[GMAT math practice question]
(Number Properties) A, B, and C are positive numbers. We have an equation A^2 + 2B^2 = C^2. What is the value of A + B + C?
1) A, B, and C are less than 11.
2) A, B, and C are integers.
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Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
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The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary.
Since A^2 + 2B^2 = C^2 , we have 2B^2 = C^2 - A^2 or 2B^2 = (C + A)(C - A).
C^2 - A^2 = (C + A)(C - A) is an even number, both A and C must either odd numbers or even numbers.
Then 2B^2 is a product of even numbers, and 2B^2 is a multiple of 4.
B is an even number.
Since we have 3 variables (A, B, and C) and 0 equations, E is most likely the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.
Conditions 1) & 2)
A, B, and C are positive integers less than or equal to 10.
Since we have 0 < C^2 - A^2 = 2B^2 < 10^2 or 0 < C^2 - A^2 = 2B^2 < 100, we have 0 < B < √50 and B = 2, 4 or 6.
We have C + A > C – A.
Case 1: B = 2
Since we have (C + A)(C - A) = 8, we have C + A = 4 and C – A = 2.
Then we have C = 3, A = 1 and A + B + C = 1 + 2 + 3 = 6.
Case 2: B = 4
Since we have (C + A)(C - A) = 32, we have C + A = 16, C - A = 2 or C + A = 8, C – A = 4.
If C + A = 16, C – A = 2, then we have A = 7, C = 9 and A + B + C = 7 + 4 + 9 = 20.
Since both conditions together do not yield a unique solution, they are not sufficient.
Therefore, E is the answer.
Answer: E
In cases where 3 or more additional equations are required, such as for original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B, or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2) when taken together. Obviously, there may be occasions on which the answer is A, B, C, or D.