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Joined: 04 Dec 2002
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GMAT Diagnostic Test Question 10  [#permalink]

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GMAT Diagnostic Test Question 10
Field: inequalities
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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Originally posted by bb on 06 Jun 2009, 21:04.
Last edited by Bunuel on 06 Oct 2013, 23:08, edited 3 times in total.
Updated
CIO  Joined: 02 Oct 2007
Posts: 1179
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Explanation:

(1) $$\frac{1}{x}>- 1$$ --> $$\frac{1+x}{x}>0$$, two cases:

A. $$x>0$$ and $$1+x>0$$, $$x>-1$$ --> $$x>0$$;

B. $$x<0$$ and $$1+x<0$$, $$x<-1$$ --> $$x<-1$$.

So, the given inequality holds true in two ranges: $$x>0$$ and $$x<-1$$, thus $$x$$ may or may not be greater than 1. Not sufficient.

(2) $$\frac{1}{x^5}> \frac{1}{x^3}$$ --> $$\frac{1-x^2}{x^5}>0$$, two cases:

A. $$x>0$$ (it's the same as $$x^5>0$$) and $$1-x^2>0$$, $$-1<x<1$$ --> $$0<x<1$$;

B. $$x<0$$ and $$1-x^2<0$$, $$x<-1$$ or $$x>1$$ --> $$x<-1$$;

We got that given inequality holds true in two ranges: $$0<x<1$$ and $$x<-1$$, ANY $$x$$ from this ranges will be less than 1. So the answer to our original question is NO. Sufficient.

Originally posted by dzyubam on 01 Jul 2009, 06:22.
Last edited by bb on 28 Sep 2013, 11:16, edited 1 time in total.
Updated
Intern  Joined: 30 Jun 2009
Posts: 33
Re: GMAT Diagnostic Test Question 9  [#permalink]

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Hi dzyubam, would you please explain again why statement 2 is sufficient

Thx again
CIO  Joined: 02 Oct 2007
Posts: 1179
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The trick to remember during solving this question is that fractions in the range $$(0,1)$$ taken to greater power get smaller, for example $$\left(\frac{1}{3}\right)^3 < \left(\frac{1}{3}\right)^2$$, etc.

$$\frac{1}{x^5} > \frac{1}{x^3}$$ is equivalent to $$x^5 < x^3$$. because if two fractions have equal numerators, the fraction with a smaller denominator is the bigger one. Now we need to find the range of values of $$x$$ in which $$x^5 < x^3$$ holds true. This is possible in $$x \in (0,1)$$ for positive $$x$$ and in $$x \in (-\infty,-1)$$ for negative $$x$$.

We can put the values of $$x$$ from these ranges to make sure it works:

positive $$x=\frac{1}{2}$$:
$$\left(\frac{1}{2}\right)^5 < \left(\frac{1}{2}\right)^3$$
$$\left(\frac{1}{32}\right) < \left(\frac{1}{8}\right)$$ -- holds true

negative $$x=-2$$:
$$(-2)^5 < (-2)^3$$
$$-32 < -8$$ -- holds true

In either range the value of $$x$$ is smaller than 1, so Statement (2) is sufficient by itself.

Hope this helps.
defoue wrote:
Hi dzyubam, would you please explain again why statement 2 is sufficient

Thx again
Manager  Joined: 08 Jul 2009
Posts: 121
Re: GMAT Diagnostic Test Question 9  [#permalink]

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why is it not correct to say
1/x>-1 => x<-1
CIO  Joined: 02 Oct 2007
Posts: 1179
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You forget that $$x$$ can be any number from the range $$(0,\infty)$$ in order to satisfy S1. $$x < -1$$ is not a complete solution. Both 2 and -2 satisfy S1, as stated in the OE. S1 is not sufficient.
sher676 wrote:
why is it not correct to say
1/x>-1 => x<-1
Intern  Joined: 30 Jun 2009
Posts: 33
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Thx dzyubam. Crystal clear
Senior Manager  Joined: 25 Oct 2008
Posts: 433
Location: Kolkata,India
Re: GMAT Diagnostic Test Question 9  [#permalink]

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Not clear.Even I got x<-1.How do you solve this question without plugging in value of x?
CIO  Joined: 02 Oct 2007
Posts: 1179
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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 $$x<-1$$?
tejal777 wrote:
Not clear.Even I got x<-1.How do you solve this question without plugging in value of x?
Manager  Joined: 22 Jul 2009
Posts: 147
Re: GMAT Diagnostic Test Question 9  [#permalink]

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Somehow I usually slow down when picking numbers in DS problems, so I prefer an algebraic approach whenever I spot that an easy one is possible.

