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# Is x > 10^10 ? (1) x > 2^34 (2) x = 2^35

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

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20 Feb 2012, 18:55
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Is x > 10^10 ?

(1) x > 2^34
(2) x = 2^35
[Reveal] Spoiler: OA

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

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20 Feb 2012, 21:41
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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.

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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.

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

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30 Jul 2012, 13:22
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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.

Dabral

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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.

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.

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]

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04 Sep 2012, 03:49
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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.

Hope it's clear.
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Re: Is x > 10^10 [#permalink]

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18 Feb 2013, 00:35
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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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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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Re: Is x > 10^10 ? (1) x > 2^34 (2) x = 2^35 [#permalink]

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11 Oct 2016, 09:10
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dvinoth86 wrote:
Is x > 10^10 ?

(1) x > 2^34
(2) x = 2^35

Statement 1: x > 2^34

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

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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.

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.

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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.

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

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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]

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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.

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.

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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Re: Is x > 10^10 [#permalink]

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18 Feb 2013, 02:22
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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Re: Is x > 10^10 [#permalink]

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18 Feb 2013, 02:25
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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Re: Is x > 10^10 ? (1) x > 2^34 (2) x = 2^35 [#permalink]

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08 Jul 2013, 00:54
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Re: Is x > 10^10 ? (1) x > 2^34 (2) x = 2^35 [#permalink]

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31 Jul 2014, 14:15
dvinoth86 wrote:
Is x > 10^10 ?

(1) x > 2^34
(2) x = 2^35

(2) is 1 number, so it is sufficient
(1) compare 2^34 > 10^10 ?
-> 2^10 * 2^24 > 2^10 * 5^10
-> 2^24 > 5^10
-> 2^12 > 5^5 ( square root)
-> 2^10 * 2^2 > 5^3 * 5^2
-> 1028*4 > 125*25
-> 4112 > 3125 ( correct)

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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.

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.

Oh Bunuel, you are always awesome. For me, your approach is always faster than Manhanttan's.
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Re: Is x > 10^10 ? (1) x > 2^34 (2) x = 2^35 [#permalink]

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27 Dec 2015, 07:10
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Re: Is x > 10^10 ? (1) x > 2^34 (2) x = 2^35   [#permalink] 27 Dec 2015, 07:10

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