Working independently at their respective constant rates, machines X

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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The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary.

Assume x and y are numbers of minutes that machines X and Y take to fill the order, respectively.
Then we have \(\frac{1}{15} = \frac{1}{x} + \frac{1}{y}\).

Since we have 2 variable and 1 equations, D is most likely to be the answer. So, we should consider each condition on its own first.

Condition 1)
We have \(\frac{1}{15} = \frac{1}{60} + \frac{1}{y}\) or \(\frac{1}{15} - \frac{1}{60} = \frac{1}{20} = \frac{1}{y}\).
Then we have \(y = 20\).
Thus, \(\frac{1}{x+y}/\frac{1}{x}= \frac{1}{80}/\frac{1}{20} = \frac{1}{4}\).


Since condition 1) yields a unique solution, it is sufficient.

Condition 2)
We have \(\frac{1}{15} = \frac{1}{x} + \frac{1}{20}\) or \(\frac{1}{15} - \frac{1}{20} = \frac{1}{x}\).
Then we have \(x = 60\).
Thus, \(\frac{1}{x+y}/\frac{1}{x}= \frac{1}{80}/\frac{1}{20} = \frac{1}{4}\).

Since condition 2) yields a unique solution, it is sufficient.

Therefore, D is the answer.

If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A,B,C or E.

Can you please explain that whilst you have put the formula as 1/X+Y / 1/X why have you used value of X as 20 min when the value of X is 60 min

Work problems can be solved by the rate - work/time concept.