GMAT Prep Algebra

[Srishti_15]

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

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


Answer : I and II

I got I. Please help me out with II.

[GMATGuruNY]

If x is positive, which of the following could be the correct ordering of 1/x, 2x, and x²?

I. x² < 2x < 1/x
II. x² < 1/x < 2x
III. 2x < x² < 1/x

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

Determine the CRITICAL POINTS by setting the expressions equal to each other:

1/x = 2x
2x² = 1
x² = 1/2
x = √(1/2) = 1/√2 ≈ 1/1.4 ≈ 10/14 ≈ 5/7.

1/x = x²
x^3 = 1
x = 1.

2x = x²
x=2
(We can divide by x because x>0.)

The critical points are x=5/7, x=1, x=2.
These critical points indicate where two of the expressions are EQUAL.
Thus, to the left and right of each critical point, the value of one expression must be GREATER than the value of another.

To determine which of I, II and II could be true, plug in values to the left and right of each critical point.
Start with the range that many test-takers will fail to consider: 5/7 < x < 1.

5/7 < x < 1:
If x = 3/4, then:
1/x = 4/3.
x² = 9/16.
2x = 3/2.
Since x² < 1/x < 2x, we know that II could be true.
Eliminate A, B and C.

In statement III, 2x<x², which implies that 2<x.
But if x>2, then 1/x cannot be the greatest of the three values.
Thus, III is not possible.
Eliminate E.

The correct answer is D.

[Brent@GMATPrepNow]

If x is positive, which of the following could be the correct ordering of 1/x, 2x and x²?
I. x² < 2x < 1/x
II. x² < 1/x < 2x
III. 2x < x² < 1/x

(A) None
(B) I only
(C) III only
(D) I and II only
(E) I II and III

Let's start by plugging in some positive values of x and see what we get.

x = 1/2
1/x = 2
2x = 1
x² = 1/4
So, we get x² < 2x < 1/x
This matches statement I.

x = 3/4
1/x = 4/3
2x = 3/2
x² = 9/16
So, we get x² < 1/x < 2x
This matches statement II

x = 3
1/x = 1/3
2x = 6
x² = 9
So, we get 1/x < 2x < x²
NO MATCHES

At this point, the correct answer is either D or E.
If you're pressed for time, you might have to guess.

Alternatively, you can use some algebra to examine statement III (2x < x² < 1/x)
Notice that there are 2 inequalities here (2x < x² and x² < 1/x)
Take 2x < x² and divide both sides by x to get 2 < x
Take x² < 1/x and multiply both sides by x to get x^3 < 1, which means x < 1
Hmmm, so x is greater than 2 AND less than 1. This is IMPOSSIBLE, so statement III cannot be true.

Answer = D

Cheers,
Brent

[Matt@VeritasPrep]

I'd think of key ranges here:

x < -1
-1 < x < 0
0 < x < 1
1 < x

I works for 0 < x < 1, so we're set there.
II works for some numbers in 0 < x < 1 (e.g. x = .75), so we're set there too.

We know that if any values satisfy III, they'd have to be positive, since x*x < 1/x. But 0 < x < 1 doesn't work, and 1 < x doesn't work, so we're out of options!

Another way to test III would be to discover that x must be positive, then multiply the entire inequality by x. This gives

2x² < x² < 1

but 2x² < x² won't work for any real value of x, so this is impossible.

[Needgmat]

Let's start by plugging in some positive values of x and see what we get.

x = 1/2
1/x = 2
2x = 1
x² = 1/4
So, we get x² < 2x < 1/x
This matches statement I.

x = 3/4
1/x = 4/3
2x = 3/2
x² = 9/16
So, we get x² < 1/x < 2x
This matches statement II

x = 3
1/x = 1/3
2x = 6
x² = 9
So, we get 1/x < 2x < x²
NO MATCHES

At this point, the correct answer is either D or E.
If you're pressed for time, you might have to guess.

Hi Brent ,

Why can't we a put any other numbers? Is there any specific reason to pic those numbers.

If I put 3 in statement 1, so this makes not sufficient.

Please explain.

Many thanks in advance.

Kavin

[[email protected]]

Hi Kavin,

The prompt asks which of the following COULD be the order of the three terms. So you have to think about whether there are ANY possible values that would put the terms in the given order(s) or not.

If you TEST X = 3 in the 1st Roman Numeral, you can clearly see that the terms do not 'fit' the inequalities. However, that does NOT mean that the 1st Roman Numeral is impossible. If you focus on the latter inequality, the easiest way to make 2X < 1/X is if X is a positive fraction, so you could TEST one (such as X = 1/2). You'd find that the 1st Roman Numeral IS possible - so the correct answer must include the 1st Roman Numeral

GMAT assassins aren't born, they're made,
Rich

[Matt@VeritasPrep]

Why can't we a put any other numbers? Is there any specific reason to pic those numbers.

If I put 3 in statement 1, so this makes not sufficient.

Please explain.

Other numbers are OK, but you can also use the inequalities to give you a sense of what's good to pick.

For instance, since (i) has x² < 2x, we know that x MUST be positive, since 0 < x² < 2x, so 0 < x.

Same idea with (ii): we know x² < 1/x, so 0 < 1/x, and again x > 0.