This is how I did this exercise:

Stat1
if x>0 then x>-1 => x>0. Not sufficient; no need to test for x<0.

Stat2
if x>0 then x<1 and x<-1 => 0<x<1. Sufficient; x is not greater than 1; no need to test for x<0.

Of course, easier explained online than done against the clock:)
Manager  Joined: 22 Jul 2009
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tejal777 wrote:
Not clear.Even I got x<-1.How do you solve this question without plugging in value of x?

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
Manager  Joined: 22 Jul 2009
Posts: 147
Re: GMAT Diagnostic Test Question 9  [#permalink]

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One last post on this thread. I see that my initial approach for the 2nd statement was not as fast as it could be. Here goes the simplified version:

Stat2
if x>0 then x^3>x^5 => x<1 => Sufficient, but I'll do x<0 for the sake of it.
if x<0 then x^3>x^5 => x<-1
Intern  Joined: 26 Aug 2009
Posts: 18
Re: GMAT Diagnostic Test Question 9  [#permalink]

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$$\frac{1}{x^5} > \frac{1}{x^3}$$ 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
CIO  Joined: 02 Oct 2007
Posts: 1179
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You have to be careful with the negative fractions. The highlighted expression doesn't hold true. While the absolute value of $$-\frac{1}{8}$$ is greater than that of $$-\frac{1}{32}$$, $$-\frac{1}{32} > -\frac{1}{8}$$ because $$-\frac{1}{32}$$ is closer to 0 on the number line than $$-\frac{1}{8}$$ is.
Justed wanted to make it clear. Hope it helps.
stilite wrote:

$$\frac{1}{x^5} > \frac{1}{x^3}$$ 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
Intern  Joined: 26 Aug 2009
Posts: 18
Re: GMAT Diagnostic Test Question 9  [#permalink]

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dzyubam wrote:
You have to be careful with the negative fractions. The highlighted expression doesn't hold true. While the absolute value of $$-\frac{1}{8}$$ is greater than that of $$-\frac{1}{32}$$, $$-\frac{1}{32} > -\frac{1}{8}$$ because $$-\frac{1}{32}$$ is closer to 0 on the number line than $$-\frac{1}{8}$$ is.
Justed wanted to make it clear. Hope it helps.
stilite wrote:

$$\frac{1}{x^5} > \frac{1}{x^3}$$ 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?
CIO  Joined: 02 Oct 2007
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Let me clarify. In the Official Explanation, I used -2 and +2 to plug into S1 because both -2 and +2 satisfy S1. So when I use different numbers that satisfy the condition (here S1) and produce a different answer to the question in the stem (if $$x$$ is greater than 1), then I can conclude that a certain Statement is NOT sufficient. However, what you did with S2 is not a good way to determine if the Statement is suff or insuff.
-2 satisfies S2, but +2 doesn't and this is why you can't use +2 to prove if S2 is sufficient or not.
So, first you have to find the range of $$x$$ that satisfies a particular Statement and only after that you can plug in different values of $$x$$ from that range. What you did was using the values of $$x$$ that worked for S1 but didn't quite work for S2.

Hope this helps.
Intern  Joined: 26 Jul 2009
Posts: 16
Re: GMAT Diagnostic Test Question 9  [#permalink]

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Hi...

I feel answer to this question must by E .

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

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
CIO  Joined: 02 Oct 2007
Posts: 1179
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Hi,
I can't see why you think E is the correct answer here. All three values that you used for $$x$$ are less than 1. If you plug in $$x=-\frac{1}{2}$$, the inequality doesn't hold true, you're right here. However, it doesn't prove that $$x$$ can be both greater than and less than 1. Plugging in $$-\frac{1}{2}$$ only shows that $$x=-\frac{1}{2}$$ is not in the range of $$x$$ that hold the inequality true.

In other words, while we consider S2, we have to use only those $$x$$ that hold the S2 true. Your example in iii) conflicts with S2 and can't be used to prove anything.

I'll have to modify the OE to use a more algebraic way of solving.

I hope my answer helps and doesn't confuse you.

sunny4frenz wrote:
Hi...

I feel answer to this question must by E .

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

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
Intern  Joined: 29 Jul 2009
Posts: 17
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1
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.

CIO  Joined: 02 Oct 2007
Posts: 1179
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We should probably revise the difficulty level of the question.
I like your approach. +1. Can't see anything wrong with it.
dpgxxx wrote:
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. Re: GMAT Diagnostic Test Question 9   [#permalink] 13 Oct 2009, 00:32

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# GMAT Diagnostic Test Question 10

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