Is x - x^2 y - y^2 ? (1) x y (2) x^2 y^2

Question

Is  x - x^2 > y - y^2  ?

(1)  x > y

(2)  x^2 > y^2

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lacktutor · Answer

Is  x —x^{2} > y—y^{2}  ?

Good approach is picking numbers

(Statement1): x > y 
—> if 3 > 2, then —6 > —2 (No)
—> if —2 > —3, then —6 > —12(yes)

Insufficient

(Statement2):  x^{2} > y^{2}

If  3^{2} > 2^{2} , then —6 > —4 (No)

If  (1.5)^{2} > (—1)^{2} , then 1.5 —2.25 > —1 —1 —> —0.75 > —2 (Yes)

Insufficient

Taken together 1&2, 
The same thing as statement2 
Insufficient

The answer is E

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minustark · Answer

x -x^2 > y -y^2  = x-y > (x+y) (x-y)

(1) x > y = x- y > 0, so we need to know the sign of x+y. insufficient.

(2)  x^2 > y^2  =  x^2- y^2 > 0 . but we don't know the value of x-y. insufficient

Together, both x - y and  x^2 -y^2  is positive. Lets take x as 1/2 and y as 1/4. so,  x - x^2  will be (1/2) which is greater than  y- y ^2  will be 3/16. Again when x=3, y =2 , then  x- x^2  becomes -6 and  y -y^2  becomes -2, so  x-x^2 < y -y^2 . insufficient.

E is the answer.

QuantMadeEasy · Answer

x - x^2 > y - y^2
x - y > x^2 - y^2 
x - y > (x - y) (x + y)
(x - y) ( x + y -1) < 0

(1) x > y 
x - y >0; ( x + y -1) can be positive or negative. Insufficient

(2) x^2 > y^2; taking square root on both the sides
|x| > |y|; signs of x and y are unknown. Insufficient

(1)+(2) still signs of x and y are unknown. Insufficient

E is correct.

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 is equivalent to (x-y)(x+y-1) < 0 for the followings. x - x^2 > y - y^2 ⇔ 0 > x^2 - y^2 - x + y ⇔ x^2 - y^2 - x + y < 0 ⇔ (x+y)(x-y) - (x-y) < 0 ⇔ (x-y)(x+y-1) < 0 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) If x = 2, y = 1 , then (x-y)(x+y+1) = 1 \cdot 2 = 2 > 0 and the answer is 'no'. If x = 2, y = -1.5 , then (x-y)(x+y-1) = 3.5 \cdot (-0.5) < 0 and the answer is 'yes'. Since both conditions together do not yield a unique solution, they are not sufficient. Therefore, E 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.

goaltop30mba · Answer

Bunuel  please help ! Is there any approach other than putting variables?

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egmat   chiranjeev  please help!

GMATinsight · Answer

INSEADIESE

Question : Is  x - x^2 > y - y^2 ?
   Question : Is  x - y > x^2 - y^2 ?

Even after combining the two statements

Case 1 : x = 1/2 and y = 1/3 i.e.   x - y > x^2 - y^2

Case 2 : x = 3 and y = 2 i.e.   x - y < x^2 - y^2

NOT SUFFICIENT

Answer: Option E

Vishalcv · Answer

can't we modify the equation to x+y<1 ?

x-y>x^2-y^2
x-y > (x-y)(x+y)
1>x+y

MBAB123 · Answer

Hey  Vishalcv , you can't simply cancel out (x-y) here as by doing so you will ignoring a lot of possible solutions. eg - if x=4 and Y=10, the equation holds true, but does that mean x+y <1? No.

Thekingmaker · Answer

fractions were my first choice 1/2 and 1/4 it's shows greater however when you add 1 and 0 it becomes equal even after eliminating negative , identifying zero is the key hence clearly insuff both statements IMO E

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