Data Sufficiency

Please help solve this exponent problem from Gmat Prep
and in general how to solve such problem.
Also how much time should this problem take

Thanks

1947 wrote:Please help solve this exponent problem from Gmat Prep
and in general how to solve such problem.
Also how much time should this problem take

Thanks

Here if X is positive: that means it can be either any positive integer or any fraction!
like x = 2, 1/2, 3,4,...etc

so, you consider various values of x for all the three expressions: 1/x,2x,X^2

you'll see for all the fractions: 1/x > 2x > X^2 but same won't be true for the integers
e.g: say X = 2, 1/2,4, 4...clearly fails!

Hence as we don't know the type of positive X values, we cannot conclude the answer as 1/x>2x>x^2!

Answer: NONE.

On something like this I would use a combination of plugging in numbers and solving the inequalities algebraically.

I. x^2<2x<1/x. This is true for x=1/2.

II. x^2<1/x<2x. This is true for x=7/8.

As x gets close to 1, 1/x gets close to 1, but 2x gets close to 2, so picking something close to 1 to plug in makes the truth of the inequality more obvious. You could also separate the inequalities out and solve them separately:

x^2<1/x AND 1/x<2x

We have the luxury of knowing x>0, so we don't have to worry about flipping around inequality signs.

x^2<1/x
x^3<1

This true only for 0<x<1 if we know for sure that x is positive.

1/x<2x

1<2x^2
1/2<x^2
x^2>1/2
x>sqrt(2)/2
This is approximately 0.7, so clearly there are some values that satisfy both inequalities.

III. Separate:

2x<x^2
2<x
x>2

x^2<1/x
x^3<1
0<x<1

to satisfy both inequalities x would have to be both greater than 2 and between 0 and 1. this is impossible, so III is impossible.

So Ans:

D

This problem should go fairly fast if you're able to make some intelligent guesses to verify I and II, and the algebra confirming it is pretty fast because we don't have to worry about x being negative.

Hi Pete,
I don't understand how you have solved this problem(Bit difficult to grasp it)
Can you please explain why don't we take a scenario where any of the conditions are satisfied with any positive number...including 0<x<1 and 1 < x < infinity.

Please explain if there is some conceptual problem in my doubt!

GmatMathPro wrote:On something like this I would use a combination of plugging in numbers and solving the inequalities algebraically.

x^2<1/x
x^3<1

This true only for 0<x<1 if we know for sure that x is positive.


x^2<1/x
x^3<1
0<x<1

D

How do you solve x^3<1 means that 0<x<1
please help on this.

i know if x^2<1 then we can factor (x-1)(x+1) but how to do this for x^3-1

Thanks

@1947

You can factor x^3-1 as (x-1)(x^2+x+1). This gets messy though with inequalities. It is simpler to use numbers. In one case, consider 0<x<1 and in the other case x>2

Yes, you can factor differences of cubes with the following formula:

a^3-b^3=(a-b)(a^2+ab+b^2). In this case a=x and b=1 and it works out to (x-1)(x^2+x+1) as sl750 said.

So you would have (x-1)(x^2+x+1)<0. x^2+x+1 is always positive and is never zero(do you see why?), so this inequality would be satisfied only if x-1<0 or x<1. In this case, we know that x is also positive, hence the additional restriction 0<x<1.

However, I do NOT recommend this approach for the GMAT. I believe, though I am not totally sure, that the GMAT does not expect you to know how to factor expressions like these.

Easier ways to solve it:

1. Think about it. If you cube a number bigger than 1, it gets bigger. If you cube a number between 0 and 1 it gets smaller. If you cube any negative number it remains negative. If you cube zero you get zero. All these taken together indicates that all numbers less than 1 satisfy the inequality x^3<1. Then add in the restriction that x is positive to get 0<x<1.

2. Take the cube root of both sides. x^3<1 -----> x<1. Add in the restriction that x is positive to get 0<x<1. You're allowed to take the cube root of both sides of an inequality without changing the inequality. However, this type of thinking does NOT apply to something like x^2<1, where the exponent is even, so be careful not to apply it too broadly.

I posted an efficient approach here:

https://www.beatthegmat.com/need-explanation-t85192.html

battlefield wrote:Hi Pete,
I don't understand how you have solved this problem(Bit difficult to grasp it)
Can you please explain why don't we take a scenario where any of the conditions are satisfied with any positive number...including 0<x<1 and 1 < x < infinity.

