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Question Number 7969 by tawakalitu last updated on 26/Sep/16
During one year in a school, (5/8) of the students had measiles. (1/2) had chickenpox, and (1/8) had Neither. What fraction of the school had both measiles and chickenpox.
$${During}\:{one}\:{year}\:{in}\:{a}\:{school},\:\frac{\mathrm{5}}{\mathrm{8}}\:{of}\:{the}\:{students} \\ $$$${had}\:{measiles}.\:\frac{\mathrm{1}}{\mathrm{2}}\:{had}\:{chickenpox},\:{and}\:\frac{\mathrm{1}}{\mathrm{8}}\:{had} \\ $$$${Neither}.\:{What}\:{fraction}\:{of}\:{the}\:{school}\:{had}\:{both} \\ $$$${measiles}\:{and}\:{chickenpox}. \\ $$
Answered by Rasheed Soomro last updated on 26/Sep/16
Let the fraction of the school who have both diseases is x fraction of, who have only measiles=(5/8)−x fraction of, who have only chickenpox=(1/2)−x (Having only measiles)+(Having only chickenpox) +(Having both)+(Having neither)=1 (Whole) ((5/8)−x)+x+((1/2)−x)+(1/8)=1 (5/8)+(1/2)+(1/8)−x=1 x=(5/8)+(1/2)+(1/8)−1=((5+4+1−8)/8)=(2/8)=(1/4)
$${Let}\:{the}\:{fraction}\:{of}\:{the}\:{school}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{who}\:{have}\:{both}\:{diseases}\:{is}\:\:\:{x} \\ $$$${fraction}\:{of},\:{who}\:{have}\:\:{only}\:{measiles}=\frac{\mathrm{5}}{\mathrm{8}}−{x} \\ $$$${fraction}\:{of},\:{who}\:{have}\:\:{only}\:{chickenpox}=\frac{\mathrm{1}}{\mathrm{2}}−{x} \\ $$$$\left({Having}\:{only}\:{measiles}\right)+\left({Having}\:{only}\:{chickenpox}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\left({Having}\:{both}\right)+\left({Having}\:{neither}\right)=\mathrm{1}\:\left({Whole}\right) \\ $$$$\left(\frac{\mathrm{5}}{\mathrm{8}}−{x}\right)+{x}+\left(\frac{\mathrm{1}}{\mathrm{2}}−{x}\right)+\frac{\mathrm{1}}{\mathrm{8}}=\mathrm{1} \\ $$$$\frac{\mathrm{5}}{\mathrm{8}}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{8}}−{x}=\mathrm{1} \\ $$$${x}=\frac{\mathrm{5}}{\mathrm{8}}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{8}}−\mathrm{1}=\frac{\mathrm{5}+\mathrm{4}+\mathrm{1}−\mathrm{8}}{\mathrm{8}}=\frac{\mathrm{2}}{\mathrm{8}}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$
Commented by tawakalitu last updated on 26/Sep/16
Thanks so much sir. God bless you.
$${Thanks}\:{so}\:{much}\:{sir}.\:{God}\:{bless}\:{you}. \\ $$
Answered by sandy_suhendra last updated on 26/Sep/16
another answer : the students had measiles = n(M) = (5/8) the students had chickenpox = n(C) = (1/2) n(M∪C)=n(M)+n(C)−n(M∩C) 1 − (1/8) = (5/8) + (1/2) − n(M∩C) (7/8) = (9/8) − n(M∩C) so had both measiles and chickenpox = n(M∩C) = (2/8) = (1/4)
$${another}\:{answer}\:: \\ $$$${the}\:{students}\:{had}\:{measiles}\:=\:{n}\left({M}\right)\:=\:\frac{\mathrm{5}}{\mathrm{8}} \\ $$$${the}\:{students}\:{had}\:{chickenpox}\:=\:{n}\left({C}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${n}\left({M}\cup{C}\right)={n}\left({M}\right)+{n}\left({C}\right)−{n}\left({M}\cap{C}\right) \\ $$$$\mathrm{1}\:−\:\frac{\mathrm{1}}{\mathrm{8}}\:=\:\frac{\mathrm{5}}{\mathrm{8}}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\:−\:{n}\left({M}\cap{C}\right) \\ $$$$\:\:\:\:\:\frac{\mathrm{7}}{\mathrm{8}}\:\:\:\:\:=\:\frac{\mathrm{9}}{\mathrm{8}}\:−\:{n}\left({M}\cap{C}\right) \\ $$$${so}\:{had}\:{both}\:{measiles}\:{and}\:{chickenpox}\:=\:{n}\left({M}\cap{C}\right)\:=\:\frac{\mathrm{2}}{\mathrm{8}}\:=\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$ \\ $$
Commented by tawakalitu last updated on 26/Sep/16
I reall appreciate sir. God bless you.
$${I}\:{reall}\:{appreciate}\:{sir}.\:{God}\:{bless}\:{you}. \\ $$

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