Question Number 177560 by aurpeyz last updated on 07/Oct/22
Commented by aurpeyz last updated on 07/Oct/22
$${Thank}\:{you}\:{Sir} \\ $$
Answered by Rasheed.Sindhi last updated on 07/Oct/22
$${Let}\:{the}\:{big}\:{list}\:{contsins}\:{x}\:{data}\:{items} \\ $$$${A}\:{contains}\:\frac{{x}}{\mathrm{2}}\:{items} \\ $$$$\because\:{Average}\:{of}\:{A}\:{is}\:\mathrm{12} \\ $$$$\therefore\:{The}\:{total}\:{of}\:{A}\:{is}\:\mathrm{12}×\frac{{x}}{\mathrm{2}}=\mathrm{6}{x} \\ $$$${B}\:{contains}\:\frac{{x}}{\mathrm{5}}\:{items} \\ $$$$\because\:{Average}\:{of}\:{B}\:{is}\:\mathrm{15} \\ $$$$\therefore\:{The}\:{total}\:{of}\:{B}\:{is}\:\mathrm{15}×\frac{{x}}{\mathrm{5}}=\mathrm{3}{x} \\ $$$${C}\:{contains}\:{x}−\left(\frac{{x}}{\mathrm{2}}+\frac{{x}}{\mathrm{5}}\right)=\frac{\mathrm{10}{x}−\mathrm{5}{x}−\mathrm{2}{x}}{\mathrm{10}}=\frac{\mathrm{3}{x}}{\mathrm{10}} \\ $$$$\because\:{Average}\:{of}\:{C}\:{is}\:\mathrm{6} \\ $$$$\therefore\:{The}\:{total}\:{of}\:{C}\:{is}\:\mathrm{6}×\frac{\mathrm{3}{x}}{\mathrm{10}}=\frac{\mathrm{9}{x}}{\mathrm{5}} \\ $$$${The}\:{total}\:{of}\:{big}\:{list}\:{is} \\ $$$$\mathrm{6}{x}+\mathrm{3}{x}+\frac{\mathrm{9}{x}}{\mathrm{5}}=\frac{\mathrm{30}{x}+\mathrm{15}{x}+\mathrm{9}{x}}{\mathrm{5}}=\frac{\mathrm{54}{x}}{\mathrm{5}} \\ $$$${The}\:{average}\:{of}\:{big}\:{list}=\frac{\left(\mathrm{54}{x}\right)/\mathrm{5}}{{x}} \\ $$$$\:\:\:\:\:\:\:=\mathrm{10}.\mathrm{8} \\ $$
Commented by aurpeyz last updated on 07/Oct/22
$${thanks} \\ $$
Answered by Rasheed.Sindhi last updated on 07/Oct/22
$${Let}\:{grand}\:{list}\:{contains}\:\mathrm{10}\:{data}\:{items} \\ $$$$\left({A}\:{number}\:{divisible}\:{by}\:\mathrm{2}\:\&\:\mathrm{5}\right) \\ $$$${A}\:{contains}\:\mathrm{5}\:{items}\left({Half}\:{of}\:\mathrm{10}\right) \\ $$$${Total}\:{of}\:{A}={average}×{number}\:{of}\:{items} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{12}×\mathrm{5}=\mathrm{60} \\ $$$${B}\:{contains}\:\mathrm{2}\:{items}\left({One}-{fifth}\:{of}\:\mathrm{10}\right) \\ $$$${Total}\:{of}\:{B}=\mathrm{15}×\mathrm{2}=\mathrm{30} \\ $$$${C}\:{contains}\:\mathrm{10}−\left(\mathrm{5}+\mathrm{2}\right)=\mathrm{3}\:{items} \\ $$$${Total}\:{of}\:{C}=\mathrm{6}×\mathrm{3}=\mathrm{18}\:{items} \\ $$$${Total}\:{of}\:{grand}\:{list}=\mathrm{60}+\mathrm{30}+\mathrm{18}=\mathrm{108} \\ $$$${Average}\:{of}\:{grand}\:{list}=\frac{\mathrm{108}}{\mathrm{10}}=\mathrm{10}.\mathrm{8} \\ $$
Commented by aurpeyz last updated on 07/Oct/22
$${thanks} \\ $$