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agree with AsterMatrix
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[quote="asterixmatrix
1. Given "n2 > 16" so n is greater than 4 or less than -4 so asnwer for question is |n| < 4 ? is no.
A is sufficient

2. lets assume n as -5 then 1/ |-5| = 1/5 > -5 and |-5| = 5 is not less than 4
and if we consider n as -3 then 1/|-3| = 1/3 > -3 and |-3| = 3 is less than 4
hence B is insuff

[/quote]

You did kill it!
OA is A
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hogann
If n is not equal to 0, is |n| < 4 ?

(1) n^2 > 16

(2) 1/|n| > n

Edit to fix exponent

What's the range for the second statement? Is it x<1?

Let us know

Cheers!
J :)
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Hi All,

This question is based around a couple of different patterns (some obvious, some not). You can TEST VALUES to answer the question...

We're told that N cannot be 0. We're asked if |N| < 4. This is a YES/NO question.

Fact 1: N^2 > 16

You probably recognize that this means that N > 4 or N < -4. This Fact gives us a consistent result; here's the proof:

IF...
N = 5, then the answer to the question is NO.

IF....
N = -5, then the answer to the question is NO.

No matter what value you use for N, under these 'restrictions', the answer to the question is ALWAYS NO.
Fact 1 is SUFFICIENT

Fact 2: 1/|N| > N

You can approach this Fact in a couple of different ways: with Algebra and Number Properties or by TESTing VALUES. Notice how the 'left side' of the inequality will ALWAYS be POSITIVE.....

IF...
N = -2, then 1/|-2| is > -2 and the answer to the question is YES.

IF...
N = -5, then 1/|-5| is > -5 and the answer to the question is NO.
Fact 2 is INSUFFICIENT

Final Answer:

GMAT assassins aren't born, they're made,
Rich
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#1. n satisfies the condition n< -4 and n>4, so mod(n) is always more than 4. Sufficient.

#2. n satisfies the condition n<0 and 0<n<1,

For n=.5, |n| < 4
But for n=-6, |n| > 4, two possibilities hence insufficient.

Only #1 Suff.

hogann
If n is not equal to 0, is |n| < 4 ?

(1) n^2 > 16

(2) 1/|n| > n
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hogann
If n is not equal to 0, is |n| < 4 ?

(1) n^2 > 16

(2) 1/|n| > n

Edit to fix exponent

What's the range for the second statement? Is it x<1?

Let us know

Cheers!
J :)

If n is not equal to 0, is |n| < 4 ?

Question basically asks is -4<n<4 true.

(1) n^2>16 --> n>4 or n<-4, the answer to the question is NO. Sufficient.

(2) 1/|n| > n, this is true for all negative values of n, hence we can not answer the question. Not sufficient.

Answer: A.

As you can see we don't really want the complete range for (2) to see that this statement is not sufficient, but still if interested:

1/|n| > n --> n*|n| < 1.

If n<0, then we'll have -n^2<1 --> n^2>-1. Which is true. So, n*|n| < 1 holds true for any negative value of n.
If n>0, then we'll have n^2<1 --> -1<n<1. So, n*|n| < 1 also holds true for 0<n<1.

Thus 1/|n| > n holds true if n<0 and 0<n<1.

Hope it's clear.
hi for stmt 2, can we also do it like this?
for n>0, 1/n > n
(1/n) - n>0
(1/n)-(n^2/n)>0
(n^2-1)/n < 0
(n+1)(n-1)/n < 0
at the end n<-1/0<n<1 (wavy line)
i understand the above solution by bunuel, but can we deduce this way too? for both ways, we are getting 0<n<1 & discard -1<n<0, since we are looking for n>0. is this correct?
thanks
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Forget the conventional way to solve DS questions.

We will solve this DS question using the variable approach.

DS question with 1 variable: Let the original condition in a DS question contain 1 variable. Now, 1 variable would generally require 1 equation for us to be able to solve for the value of the variable.

We know that each condition would usually give us an equation, and Since we need 1 equation to match the numbers of variables and equations in the original condition, the logical answer is D.

To master the Variable Approach, visit https://www.mathrevolution.com and check our lessons and proven techniques to score high in DS questions.

Let’s apply the 3 steps suggested previously. [Watch lessons on our website to master these 3 steps]

Step 1 of the Variable Approach: Modifying and rechecking the original condition and the question.

We have to find whether |n| < 4? where n ≠ 0.


Second and the third step of Variable Approach: From the original condition, we have 1 variable (n).To match the number of variables with the number of equations, we need 1 equation. Since conditions (1) and (2) will provide 1 equation each, D would most likely be the answer.

Let’s take look at each condition separately.

Condition(1) tells us that \(n^2 > 16\) .

=> 'n^2' is greater than 16 means n > 4 or n < -4 hence is |n| < 4 - NO

Since the answer is a unique NO , condition(1) alone is sufficient by CMT 1.

Condition(2) tells us that \(\frac{1}{|n|} > n\) .

=> If n = -3 , then \(\frac{1}{|-3|}\) > -3 = 1/3 > -3 - But |-3| = 3 is less than 4 - YES

=> But if n = -6 , then \(\frac{1}{|-6|}\) > -6 = 1/6 > -6 - But |-6| = 6 is greater than 4 - NO

Since the answer is not a unique YES or NO , condition(2) alone is not sufficient by CMT 1.

Condition (1) alone is sufficient.

So, A is the correct answer.

Answer: A


SAVE TIME: By Variable Approach, when you know that we need 1 equation, we will directly check each conditions to be sufficient. We will save time in checking the conditions individually.
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If n is not equal to 0, is |n| < 4 ?

Is -4 < n < 4?

(1) n^2 > 16

n > 4 or n < -4

We can answer with certainty that n does not fall between -4 < n < 4. Sufficient.

(2) 1/|n| > n

We can plug in values for this statement

\(\frac{1}{|-5|} > -5\)

\(\frac{1}{|-3|} > -3\)

We can get a yes and no answer. Insufficient.

Answer is A.
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Bunuel

2) 1/|n| > n

if we square on both sides

1/n^2 > n^2
1 > n^4

therefore, if n^4 is less than 1, this means n is a fraction and we get the answer, don't we? n lies between 4 and -4
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Bunuel

2) 1/|n| > n

if we square on both sides

1/n^2 > n^2
1 > n^4

therefore, if n^4 is less than 1, this means n is a fraction and we get the answer, don't we? n lies between 4 and -4

We can raise both parts of an inequality to a positive even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality). Here we cannot be sure that n (the right hand side) is non-negative, so we cannot square.

Adding, subtracting, squaring etc.: Manipulating Inequalities.
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