Question Number 45905 by peter frank last updated on 18/Oct/18
Answered by math1967 last updated on 18/Oct/18
$${let}\:{no}\:{of}\:{boys}={x}\:{and}\:{no}\:{of}\:{girls}={y} \\ $$$$\therefore\frac{\mathrm{50}{x}+\mathrm{54}{y}}{{x}+{y}}=\mathrm{52} \\ $$$$\mathrm{50}{x}+\mathrm{54}{y}=\mathrm{52}{x}+\mathrm{52}{y} \\ $$$$\mathrm{2}{x}=\mathrm{2}{y}\Rightarrow\frac{{x}}{{y}}=\frac{\mathrm{1}}{\mathrm{1}} \\ $$$${boys}:{girls}=\mathrm{1}:\mathrm{1} \\ $$
Answered by math1967 last updated on 18/Oct/18
$${total}\:{reduce}\:\mathrm{15}×\mathrm{1}=\mathrm{15}{kg} \\ $$$${wt}.\:{of}\:{new}\:{pupil}=\mathrm{40}−\mathrm{15}=\mathrm{25}{kgans} \\ $$