| 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. \[3{{y}^{2}}-19y+28=0\] |
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: 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. \[3{{y}^{2}}-19y+28=0\] |
| \[\Rightarrow \] \[3{{y}^{2}}-12y-7y+28=0\] |
| \[\Rightarrow \] \[3y\,(y-4)-7\,(y-4)=0\] |
| \[\Rightarrow \] \[y=\frac{7}{3},\]4 |
| \[\therefore \] \[x\ge y\] |
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