GMAT Diagnostic Test Question 9
Field: number properties, fractions
Difficulty: 750



Is x greater than 1?

(1) \frac{1}{x} > -1
(2) \frac{1}{x^5} > \frac{1}{x^3}

A. Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient
B. Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
D. EACH statement ALONE is sufficient
E. Statements (1) and (2) TOGETHER are NOT sufficient
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Hi dzyubam, would you please explain again why statement 2 is sufficient

Thx again

why is it not correct to say
1/x>-1 => x<-1

Thx dzyubam. Crystal clear

Plugging in a couple of numbers explained in the OE is fast enough. Do you think plugging in values is not fast enough for this question?

How did you get x<-1?

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Stat1
if x>0 then x>-1 => x>0
if x<0 then x<-1 => x<-1

Stat2
if x>0 then x^3/x^5>1 => 1/x^2>1 => 1>x^2 => |x|<1 => x<1 or x>-1 => 0<x<1
if x<0 then x^3/x^5<1 => 1/x^2<1 => (as x^2 is always positive) 1<x^2 => |x|>1 => x>1 or x<-1 => x<-1

Since I didn't know about this rule:

\frac{1}{x^5} > \frac{1}{x^3} is equivalent to x^5 < x^3. because if two fractions have equal numerators, the fraction with a smaller denominator is the bigger one.

...I simply went ahead and pluggeed-in "2" and "-2" for the the second expression (just like the first). In this way, I concluded a "NO" answer for both 1/32 > 1/8 and -1/32 > -1/8

Let me see if I understand this question correctly - and DS problems in general since they are throwing me off left and right simply because of their demand of figuring out "yes" or "no" "yes + no" for every possibility. **So please at least tell me if I have approached this question in the correct way**



#1 The question is asking if x is greater than 1. Now I have to look at each possibility individually and plug-in numbers until I can justify whether the possibility is, in itself, enough to answer the original question stem.


#2 So, the first possibility gives us \frac{1}{x} > -1

Now I have to choose two numbers ("2" for ease) and both + and - to see if this possibility will prove the stem.

For a +2: \frac{1}{2} > -1 *This holds true as +0.5 is > -1

However,

For a -2: \frac{1}{-2} > -1 *This is also a confirmation of the expression itself, but creates a conflict with the stem which is only looking for the confirmation of x being greater than 1.

Therefore, the answer "A" can be ruled out and we are left with the possibilities of B, C or E.


#2 Now if I am to look at the second possibility, \frac{1}{x^5} > \frac{1}{x^3}, and have not "simplified it" for whatever reason (i.e. I didn't know that rule)... I will plug in the same two numbers to see what happens.

For a +2: \frac{1}{2^5} > \frac{1}{2^3}

=\frac{1}{32} > \frac{1}{8} *This does not hold true as +3.125% is not > +12.5%

However, now...

For a -2: \frac{1}{-2^5} > \frac{1}{-2^3}

=\frac{1}{-32} > \frac{1}{-8} *This does hold true as -3.125% is > -12.5%


Now I am thoroughly confused, because I'm being tricked with the second possibility (#2) with a +# as "not true" and a -# as "true." This is the main reason I think that I am failing at these DS q's because I thought that any possibility that presents both a "Yes" and "No" answer is therefore "IN SUFFICIENT"...


Where have I gone wrong here?

Hi...

I feel answer to this question must by E .

Statement (1) is obvious why it is not sufficient to find the answer

About Statement (2)

i) x = 1/2
( 1/x(power 5) = 32 ) > (1/x(power3) = 8 )

ii) x = -2
(1/x(power 5) = -1/32 ) > ( 1/x(power3) = -1/8 )

iii) x = -1/2
(1/x(power 5) = -32) < (1/x(power3) = -8 )

So that statement is also not sufficient . Please respond to my reasoning !!!

Thanks

Here is my way, which worries me by its simplicity in comparison to a "750" level problem. Please let me know if there are any errors in the following approach. I am not the best with these question types and am not sure if plugging in numbers, like others did, considers an element my approach does not.

S1) 1/X > - 1 ---> 1>-X ---> (multiply by -1/turn sign)---> -1<X ..... This means X can be -.5, 0, 1, 2..insufficient

S2) 1/X^5 > 1/X^3 --->(multiply both sides by X^3)---> 1/x^2 > 1 ---> 1>X2 ---> +-1 > X ...if X is less than 1 or -1, we know its not greater than 1; sufficient.

Answer: B