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Is x > 10^10 ? (1) x > 2^34 (2) x = 2^35 [#permalink]
 20 Feb 2012, 18:55
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Question Stats:
58% (02:15) correct
41% (01:14) wrong based on 1 sessions
Is x > 10^10 ? (1) x > 2^34 (2) x = 2^35
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Re: Is x > 10^10 ? (1) x > 2^34 (2) x = 2^35 [#permalink]
20 Feb 2012, 21:41
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Re: Is x > 10^10 ? (1) x > 2^34 (2) x = 2^35 [#permalink]
21 Feb 2012, 20:36
u r awesome Bunuel 
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Re: Is x > 10^10 ? (1) x > 2^34 (2) x = 2^35 [#permalink]
29 Jul 2012, 21:58
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.
OR: 2^{34}=(2^{10})^{3.4}=(1,024)^{3.4}>(10^3)^{3.4}=10^{(3*3.4)}=10^{10.2}>10^{10}.
(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.
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Re: Is x > 10^10 ? (1) x > 2^34 (2) x = 2^35 [#permalink]
30 Jul 2012, 03:09
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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.
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Re: Is x > 10^10 ? (1) x > 2^34 (2) x = 2^35 [#permalink]
30 Jul 2012, 06:19
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.
OR: 2^{34}=(2^{10})^{3.4}=(1,024)^{3.4}>(10^3)^{3.4}=10^{(3*3.4)}=10^{10.2}>10^{10}.
(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.
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Re: Is x > 10^10 ? (1) x > 2^34 (2) x = 2^35 [#permalink]
30 Jul 2012, 13:22
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. Dabral
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Re: Is x > 10^10 ? (1) x > 2^34 (2) x = 2^35 [#permalink]
04 Sep 2012, 03:43
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.
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Re: Is x > 10^10 ? (1) x > 2^34 (2) x = 2^35 [#permalink]
04 Sep 2012, 03:49
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Re: Is x > 10^10 ? (1) x > 2^34 (2) x = 2^35 [#permalink]
04 Sep 2012, 03:54
Crystal.
Classic mistake, which is what is probably REALLY being tested, realised it when re-reading my post.
Thanks for response though.
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Re: Is x > 10^10 ? (1) x > 2^34 (2) x = 2^35 [#permalink]
16 Feb 2013, 14:19
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.
OR: 2^{34}=(2^{10})^{3.4}=(1,024)^{3.4}>(10^3)^{3.4}=10^{(3*3.4)}=10^{10.2}>10^{10}.
(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 (2) X = 2^35 - SUFFICIENT Is this approach correct?
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Is x > 10^10 ? (1) x > 2^34 (2) x = 2^35 Is my approach for solving this question correct?
10^10 = 2^10 * 5^10 = 2^10 * 2^20 (approx)= 2^30.
Therefore, st1 & st2 both are sufficient to ans this question. Thus, Ans D.
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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 a pretty quick one. I don't think it is a 700+ level question.
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vinaymimani wrote: 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?
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greatps24 wrote: vinaymimani wrote: 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!
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