If x and y are positive integers and x>y, is x/y an integ

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[GMAT math practice question]

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

1) x is a multiple of 10
2) y is a multiple of 2

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by Brent@GMATPrepNow » Fri May 04, 2018 5:48 am

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Max@Math Revolution wrote: If x and y are positive integers and x>y, is x/y an integer?

1) x is a multiple of 10
2) y is a multiple of 2
Target question: Is x/y an integer?

Given: x and y are positive integers and x>y

Statement 1: x is a multiple of 10
No information about y.
Statement 1 is NOT SUFFICIENT

Statement 2: y is a multiple of 2
No information about x.
Statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
There are several values of x and y that satisfy BOTH statements. Here are two:
Case a: x = 10 and y = 2. In this case, x/y = 10/2 = 5 (which is an integer). So, the answer to the target question is YES, x/y IS an integer
Case b: x = 10 and y = 4. In this case, x/y = 10/4 = 2.5 (which is not an integer). So, the answer to the target question is NO, x/y is NOT an integer
Since we cannot answer the target question with certainty, the combined statements are NOT SUFFICIENT

Answer: E

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Brent
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by Max@Math Revolution » Mon May 07, 2018 1:06 am

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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.
Since we have 2 variables (x and y) and 0 equations, C is most likely to be 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):
If x = 10 and y = 2, x / y = 5 is an integer.
If x = 10 and y = 4, x / y = 10 / 4 = 5 / 2 is not an integer.
Since we don't have a unique solution, both conditions together are not sufficient.

Therefore, E is the answer.
Answer: E

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.