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Re: Is x > 10^10 ? (1) x > 2^34 (2) x = 2^35 [#permalink]
20 Feb 2012, 20:41

19

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Expert's post

5

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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.

(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

5

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

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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!!!

gmatclubot

Re: Is x > 10^10 ? (1) x > 2^34 (2) x = 2^35
[#permalink]
08 Dec 2014, 00:34

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