SATHEE: Quantitative Aptitude Ques 265

Quantitative Aptitude Ques 265

Question: Directions: In the given questions, two equations numbered I and II are given. Solve both the equations and mark the appropriate answer.

I. $ 6x^{2}+25x+24=0 $ II. $ 12y^{2}+13y+3=0 $

Options:

A) $ x>y $

B) $ x\ge y $

C) $ x<y $

D) Relationship between a; and y cannot be determined

E) $ x\le y $

Show Answer

Answer:

Correct Answer: C

Solution:

I. $ 6x^{2}+25x+24=0 $ $ D=\sqrt{b^{2}-4ac} $

$ \Rightarrow $ $ D=\sqrt{625-4\times 24\times 6}=\sqrt{49}=7 $ $ x _1=\frac{-,b+7}{12}=\frac{-,25+7}{12}=\frac{-18}{12}=-\frac{3}{2} $ $ x _2=\frac{-,b-7}{12}=\frac{-,25-7}{12}=\frac{-,32}{12}=-\frac{8}{3} $

$ \Rightarrow $ $ x=\frac{-,3}{2},, $ $ \frac{-,8}{3} $ II. $ 12y^{2}+13y+3=0 $ $ y _1=\frac{-13+\sqrt{169-144}}{24} $ $ =\frac{-13+5}{24}=\frac{-,8}{24}=\frac{-1}{3} $ $ y _2=\frac{-13-\sqrt{169-144}}{24} $ $ =\frac{-13-5}{24}=\frac{-18}{24}=\frac{-,3}{4} $

$ \Rightarrow $ $ y=\frac{-1}{3}, $ $ \frac{-,3}{4} $

$ \therefore $ $ x<y $

Quantitative Aptitude Ques 264

Quantitative Aptitude Ques 266

Quantitative Aptitude Ques 265

Question: Directions: In the given questions, two equations numbered I and II are given. Solve both the equations and mark the appropriate answer.

Options:

Answer:

Solution:

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