I can not solve but answer is either B or D. I would have guessed B.

Anurag@Gurome wrote:

goyalsau wrote:Is x > 10^10 ?

(1) x > 2^34

(2) x = 2^35

Statement 1: x > 2^34

Now, (2^34) = (2^4)*(2^30) = (2^4)*((2^10)^3) = (2^4)*(1024)^3
Now, 1024 > 1000 => (1024)^3 > (1000)^3 = (10^9)
Now, (16)*(10^9) > (10)*(10^9) => (2^4)*(10^9) > (10^10) => (2^4)*(1024)^3 > (10^10)

Thus, x > (2^34) > (10^10)

Sufficient.

Statement 2: x = 2^35
Without applying the same analysis as of statement 1, we can say this is sufficient as we have a definite value for x.

Sufficient.

The correct answer is D.

HI! Anurag,

Well i able to understand your explanation, I think the important take way is to make both sides similar, Like you 1024 = 2^10 = 10 ^3 ... I know equal to is not the right sign over here But actually you were trying to reach a position from where you can compare both sides.

This time i know 2^10 = 1024, but what if they replace 2 with 3 or 4 , I have no idea what 3^10 or 4^10. is??? , I am sure by the time i end up calculating my 2 minutes will already be over.

So please if you have any other approach : : :, Do share that otherwise, God Help Me,......

Anurag@Gurome wrote:

goyalsau wrote:This time i know 2^10 = 1024, but what if they replace 2 with 3 or 4 , I have no idea what 3^10 or 4^10. is??? , I am sure by the time i end up calculating my 2 minutes will already be over.

Don't be so stressed.
If you are able to understand how I did it (which I assume you are as you've noted the basic idea is "trying to reach a position from where you can compare both sides") then you'll be able do it if they replace the numbers. Because if they do give 3 or 4 in place of 2, they'll change the other side accordingly.

And you know what 4^10 is! It's (2^10)^2.
Most of the times these inequalities involve power of primes or numbers which are easy to deal with like 10. So don't worry. If you feel comfortable with the above method you can do it as the above method is not so easy one.

Thanks Anurag..

goyalsau wrote:

Anurag@Gurome wrote:

goyalsau wrote:Is x > 10^10 ?

(1) x > 2^34

(2) x = 2^35

Statement 1: x > 2^34

Now, (2^34) = (2^4)*(2^30) = (2^4)*((2^10)^3) = (2^4)*(1024)^3
Now, 1024 > 1000 => (1024)^3 > (1000)^3 = (10^9)
Now, (16)*(10^9) > (10)*(10^9) => (2^4)*(10^9) > (10^10) => (2^4)*(1024)^3 > (10^10)

Thus, x > (2^34) > (10^10)

Sufficient.

Statement 2: x = 2^35
Without applying the same analysis as of statement 1, we can say this is sufficient as we have a definite value for x.

Sufficient.

The correct answer is D.

HI! Anurag,

Well i able to understand your explanation, I think the important take way is to make both sides similar, Like you 1024 = 2^10 = 10 ^3 ... I know equal to is not the right sign over here But actually you were trying to reach a position from where you can compare both sides.

This time i know 2^10 = 1024, but what if they replace 2 with 3 or 4 , I have no idea what 3^10 or 4^10. is??? , I am sure by the time i end up calculating my 2 minutes will already be over.

So please if you have any other approach : : :, Do share that otherwise, God Help Me,......

goyalsau,

I think we should remember the powers of 2 upto 10 (which you already do... And 2^10 almost equal to 1000 is very handy and useful in other fields where we deal with computer memory or bytes of data, etc..), and power of 3 to 5 upto 5 . Apart from that GMAT question wouldn't test us remembering these things. They may twist it in a way so that we use prime factorization technique to deal with the case.
For example they may ask where you need to find 6^5, But we may get away with 6^5 = 2^5*3^5.

Here is how I solved this question (It is exactly in theory as Anurag solved, but slight differently represented ) :

Lets calculate the fraction :

2^34/10^10 = (2^10)*(2^24)/(10^3)*(10^7) > (2^24)/(10^7) = (2^10)*(2^14)/(10^3)*(10^4) > (2^14)/(10^4)
= (2^10)*(2^4)/(10^3)*10 > 16/10 > 1

So, Sufficient.

Using this approach, you can see that we can solve the problem if it uses 3 or 5 (And believe me they wouldn't choose a value which will lead to tiresome calculation).

Thanks