Question

Divide 40 into two parts such that the sum of their reciprocals is $\frac{8}{75}$.

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{first}\mathrm{part}\mathrm{of}40\mathrm{be}x.\text{T}\mathrm{hen},\mathrm{its}\mathrm{second}\mathrm{part}\text{will be}\left(40-x.\right) \\ \mathrm{Thus},\mathrm{by}\mathrm{the}\mathrm{given}\mathrm{condition},\mathrm{we}\mathrm{get}: \\ \frac{1}{x}+\frac{1}{40-x}=\frac{8}{75} \\ \Rightarrow \frac{40-x+x}{x\left(40-x\right)}=\frac{8}{75} \\ \Rightarrow \frac{40}{x\left(40-x\right)}=\frac{8}{75} \\ \Rightarrow \frac{5}{x\left(40-x\right)}=\frac{1}{75} \\ \Rightarrow x\left(40-x\right)=375 \\ \Rightarrow 40x-x^2=375 \\ \Rightarrow x^2-40x+375=0 \\ \mathrm{On}\mathrm{splitting}\mathrm{the}\mathrm{middle}\mathrm{term}40x\mathrm{as}-15x-25x,\mathrm{we}\mathrm{get}: \\ {x}^2-15x-25x+375=0 \\ \Rightarrow x\left(x-15\right)-25\left(x-15\right)=0 \\ \Rightarrow \left(x-15\right)\left(x-25\right)=0 \\ \Rightarrow x-15=0orx-25=0 \\ \Rightarrow x=15orx=25 \\ \mathrm{Now},\mathrm{if}x=15,\mathrm{then}\left(40-x\right)=40-15=25 \\ \mathrm{If}x=25,\mathrm{then}\left(40-x\right)=40-25=15 \\ \mathrm{Thus},40\mathrm{is}\mathrm{divided}\mathrm{into}15\mathrm{and}25. \end{gathered}$$