Stuart Kovinsky wrote:

Nijeesh wrote:Is x > k?

(1) 2^x "¢ 2^k = 4

(2) 9^x "¢ 3^k = 81

Step 1: Analyze the Problem

We see "is", we think "yes/no question: if I can answer with a definite yes or definite no, sufficient; if I answer with a maybe or sometimes or not sure, insufficient".

What do I know? Absolutely nothing! We have 2 variables and 0 equations, so 2 equations would be nice.

Step 2: Evaluate the Statements

(1) 2^x * 2^k = 4

Using the rules of exponents, we now know that:

2^(x+k) = 2^2

so:

x + k = 2

No clue which one is bigger, so insufficient. Eliminate A and D.

(2) 9^x * 3^k = 81

By the same process as above, we know that:

(3^2)^x * 3^k = 3^4

3^2x * 3^k = 3^4

3^(2x + k) = 3^4

2x + k = 4

Again, no clue which one is bigger, so insufficient. Eliminate B.

Together:

2 equations, 2 unknowns, we can solve for x and k: sufficient. Choose C.

* * *

This question is a great illustration of the power of the "number of equations for number of unknowns" rule; the better you understand and apply that rule, the less math you'll have to do.

For example, if on this question we quickly realized that each statement would give us one distinct linear equation, and that each equation contained our two variables (and no others), we could have gone directly to "C" without actually solving each statement.

tgou008 wrote:I got to C, but used a different technique; testing numbers.

Is X>K? = Y/N question

(1) 2^x.2^k = 4

Possible value combinations for x & k that total to four include:
x = 1, k = 1. Here x = k
x = 0, k = 2. Here x < k
x = 2, k = 0 . Here x > k

As there are multiple possibilities for X>K, we can eliminate answers A and D.

(2) 9^x.3^k = 81

Possible value combinations for x & k that total to 81 include:
x = 2, k = 0. Here x > k
x = 0, k = 4. Here x < k
x = 1, k = 2 . Here x < k

Once again there are multiple possibilities for X>K, so we can eliminate B. Must now choose between C and E.

For it to be C, there must be 1 X vs K combination that is common across both (1) and (2). We can see that there is, the only combination common across both (1) and (2) is x= 2, k=0, which is sufficient as we can now see that x > k.

Therefore answer = C

That took me ~2mins to do. Not sure whether this or the other approach is quicker. What level difficultly would you say this question is?

Thanks