| Directions: In each of the following questions two equations are given, solve these equations and give answer. [IBPS (PO) 2013] |
| I. \[{{x}^{2}}+20=9x\] |
| II. \[{{y}^{2}}+42=13y\] |
A) If \[x\ge y\]
B) If \[x>y\]
C) If \[x\le y\]
D) If \[x<y\]
E) If \[x=y\]
Correct Answer: D
Solution :
| I. \[{{x}^{2}}+20=9x\] |
| \[\Rightarrow \]\[{{x}^{2}}-9x+20=0\] |
| \[\Rightarrow \]\[{{x}^{2}}-5x-4x+20=0\] |
| \[\Rightarrow \]\[x\,\,(x-5)-4\,\,(x-5)=0\] |
| \[\Rightarrow \] \[(x-4)(x-5)=0\] |
| \[\therefore \] \[x=4,\]\[5\] |
| II. \[{{y}^{2}}+42=13y\] |
| \[\Rightarrow \]\[{{y}^{2}}-13y+42=0\] |
| \[\Rightarrow \]\[{{y}^{2}}-7y-6y+42=0\] |
| \[\Rightarrow \]\[y\,\,(y-7)-6\,\,(y-7)=0\] |
| \[\Rightarrow \] \[(y-6)(y-7)=0\] |
| \[\therefore \] \[y=6,7\] |
| Hence, \[x<y\] |
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