If x, y, and z are distinct positive integers, is x(y – z) less than y

Q. x(y-z) < y(x-z)?
or, -xz < -yz
or, (since z is a distinct +ve integer, we can cancel z from both sides)
or, y<x? -> this is the question we need to answer.

Stmt1: suffices.
Stmt2: not sufficient.

Hence, A.

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

The first step of VA(Variable Approach) method is modifying the original condition and the question, and rechecking the number of variables and the number of equations.
We can modify the original condition and question as follows.

The question is asks x(y-z) < y(x-z) or x(y-z) - y(x-z) < 0.
It is equivalent to -xz +yz < 0 or z(y-x) < 0
Since z > 0, the question asks if y - x < 0.

Thus, the condition 1) is sufficent, since the answer is always no.

Therefore, the answer is A.

For cases where we need 3 more equations, such as original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 80 % chance that E is the answer, while C has 15% chance and A, B or D has 5% chance. Since E is most likely to be the answer using 1) and 2) together according to DS definition. Obviously there may be cases where the answer is A, B, C or D.