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Bunuel
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reatinus
Why isn't A enough?
Y can also be negative. You need to know know the sign of the inequality before bringing it to the other side of the inequality.

This is a basic rule for inequality.

Statement 2 gives the sign of y and hence you need both.

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reatinus
Why isn't A enough?

You can check using 2 different scenarios

if x=4, y=1 then x/y>3 and 4 > 3(1) (i.e. Is x > 3y)

but if x=-4 and y=-1 then x/y>3 (per statement 1) but -4 < (-1)3 (i.e. the stem fails)

Hence the learning is that you can cross multiply only if you know the sign because the inequality changes accordingly.

Hope that helps!
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reatinus
Why isn't A enough?

x>3y? Yes/No?
a) x/y > 3
Here we can't multiply y on right side as we don't know the polarity of y.
So there are two cases -
1) if y is positive:
x/y > 3
x > 3y
2) if y is negative:
x/y>3
x < 3y
So, there are two possibilities and there is no definite Yes/No answer - so A is not sufficient.

b) y>0
We don't know about x.
x could be greater than y, equal to y, less than y - so, B is not sufficient.
Now, combine a) and b)
We know from a) y could be positive or y could be negative,
from b) we know y is positive
so, y is positive and if y is positive x>3y
C)
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Bunuel
Is x > 3y?

(1) x/y > 3
(2) y > 0

(1) x/y > 3

Let x =12 , y =3 .............x/y > 3................12 > (3) (3).....12 > 9.........Answer is Yes

Let x =-12 , y =-3 ............x/y > 3................12 > (3) (3)....-12 >- 9.........Answer is No

Insufficient

(2) y > 0

No info about x

Insufficient

Combine 1 & 2

From 2..........y is positive

From 1.......... x must be larger than 3y

Answer is always Yes

Answer: C
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Bunuel
Is x > 3y?

(1) x/y > 3
(2) y > 0


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)
x/y > 3
⇔ x > 3y since y > 0 from the condition 2)
Both conditions together 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 = 4 and y = 1, then the answer is yes.
If x = -4 and y = -1, then the answer is no.
The condition 1) only is not sufficient.

Condition 2)
Since we don't have any information about the variable x, the condition 2) only is not sufficient.

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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