If n is positive and less than 1, then which of the following is true?

Question

I. n^2 - n < 0
II. n^3 < n
III. n + 1 < 1

A. I only
B. II only
C. III only
D. I and II only
E. II and III only

lmuenzel · Answer

Plug in numbers: my example =  \frac{1}{2}

(1)  \frac{1}{2} ^2 -  \frac{1}{2}  < 0

=>  \frac{1}{4}  -  \frac{1}{2}  < 0  correct

(2)  \frac{1}{2} ^3 <  \frac{1}{2}

\frac{1}{8}  <  \frac{1}{2}   correct

(3)  \frac{1}{2}  + 1 < 1  wrong!

EMPOWERgmatRichC · Answer

Hi All,

When dealing with Roman Numeral questions, you can work through the 3 Roman Numerals in any order (and eliminate answers according to your results). For this question though, I'll work through all 3 options in order:

We're told that N is POSITIVE and LESS than 1. This means that 0 < N < 1. We're asked which of the Roman Numerals is true (meaning which is ALWAYS true no matter how many different examples we can come up with). This question can be solved with a mix of logic and TESTing VALUES.

I. N^2 - N < 0

Since N is a POSITIVE FRACTION - and squaring a positive fraction will always lead to a SMALLER fraction - we know that N^2 will always be less than N. 
For example, (1/2)^2 = 1/4. 
Thus N^2 - N will ALWAYS be less than 0. 
Roman Numeral 1 IS always true.

II. N^3 < N

The same logic we used in Roman Numeral 1 applies to Roman Numeral 2. Cubing a positive fraction will ALWAYS lead to a SMALLER fraction, so N^3 will always be less than N. 
For example, (1/2)^3 = 1/8 
Roman Numeral 2 IS always true.

III. N+1 < 1

We're told that N is POSITIVE, so N+1 will be greater than 1. 
Roman Numeral 3 is NOT true.

Final Answer:  D

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

ScottTargetTestPrep · Answer

We are given that n is a number between 0 and 1. Since, when we raise such a number to a positive integer exponent (greater than 1), the value decreases, (1) and (2) are always true. In (3), since we can simplify the inequality to n < 0, we see that (3) is not true.

Answer: D

Array_rayray · Answer

Hi, everyone. I'm new to the forum, though been a silent lurker here for a whole while. Can someone explain why the answer is D and not A? I used the approach of rules of sets and functions: 0 n(n-1)<0 (through factorisation). Then the domain of the function will be (0,1) making A and D the only remaining options to check for. So for (II), n^{3}n^{3}-n<0 => n(n^{2}-1)<0 => n(n+1)(n-1) <0. In this case, domain of the function will be (-infinity, -1) U (0, 1) . Here's my doubt, they have mentioned in the text that n is positive. Shouldn't this rule out (D)?

Bunuel · Answer

Yes, n^3 < n means n < -1 or 0 < n < 1. However, the question asks the following: if 0 < n < 1, which of the following must be true? Since given that 0 < n < 1, then for any n from that range n^3 < n would be true. Hence, II must betrue.

This is a trap many fall in. Check other similar questions from  Trickiest Inequality Questions Type: Confusing Ranges  (part of our  Special Questions Directory ).

Hope it helps.

Array_rayray · Answer

It makes sense now, question is asking whether for the given range it is true, making the answer (D). Thanks a lot!

If n is positive and less than 1, then which of the following is true?

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If n is positive and less than 1, then which of the following is true?

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