| 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. \[88{{x}^{2}}-19x+1=0\] |
| II. \[132{{y}^{2}}-23y+1=0\] |
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: A
Solution :
| I. \[88{{x}^{2}}-19x+1=0\] |
| \[\Rightarrow \]\[88{{x}^{2}}-11x-8x+1=0\] |
| \[\Rightarrow \]\[11x\,\,(8x-1)-1\,\,(8x-1)=0\] |
| \[\Rightarrow \]\[(11x-1)(8x-1)=0\] |
| \[\therefore \]\[x=\frac{1}{8},\]\[\frac{1}{11}\] |
| II. \[132{{y}^{2}}-23y+1=0\] |
| \[\Rightarrow \]\[132{{y}^{2}}-12y-11y+1=0\] |
| \[\Rightarrow \]\[12y\,\,(11y-1)-1(11y-1)=0\] |
| \[\Rightarrow \]\[(12y-1)(11y-1)=0\] |
| \[\therefore \]\[y=\frac{1}{12},\]\[\frac{1}{11}\] |
| Hence, \[x\ge y\] |
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