| Directions: In each of these questions, two equations I and II are given. You have to solve both the equations and give answer. [IBPS (PO) 2013] |
| I. \[{{x}^{2}}-24x+144=0\] |
| II. \[{{y}^{2}}-26y+169=0\] |
A) If \[x<y\]
B) If \[x>y\]
C) If \[x=y\]
D) If \[x\ge y\]
E) If \[x\le y\]or no relationship can be established between x and y
Correct Answer: A
Solution :
| I. \[{{x}^{2}}-24x+144=0\] |
| \[\Rightarrow \]\[{{x}^{2}}-12x-12x+144=0\] |
| \[\Rightarrow \]\[x\,\,(x-12)-12\,\,(x-12)=0\] |
| \[\Rightarrow \] \[(x-12)(x-12)=0\] |
| \[\Rightarrow \] \[{{(x-12)}^{2}}=0\] |
| \[\therefore \] \[x=12\] |
| II. \[{{y}^{2}}-26y+169=0\] |
| \[\Rightarrow \]\[{{y}^{2}}-13y-13y+169=0\] |
| \[\Rightarrow \]\[y\,\,(y-13)-13\,\,(y-13)=0\] |
| \[\Rightarrow \]\[{{(y-13)}^{2}}=0\] |
| \[\therefore \] \[y=13\] |
| Hence, \[y>x\] |
| Alternate Method |
| I. \[{{x}^{2}}-24x+144=0\] |
| \[\Rightarrow \]\[{{x}^{2}}-2\,\,(12)(x)+{{(12)}^{2}}=0\] |
| \[\Rightarrow \]\[{{(x-12)}^{2}}=0\] |
| \[\therefore \] \[x=12\] |
| II. \[{{y}^{2}}-26y+169=0\] |
| \[\Rightarrow \]\[{{y}^{2}}-2\,\,(13)(y)+{{(13)}^{2}}=0\] |
| \[\Rightarrow \] \[{{(y-13)}^{2}}=0\] |
| \[\therefore \] \[y=13\] |
| Hence, \[y>x\] |
You need to login to perform this action.
You will be redirected in
3 sec
