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In the fo1lowing question two equations numbered I and II are given. You have to solve both the equations and Give answer

I. 20 \( \Large x^{2} \) - x - 12 = 0

II. 20\( \Large y^{2} \) + 27y + 9 = 0

A) x > y
B) x \( \Large \geq \) y
C) x < y
D) x \( \Large \leq \) y

Correct Answer:


B) x \( \Large \geq \) y

Description for Correct answer:
I. 20\( \Large x^{2} \) - x - 12 = 0

=> 20\( \Large x^{2} \) - 16x + 15x - 12 = 0

=> 4x ( 5x - 4 ) + 3 ( 5x - 4 ) = 0

=> ( 5x - 4 ) ( 4x + 3 ) = 0

=> 5x - 4 = 0 or 4x + 3 = 0

=> x = \( \Large \frac{4}{5} \ or \frac{-3}{4} \)

II. 20\( \Large y^{2} \) + 27y + 9 = 0

=> 20\( \Large y^{2} \) + 15y + 12y + 9 = 0

=> 5y ( 4y + 3 ) + 3 ( 4y + 3 ) = 0

=> ( 5y + 3 ) ( 4y + 3 ) = 0

=> y = \( \Large \frac{-3}{5} \ or \frac{-3}{4} \)

Clearly, x \( \Large \geq \) y

Part of solved Linear Equations questions and answers : >> Aptitude >> Linear Equations

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