Quantitative Aptitude Ques 266
Question: Directions: In the given questions, two equations numbered I and II are given. Solve both the equations and mark the appropriate answer.
I. $ 12x^{2}-x-1=0 $
II. $ 20y^{2}-41y+20=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. $ 12x^{2}-x-1=0 $ $ D=\sqrt{b^{2}-4ac}=\sqrt{1-4\times 12\times (-1)} $ $ =\sqrt{1+48}=\sqrt{49}=7 $ $ x _1=\frac{-,b+D}{2a}=\frac{1+7}{24}=\frac{8}{24}=\frac{1}{3} $ $ x _2=\frac{-,b-D}{2a} $
$ \Rightarrow $ $ x^{2}=\frac{1-7}{24}=\frac{-,6}{24}=\frac{-1}{4} $
$ \Rightarrow $ $ x=\frac{1}{3}, $ $ -\frac{1}{4} $ II. $ 20y^{2}-41y+20 $ $ y _1=\frac{41+\sqrt{1681-1600}}{40} $
$ \Rightarrow $ $ y _1=\frac{41+9}{40}=\frac{50}{40}=\frac{5}{4} $ and $ y _2=\frac{41-\sqrt{1681-1600}}{40} $
$ \Rightarrow $ $ y _2=\frac{32}{40}=\frac{4}{5} $
$ \Rightarrow $ $ y=\frac{5}{4}, $ $ \frac{4}{5} $
$ \therefore $ $ x<y $
BETA
