Question Number 44806 by jasno91 last updated on 05/Oct/18
Answered by Kunal12588 last updated on 05/Oct/18
$${let}\:{the}\:{numbers}\:{be}\:{x}\:{and}\:{y} \\ $$$${x}:{y}=\mathrm{3}:\mathrm{5} \\ $$$$\Rightarrow\frac{{x}}{{y}}=\frac{\mathrm{3}}{\mathrm{5}}\Rightarrow{x}=\frac{\mathrm{3}{y}}{\mathrm{5}}\:\:\:….\left(\mathrm{1}\right) \\ $$$$\left({x}+\mathrm{10}\right):\left({y}+\mathrm{10}\right)=\mathrm{5}:\mathrm{7} \\ $$$$\Rightarrow{x}=\frac{\mathrm{5}}{\mathrm{7}}\left({y}+\mathrm{10}\right)−\mathrm{10}\:\left(\mathrm{2}\right) \\ $$$$\frac{\mathrm{3}{y}}{\mathrm{5}}=\frac{\mathrm{5}{y}+\mathrm{50}−\mathrm{70}}{\mathrm{7}}\:\:\:\left[{from}\left(\mathrm{1}\right)\:{and}\:\left(\mathrm{2}\right)\right] \\ $$$$\Rightarrow\mathrm{21}{y}=\mathrm{25}{y}−\mathrm{100} \\ $$$$\Rightarrow{y}=\mathrm{25} \\ $$$${x}=\mathrm{3}\left(\mathrm{25}\right)/\mathrm{5}=\mathrm{15}\:\:\:\:\:\left[{from}\:\left(\mathrm{1}\right)\right] \\ $$