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Is X^4 + Y^4 > Z^4? (1) X^2 + Y^2 > Z^2 (2) X + Y > [#permalink]
 17 Dec 2005, 10:46
Question Stats:
0% (00:00) correct
100% (01:03) wrong based on 1 sessions
Is X^4 + Y^4 > Z^4?
(1) X^2 + Y^2 > Z^2
(2) X + Y > Z
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Current Student
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I`d say (A)
1. Suff for neg, pos, fractions, and sqrts.
2. Insuff for neg.
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SVP
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if x, y and z are fraction then 1 or 2 is not sufficient.
so E is the answer.
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hey ya......
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Director
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I agree that A is suff but shouldnt B is also be suff.
x+y>z is nothng but x+y-z>0 and if x,y or z is negative or positive in this particular condition , when each of them is raised to the power of 4, the condition of x^4+y^4>z^4 is satisfied.
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VP
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shahnandan wrote: I agree that A is suff but shouldnt B is also be suff.
x+y>z is nothng but x+y-z>0 and if x,y or z is negative or positive in this particular condition , when each of them is raised to the power of 4, the condition of x^4+y^4>z^4 is satisfied.
If x = 1, y = 1, and z = -4, then (x+y) > z, but x^4 + y^4 < z^4.
If x = 2, y = 2, and z = 2, then (x+y) > z, and x^4 + y+4 > z^4.
Statement 2 alone is insufficient.
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Director
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Do we need a statement at all to answer the question?
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Manager
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naki77 can you explain why, E is the OA, but I don't understand why
I can't find any values for which if 1 holds then the statement in the question does not hold
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Manager
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Ok I found a combination for which X^2 + Y^2 > Z^2 but X^4 + y^4 is not > Z^4
if X = 1/sqrt(2), Y = 1/2, Z = Sqrt (3/5)
then X^2 + Y^2 = 3/4 (>) Z^2 or 3/5
but X^4 + Y^4 = 1/4 + 1/16 = 5/16 which is less than 16/25 or Z^4
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Director
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I thought the answer for this was E also, and was surprised when so many people said A.
Is X^4 + Y^4 > Z^4?
(1) X^2 + Y^2 > Z^2 -
By squaring both sides x^4+y^4 +2x^2*y^2 > z^4 - the inequality holds good for integers as well as fractions. they are always positive since square.
So with what guarantee can we say that x^4+y^4 > z^4?
(2) X + Y > Z - this we all agree in insuff.
I rarely pick numbers, and that seems to be helping me here.
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Manager
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i ended up squaring inequality in another question and made mistake, can we square inequalities in general
your approach doe snot look worng but my examples of numbers picked show that given statement is not true
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Director
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jainan24 wrote: i ended up squaring inequality in another question and made mistake, can we square inequalities in general
your approach doe snot look worng but my examples of numbers picked show that given statement is not true
To answer your question, no we cannot square inequalities in general, because if they are both negative, squaring will not hold the inequality.
However, in this case, I think it's ok because we have x^2, y^2 and z^2, which are all positive since they are squares. For all positives, squaring should hold the inequality.
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Intern
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I would go with E
whats the correct Answer?
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Director
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Will someone explain why E is the OA?
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Senior Manager
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(1): Pick 1, 1, 1: 1^2 + 1^2 > 1^2. Also 1^4 + 1^4 > 1^4. Nevertheless, if we pick 3, 3, 4: 3^2 + 3^2 > 4^2, but 3^4 + 3^4 < 4^4. Insuff.
(2) Insuff as stated in previous posts.
(1&2): Pick again 1, 1, 1 and 3, 3, 4. The inequalities hold for the former, but don´t hold for the latter.
E.
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Re: DS - Inequality (GMATPrep) [#permalink]
06 Jan 2007, 01:24
ffgmat wrote: Is X^4 + Y^4 > Z^4?
(1) X^2 + Y^2 > Z^2
(2) X + Y > Z
My Very simple reasoning...
From 1:
Squaring both sides
X^4 + Y^4 + X^2*Y^2 > Z^4
But from 1 and 2 we have no idea what X^2*Y^2 is so has to be E
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Manager
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Re: DS - Inequality (GMATPrep) [#permalink]
08 Dec 2009, 12:04
please post the OA..
Rohit
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Re: DS - Inequality (GMATPrep)
[#permalink]
08 Dec 2009, 12:04
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