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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] New post 20 Feb 2012, 17:55
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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] New post 20 Feb 2012, 20:41
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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.

Answer: D.
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Re: Is x > 10^10 ? (1) x > 2^34 (2) x = 2^35 [#permalink] New post 21 Feb 2012, 19:36
u r awesome Bunuel :-D
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Re: Is x > 10^10 ? (1) x > 2^34 (2) x = 2^35 [#permalink] New post 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.

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] New post 30 Jul 2012, 02:09
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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] New post 30 Jul 2012, 05:19
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] New post 30 Jul 2012, 12:22
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.

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Re: Is x > 10^10 ? (1) x > 2^34 (2) x = 2^35 [#permalink] New post 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.
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Re: Is x > 10^10 ? (1) x > 2^34 (2) x = 2^35 [#permalink] New post 04 Sep 2012, 02:49
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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 ? (1) x > 2^34 (2) x = 2^35 [#permalink] New post 04 Sep 2012, 02:54
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] New post 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.

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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Re: Is x > 10^10 [#permalink] New post 17 Feb 2013, 23:35
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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 [#permalink] New post 18 Feb 2013, 01:22
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] New post 18 Feb 2013, 01:25
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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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Re: Is x > 10^10 ? (1) x > 2^34 (2) x = 2^35 [#permalink] New post 07 Jul 2013, 23:54
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Re: Is x > 10^10 ? (1) x > 2^34 (2) x = 2^35 [#permalink] New post 08 Jul 2013, 14:25
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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] New post 31 Jul 2014, 13:15
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)

answer D
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Re: Is x > 10^10 ? (1) x > 2^34 (2) x = 2^35   [#permalink] 31 Jul 2014, 13:15
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