SSC CGL Model Paper 7

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Question : 62 of 100

Marks: +1, -0

If

xy =r, xz = r ^2, yz = r ^3, x+y+z = 14

and

x^2 + y ^2 + z^2 =28

then find

z/y \ ?

3
4
$7 /3$
$14/3$

Solution:

Given,

x + y + z = 14

Squaring both sides, we get

x^ 2 + y ^2 + z^2 + 2(xy + yz + zx) = 196

⇒

28 + 2(xy + yz + zx) = 196

⇒

2(r + r ^2 + r^ 3) = 168

⇒

r + r^ 2 + r ^3 = 84

⇒

r + r ^2 + r ^3 = 4 + 16 + 64

⇒

r = 4

∴ xy = 4 and xz = 16
⇒

{xz }/{xy}=z/y=16/4=4

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