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:
Re: Is x > 10^10 ? (1) x > 2^34 (2) x = 2^35 [#permalink]
20 Feb 2012, 20:41
24
This post received KUDOS
Expert's post
8
This post was BOOKMARKED
Is x > 10^10 ?
(1) x > 2^34 --> we should compare \(2^{34}\) and \(10^{10}\) --> take the square root from both: we should compare \(2^{17}\) and \(10^5=100,000\). Now, \(2^{17}=2^{10}*2^7=1,024*128>100,000\). Sufficient.
(2) x = 2^35. Since we have the exact numerical value of x we should be able compare it to 10^10 and answer the question. It really doesn't matter whether 2^35>10^10, the main point is that we have sufficient information to get the answer. Sufficient.
Re: Is x > 10^10 ? (1) x > 2^34 (2) x = 2^35 [#permalink]
30 Jul 2012, 02:09
6
This post received KUDOS
dvinoth86 wrote:
Is x > 10^10 ?
(1) x > 2^34 (2) x = 2^35
(1) Let's see if \(2^{34}>10^{10}\).
\(2^{34}>2^{10}*5^{10}\). Divide through by \(2^{10}\), and we get \(2^{24}>5^{10}\). Take the square root of both sides: \(2^{12}>5^5\). This we can compute quite easily, and we find that 1024*4=4096 > 625*5= 3125, TRUE. Sufficient.
(2) We have already seen that (1) is sufficient, obviously (2) is also sufficient. Just to play with powers, we can check that \(2^{35}>10^{10}\): Start again with \(2^{35}>2^{10}*5^{10}\), divide through by \(2^{10}\), then \(2^{25}>5^{10}\). Now we can take the 5th order root of both sides and obtain \(2^5>5^2\) or 32 > 25, TRUE.
Thus, answer D. _________________
PhD in Applied Mathematics Love GMAT Quant questions and running.
Re: Is x > 10^10 ? (1) x > 2^34 (2) x = 2^35 [#permalink]
30 Jul 2012, 05:19
1
This post received KUDOS
venmic wrote:
Bunuel rocks this is a very nice solution compared to the other ones I ve seen
Bunuel wrote:
Is x > 10^10 ?
(1) x > 2^34 --> we should compare \(2^{34}\) and \(10^{10}\) --> take the square root from both: we should compare \(2^{17}\) and \(10^5=100,000\). Now, \(2^{17}=2^{10}*2^7=1,024*128>100,000\). Sufficient.
(2) x = 2^35. Since we have the exact numerical value of x we should be able compare it to 10^10 and answer the question. It really doesn't matter whether 2^35>10^10, the main point is that we have sufficient information to get the answer. Sufficient.
Answer: D.
Bunuel's logic for testing assumption (2) is excellent. In the present form of the question, one can use that \(2^{35}>2^{34}\), so once (1) turns out sufficient and necessarily provides the info that \(2^{34}>10^{10}\), testing (2) is very easy. Maybe, it would have been somehow more challenging to choose a smaller exponent in statement (2), like 33, with which direct comparison would have been not so straightforward, and a time saving approach would need similar logic to Bunuel's. _________________
PhD in Applied Mathematics Love GMAT Quant questions and running.
Re: Is x > 10^10 ? (1) x > 2^34 (2) x = 2^35 [#permalink]
30 Jul 2012, 12:22
1
This post received KUDOS
In my personal opinion, I just don't see this question following the style of the GMAT test writers. Unless, whoever wrote this can confirm that they recently saw a similar idea on the exam.
Re: Is x > 10^10 ? (1) x > 2^34 (2) x = 2^35 [#permalink]
04 Sep 2012, 02:49
1
This post received KUDOS
Expert's post
reklaw wrote:
Why would you need to compute anything in this problem?
Statement 1 gives us a minimum value of x. It doesn't matter if 10^10 is smaller or larger, it is sufficient to answer the question. Statement 2 needs no computation either, which Bunuel already pointed out.
We do need to compare 10^10 and 2^34 in (1). Because if 2^34 were less than 10^10, then the statement wouldn't b sufficient. Consider this:
Is x>10?
(1) x>2. If x=5 then the answer is NO but if x=15, then the answer is YES. Not sufficient.
Re: Is x > 10^10 ? (1) x > 2^34 (2) x = 2^35 [#permalink]
08 Jul 2013, 14:25
1
This post received KUDOS
dvinoth86 wrote:
Is x > 10^10 ?
(1) x > 2^34 (2) x = 2^35
hi,
statement 2 is clear we are able to calculate...HENCE SUFFICIENT for statement 1 2^34=2^10*8^8===>(1 10^10=2^10*5^10===>(2
actually we have to compare 2^34..and 10^10 both of them have 2^10 common...so we have to compare actually 8^8 and 5^10...
