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(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]

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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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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]

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30 Jul 2012, 06:19

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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.
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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]

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08 Jul 2013, 15:25

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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
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Comparing 2^34 with 10^10 Comparing 2^34 with 2^10 * 5^10 [Now we can cancel out common terms 2^10 from both sides] Comparing 2^24 with 5^10 Comparing (2^2)^12 with 5^10 Comparing (4)^12 with (5)^10 Comparing (4^6)^2 with (5^5)^2 [Now we can cancel out common powers 2 from both sides] Comparing (4)^6 with (5)^5 Comparing (4096) with (3125)

Since 4096 > 3125 therefore, 2^34 > 10^10 SUFFICIENT

Statement 2: x = 2^35

Since I know the exact value of x so a comparison can be established hence the statement is sufficient SUFFICIENT

Answer: Option D
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Re: Is x > 10^10 ? (1) x > 2^34 (2) x = 2^35 [#permalink]

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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]

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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.

(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]

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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.

Re: Is x > 10^10 ? (1) x > 2^34 (2) x = 2^35 [#permalink]

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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.

(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!
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Re: Is x > 10^10 ? (1) x > 2^34 (2) x = 2^35 [#permalink]

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08 Dec 2014, 01:34

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]

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