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

Bunuel rocks this is a very nice solution compared to the other ones I ve seen

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.

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

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.

Crystal.

Classic mistake, which is what is probably REALLY being tested, realised it when re-reading my post.

Thanks for response though.

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?

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.

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!

Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

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

(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

Oh Bunuel, you are always awesome. For me, your approach is always faster than Manhanttan's.
Highly recommend GMAT club for all GMAT learner!!!

whenever there is a 10 (or its multiples) involved, i use log.

is x > 10^10
or log x > 10 log 10
or log x > 10 ?

1) x > 2^34
log x > 34*log 2
log x > 10.23 (sufficient)

2) has to be sufficient

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Is x > 10^10 ?

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

In case of inequality questions, it is important to note that the conditions are sufficient if the range of the question includes the range of the conditions.
There is 1 variable (x) in the original condition. In order to match the number of variables and the number of equations, we need 1 equation. Since the condition 1) and 2) each has 1 equation, there is high chance D is going to be the answer.

In the case of the condition 1), if we do the calculation, we get 10^10<10*(10^3)^3=10*(1000)^3<16*(1024)^3=(2^4)(2^10)^3=2^34. Since the range of the question includes the range of the conditions, the condition 1) is sufficient.

In the case of the condition 2), if x=2^35, the answer can always be either ‘yes’ or ‘no’. Therefore, the condition is sufficient, and the correct answer is D.


For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously, there may be cases where the answer is A, B, C or E.

My working is just a tad different.

Statement 2, enough
Statement 1,
x > 10^10
x > 2^10.5^10

So x > 2^34
x > 2^10.2^24

Now, my thinking was 5 is 2.5 times more than 2, so 5^10 will be closer to 2^20 but w/o working i cant say higher or lower , whereas 2^21 > 5^10 (definite as its 1 more than double), we have 2^24. So sufficient.
I know its not perfect but saved time for me, of course Bunuel's method is flawless !!.