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Need shortcut for Statement 1

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Source: — Data Sufficiency |

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by Brent@GMATPrepNow » Tue Aug 05, 2014 12:18 pm
Is x > 10^10?

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

Statement 1: x > 2^34
NOTE: if 2^34 > 10^10, then x will be greater than 10^10, in which case the statement is sufficient.
If 2^34 < 10^10, then x may or may not be greater than 10^10, in which case the statement is not sufficient

Since the statement is sufficient only if 2^34 > 10^10, we can reword the target question as Is 2^34 > 10^10?
- Apply exponent laws to rewrite both sides: Is (2^10)(2^24) > (2^10)(5^10)?
- Divide both sides by 2^10 to get: Is 2^24 > 5^10?
- Apply exponent laws to rewrite both sides: Is (2^12)² > (5^5)²?
- Take the square root of both quantities to get: Is 2^12 > 5^5?

This is pretty manageable.
- Rewrite both sides to get: Is (2^6)(2^6) > (5^4)(5)?
- Evaluate to get: Is (64)(64) > (625)(5)?
- Estimate to get: Is 3600+ > ≈3000?
The answer is a definite YES
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: x = 2^35
If I wanted to, I could evaluate 2^35, and then determine whether or not x is greater 10^10
So, since I could use statement 2 to answer the target question with certainty, statement 2 is SUFFICIENT

Answer = D

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by [email protected] » Tue Aug 05, 2014 4:44 pm
Hi nandinitaneja,

Brent has shown one of the ways to re-organize the given values for comparison purposes. Here's another way to re-organize the numbers:

Once we get down to 2^12 vs. 5^5, to figure out which is bigger, we need to find a point at which the values are "pretty close"

2^7 = 128
3^5 = 125

With a little "playing around", it's not tough to get these values. So now we have....

2^12 = (2^7)(2^5) = (128)(32)
5^5 = (5^3)(5^2) = (125)(25)

Notice how each "piece" of 2^12 is bigger than the corresponding "piece" in 5^5. Those numbers prove that 2^12 is greater than 5^5.

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by GMATGuruNY » Wed Aug 06, 2014 3:07 am
nandinitaneja wrote:In the question below, what is the fastest way of finding out that 2^34 > 10^10 to prove statement 1 sufficient?

Is x > 10^10 ?

(1) x > 2^34

(2) x = 2^35

Thanks!
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�
Determine whether 2³� > 10¹�:
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:
2³� > 10¹�.
Result:
x > 2³� > 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.
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by Matt@VeritasPrep » Wed Aug 06, 2014 4:37 pm
One last trick: if you remember that 10^(.3) is approximately 2, you can do these pretty quickly.

For instance, 2³� becomes (10^.3)³�, or 10^(.3 * 34) or about 10^(10.3). This is bigger than 10¹�, so we're done.
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by GMATinsight » Wed Aug 06, 2014 6:27 pm
nandinitaneja wrote:In the question below, what is the fastest way of finding out that 2^34 > 10^10 to prove statement 1 sufficient?

Is x > 10^10 ?

(1) x > 2^34

(2) x = 2^35

Thanks!
Hi Nandita,

CONCEPT : We can compare the exponents when the Bases or the Powers among the exponents are same

i.g. If I ask Which one is greater between 2^2 and 2^3 then Instant answer will be 2^3 because Bases are same in both Numbers

and If I ask Which one is greater between 2^3 and 3^3 then Instant answer will be 3^3 because Powers are same in both Numbers

But the problem arises when Questions asks comparison between number with different bases and different powers

COURSE OF ACTION: Either make their bases same or Powers same and Compare the numbers in its parts

To Compare 2^34 and 10^10
We can write 2^34 = (2^10)^3 * 2^4 [We fine a power of 2 which is closer to some power of 10 and 2^10 = 1024 which is close to 10^3
i.e. 2^34 = (1024)^3 * 16 = (10^3)^3 * 16 = 10^9 * 16
on the other part 10^10 = (10^9) * 10

Blue Parts are same therefore now we have to compare 16 and 10
Now we know which one is bigger :)
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