| Directions: In the following questions, two equations numbered I and II are given. You have to solve both the equations and give answer. |
| I.\[\sqrt{500}x+\sqrt{402}=0\] |
| II. \[\sqrt{360}y+{{(200)}^{1/2}}=0\] |
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. \[\sqrt{500}x+\sqrt{402}=0\] |
| \[\Rightarrow \]\[10\sqrt{5}x+\sqrt{402}=0\] |
| \[\Rightarrow \]\[10\sqrt{5x}+20.04=0\] |
| \[\Rightarrow \] \[x=\frac{-\,\,20.04}{22.36}=-\,\,0.896\] |
| II.\[\sqrt{360}y+\sqrt{200}=0\] |
| \[\Rightarrow \]\[6\sqrt{10}y+10\sqrt{2}=0\] |
| \[\Rightarrow \]\[y=\frac{-10\sqrt{2}}{6\times \sqrt{5}\times \sqrt{2}}=\frac{-10}{13.41}=-\,\,0.745\] |
| \[\therefore \]\[x<y\] |
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