Hi jain2016,
We're given the function f(A) = A^2 + 3A. We're then asked how many times greater the value of 3f(A-B) is than B?
This question can be solved by TESTing VALUES.
1) B = 3
Without the value of A, there's no way to answer this question, but here's proof that Fact 1 is insufficient:
IF....
A = 3
3(f(3-3)) = 3(f(0) = 3(0 + 0) = 0
0 is 0 times greater than 3.
IF....
A = 4
3(f(4-3)) = 3(f(1) = 3(1 + 3) = 12
12 is 4 times greater than 3.
Fact 1 is INSUFFICIENT
2) A is a positive integer
We can use the same two TESTs (from above) here:
IF....
A = 3
B = 3
3(f(3-3)) = 3(f(0) = 3(0 + 0) = 0
0 is 0 times greater than 3.
IF....
A = 4
B = 3
3(f(4-3)) = 3(f(1) = 3(1 + 3) = 12
12 is 4 times greater than 3.
Fact 2 is INSUFFICIENT
Combined, we have the same two TESTs with different results.
Combined, INSUFFICIENT
Final Answer: E
GMAT assassins aren't born, they're made,
Rich
3f(a-b)
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Here's a succinct approach:
We want 3*f(a-b) - b, or
3*((a - b)² + 3(a - b)) - b, or
3*(a² - 2ab + b² + 3a - 3b) - b, or
3a² - 6ab + 3b² + 9a - 10b
To solve this, we'll need the value of a. Neither statement helps us here, though, so the answer must be E.
We want 3*f(a-b) - b, or
3*((a - b)² + 3(a - b)) - b, or
3*(a² - 2ab + b² + 3a - 3b) - b, or
3a² - 6ab + 3b² + 9a - 10b
To solve this, we'll need the value of a. Neither statement helps us here, though, so the answer must be E.
