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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.

Condition 2):

If \(p = 3\), then \(q ≠ 4\) since \((p,q)\) is different from \((3,4).\)

If \(s = 4\), then \(r ≠ 3\) since \((r,s)\) is different from \((3,4)\).

So, \((p,q)\) is on the line \(x = 3\) and \((r,s)\) is on the line \(y = 4.\)

As these lines are perpendicular AB and AC are perpendicular, and triangle ABC is a right triangle.

Thus, condition 2) is sufficient.

Condition 1) is complicated. If you can’t figure out how to apply condition 1), CMT4(B) tells you to choose D as the answer.

Condition 1)

\((p-3)(r-3)+(q-4)(s-4)=0\)

\(=> (p-3)(r-3) = -(q-4)(s-4)\)

\(=> \frac{(p-3)(r-3)}{(q-4)(s-4)} = -1\) or \((q-4)(s-4) = 0\)

\(=> {\frac{(p-3)}{(q-4)}} {\frac{(r-3)}{(s-4)}} = -1\) or \(q = 4\) or \(s = 4\)

Case 1: \(\frac{(p-3)}{(q-4)} * {\frac{(r-3)}{(s-4})} = -1\)

\(\frac{(p-3)}{(q-4)}\) and \(\frac{(r-3)}{(s-4)}\) are the slopes of two sides of the triangle.

Since the product of these slopes is \(-1\), the two sides are perpendicular and the triangle is a right triangle.

Case 2: \(q = 4\)

If \(q = 4\), then \(p\) is not \(3\). Also, we must have \((p-3)(r- 3) = 0\). So, \(r = 3\) since \((p,q)\) is different from \((3,4)\).

Thus, \((p,q)\) lies on the line \(y = 4\) and \((r,s)\) lies on the line \(x = 3\).

As these lines are perpendicular, AB and AC are perpendicular, and triangle ABC is a right triangle.

Case 3: \(s = 4.\)

A similar argument to the one used for case 2 shows that triangle ABC is a right triangle.

Thus, condition 1) is also sufficient.

Therefore, D is the answer.

Answer: D

In cases where 3 or more additional equations are required, such as for original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2), when taken together. Obviously, there may be occasions on which the answer is A, B, C or D.

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