| Directions: In each of the following questions two equations I and II are given. You have to solve both the equations and find out values of x, y and give answer. [LIC (AAO) 2014] |
| I. \[\frac{12}{\sqrt{(x)}}+\frac{8}{\sqrt{(x)}}=\sqrt{(x)}\] |
| II. \[{{y}^{4}}-\frac{{{(18)}^{9/2}}}{\sqrt{(y)}}=0\] |
A) If \[x>y\]
B) If \[x\le y\]
C) If \[x<y\]
D) If \[x\ge y\]
E) If relationship cannot be established
Correct Answer: A
Solution :
| I. \[\frac{12}{\sqrt{x}}+\frac{8}{\sqrt{x}}=\sqrt{x}\]\[\Rightarrow \]\[\frac{12\times 8}{\sqrt{x}}=\sqrt{x}\] |
| \[\Rightarrow \] \[12+8=x\]\[\Rightarrow \]\[x=20\] |
| II. \[{{y}^{4}}-\frac{{{(18)}^{9/2}}}{\sqrt{y}}=0\]\[\Rightarrow \]\[{{y}^{4}}=\frac{{{(18)}^{9/2}}}{\sqrt{y}}\] |
| \[\Rightarrow \] \[{{(y)}^{9/2}}={{(18)}^{9/2}}\]\[\Rightarrow \]\[y=18\] |
| Hence,\[x>y\] |
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