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23 Jul 2009, 20:29
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1) x^2 + y^2 > Z^2
X^2, Y^2, Z^2 are always positive numbers, so squaring on both sides does not have any effect on the inequality
i.e
(x^2 + y^2)^2 > (Z^2)^2
X^4 + Y^4 + 2X^2Y^2 > Z^4
All the values mentioned above are positive numbers. So obviously
X^4 + Y^4 > Z^4
2) X + Y > Z
In this scenario nothing is mentioned about X, Y & Z. They might be either positive or negative numbers.
Lets suppose X= 1, Y = 1, Z = -50
2> -50
squaring both sides, 4 Z^4 holds true.
So 1 alone is sufficient but 2 is not required and the answer is "A".
X^2, Y^2, Z^2 are always positive numbers, so squaring on both sides does not have any effect on the inequality
i.e
(x^2 + y^2)^2 > (Z^2)^2
X^4 + Y^4 + 2X^2Y^2 > Z^4
All the values mentioned above are positive numbers. So obviously
X^4 + Y^4 > Z^4
2) X + Y > Z
In this scenario nothing is mentioned about X, Y & Z. They might be either positive or negative numbers.
Lets suppose X= 1, Y = 1, Z = -50
2> -50
squaring both sides, 4 Z^4 holds true.
So 1 alone is sufficient but 2 is not required and the answer is "A".
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23 Jul 2009, 23:08
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Aleehsgonji
No offence but neither is the approach to this problem correct nor is the solution.
Lets start with the first statement:
I. x^2+y^2>z^2
Squaring both sides we get:
x^4+y^4+2x^2y^2 > z^4
Since 2x^2y^2 is a positive quantity, on it's own this expression is insufficient to determine whether x^4+y^4>z^4
Alternatively,
Let's try and plug-in certain values in this expression:
a. x=1; y=0; z=0
x^2+y^2>z^2
Clearly, x^4+y^4>z^4 holds true here since [((0+1)^2)^2]>[(0)^2]^2
TRUE
b. x=1; y=3; z=2
z^2=4
x^2+y^2=(1+9)=10
Clearly x^2+y^2>z^2
x^4=1; y^4=81; z^4=16
x^4+y^4 > z^4
TRUE
c. x=4; y=5; z=6
z^2=36
x^2+y^2=(16+25)=41
Clearly x^2+y^2>z^2
x^4=256; y^4=625; z^4=1296
x^4+y^4 < z^4
FALSE
Hence, Statement I is INSUFFICIENT
Let's look at the second statement:
II. x+y>z
Squaring both sides
(x+y)^2>z^2
x^2+y^2+2xy > z^2
Squaring again we get:
x^4+4x^3y+6x^2y^2+4xy^3+y^4 > z^4
This expression is Insufficient on it's own to determine whether x^4+y^4 > z^4 since don't know whether x,y,z are positive or negative integers.
Alternatively,
Lets plug-in certain values for x,y and z:
a. x=1; y=1; z=0
(1^4 + 1^4) > 0^4
Hence, x^4+y^4 > z^4
TRUE
b. x=2; y=3; z=4
(2^4+3^4)= (16+81)=97
4^4=256
x^4+y^4 < z^4
FALSE
c. x=1; y=1; z=-10
(x^4+y^4)=2
z^4= 10000
x^4+y^4 < z^4
FALSE
Hence, Statement II is INSUFFICIENT
Combining both the statements doesn't help either as there is no common ground between the two statements which could lead to a solution to the question.
Hence the Answer is E
24 Jul 2009, 02:19
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x=3, y=3, z=4
x+y>z
x^2+y^2>z^2
But
x^4+y^4<z^4
x=10, y=10, z=1
x+y>z
x^2+y^2>z^2
And
x^4+y^4>z^4
So E
x+y>z
x^2+y^2>z^2
But
x^4+y^4<z^4
x=10, y=10, z=1
x+y>z
x^2+y^2>z^2
And
x^4+y^4>z^4
So E
