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Is X^4 + Y^4 > Z^4? (1) X^2 + Y^2 > Z^2 (2) X + Y >

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Is X^4 + Y^4 > Z^4? (1) X^2 + Y^2 > Z^2 (2) X + Y > [#permalink] New post 17 Dec 2005, 09:46
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Is X^4 + Y^4 > Z^4?

(1) X^2 + Y^2 > Z^2

(2) X + Y > Z
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 [#permalink] New post 17 Dec 2005, 09:54
I`d say (A)

1. Suff for neg, pos, fractions, and sqrts.
2. Insuff for neg.
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 [#permalink] New post 17 Dec 2005, 13:16
if x, y and z are fraction then 1 or 2 is not sufficient.

so E is the answer.
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 [#permalink] New post 20 Dec 2005, 21:42
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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 [#permalink] New post 20 Dec 2005, 22:02
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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 [#permalink] New post 21 Dec 2005, 02:41
Do we need a statement at all to answer the question?
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 [#permalink] New post 31 Dec 2006, 16:59
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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 [#permalink] New post 31 Dec 2006, 17:12
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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 [#permalink] New post 01 Jan 2007, 06:06
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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 [#permalink] New post 01 Jan 2007, 21:55
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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 [#permalink] New post 02 Jan 2007, 02:03
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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 [#permalink] New post 02 Jan 2007, 12:19
I would go with E

whats the correct Answer?
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 [#permalink] New post 02 Jan 2007, 15:33
Will someone explain why E is the OA?
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 [#permalink] New post 03 Jan 2007, 13:31
(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] New post 06 Jan 2007, 00:24
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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Re: DS - Inequality (GMATPrep) [#permalink] New post 08 Dec 2009, 11:04
please post the OA..

Rohit
Re: DS - Inequality (GMATPrep)   [#permalink] 08 Dec 2009, 11:04
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