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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.
Duringoneyearinaschool,58ofthestudentshadmeasiles.12hadchickenpox,and18hadNeither.Whatfractionoftheschoolhadbothmeasilesandchickenpox.
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)
Letthefractionoftheschoolwhohavebothdiseasesisxfractionof,whohaveonlymeasiles=58xfractionof,whohaveonlychickenpox=12x(Havingonlymeasiles)+(Havingonlychickenpox)+(Havingboth)+(Havingneither)=1(Whole)(58x)+x+(12x)+18=158+12+18x=1x=58+12+181=5+4+188=28=14
Commented by tawakalitu last updated on 26/Sep/16
Thanks so much sir. God bless you.
Thankssomuchsir.Godblessyou.
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)
anotheranswer:thestudentshadmeasiles=n(M)=58thestudentshadchickenpox=n(C)=12n(MC)=n(M)+n(C)n(MC)118=58+12n(MC)78=98n(MC)sohadbothmeasilesandchickenpox=n(MC)=28=14
Commented by tawakalitu last updated on 26/Sep/16
I reall appreciate sir. God bless you.
Ireallappreciatesir.Godblessyou.

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