Question Number 69431 by azizullah last updated on 23/Sep/19
Answered by Kunal12588 last updated on 23/Sep/19
$$\mathrm{120}={x}+{y}…\left(\mathrm{1}\right) \\ $$$${let}\:{x}>{y} \\ $$$$\mathrm{3}{y}=\mathrm{2}{x}…\left(\mathrm{2}\right) \\ $$$${from}\:\left(\mathrm{1}\right) \\ $$$$\mathrm{240}=\mathrm{2}{x}+\mathrm{2}{y} \\ $$$$\Rightarrow\mathrm{240}=\mathrm{3}{y}+\mathrm{2}{y}=\mathrm{5}{y}\:\:\:\:{from}\left(\mathrm{2}\right) \\ $$$$\Rightarrow{y}=\frac{\mathrm{240}}{\mathrm{5}}=\mathrm{48} \\ $$$${x}=\mathrm{120}−\mathrm{48}=\mathrm{72} \\ $$$${Don}'{t}\:{you}\:{really}\:{understand}\:{the}\:{question}? \\ $$
Commented by azizullah last updated on 24/Sep/19
$$\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{thanks}}\:\boldsymbol{\mathrm{sir}} \\ $$
Answered by $@ty@m123 last updated on 23/Sep/19
$$\mathrm{3}{x}=\mathrm{2}\left(\mathrm{120}−{x}\right) \\ $$$$\mathrm{5}{x}=\mathrm{240} \\ $$$${x}=\mathrm{48} \\ $$
Commented by azizullah last updated on 24/Sep/19
$$\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{Sir}}\:\boldsymbol{\mathrm{thanks}}\:\boldsymbol{\mathrm{alot}}. \\ $$