Bunuel wrote:

If x and y are integers, is \(\frac{y}{x} = \frac{17}{9}\) ?

(1) \(\frac{13}{7} < \frac{y}{x} < \frac{19}{10}\)

(2) \(x = 9\)

Are You Up For the Challenge: 700 Level QuestionsForget 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.

Since we have 2 variables (x and y) and 0 equations, C is most likely 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)

Since \(x = 9\), we should find the value of \(y\) such that \(13/7 < y/9 < 19/10\).

When we multiply all sides of the inequality by \(9\), we have \(16 < 117/7 < y < 171/10 < 18\) or\( y = 17\).

Since both conditions together yield a unique solution, they are sufficient.

Since this question is a statistics 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)

If\( x = 9, y = 17\), then the answer is 'yes'.

If \(x = 17 = 7 + 10, y = 32 = 13 + 19\), then the answer is 'no'.

Note. When we have \(\frac{a}{b} < \frac{c}{d}\) for positive integers \(a, b, c\) and \(d\), we have \(\frac{a}{b} < \frac{a+c}{b+d} < \frac{c}{d}\).

Since condition 1) does not yield a unique solution, it is not sufficient

Condition 2)

Since we don't have any information about y, condition 2) is not sufficient obviously.

Therefore, C is the answer.

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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