Is x/y 0? (1) x/(x + y) 0 (2) y/(x + y) 0

Question

Is  \frac{x}{y} > 0 ?

(1)  \frac{x}{x + y} > 0

(2)  \frac{y}{x + y} > 0

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IanStewart · Answer

Neither Statement is sufficient alone -- for Statement 1, of course x/y could be positive, but we could have x = 2 and y = -1, and then x/y is negative. Similarly for Statement 2, we could have y = 2 and = -1, and x/y can be negative.

Using both Statements, Statement 1 tells us that x and x+y have the same sign, and Statement 2 tells us that y and x+y have the same sign. If x and y both have the same sign as x+y, they have the same sign as each other, which ensures x/y is positive, so the answer is C.

MathRevolution · Answer

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. Visit https://www.mathrevolution.com/gmat/lesson for details. 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. The question \frac{x}{y} > 0 is equivalent to xy > 0 by multiplying both sides by the square of y . 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) Condition 1) \frac{ x }{ {x+y}} > 0 is equivalent to x(x+y) > 0 by multiplying both sides by the square of x+y , which means x and x+y have same signs. Condition 2) \frac{ y }{ {x+y}} > 0 is equivalent to y(x+y) > 0 by multiplying both sides by the square of x+y , which means x and x+y have same signs. Thus x and y have same signs and we have xy > 0 . Since both conditions together yield a unique solution, they are sufficient. Since this question is an inequality 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) (1,1) and (2,-1) are possible pairs of (x,y) . If x=1 and y=1 , then \frac{x}{y} = 1 > 0 and the answer is 'yes'. If x=2 and y=-1 , then \frac{x}{y} = -2 < 0 and the answer is 'no'. Since condition 1) does not yield a unique solution, it is not sufficient. Condition 2) (1,1) and (2,-1) are possible pairs of (x,y) . If x=1 and y=1 , then \frac{x}{y} = 1 > 0 and the answer is 'yes'. If x=-1 and y=2 , then \frac{x}{y} = -2 < 0 and the answer is 'no'. Since condition 2) does not yield a unique solution, it 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.

Andrewcoleman · Answer

Hi  MathRevolution  !

I have been employing the variable approach method! I am still learning about it. Why does the stem have 0 equations instead of the one that is there?

I want to learn more about the VA method and am going to sign up for your course

Thanks!

TestPrepUnlimited · Answer

The question is asking if x and y have the same sign. We can check this by plugging in numbers. Statement 1: We can have both x and y positive for the same signs, or x = 5 and y = -1 for example for opposite signs so insufficient. Statement 2: We can have both x and y positive for the same signs, or y = 5 and x = -1 for example for opposite signs so insufficient. Combined: Now let's note x + y is on both equations. If x + y is positive we get x > 0 and y > 0. If x + y is negative we get x < 0 and y < 0. In either case, x and y have the same sign so combined this is sufficient.

Papist · Answer

I realized this question is actually pretty easy after wasting ~3 minutes plugging in test values.

In order for this to be true, x and y must either both be positive or both be negative. S1: Multiply both sides of the equation by the denominator and you get a new equation of x>0. However, we have no information about y. NOT SUFFICIENT. S2: Multiply both sides of the equation by the denominator and you get a new equation of y>0. However, we have no information about x. NOT SUFFICIENT. S1+S2: We know that both x and y are positive, so the question will always be true. SUFFICIENT. ANSWER: C

kungfury42 · Answer

One of the more challenging problems I've done in a while

Statement 1: \frac{x}{x+y}>0 Simple knowledge of numbers tells us that if a number is positive its reciprocal too must be positive Since, \frac{x}{x+y}>0 therefore, \frac{x+y}{x}>0 \frac{x+y}{x}>0 is equivalent to saying 1+\frac{y}{x}>0 or \frac{y}{x}>-1 Now since \frac{y}{x}>-1 it can be both negative or positive in value. For ex: \frac{y}{x} can be both -0.5 and 5 If \frac{y}{x}=-0.5 then \frac{x}{y}=-2 If \frac{y}{x}=5 then \frac{x}{y}=\frac{1}{5} Clearly, we cannot tell whether \frac{x}{y}>0 Statement 2: \frac{y}{x+y}>0 Using a similar analysis as above we will arrive at \frac{x}{y}>-1 and we again would not be able to tell whether \frac{x}{y}>0 Clearly, even statement 2 by itself is insufficient. Combining the two statements we get \frac{y}{x}>-1 and \frac{x}{y}>-1 Now we know that for both of these statements to co-exist, neither of \frac{x}{y} or \frac{y}{x} can lie between -1 and 0 because any number that lies between -1 and 0 has its reciprocal <-1 but here we know that both the numbers are reciprocal of each other and both are >-1 Hence, we can safely say that using both the statements together \frac{x}{y}>0 and option C is our answer.

Is x/y > 0? (1) x/(x + y) > 0 (2) y/(x + y) > 0

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