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 and y and give answer. |
I. \[\frac{25}{\sqrt{(x)}}+\frac{9}{\sqrt{(x)}}=17\sqrt{(x)}\] |
II. \[\frac{\sqrt{(y)}}{3}+\frac{5\sqrt{(y)}}{6}=\frac{3}{\sqrt{(y)}}\] |
A) If \[x>y\]
B) If \[x\le y\]
C) If \[x<y\]
D) If relationship between x and y cannot be determined
E) If \[x\ge y\]
Correct Answer: C
Solution :
I. \[\frac{25}{\sqrt{x}}+\frac{9}{\sqrt{x}}=17\sqrt{x}\] |
\[\Rightarrow \] \[\frac{25+9}{\sqrt{x}}=17\sqrt{x}\]\[\Rightarrow \]\[34=17x\] |
\[\Rightarrow \] \[x=2\] II. \[\frac{\sqrt{y}}{3}+\frac{5\sqrt{y}}{6}=\frac{3}{\sqrt{y}}\] |
\[\Rightarrow \] \[\frac{2\sqrt{y}+5\sqrt{y}}{6}=\frac{3}{\sqrt{y}}\]\[\Rightarrow \]\[\frac{7\sqrt{y}}{6}=\frac{3}{\sqrt{y}}\] |
\[\Rightarrow \] \[7y=18\]\[\Rightarrow \]\[y=\frac{18}{7}\] |
\[\therefore \] \[x<y\] |
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