If n is an integer, x=?

[veenu08]

If n is an integer, x=?
1. 9^n=100,000*x
2. 1/10<x<1

OA C

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Hi veenu08,

The prompt for this DS question is a bit thin. You're told that N is an integer, but you're asked about X (which we know NOTHING about).

Fact 1 gives us a relationship between N and X. As N changes, X changes.

Different X values = inconsistent = INSUFFICIENT.

Fact 2 gives us a range of values for X, but we still don't know what X is. INSUFFICIENT.

Combining Facts, we have a real SMALL range for X and a relationship between X and Y

If N = 1, then 9 = 100,000X so X = .00009 This is NOT POSSIBLE because it's outside the range in Fact 2
If N = 2, then 81 = 100,000X so X = .00018 Also NOT POSSIBLE
If N = 3, then 729 = 100,000X so X = .00729 Also NOT POSSIBLE
If N = 4, then 6561 = 100,000X so X = .06561 Also NOT POSSIBLE
If N = 5, then 59049 = 100,000X so X = .59049 This IS A POSSIBLE VALUE FOR X
As N gets bigger, X becomes greater than 1, which is NOT POSSIBLE.

Notice how the decimal shifted with every additional "power of 9"? If you spot the math pattern, then you wouldn't need to calculate all of the above values. You'd just need to figure out which value(s) would put you in range defined by Fact 2.

There is ONLY 1 POSSIBLE ANSWER, which makes the final answer C (Combined Sufficient)

GMAT assassins aren't born, they're made,
Rich

[Matt@VeritasPrep]

S1 isn't sufficient: x could be almost anything.
S2 isn't sufficient: x could be an infinite number of real numbers.

With the two together, we know 10,000 < 9^n < 100,000. Quickly finding our powers of 9, we have 9, 81, 729, and 6561 for the first four. The next number will be 6561 * 9, which IS between 10,000 and 100,000. The number after that is 6561 * 81, which is clearly greater than 100,000. So n = 5, and the answer is C.