A) 19 : 30
B) 30 : 19
C) 19 : 39
D) 39 : 19
Correct Answer: B
Solution :
\[x:y=3:2;\] or \[{{x}^{2}}:{{y}^{2}}=9:4\] \[\therefore \] \[\frac{2{{x}^{2}}+3{{y}^{2}}}{3{{x}^{2}}-2{{y}^{2}}}=\frac{2\frac{{{x}^{2}}}{{{y}^{2}}}+3}{3\frac{{{x}^{2}}}{{{y}^{2}}}-2}\] \[=\frac{2\times \frac{9}{4}+3}{3\times \frac{9}{4}-2}=\frac{\frac{15}{2}}{\frac{19}{4}}=\frac{15}{2}\times \frac{4}{19}=\frac{30}{19}\] \[\therefore \] \[2{{x}^{2}}+3{{y}^{2}}:3{{x}^{2}}-2{{y}^{2}}=30:19\]
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