A) 12 : 5
B) 6 : 5
C) 30 : 19
D) 5 : 3
Correct Answer: C
Solution :
Here\[\frac{x}{y}=\frac{3}{2}\] \[\therefore \] \[\frac{{{x}^{2}}}{{{y}^{2}}}={{\left( \frac{3}{2} \right)}^{2}}=\frac{9}{4}\] Now, \[\frac{2{{x}^{2}}+3{{y}^{2}}}{3{{x}^{2}}-2{{y}^{2}}}=\frac{2\left( \frac{{{x}^{2}}}{{{y}^{2}}} \right)+3}{3\left( \frac{{{x}^{2}}}{{{y}^{2}}} \right)-2}\] [on dividing \[{{N}^{r}}\]and \[{{D}^{r}}\]by \[{{y}^{2}}\]] \[=\frac{\left( 2\times \frac{9}{4} \right)+3}{\left( 3\times \frac{9}{4} \right)-2}=\frac{\frac{9}{2}+3}{\frac{27}{4}-2}\] \[=\frac{\frac{9+6}{2}}{\frac{27-8}{4}}=\frac{15}{2}\times \frac{4}{19}=\frac{30}{19}=30:19\]
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