Question

If $\frac{a}{b}=\frac{4}{3},$ then the value of $\frac{6a+4b}{6a-5b}$ is

(a) −1
(b) 3
(c) 4
(d) 5

Answer 1

(c) 4

$$\begin{gathered} \frac{a}{b}=\frac{4}{3} \\ a=\frac{4b}{3} \\ \mathrm{On}\mathrm{putting}\mathrm{the}\mathrm{value}\mathrm{of}a=\frac{4b}{3}\mathrm{in}\frac{6a+4b}{6a-5b},\mathrm{we}\mathrm{get}: \\ \frac{6a+4b}{6a-5b}=\frac{6\left({\displaystyle \frac{4b}{3}}\right)+4b}{6\left({\displaystyle \frac{4b}{3}}\right)-5b}=\frac{{\displaystyle \frac{24b}{3}}+4b}{{\displaystyle \frac{24b}{3}}-5b} \\ \mathrm{LCM}\mathrm{of}3\&1\mathrm{is}3. \\ \frac{{\displaystyle \frac{24b}{3}}+{\displaystyle \frac{4b\times 3}{1\times 3}}}{{\displaystyle \frac{24b}{3}}-{\displaystyle \frac{5b\times 3}{1\times 3}}}=\frac{{\displaystyle \frac{24b}{3}}+{\displaystyle \frac{12b}{3}}}{{\displaystyle \frac{24b}{3}}-{\displaystyle \frac{15b}{3}}}=\frac{{\displaystyle \frac{24b+12b}{3}}}{{\displaystyle \frac{24b-15b}{3}}}=\frac{36b}{9b} \\ \mathrm{On}\mathrm{dividing}\mathrm{the}\mathrm{numerator}\&\mathrm{denominator}\mathrm{by}\mathrm{the}\mathrm{HCF}\mathrm{of}36\mathrm{b}\&9\mathrm{b},\mathrm{we}\mathrm{get}: \\ \frac{36\mathrm{b}÷9\mathrm{b}}{9\mathrm{b}÷9\mathrm{b}}=4 \end{gathered}$$