moneyman wrote:Is x^4+y^4 > z^4 ?
(1) x^2+y^2 > z^2
(2) x+y > z
Ans E
This problem freaked me out..looks simple but I dont know what numbers to pick to test the answers..Guys pls tell me the best approach for problems like these.Thanks!!
Picking numbers is about making sure you have covered all possible ranges of number. Broadly these are the numbers you should check for. 1.) negative numbers 2.) positive numbers 3.) numbers between 0 and 1. When we have variables with even powers we usually need not check for negative numbers as a negative number raised to an even number is positive.
In the above question we get nothing from the question stem. So let us look at the first statement, the first statement says x^2+y^2>z^2. As I said before since x,y is raised to 2 an even number we dont have to test for negative numbers. So let us test it for numbers between 0 and 1 and for positive numbers greater than 1.
If x^2 is 0.6 and y^2 = 0.5 and z^2=0.9 , then as we can see 0.6+0.5>0.9 i.e. x^2+y^2>z^2 but x^4+y^4 = 0.36+0.25= 0.61<z^2=0.81. Now, let x^2=20 and y^2=6 and z^2=10 in which case x^2+y^2>z^2 and x^4+y^4= 436 > z^4=100. So we do not get a definite answer and hence this statement is insufficient.
The second statement says x+y>z, over here you will have to check for negative numbers too. So let x= 1 and y=2 and z = -10 here x+y>z but x^4+y^4 is clearly <z^4=100. Now if x=12 and y=16 and z=15 then x+y>z and x^4+y^4>z^4. Since we do not get a definite answer, hence this statement is also not sufficient.
Combining the 2 will also not give a definite answer hence the answer is E.
