Question: If a:b=2:3, find \displaystyle \frac{3a^3+4b^2a}{5a^2b+5b^3}
(a) 7 : 10
(b) 32 : 65
(c) 15 : 41
(d) 19 : 63
Ans: (b) 32 : 65
Explanation:
\displaystyle \frac{a}{b}=\displaystyle\frac{2}{3}
\begin{aligned}
\\
\frac{3a^3+4b^2a}{5a^2b+5b^3}&=\frac{3(\frac{a}{b})^3 +4(\frac{a}{b})}{5(\frac{a}{b})^2 +5}\\[1.5em]
&=\frac{3(\frac{2}{3})^3 +4(\frac{2}{3})}{5(\frac{2}{3})^2 +5}\\[1.5em]
&=\frac{3\times \frac{8}{27} +\frac{8}{3}}{5\times \frac{4}{9} +5}\\[1.5em]
&=\frac{\frac{8}{9} +\frac{8}{3}}{\frac{20}{9} +5}\\[1.5em]
&=\frac{8+24}{20+45}\\[1.5em]
&=\frac{32}{65}
\end{aligned}
Aliter:
If you are only interested in answer try solving in the following way:
\displaystyle \frac{a}{b}=\displaystyle\frac{2}{3} \Rightarrow put a = 2, b = 3 in the expression
\begin{aligned}
\frac{3a^3+4b^2a}{5a^2b+5b^3}&=\frac{3\times 8 + 4\times 9\times 2}{5 \times 4\times 3 + 5\times 27}\\[1.6em]
&=\frac{8+24}{20+45}\quad \text{[Dividing both numerator and denominator by 3]}\\[1.6em]
&=\frac{32}{65}
\end{aligned}
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