Is x^3 + y^3 x^2 + y^2? (1) x + y x^2+y^2 (2) x^4 + y^4 x^2 + y^2

Question

Is  x^3+y^3>x^2 +y^2 ?

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

(2)  x^4 + y^4 > x^2+y^2

GMATGuruNY · Answer

The problem can be rephrased as follows:

Let x=1.1.
Substituting x=1.1 into the rephrased problem, we get:

Case 1:  y = 0.2 
Statement 1:  y^2-y=-0.16 , with the result that  y^2-y<-0.11 
Statement 2:  y^2-y^4 = 0.0384 , with the result that  y^2-y^4 < 0.2541 
Is  y^2-y^3<0.121 ?
In this case,  y^2-y^3=0.032 , so the answer to the rephrased question stem is YES.

Case 2:  y = 0.5 
Statement 1:  y^2-y=0.5^2-0.5 = -0.25 , with the result that  y^2-y<-0.11 
Statement 2:  y^2-y^4 = 0.1875 , with the result that  y^2-y^4 < 0.2541 
Is  y^2-y^3<0.121 ?
In this case,  y^2-y^3=0.125 , so the answer to the rephrased question stem is NO.

Since the answer is YES in Case 1 but NO in Case 2, the two statements combined are INSUFFICIENT.

E

This problem seems WAY beyond what the GMAT would expect of a test-taker -- even one performing at the very highest level.

hellothere88 · Answer

GMATGuruNY · Answer

The values for y were chosen through trial and error. 
I tested 0.1 through 0.5.
y=0.2 satisfied both statements and yielded an answer of YES when inserted into the rephrased question stem.
y=0.5 satisfied both statements and yielded an answer of NO when inserted into the rephrased question stem.

satyaki23 · Answer

Shouldn't the first option be sufficient to answer the question?
if x + y > x^2 + y^2, both x and y need to be less than 1 and non-negative. (as the right hand side is positive and the only way the sum of squares of two numbers is less than the sum of the numbers is if they lie between 0 and 1). 
Using this logic, x^3 + y^3 will always be less than x^2 + y^2 (as x and y are positive numbers between 0 and 1). Hence the answer will be NO for all the cases.
That's why i believe that option A should be the answer.

chetan2u · Answer

No, you can have x slightly more than 1 and y lesser than 1, closer to 0.5

x=1.2 and y = 0.5
 1.2+0.5>1.2^2+0.5^2……1.7>1.44+0.25 ….Yes
 1.2^3+0.5^3>1.2^2+0.5^2……1.728+0.125>1.44+0.25 ….Yes

Bp36mafia · Answer

Hello team , There seems to be some error in this question .
in 1st condition :- x +y > x^2 + y^2 , we can also manipulate this and rewrite it as x(x-1) + y (y-1) <0 and here we can see that this situation is only true when 0 < x,y <1 , other wise it will be positive and not true.
and if we reanalyze the question statement x^2(x-1) + y^2(y-1) >0 this is what we need to prove , and this can only be true if x,y >1.
So thus condition 1 gives us the inference that x , y is not > 1 but it 0 < x , y <1. ( Hence Sufficient)

in 2nd condition :- x^2 ( x^2 - 1) + y^2 ( y^2 - 1) > 0 this situation is only true when x , y >1 or x , y <-1 otherwise not . so using this we'll be able to state a yes or a no to our question statement.

Thus A is sufficient

StacyArko · Answer

Dear Experts,
I understand that a mix of fractions (for one of the variables) and improper fractions (for the other variable) works out well for this question. But how do you know which numbers/how should you process your thoughts to choose the numbers to test these cases, without wasting time?

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Is x^3 + y^3 > x^2 + y^2? (1) x + y >x^2+y^2 (2) x^4 + y^4 > x^2 + y^2

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