Question Number 185224 by alcohol last updated on 18/Jan/23
Answered by hafiz last updated on 18/Jan/23
Answered by Rasheed.Sindhi last updated on 19/Jan/23
$${Let}\:{there}\:{are}\:\mathrm{100}\:{families}\:{in}\:{the} \\ $$$${town}.{Let}\:{y}\:{families}\:{have}\:{both}\:{car}\: \\ $$$$\&\:{phone} \\ $$$$\mathrm{60}\:{families}\:{have}\:{neither}.. \\ $$$$\mathrm{25}−{y}\:{have}\:{only}\:{phones} \\ $$$$\mathrm{20}−{y}\:{have}\:{only}\:{cars} \\ $$$${y}\:{have}\:{both} \\ $$$$\mathrm{60}+\left(\mathrm{25}−{y}\right)+\left(\mathrm{20}−{y}\right)+{y}=\mathrm{100} \\ $$$$\mathrm{105}−{y}=\mathrm{100}\Rightarrow{y}=\mathrm{5} \\ $$$$\mathrm{5\%}\:{have}\:{both} \\ $$$${Let}\:{there}\:{are}\:{x}\:{families}\:{in}\:{the}\:{town} \\ $$$$\mathrm{5\%}\:{of}\:{x}=\mathrm{2000} \\ $$$$\frac{\mathrm{5}}{\mathrm{100}}{x}=\mathrm{2000} \\ $$$${x}=\frac{\mathrm{2000}×\mathrm{100}}{\mathrm{5}}=\mathrm{40000} \\ $$$$\boldsymbol{{a}}.\mathrm{40000}\:{families}\:{in}\:{the}\:{town} \\ $$$$\boldsymbol{{b}}\left(\mathrm{1}\right)\:\mathrm{P}\cap\mathrm{C}=\mathrm{5\%}\:\:\left(\mathrm{2}\right)\:\mathrm{24000}\:\: \\ $$$$\:\:\:\left(\mathrm{3}\right)\:\mathrm{6000}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{4}\right)\:\mathrm{8000} \\ $$$$ \\ $$$$ \\ $$