What is the value of (a+b)^2?

Each statement alone is clearly insufficient.

When we take the statements together we have:
\(ab=0\) and \(a^2-2ab+b^2=36\), thus \(a^2+b^2=36\) --> \((a+b)^2=a^2+2ab+b^2=36+0=36\). Sufficient.

Answer: C.
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Hi All,

This question can be solved with Algebra (and recognizing Classic Quadratics) or by TESTing VALUES.

We're asked for the value of (A+B)^2

Fact 1: AB = 0

This means that either A or B or Both = 0

IF...
A = 0
B = 0
(0+0)^2 = 0

IF....
A = 1
B = 0
(1+0)^2 = 1
Fact 1 is INSUFFICIENT

Fact 2: (A-B)^2 = 36

This ultimately means that (A-B) = 6 or -6

IF.....
A = 6
B = 0
(6+0)^2 = 36

IF....
A = 7
B = 1
(7+1)^2 = 64
Fact 2 is INSUFFICIENT

Combined, we know....
A or B or Both = 0
(A-B) = 6 or -6

IF....
A = 0
B = 6 OR -6
(0+6)^2 = (0-6)^2 = 36

IF....
B = 0
A = 6 OR -6
(0+6)^2 = (0-6)^2 = 36

The answer is ALWAYS 36.
Combined, SUFFICIENT

Final Answer:

GMAT assassins aren't born, they're made,
Rich

\((a+b)^2= a^2+b^2+2ab\)

St1
ab=0
=> either a=0 or b=0 or both.
=> \((a+b)^2= a^2+b^2\)
Since we cant determine this value. Insufficient

St2
(\(a-b)^2=36\)
=>\(a^2+b^2-2ab=36\)
Insufficient

St1+St2
since ab=0, \((a-b)^2=a^2+b^2=36 = (a+b)^2\)
Sufficient

Answer C.

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.
Visit https://www.mathrevolution.com/gmat/lesson for details.

Since we have 2 variables (a and b) and 0 equations, C is most likely to be 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)
\((a+b)^2 = a^2 + 2ab + b^2 = a^2 - 2ab + b^2 + 4ab = (a-b)^2 + 4ab = 36 + 4*0 = 36\).
Thus, both conditions together are sufficient.

Therefore, C is the answer.

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.

Asked: What is the value of (a+b)^2?
\((a+b)^2 = a^2 + b^2 + 2ab = (a-b)^2 + 4ab\)
(1) ab = 0
since a or b=0
But values of a & b are unknown
NOT SUFFICIENT

(2) (a-b)^2 = 36
\((a+b)^2 = a^2 + b^2 + 2ab = (a-b)^2 + 4ab\)
Since value of ab is unknown
NOT SUFFICIENT

Combining (1) & (2)
(1) ab = 0
(2) (a-b)^2 = 36
\((a+b)^2 = a^2 + b^2 + 2ab = (a-b)^2 + 4ab = 36 + 0 = 36\)
SUFFICIENT

IMO C

IMO A

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