5^10=(8-3)^8*25=25*(8-3)^8 lets divide 5^10..with 8^8..==>25*((8-3)/8)^8==>clearly we can see that 25 is multiplied to a very small no.(as bracket number is less than 1,and it has been raised to power 8)==>end resul will be less than 1...==>this proves 8^8 is greater than 5^10....hence...2^34>10^10==>sufficient.
both statements are sufficient..hence D _________________
When you want to succeed as bad as you want to breathe ...then you will be successfull....
Re: Is x > 10^10 ? (1) x > 2^34 (2) x = 2^35 [#permalink]
29 Jul 2012, 20:58
Bunuel rocks this is a very nice solution compared to the other ones I ve seen
Bunuel wrote:
Is x > 10^10 ?
(1) x > 2^34 --> we should compare \(2^{34}\) and \(10^{10}\) --> take the square root from both: we should compare \(2^{17}\) and \(10^5=100,000\). Now, \(2^{17}=2^{10}*2^7=1,024*128>100,000\). Sufficient.
(2) x = 2^35. Since we have the exact numerical value of x we should be able compare it to 10^10 and answer the question. It really doesn't matter whether 2^35>10^10, the main point is that we have sufficient information to get the answer. Sufficient.
Re: Is x > 10^10 ? (1) x > 2^34 (2) x = 2^35 [#permalink]
04 Sep 2012, 02:43
Why would you need to compute anything in this problem?
Statement 1 gives us a minimum value of x. It doesn't matter if 10^10 is smaller or larger, it is sufficient to answer the question. Statement 2 needs no computation either, which Bunuel already pointed out.
Re: Is x > 10^10 ? (1) x > 2^34 (2) x = 2^35 [#permalink]
16 Feb 2013, 13:19
Bunuel wrote:
Is x > 10^10 ?
(1) x > 2^34 --> we should compare \(2^{34}\) and \(10^{10}\) --> take the square root from both: we should compare \(2^{17}\) and \(10^5=100,000\). Now, \(2^{17}=2^{10}*2^7=1,024*128>100,000\). Sufficient.
(2) x = 2^35. Since we have the exact numerical value of x we should be able compare it to 10^10 and answer the question. It really doesn't matter whether 2^35>10^10, the main point is that we have sufficient information to get the answer. Sufficient.
Answer: D.
Bunuel,
I solved it a different way- would you mind checking my approach?
I restructured the qstem to is x =,> 2^11 * 5^11?
(1) x > 2^34
- x has 2^11 therefore 2^35 - 2^11 = 2^24 - estimated 2^2 to be 5 and divided 24 by 2 and got 5^12 - x = 2^11 * 5^12 -------- SUFFICIENT
A very coarse method but I would do this problem by log.
F.S.1 gives logx > 34 log2 = 34*0.3 = 10.2(approx). As problem statement is asking about whether x>10^10, it boils down to logx>10. Thus suficient.
F.S.2 anyways gives the value of logx = 35*0.3. Again logx>10. Sufficient.
D.
Not the best method but pretty quick. I don't think it is a 700+ level question.
Good one. Can you please share some source/ tutorial of this method?
No tutorial as such. Just that they gave a power of 10 in this question and also 2^something. So it just struck me. Worked in this question, might not work always! _________________
Re: Is x > 10^10 ? (1) x > 2^34 (2) x = 2^35 [#permalink]
08 Dec 2014, 00:34
Bunuel wrote:
Is x > 10^10 ?
(1) x > 2^34 --> we should compare \(2^{34}\) and \(10^{10}\) --> take the square root from both: we should compare \(2^{17}\) and \(10^5=100,000\). Now, \(2^{17}=2^{10}*2^7=1,024*128>100,000\). Sufficient.
(2) x = 2^35. Since we have the exact numerical value of x we should be able compare it to 10^10 and answer the question. It really doesn't matter whether 2^35>10^10, the main point is that we have sufficient information to get the answer. Sufficient.
Answer: D.
Oh Bunuel, you are always awesome. For me, your approach is always faster than Manhanttan's. Highly recommend GMAT club for all GMAT learner!!!
Re: Is x > 10^10 ? (1) x > 2^34 (2) x = 2^35 [#permalink]
27 Dec 2015, 06:10
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________
You know what’s worse than getting a ding at one of your dreams schools . Yes its getting that horrid wait-listed email . This limbo is frustrating as hell . Somewhere...
As I’m halfway through my second year now, graduation is now rapidly approaching. I’ve neglected this blog in the last year, mainly because I felt I didn’...
Wow! MBA life is hectic indeed. Time flies by. It is hard to keep track of the time. Last week was high intense training Yeah, Finance, Accounting, Marketing, Economics...