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

Thanks

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