M11-10

From (1) a = -4 and b = -3, so ab = 12, which does NOT contradict (2).

I think this is a high-quality question and I agree with explanation.

Can't value here be a=-4, b=-3

From (1) we got that a - b= -1 and a + b= -7. This gives a = -4 and b = -3 but we don't need to solve for a and b because we asked to find the value of a - b, not individual values of a and b.

They are the definite values solved from statement 1

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I chose a different approach:

Statement 1:
\(a^2=b^2+7\)

Since we know that a & b are negative integers, \(a^2\) & \(b^2\) are perfect squares
Lets list up the first perfect squares: 1, 4, 9, 16, 24, 36, 49, 64, ...
We know \(a^2\) is a perfect square that is 7 bigger than another perfect square.
If we look at our squares, we can quickly see that this is only the case when \(a^2 = 16\) and \(b^2 = 9\)
The distances between the squares will get bigger and bigger if you would continue to write them down.
Therefore, there will be no other case in which statement 1 is true.

This means that a = -4 and b = -3 (which fits to the question stem).
a-b = -4 - -3 = -1

SUFFICIENT

Statement 2:
\(ab=12\)

All possible negative factor pairs of 12 are:
-1 * -12
-2 * -6
-3 * -4

We know that a < b, but we still have several possibilities:
a = -12 and b = -1 --> a-b = -12 - -1 = -11
a = -6 and b = -2 --> a-b = -6 - -2 = -4

INSUFFICIENT

Answer A

wow, I'm just gonna drop this here. this is an absolutely phenomenal question !!! the level of prethinking that one has to do is respectable !!

in my case, when I noticed that we have a difference of squares and the difference of squares is given to us (which we can infer that there's only a unique combination for a and b), I stopped there and just assumed that yes we can find the value of a and b.

do you think during the exam this thinking process is good?

Your thinking process during the exam should be more thorough in order to avoid mistakes.

a^2 - b^2 = 7 has infinitely many solutions for a and b. When we restrict the solutions to integers, we still have four possible pairs of (a, b): (4, 3), (4, -3), (-4, 3), and (-4, -3). Only by further limiting the solutions to non-negative integers will you get a unique solution: (-4, -3).

So, during the exam, make sure to consider all given constraints and systematically narrow down the possibilities to avoid overlooking important information or making incorrect assumptions.

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

Similar PS question to practice:

https://gmatclub.com/forum/m41-429501.html­