Bunuel wrote:

What is the value of |a + b|?

(1) (a + b + c + d) (a + b – c – d) = 16

(2) c + d = 3

Kudos for a correct solution.

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.

Since we have 2 variables ( a and b ) and 0 equations, C is most likely to be the answer.

Thus, we need to consider both conditions together first.

\((a+b+c+d) (a+b–c–d) = 16\)

\({(a+b)+(c+d)}{(a+b)–(c+d)} = 16\)

\((a+b)^2 - (c+d)^2 = 16\)

\((a+b)^2 = 25\) since \(c + d = 3\)

\(|a+b| = 5\)

They are sufficient.

By CMT(Common Mistake Type)4, we need to check A or B and consider each condition only.

Condition 1)

\((a+b+c+d) (a+b–c–d) = 16\)

\({(a+b)+(c+d)}{(a+b)–(c+d)} = 16\)

\((a+b)^2 - (c+d)^2 = 16\)

This is not sufficient, since we don't know \(c+d\).

Condition 2)

We don't have anything about \(a\) and \(b\).

This is not sufficient either.

Therefore, the answer is C.

Normally, in problems which require 2 or more additional 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).

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