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x^2 + y^2 + z^2| < 4, is |x^2 + z^2| < 1 1. |x^2 +

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x^2 + y^2 + z^2| < 4, is |x^2 + z^2| < 1 1. |x^2 + [#permalink] New post 01 May 2004, 01:02
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A
B
C
D
E

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0% (00:00) correct 100% (01:02) wrong based on 6 sessions
|x^2 + y^2 + z^2| < 4, is |x^2 + z^2| < 1

1. |x^2 + y^2| < 1
2. |y^2 + z^2 | < 1
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 [#permalink] New post 01 May 2004, 06:26
E for me as well

A) alone is not sufficient
B) alone is not sufficient

The interesting part is combining these two

let us assume x^2+z^2 = 1.5 where x^2= 0.75 and z^2 = 0.75
and y^2 = 0 The both A and B are satisfied and also the stem is satisfied.

let us assume x^2+z^2 = 0.5 where z^2 = 0 and x^2 = 0.5 and y^2 = 0
Even then stem and conditions A) and B) are satisfied.

So nothing can be concluded.
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 [#permalink] New post 08 May 2004, 17:45
can you give us the answer for this?
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 [#permalink] New post 09 May 2004, 03:09
The answer seems to be E, for only one line in 3D satisfies 3 inequalities. However, only one point on that line satisfies the inequality in the question.
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Re: DS-77 [#permalink] New post 12 May 2004, 05:05
hallelujah1234 wrote:
|x^2 + y^2 + z^2| < 4, is |x^2 + z^2| < 1

1. |x^2 + y^2| < 1
2. |y^2 + z^2 | < 1


Neither is sufficient. Together they are also NOT sufficient.

Example: z = 0.99, x = 0.1, y = 0.
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Re: DS-77 [#permalink] New post 28 Jul 2010, 02:43
Hey I agree that E is the answer, but do not understand the solution. Can someone explain without plugging values please ?
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Re: DS-77 [#permalink] New post 28 Jul 2010, 18:51
devashish wrote:
Hey I agree that E is the answer, but do not understand the solution. Can someone explain without plugging values please ?

This is how i tried to solve it

Since squares of numbers are +ve we can remove the absolute symbol
|x^2 + y^2 + z^2| < 4 So x^2+y^2+z^2<4 and 0<= x^2 < 4, 0<= y^2 < 4, 0<= z^2 <4

1. |x^2 + y^2| < 1
x^2+y^2<1 and also 0<=X^2<1 , 0<=y^2<1. Z^2 can be still be between 0 and 4
So 0<= x^2+z^2 < 4 . So we can't determine if |x^2+z^2|< 1

2. y^2 + z^2 < 1. Similar reasoning as statement 1. Stmt2 not sufficient

Combining (1) and (2) 0<=x^2<1 , 0<=y^2<1, 0<=z^2<1.
So 0<= x^2+z^2 < 2 .Still can't determine if |x^2+z^2|< 1.So answer E

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Re: DS-77 [#permalink] New post 29 Jul 2010, 06:56
crack700 wrote:
devashish wrote:
Hey I agree that E is the answer, but do not understand the solution. Can someone explain without plugging values please ?


If you're looking for a literal explanation, the reason you can't solve for |x^2 + z^2| < 1 is because statement 1 tells us about x and y, but we can't pinpoint z. And statement two tells us about y and z, but we can't pinpoint x. From either statement we can tell that the third term is most certainly less than 4, because that's stated in the original question. But we can't determine what that third term is, so we can't determine if |x^2 + z^2| < 1 is true.

Even taken together we can't solve because we still can't pinpoint which term stands for which number. They could all be less than 1. But one of them could also be great than 1.
Re: DS-77   [#permalink] 29 Jul 2010, 06:56
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