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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.
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Since we have 3 variables (x, y, and z) and 0 equations, E is most likely to be the answer. Let’s look at both conditions together. However, since the value of condition (1) is equal to the value of condition (2), by Tip 1, we get D as the most likely answer. Let’s look at each condition separately
Let’s look at the condition 1). It tells us that since we have z ≥ 1, y ≥ 2, and x ≥ 3, we have 1/z ≤ 1, 1/y ≤ 1/2, and 1/x ≤ 1/3.
1/x + 1/y + 1/z ≤ 1 + 1/2 + 1/3 = 1 + 3/6 + 2/6 = 1 + 5/6 = 6/6 + 5/6 = 11/6 < 2 from condition 1).
Since the unique positive integer less than 2 is 1, we have 1/x + 1/y + 1/z = 1.
The actual values of x, y, and z are 6, 3, and 2, respectively.
The answer is unique, so the condition is sufficient according to Common Mistake Type 2, which states that the number of answers must be only one.
Let’s look at the condition 2). It tells us that
Since 2 and 3 are unique consecutive integers and prime numbers, we have y = 3 and z = 2.
If x = 6, y = 3 and z = 2, then 1/x + 1/y + 1/z = 1/6 + 1/3 + 1/2 = 1/6 + 2/6 + 3/6 = 6/6 = 1.
The answer is unique, so the condition is sufficient according to Common Mistake Type 2, which states that the number of answers must be only one.
Each condition ALONE is sufficient.
Therefore, D is the correct 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.
