avohden wrote:

If x and y are integers, is x/3 an integer?

(1) x + 14 = 2y

(2) (2y+1)/3 is an integer.

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.

There are 2 variables and 0 equation. By Variable Approach method, the answer is C most likely.

The question asks if \(x\) is a multiple of \(3\).

Condition 1) : \(x + 14 = 2y\) is equivalent to \(x = 2y - 14\).

If \(y = 0\), then \(x = -14\) is not a multiple of \(3\).

If \(y = 7\), then \(x = 0\) is a multiple of \(3\).

Thus the solutions are not unique.

Hence this condition is not sufficient.

Condition 2) : We don't have any information about \(x\).

Thus this is not sufficient.

Condition 1) & 2) The second condition that \(\frac{2y + 1}{3}\) is an integer is equivalent to the statement that \(2y + 1\) is a multiple of \(3\).

If \(2y + 1\) is a multiple of \(3\), then \(2y\) has a remainder \(2\) when it is divided by \(3\), that is \(2y = 3k + 2\) for some an integer \(k\).

Then \(x = 2y - 14 = 3k + 2 - 14 = 3k - 12 = 3(k-4)\).

Thus \(x\) is a multiple of \(3\).

Therefore, C is the answer as expected.

Normally for cases where we need 2 more equations, such as original conditions with 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, we have 1 equation each in both 1) and 2). Therefore C has a high chance of being the answer, which is why we attempt to solve the question using 1) and 2) together. Here, there is 70% chance that C is the answer, while E has 25% chance. These two are the key questions. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer according to DS definition, we solve the question assuming C would be our answer hence using 1) and 2) together. (It saves us time). Obviously there may be cases where the answer is A, B, D or E.

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