| Directions: In each of these questions two equations I and II are given. You have to solve both the equations and give answer. [IBPS (SO) 2014] |
| I. \[6{{x}^{2}}-7x+2=0\] |
| II. \[20{{y}^{2}}-31y+12=0\] |
A) If \[x\ge y\]
B) If \[x>y\]
C) If \[x\le y\]
D) If \[x<y\]
E) If relationship between x and y cannot be established
Correct Answer: D
Solution :
| I. \[6{{x}^{2}}-7x+2=0\] |
| \[\Rightarrow \]\[6{{x}^{2}}-4x-3x+2=0\] |
| \[\Rightarrow \]\[2x\,\,(3x-2)-1\,\,(3x-2)=0\] |
| \[\Rightarrow \]\[(2x-1)(3x-2)=0\]\[\Rightarrow \]\[x=\frac{1}{2},\]\[\frac{2}{3}\] |
| II. \[20{{y}^{2}}-31y+12=0\] |
| \[\Rightarrow \]\[20{{y}^{2}}-15y-16y+12=0\] |
| \[\Rightarrow \]\[5y\,\,(4y-3)-4\,\,(4y-3)=0\] |
| \[\Rightarrow \] \[(5y-4)(4y-3)=0\] |
| \[\therefore \]\[y=\frac{4}{3},\]\[\frac{3}{4}\] |
| Hence, \[y>x\] |
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