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IBPS RRBs Officer Scale I Sep 2013 Paper

Directions (76-80) :

In the following questions two equations numbered I and II are given. You have to solve both equations and

Give answer if(1) x > y

Give answer (2) if x > y

Give answer (3)if x < y

Give answer (4)if x < y

Give answer (5)if x = yor the relationship cannot be established

Question : 77 of 200

Marks: +1, -0

x^2 + 28x + 192 = 0

y^2 + 16y + 48 = 0

Solution:

(4)

I. x^2 + 28x + 192 = 0

\Rightarrow x^2 + 16x + 12x + 192 = 0

⇒x (x + 16) + 12 (x + 16) = 0

⇒(x + 12) (x + 16) = 0

⇒x = –12 or –16

II. y^2 + 16y + 48 = 0

\Rightarrow y^2 + 12y + 4y + 48 = 0

⇒y (y + 12) + 4 (y + 12) = 0

⇒(y + 12) (y + 4) = 0

⇒y = – 12 or– 4

Clearly, x < y

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