Is x^4+y^4 > z^4 ?
(1) x^2+y^2 > z^2
(2) x+y > z
How to quickly arrive at answers for such problems?
Inequality DS
This topic has expert replies
-
- Master | Next Rank: 500 Posts
- Posts: 119
- Joined: Tue Jan 20, 2009 10:16 pm
- Thanked: 9 times
- GMAT Score:730
IMO E
you can solve the question by picking up numbers. I think is easy to see that if x,y and z are greater than 1 the statement is true and both statements will be sufficient
for instance x = 3, y = 4 and z = 2
we know that if a number is between 0 and 1 when you square it gets smaller so that is the kind of number you have to look for.
for z pick 1 since it does not matter how many times you square 1, the number will always be 1
statement 2)
you need to find two numbers which sum will be slightly greater than 1
let's say x=0.51, y=0.51, z =1
x + y > z ---> (0.51)^4 + (0.51)^4 > 1
I think is clear that (0.51)^4 is less than 0.5 so statement 2 is not sufficient
statement 1)
x^2 + y ^2 > z
same procedure, based on the previous example pick
x = sqrt(0.51) y = sqrt(0.51) z = 1
not sufficient
statement 1 and 2
this is the difficult part let's pick z = 1 as always and now two numbers that when square and sum will be greater than 1
x = 0.8 and Y = 0.8 z = 1
II) x + y > z
I) x^2 + y^2 > z^2 --> 0.64 + 0.64 > 1
x^4 + y^ 4 > z^4 -->2*(0.64)^2 > 1 --> 2 * 0.4 > 1
not sufficient
you can solve the question by picking up numbers. I think is easy to see that if x,y and z are greater than 1 the statement is true and both statements will be sufficient
for instance x = 3, y = 4 and z = 2
we know that if a number is between 0 and 1 when you square it gets smaller so that is the kind of number you have to look for.
for z pick 1 since it does not matter how many times you square 1, the number will always be 1
statement 2)
you need to find two numbers which sum will be slightly greater than 1
let's say x=0.51, y=0.51, z =1
x + y > z ---> (0.51)^4 + (0.51)^4 > 1
I think is clear that (0.51)^4 is less than 0.5 so statement 2 is not sufficient
statement 1)
x^2 + y ^2 > z
same procedure, based on the previous example pick
x = sqrt(0.51) y = sqrt(0.51) z = 1
not sufficient
statement 1 and 2
this is the difficult part let's pick z = 1 as always and now two numbers that when square and sum will be greater than 1
x = 0.8 and Y = 0.8 z = 1
II) x + y > z
I) x^2 + y^2 > z^2 --> 0.64 + 0.64 > 1
x^4 + y^ 4 > z^4 -->2*(0.64)^2 > 1 --> 2 * 0.4 > 1
not sufficient
- rahulg83
- Legendary Member
- Posts: 575
- Joined: Tue Nov 04, 2008 2:58 am
- Location: India
- Thanked: 18 times
- Followed by:4 members
- GMAT Score:710
IMO E, I simply plugged in x^2=2, y^2=3, and z^3=4 for statement 1 and x=2,y=3,z=4 for statement 2..in both the cases the inequality in the question gets REVERSED
Remember, the tricky part was this one. In any case, you can have a huge pool of (x,y,z) that SATISFIES the above inequality..
therefore E
Remember, the tricky part was this one. In any case, you can have a huge pool of (x,y,z) that SATISFIES the above inequality..
therefore E