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
