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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 have to find the value of a + b.
Follow the second and the third step: From the original condition, we have 2 variables (a and b). To match the number of variables with the number of equations, we need 2 equations. Since conditions (1) and (2) will provide 1 equation each, C would most likely be the answer.
Recall 3 Principles and choose C as the most likely answer.
Let’s look at both conditions together. However, since the value of condition (1) is equal to the value of condition (2), by Tip 1, we get D as the most likely answer. Let’s look at each condition separately.
Condition 1) tells us that a = -2 and b = 2. In order for the equation to have more than one solution, the corresponding coefficients on both sides must be equal, respectively. Then we have the left-hand side 2(x + a) = 2x + 2a and we have 2x + 2a = bx – 4. Since the equation 2x + 2a = bx – 4 has more than one solution, we have 2 = b and 2a = -4. Thus, condition 1) tells us that a = -2 and b = 2.
Then we have a + b = -2 + 2 = 0.
The answer is unique, so the condition is sufficient according to Common Mistake Type 2, which states that the number of answers must be only one.
Condition 2) tells us that a = -b. Since we have |a| = |b| and ab < 0, a and b have different signs and a = -b. Thus, we have a + b = 0.
The answer is unique, so the condition is sufficient according to Common Mistake Type 2, which states that the number of answers must be only one.
Each condition alone is sufficient.
Therefore, D is the correct answer.
Answer: D
Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B, or D is the answer. This occurs in Common Mistake Types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D, or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
