| Directions: In each of the following questions there are two equations. You have to solve both equations and give answer. [SBI (PO) 2015] |
| I. \[{{x}^{2}}-13x+40=0\] |
| II. \[{{y}^{2}}-11y+28=0\] |
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
Correct Answer: A
Solution :
| I. \[{{x}^{2}}-13x+40=0\] |
| \[\Rightarrow \] \[{{x}^{2}}-8x-5x+40=0\] |
| \[\Rightarrow \] \[x\,(x-8)-5\,(x-8)=0\] |
| \[\Rightarrow \] \[x=5,\]\[8\] |
| II. \[{{y}^{2}}-11y+28=0\] |
| \[\Rightarrow \] \[{{y}^{2}}-7y-4y+28=0\] |
| \[\Rightarrow \] \[y\,(y-7)-4\,(y-7)=0\] |
| \[\Rightarrow \] \[y=4,\]7 |
| \[\therefore \] \[x>y\] |
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