I chose C and used the number 2 and 0 to test each equation. I like your method better though!Stuart Kovinsky wrote:Step 1: Analyze the ProblemNijeesh wrote:Is x > k?
(1) 2^x "¢ 2^k = 4
(2) 9^x "¢ 3^k = 81
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.
Is x > k?
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(1) 2^x "¢ 2^k = 4
4 = 2^2 = 2^(x + k)
so x + k = 2
so k = 2 - x (A)
(2) 9^x "¢ 3^k = 81
81 = 3^4 = 3^(2x + k)
so 2x + k = 4 (B)
Substitute A into B:
2x + 2 - x = 4
so x = 2
Subsitute into A, gives k = 0
2 > 0
Therefore x > k TRUE
4 = 2^2 = 2^(x + k)
so x + k = 2
so k = 2 - x (A)
(2) 9^x "¢ 3^k = 81
81 = 3^4 = 3^(2x + k)
so 2x + k = 4 (B)
Substitute A into B:
2x + 2 - x = 4
so x = 2
Subsitute into A, gives k = 0
2 > 0
Therefore x > k TRUE
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My answer is C
if we take x=2, k=0 then for
1st case x>k
but if we take x=0 and k=2 then also
1st case satisfy but k>x
cant be sure with only 1 so insufficient
taking same process in 2 we get the same result
X>k but also K>x Insufficient option
But we take both 1 and 2 then only for
x=2 and k=0 they are valid
thus both are Sufficient
Hence option C
if we take x=2, k=0 then for
1st case x>k
but if we take x=0 and k=2 then also
1st case satisfy but k>x
cant be sure with only 1 so insufficient
taking same process in 2 we get the same result
X>k but also K>x Insufficient option
But we take both 1 and 2 then only for
x=2 and k=0 they are valid
thus both are Sufficient
Hence option C
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