Problem Solving

If -1 < x < 0, which of the following must be true?

I. x³ < x²
II. x� < 1 - x
III. x� < x²

A)I only
B)I and II only
C) II and III only
D) I and III only
E) I, II and III

USEFUL RULES:
(negative)^(ODD integer) = some negative value
(negative)^(EVEN integer) = some positive value

I. x³ < x²
x is NEGATIVE. So, the above rule tells us that x³ is negative, and x² is positive
So, it must be true that x³ < x²

II. x� < 1 - x
x is NEGATIVE. So, the above rule tells us that x� is negative.
Also, since x is negative, 1 - x must be positive
So, it must be true that x� < 1 - x

III. x� < x²
We can take this inequality and divide both sides by x² (since we can be certain that x² is positive).
When we do so, we get: x² < 1, and this is definitely true.
We know that x² is less than 1 because we're told that -1 < x < 0
So, it must be true that x� < x²

So all 3 statements must be true.

Cheers,
Brent

Hi Anaira Mitch,

Both Brent and Mitch have discussed the Number Properties involved with this question. In many cases, it helps to TEST VALUES, so you can 'see' the results and recognize any patterns involved.

Here, we're told that -1 < X < 0, so X is a NEGATIVE FRACTION. We're asked which of the three Roman Numerals MUST be true (meaning which is ALWAYS TRUE no matter how many different examples we try).

I. X^3 < X^2

Since X is a negative fraction, let's TEST something obvious: X = -1/2

(-1/2)^3 = -1/8
(-1/2)^2 = +1/4

Notice how when you cube a negative value you end up with a NEGATIVE value.
Also notice how when you square a negative value you end up with POSITIVE value.
You can do additional calculations with other negative fractional values, but those patterns prove that, when dealing with negative fractions, X^3 will ALWAYS be less than X^2.
Roman Numeral 1 is ALWAYS TRUE.

II. X^5 < (1 - X)

Again. let's TEST X = -1/2
(-1/2)^5 = -1/32
(1 - -1/2) = + 1.5

Notice how when you take a negative value to the 5th power, you end up with a NEGATIVE value.
Also notice how when you subtract a negative value from 1 you end up with POSITIVE value.
Those patterns prove that, when dealing with negative fractions, X^5 will ALWAYS be less than (1 - X).
Roman Numeral 2 is ALWAYS TRUE.

III. X^4 < X^2

Again. let's TEST X = -1/2
(-1/2)^4 = +1/16
(-1/2)^2= +1/4

This Roman Numeral requires a bit more attention to detail. Raising a negative value to the 2nd and 4th powers will always lead to a POSITIVE value. However, you have to think about how those values relate to one another. Notice how when we INCREASE the EVEN exponent, the result is a smaller positive value. You can try it with other negative fractions and you'll end up with the same result. Thus, when dealing with negative fractions, X^4 will ALWAYS be less than X^2.
Roman Numeral 3 is ALWAYS TRUE.

Final Answer: E

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

Anaira Mitch wrote:If -1 < x < 0, which of the following must be true?
I. x^3 < x^2
II. x^5 < 1 - x
III. x^4 < x^2

a)I only
b)I and II only
c) II and III only
d) I and III only
e) I, II and III

Since x is between -1 and 0, we know that x is a negative proper fraction. We can test the Roman numerals using an actual negative proper fraction for x such as -½ (when necessary).

I. x^3 < x^2

We don't need to plug in for the above inequality because x^3 will always be negative and x^2 will always be positive. Thus, Roman numeral I is true.

II. x^5 < 1 - x

Once again we do not need to plug, because 1 - x will always be positive (since 1 - (negative number) equals to 1 + (positive number)) and x^5 will always be negative. Thus, Roman numeral II is true.

III. x^4 < x^2

x^4 = (-1/2)^4 = 1/16

x^4 = (-1/2)^2 = 1/4

If we did not want to plug in a value for x, we could have proved Roman numeral III as follows:

Since -1 < x < 0, x^2 is a positive fraction between 0 and 1. Moreover, x^4 is equal to the square of x^2; that is, x^4 = (x^2)^2. Since 0 < x^2 < 1, the square of x^2 is actually less than x^2.

Roman numeral III must also always be true.

Answer: E