Problem Solving

If x is positive, which of the following could be the correct ordering of 1/x,2x, and x^2

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

a. None
b. I only
c. III only
d. I and II
e I, II and III

I understand how one works with fractions but I'm confused that the answer is D I and II. I'm sure its something simple that i'm over thinking. Thanks

davidnodine wrote:If x is positive, which of the following could be the correct ordering of 1/x,2x, and x^2

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

a. None
b. I only
c. III only
d. I and II
e I, II and III

I understand how one works with fractions but I'm confused that the answer is D I and II. I'm sure its something simple that i'm over thinking. Thanks

Are you sure II = x^2<1/x<2x ?

Coz if you take a fraction like 3/2,
x^2 = 9/4
1/x = 2/3
2x = 3

So i'm getting the expression as 1/x<x^2<2x.

ratindasgupta wrote:

davidnodine wrote:If x is positive, which of the following could be the correct ordering of 1/x,2x, and x^2

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

a. None
b. I only
c. III only
d. I and II
e I, II and III

I understand how one works with fractions but I'm confused that the answer is D I and II. I'm sure its something simple that i'm over thinking. Thanks

Are you sure II = x^2<1/x<2x ?

Coz if you take a fraction like 3/2,
x^2 = 9/4
1/x = 2/3
2x = 3

So i'm getting the expression as 1/x<x^2<2x.

Just because one example works, doesn't mean he typed the problem incorrectly...

x=.9 would satisfy criterion II.

It is written correctly. Is there any way to look at this question without plugging in numbers?

Source of the question is: GMAT Prep Test 1

If x is positive which of the following could be correct ordering of 1/x, 2x, x^2.
1) x^2 < 2x < 1/x

2) x^2 < 1/x < 2x

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

Choices are:

1) None
2) 1 Only
3) 3 Only
4) I and 2 only
5) 1, 2, 3

How to proceed without plugging in?I plugged in in the exam and got only choice 1 as correct and hence marke db which is wrong:(

uptowngirl92 wrote:Source of the question is: GMAT Prep Test 1

If x is positive which of the following could be correct ordering of 1/x, 2x, x^2.
1) x^2 < 2x < 1/x

2) x^2 < 1/x < 2x

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

Choices are:

1) None
2) 1 Only
3) 3 Only
4) I and 2 only
5) 1, 2, 3

How to proceed without plugging in?I plugged in in the exam and got only choice 1 as correct and hence marke db which is wrong:(

Hybrid approach to make things smooth

"Could be true" Find one positive instance for each case. Proving that there are no positive instances requires algebraic way.

"must be true" Find one negative instance for each case. Proving that there are all positive instances requires algebraic way.

Lets focus on algebraic way.

1. x^2 < 2x < 1/x
x^2 - 2x < 0 --> (0,2)
2x < 1/x
x^2 - (1/2) < 0 --> (-1/sqrt(2), 1/sqrt(2))

Combining together: (0, 1/sqrt(2))

The domain of this inequality is the above.

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

x^3 < 1
domain: (-inf, 1)

and x^2 > 1/2
domain: (-inf, 1/sqrt(2)) U (1/sqrt(2), +inf)

Find the intersection of both domains: (-inf, 1/sqrt(2). Note that x is positive.
(0, 1/sqrt(2)) is the domain.

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

x^2 - 2x > 0
domain: (-inf, 0) U (2, +inf)


x^3 < 1
domain: (-inf, 1)

Intersection: (-inf, 0)

But x is positive.


Some key things to remember:

(x-a)(x-b) > 0, (-inf, a) U (b, +inf), assumin a < b
(x-a)(x-b) < 0 (a, b), assumin a < b
(x^3-a^3) > 0 (a, +inf)
(x^3+a^3) < 0 (-inf, a)

Being cool and taking the restctions (like x being +ve) into account help your test.

Smart numbers help you to get rid of some choices, thereby ending up with fewer choices. There, focus on algebraic approach.

Since we know X is positive, cant we multiply by X to get rid of the fraction and then compare?

x = 1/2 satisfies I

x = 3/4 satisfies II

If x = 3/4, then x^2 = 9/16, and 1/x = 4/3, and 2x = 3/2

9/16 < 4/3 < 3/2

So, both I and II are possible orders. If x is positive, however, it is impossible that x^2 < 1/x while simultaneously 2x < x^2.

Choose D.

Testluv wrote:x = 1/2 satisfies I

x = 3/4 satisfies II

If x = 3/4, then x^2 = 9/16, and 1/x = 4/3, and 2x = 3/2

9/16 < 4/3 < 3/2

So, both I and II are possible orders. If x is positive, however, it is impossible that x^2 < 1/x while simultaneously 2x < x^2.

Choose D.

But in order for a choice such as II to be correct, the choice has to be consistent with any value for x. Clearly, x=1 will destroy the equation and x = 3/2 will destroy it. How can we chose based on one value only. I haven't seen this yet ?

heshamelaziry wrote:

Testluv wrote:x = 1/2 satisfies I

x = 3/4 satisfies II

If x = 3/4, then x^2 = 9/16, and 1/x = 4/3, and 2x = 3/2

9/16 < 4/3 < 3/2

So, both I and II are possible orders. If x is positive, however, it is impossible that x^2 < 1/x while simultaneously 2x < x^2.

Choose D.

But in order for a choice such as II to be correct, the choice has to be consistent with any value for x. Clearly, x=1 will destroy the equation and x = 3/2 will destroy it. How can we chose based on one value only. I haven't seen this yet ?

No, it doesn't have to be consistent with any value of x. The question asks which of them COULD be a possible correct ordering. So, we only need to find a value of x that will work. As soon, as we do, we know that that COULD be the possible order.