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Question-185224




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 100 families in the  town.Let y families have both car   & phone  60 families have neither..  25−y have only phones  20−y have only cars  y have both  60+(25−y)+(20−y)+y=100  105−y=100⇒y=5  5% have both  Let there are x families in the town  5% of x=2000  (5/(100))x=2000  x=((2000×100)/5)=40000  a.40000 families in the town  b(1) P∩C=5%  (2) 24000       (3) 6000              (4) 8000
$${Let}\:{there}\:{are}\:\mathrm{100}\:{families}\:{in}\:{the} \\ $$$${town}.{Let}\:{y}\:{families}\:{have}\:{both}\:{car}\: \\ $$$$\&\:{phone} \\ $$$$\mathrm{60}\:{families}\:{have}\:{neither}.. \\ $$$$\mathrm{25}−{y}\:{have}\:{only}\:{phones} \\ $$$$\mathrm{20}−{y}\:{have}\:{only}\:{cars} \\ $$$${y}\:{have}\:{both} \\ $$$$\mathrm{60}+\left(\mathrm{25}−{y}\right)+\left(\mathrm{20}−{y}\right)+{y}=\mathrm{100} \\ $$$$\mathrm{105}−{y}=\mathrm{100}\Rightarrow{y}=\mathrm{5} \\ $$$$\mathrm{5\%}\:{have}\:{both} \\ $$$${Let}\:{there}\:{are}\:{x}\:{families}\:{in}\:{the}\:{town} \\ $$$$\mathrm{5\%}\:{of}\:{x}=\mathrm{2000} \\ $$$$\frac{\mathrm{5}}{\mathrm{100}}{x}=\mathrm{2000} \\ $$$${x}=\frac{\mathrm{2000}×\mathrm{100}}{\mathrm{5}}=\mathrm{40000} \\ $$$$\boldsymbol{{a}}.\mathrm{40000}\:{families}\:{in}\:{the}\:{town} \\ $$$$\boldsymbol{{b}}\left(\mathrm{1}\right)\:\mathrm{P}\cap\mathrm{C}=\mathrm{5\%}\:\:\left(\mathrm{2}\right)\:\mathrm{24000}\:\: \\ $$$$\:\:\:\left(\mathrm{3}\right)\:\mathrm{6000}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{4}\right)\:\mathrm{8000} \\ $$$$ \\ $$$$ \\ $$

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