If xy/5 is an integer, is x divisible by 5?

Case 1 ->
y = 30
x = 1

NO, x is not divisible by 5

Case 2 -->
y=30
x=5
YES, x is divisible by 5


Use the above test case to reject A,B,C

and then off-course -> Smash that E

Answer E.

A few ways to solve this:

Using logic:
Since we know xy is divisible by 5, The only way to know for sure whether x is divisible by 5 or not is either information about x, or knowing that y definitely isn't divisible by 5 (in which case x definitely will be). Neither statement gives us any of these: (1) doesn't rule out y being divisible by 5 as well, and (2) tells us y is - which tells us nothing about x. Combining them doesn't help, for the same reason that (2) is insufficient. Not enough data.


Using the alternative method of using numbers:
(1) say y=3 > 3x/5 > x must be divisible by 5. But say y equals 30 > 30x/5 > x could equal 1 - insufficient!
2) As discussed, in 30x/5 we can't know whether x is or isn't divisible by 5. Insufficient!
Combined: once again, 30x/5 fits the criteria and can go both ways. Insufficient!

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 1) & 2) first.

Conditions 1) & 2):

x = 5, y = 30 : Yes
x = 1, y = 30 : No

Since the answer is not unique, both conditions together are not sufficient.
Therefore, the answer is 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.