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indonesianfintech
how did you choose the numbers for test case?

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.
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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.
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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.

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
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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
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satyaki23
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.

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


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?

Posted from my mobile device
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