Quantitative Aptitude Ques 1426

Question: Directions: In these questions two equations numbered I and II are given.

You have to solve both the equations and give answer. I. $ 24x^{2}+38x+15=0 $ II. $ 12y^{2}+28y+15=0 $

Options:

A) If $ x\le y $

B) If $ x\ge y $

C) If $ x<y $

D) If $ x>y $

E) If relationship between x and .y cannot be established

Show Answer

Answer:

Correct Answer: B

Solution:

  • I. $ 24x^{2}+38x+15=0 $

$ \Rightarrow $ $ 24x^{2}+20x+18x+15=0 $

$ \Rightarrow $ $ 4x(6x+5)+3(6x+5)=0 $

$ \Rightarrow $ $ (4x+3)(6x+5)=0 $

$ \therefore $ $ x=-\frac{3}{4}, $ $ -\frac{5}{6} $ II. $ 12y^{2}+28y+15=0 $

$ \Rightarrow $ $ 12y^{2}+18y+10y+15=0 $

$ \Rightarrow $ $ 6y(2y+3)+5(2y+3)=0 $

$ \Rightarrow $ $ (6y+5)(2y+3)=0 $

$ \therefore $ $ y=-\frac{3}{2}, $ $ -\frac{5}{6} $ Hence, $ x\ge y $