Tricky option A.

[bryan88]

Is x> 10^10?

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

Quick solution to verify A?

[Mike@Magoosh]

Hi, there. I'll give my 2 cents on this.

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

The first thing I'll say is --- this seem to me well beyond what the real GMAT expects of students. This kind of exponential estimation is not something I have seen asked on the real GMAT.

So, without a calculator, what can we do?

We know 2^10 = 1024 ~ 1000 = 10^3, so
2^20 ~ 10^6
2^30 ~ 10^9

(BTW, the "kilo-" and "Mega-" they use in computer science, as in kb or Mb, are actually 2^10 and 2^20, not 1000 and 100000.)

Technically, the powers of two are slightly larger than the powers of 10, so 2^30 > 10^9.

Well, 2^4 = 16 > 10, so 2^34 = (2^30)*(2^4) > (10^9)*(10) = 10^10

So, this implies that if x is 2^34 or anything greater than that, then it's greater than 10^10. The calculator verifies that 17,179,869,184, which is slightly larger than 10^10.

Both statements are sufficient. Answer = D

Again, this kind of estimation is beyond what I have seen the GMAT asking of students.

Here's an example of an estimation question the GMAT is much more likely to ask.
https://gmat.magoosh.com/questions/50
When you submit your answer to that, the next page will have the video explanation of the problem.

Does all this make sense? Please let me know if you have any questions.

Mike

[GMATGuruNY]

Is x> 10^10?

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

Quick solution to verify A?

To compare exponents, try to get SIMILAR BASES.
It is helpful to have memorized the powers of 2 up to 2¹�.
2¹� = 1024 ≈ 10³.

Statement 1: x > 2³�
2³� > 10¹�
2¹� * 2¹� * 2¹� * 2� > 10¹�
10³ * 10³ * 10³ * 16 > 10¹�
10� * 16 > 10� * 10
The lefthand side is greater than the righthand side.
Thus, x > 10¹�.
SUFFICIENT.

Statement 2: x = 2³�
Since the value of x is known, we can determine whether x > 10¹�.
SUFFICIENT.

The correct answer is D.

[bx]

Is x> 10^10?

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

Quick solution to verify A?

10^10 = (2*5)^10 = (2^10)*(5^10)

I use the following approximation:
5^3 = 125 ≈ 128 = 2^7 and 5 ≈ 2^2

Thus,
10^10 = 2^10 * (5^3)^3 * 5 ≈ 2^10 * 2^21 * 2^2 ≈ 2^33

Hence, both statements are sufficient