Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized for You
we will pick new questions that match your level based on your Timer History
Track Your Progress
every week, we’ll send you an estimated GMAT score based on your performance
Practice Pays
we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
It appears that you are browsing the GMAT Club forum unregistered!
Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club
Registration gives you:
Tests
Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.
Applicant Stats
View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more
Books/Downloads
Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
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 _________________
(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.
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:
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.
Re: GMAT Diagnostic Test Question 9 [#permalink]
28 Aug 2009, 05:03
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
Re: GMAT Diagnostic Test Question 9 [#permalink]
28 Aug 2009, 06:48
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:
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
Re: GMAT Diagnostic Test Question 9 [#permalink]
28 Aug 2009, 08:17
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:
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"...
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.
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
Re: GMAT Diagnostic Test Question 9 [#permalink]
12 Oct 2009, 16:01
5
This post received KUDOS
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
2
This post received KUDOS
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.