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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.

Since we have 2 variables (x and y) 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):

\(x^2 + y^2 < 1\)

\(=> x^2< 1 - y^2 ≤ 1\) since \(y^2 ≥ 0\)

\(=> x^2< 1\)

\(=> -1 < x < 1\)

and

\(x^2 + y^2 < 1\)

\(=> y^2< 1 - x^2 ≤ 1\) since \(x^2 ≥ 0\)

\(=> y^2< 1\)

\(=> -1 < y < 1\)

Combining these two inequalities yields \(-1 < xy < 1\), so \(xy < 1\). Both conditions are sufficient, when taken together.

Since this question is an inequality question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.

Condition 1)

The argument showing that conditions 1 and 2 are sufficient, when taken together, only used condition 1. Therefore, condition 1 is sufficient.

Using the same argument as above,

\(x^2 + y^2 < 1\)

\(=> x^2< 1 - y^2 ≤ 1\) since \(y^2 ≥ 0\)

\(=> x^2< 1\)

\(=> -1 < x < 1\)

and

\(x^2 + y^2 < 1\)

\(=> y^2< 1 - x^2 ≤ 1\) since \(x^2 ≥ 0\)

\(=> y^2< 1\)

\(=> -1 < y < 1\).

So, \(-1 < xy < 1\), and condition 1) is sufficient.

Condition 2)

If \(x = \frac{1}{3}\) and \(y = \frac{1}{3,}\) then \(xy = \frac{1}{9} < 1\) and the answer is ‘yes’

If \(x = -2\) and \(y = -2\), then \(xy = 4 > 1\) and the answer is ‘no’

Since we don’t have a unique solution, condition 2) is not sufficient.

Therefore, A is the answer.

Answer: A

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.

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