, 1<x is derived from x<x^3(the direction of the sign doesn’t change when x^2 is multiplied to the both equations. When it comes to an inequality, square only matters.), 0<x^3-x, 0<x(x^2-1), 0<x(x-1)(x+1). Since the range of que doesn’t contain the range of con, it is not sufficient. When 1) & 2), from x>1, the range of que contains the range of con, which is sufficient. Therefore the answer is C.

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.
_________________

Is x^5 > x^4? Is x^5 - x^4 > 0? Is x^4(x - 1) > 0?

We ignore the even powers because x^4 will never be negative. But we need to ensure that x should not be 0 for the inequality to hold.

Is (x - 1) > 0? Is x > 1?

Statement 1: x^3 > −x

x^3 + x > 0 x(x^2 + 1) > 0 x^2 + 1 will always be positive. So we see that x > 0. But we need to know whether it is greater than 1. Greater than 0 could be 0.5 or it could be 2 (or any other number). So this is not sufficient to say whether x will eb greater than 1. Not sufficient.

(2) 1/x < x When the sign of x is not known, we do not multiply by x. We bring it to the o there side. (1/x) - x < 0 (1 - x^2)/x < 0 (x^2 - 1)/x > 0 (x + 1)(x - 1)/x > 0 The transition points on the number line are -1, 0, and 1. The expression will be positive when x > 1 or -1 < x < 0. This alone is also not sufficient.

Both together we know that x is positive and x > 1 or -1 < x < 0. So x > 1 is a must. Sufficient.

Is \(x^5 > x^4\)? --> reduce by x^4: is \(x > 1\)?

(1) x^3 > −x --> \(x(x^2 + 1) > 0\). Since \(x^2 + 1\) is always positive, then the second multiple, x, must also be positive. So, this statement implies that x > 0. Not sufficient.

(2) 1/x < x --> multiply by x^2 (it has to be positive, so we can safely do that): \(x < x^3\) --> \(x(x^2 -1) > 0\) --> \((x + 1)x(x - 1) > 0\) --> \(-1 < x < 0\) or \(x > 1\). Not sufficient.

(1)+(2) Intersection of ranges from (1) and (2) is \(x > 1\). Sufficient.

How can you divide the equation by x^4 throughout ? As we don't know the value of x. We can't do that right? what if x was 0 , then we can't reduce a equation by 0

Please let me know if I am wrong. Eager to hear back from you

Is \(x^5 > x^4\)? --> reduce by x^4: is \(x > 1\)?

(1) x^3 > −x --> \(x(x^2 + 1) > 0\). Since \(x^2 + 1\) is always positive, then the second multiple, x, must also be positive. So, this statement implies that x > 0. Not sufficient.

(2) 1/x < x --> multiply by x^2 (it has to be positive, so we can safely do that): \(x < x^3\) --> \(x(x^2 -1) > 0\) --> \((x + 1)x(x - 1) > 0\) --> \(-1 < x < 0\) or \(x > 1\). Not sufficient.

(1)+(2) Intersection of ranges from (1) and (2) is \(x > 1\). Sufficient.

How can you divide the equation by x^4 throughout ? As we don't know the value of x. We can't do that right? what if x was 0 , then we can't reduce a equation by 0

Please let me know if I am wrong. Eager to hear back from you

If it were x^5 < x^4, then after dividing by x^4 we'd get x < 1 but x cannot be 0, so it would be x< 0 or 0 < x < 1.

x^5 > x^4 gives x > 1, so it's redundant to mention that x cannot be 0 because we got x > 1.
_________________

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