Please explain if there is some conceptual problem in my doubt!

Hi. I'm not 100% sure I understand your question, but let me know if this helps. The problem with just checking if numbers from 0<x<1 and 1<x<infinity work is that not all numbers from those regions will work out the same way. x=1/2 gives us one way of ordering the expressions, whereas x=7/8 gives us another way. So if you just plug in one number from 0<x<1, you might think that whatever answer you get will be true for EVERY value of 0<x<1, but that is not the case.

Look at the following graph where I have plotted y=x^2 in blue, y=1/x in red, and y=2x in green and we are looking only at the part of the graph where x>0:


Blue(B)=x^2, Red(R)=1/x, and Green(G)=2x. Notice that when x is just slightly bigger than 0, the blue graph is the lowest, green is in the middle, and red is highest. B<G<R or x^2<2x<1/x. Then, the green and red graphs cross a little short of x=1 and it becomes B<R<G or x^2<1/x<2x. Then, at x=1, the blue and red graphs cross and it becomes R<B<G or 1/x<x^2<2x. Finally, the green and blue graphs cross at x=2 and it becomes R<G<B or 1/x<2x<x^2. To summarize, the possibilities, if x is positive, are:

x^2<2x<1/x
x^2<1/x<2x
1/x<x^2<2x
1/x<2x<x^2.

The key thing to realize is that you shouldn't assume that 0<x<1 and 1<x<infinity are the ONLY regions you need to consider. Notice that the inequalities changed every time the graphs intersect. They intersect when the expressions are equal to each other. If you want to solve this by plugging in numbers from different zones on the number line, you could do the following:

Set each pair of equations equal to each other and determine intersection points(keeping in mind that we are only interested in solutions where x>0:

x^2=1/x ---> x^3=1 ---->
x=1

x^2=2x
x^2-2x=0
x(x-2)=0
x=2

2x=1/x
2x^2=1
x^2=1/2
x=sqrt(1/2)
x=sqrt(2)/2

Let these three values determine the regions from which you are choosing numbers to plug in:

0<x<sqrt(2)/2, sqrt(2)/2<x<1, 1<x<2, 2<x<infinity.

Plug in numbers from each of these zones to the expressions listed, and see which of the choices are valid.

GMATGuruNY wrote:I posted an efficient approach here:

https://www.beatthegmat.com/need-explanation-t85192.html

Hi Mitch,

The solution provided is fine.
But it will surely take atleast 3 mins.
Is there any other easier way of dealing with this one?

Many thanks..

Look at each choice and determine if all three inequalities lead to x ranges that overlap. This can be done because the questions ask which could be true, meaning they aren't always true but are for at least one x.

Writing this takes time. Doing this with a pencil and paper takes a few seconds.

I. x^2<2x<1/x

A. x^2 < 2x .....divide by x....x < 2

B. x^2<1/x ....multiply by x... x^3 < 1 ...cube root...x < 1 (contains some vales of x contained in A., so both can be true.)

C. 2x < 1/x (multiply by x) 2x^2 < 1 ... divide by 2 ... x^2 < 1/2 ...square root... x < sqrt (1/2) that could be + or - 1.4 and change, and x < 1.4... (contains some values from A and B, so I is true.)

II. x^2 < 1/x < 2X

A. x^2 < 1/x multiply by x x^3 < 1 cube roots x < 1

B. x^2 < 2x divide by x x < 2 (contains some vales of x contained in A., so both can be true.)

C. 1/x < 2x multiply by x 1 < 2x^2 divide by 2 x^2 > 1/2 square root x > + 1.41 and x > - 1.41 (contains some values from A and B, so II is true.)

III. 2x < x^2 < 1/x

A. 2x < x^2 divide by x 2 < x
B. 2x < 1/x multiply by x 2x^2 < 1 divide by 2 x^2 < 1/2 square root x < +1.41 and x < -1.41 (x cant be greater than 2, while at the same time less than 1.41 and less than -1.41. III. is not possible.)

I noticed some people were able to insert math code in their posts to format the math. I tried using LaTeX, but that didn't work. How is it done in this forum?

thats a fast way to solve. Thanks