SSC CGL 10 Jun 2019 Shift 1 Paper

© examsiri.com

Question : 70 of 100

Marks: +1, -0

If

\ x+y+z=0,

then the value of

\ (x^{2}+y^{2}+z^{2}\ ) \÷

\ (z^{2}-x y\ )

is :

1
2
-2
-1

Solution:

x+y+z=0

Put

z=0, x=1, y=-1

Now

\ (x^{2}+y^{2}+z^{2}\ ) \÷\ (z^{2}-x y\ )

\⇒(1+1+0) \÷(0+1) \⇒ 2

© examsiri.com

Go to Question:

Prev Question

Next Question