| Directions: In the following questions, two equations numbered I and II are given. You have to solve both the equations and give answer. [IBPS (RRB) Grade A 2012] |
| I. \[{{(169)}^{1/2}}x+\sqrt{289}=134\] |
| II. \[{{(361)}^{1/2}}{{y}^{2}}-270=1269\] |
A) If \[x>y\]
B) If \[x\ge y\]
C) If \[x<y\]
D) If \[x\le y\]
E) If \[x=y\] or the relationship cannot be established
Correct Answer: B
Solution :
| \[{{(169)}^{1/2}}x+\sqrt{289}=134\] (i) |
| \[{{(361)}^{1/2}}{{y}^{2}}-270=1269\] (ii) |
| Above equations can be rewritten as, |
| I. \[13x+17=134\] |
| \[\Rightarrow \] \[13x=134-17\] |
| \[\Rightarrow \] \[13x=117\]\[\Rightarrow \]\[x=\frac{117}{13}=9\] |
| \[\therefore \] \[x=9\] |
| II. \[19{{y}^{2}}-270=1269\] |
| \[\Rightarrow \] \[19{{y}^{2}}=1269+270\] |
| \[\Rightarrow \] \[19{{y}^{2}}=1539\] |
| \[\Rightarrow \] \[{{y}^{2}}=\frac{1539}{19}\]\[\Rightarrow \]\[{{y}^{2}}=81\] |
| \[\therefore \] \[y=\pm \,\,9\] |
| Hence, \[x\ge y\] |
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