| Directions: In the following questions two equations numbered I and II are given. You have to solve both the equations and give answer. [Allahabad Bank (PO) 2011] |
| I. \[\frac{5}{7}-\frac{5}{21}=\frac{\sqrt{x}}{42}\] |
| II. \[\frac{\sqrt{y}}{4}+\frac{\sqrt{y}}{16}=\frac{250}{\sqrt{y}}\] |
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: C
Solution :
| I. \[\frac{5}{7}-\frac{5}{21}=\frac{\sqrt{x}}{42}\]\[\Rightarrow \]\[\frac{15-5}{21}=\frac{\sqrt{x}}{42}\] |
| \[\Rightarrow \]\[\frac{10}{21}=\frac{\sqrt{x}}{42}\]\[\Rightarrow \]\[\sqrt{x}=20\]\[\Rightarrow \]\[x=400\] |
| II. \[\frac{\sqrt{y}}{4}+\frac{\sqrt{y}}{16}=\frac{250}{\sqrt{y}}\]\[\Rightarrow \]\[\frac{4\sqrt{y}+\sqrt{y}}{16}=\frac{250}{\sqrt{y}}\] |
| \[\Rightarrow \]\[\frac{5\sqrt{y}}{16}=\frac{250}{\sqrt{y}}\]\[\Rightarrow \]\[5y=250\times 16\] |
| \[\Rightarrow \]\[y=50\times 16=800\] |
| \[\therefore \] \[x<y\] |
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