Quantitative Aptitude Ques 458
Question: Directions: In each of the following questions there are two equations. You have to solve both equations and give answer. [SBI (PO) 2015]
I. $ x^{2}-10x+24=0 $ II. $ 3y^{2}-19y+28=0 $
Options:
A) If $ x>y $
B) If $ x\ge y $
C) If $ x<y $
D) If $ x\le y $
E) If $ x=y $ of relation cannot be established
Show Answer
Answer:
Correct Answer: B
Solution:
- I. $ x^{2}-10x+24=0 $
$ \Rightarrow $ $ x^{2}-6x-4x+24=0 $
$ \Rightarrow $ $ x,(x-6)-4,(x-6)=0 $
$ \Rightarrow $ $ x=4, $ 6 II. $ 3y^{2}-19y+28=0 $
$ \Rightarrow $ $ 3y^{2}-12y-7y+28=0 $
$ \Rightarrow $ $ 3y,(y-4)-7,(y-4)=0 $
$ \Rightarrow $ $ y=\frac{7}{3}, $ 4
$ \therefore $ $ x\ge y $
BETA
