at break,in a school123 students go to canteen which sells cakes,ice cream and buns. 42 students buy ice-cream, 36 buy bunsand 10 buy only cakes, 15 students buy ice-cream and buns, 10 buy ice-cream and cakes 4 buy cakes and buns but not ice creamand 11 buy ice cream and buns but no cakes draw venn diagram to illustratethe above information and find 1)how many students buy nothing at all 2)how many students buy at least two items. 3)how many students buy all three items. please can u ans this..
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The sets should be denoted as follows:
A:student who ate cake only.
B:Student who ate ice-cream only.
C:students who ate buns only.
D:Students who ate ice cream & cake but not buns.
E:Students who ate cake & buns but not ice-cream.
F:Students who ate ice cream & buns but not cake.
G:students who ate cake,buns& ice cream.
Then according to the given information we have:
n(D)+n(B)+n(G)+n(F)=42
n(E)+n(G)+n(F)+n(C)=36
n(A)=10
n(G)+n(F)=15
n(D)+n(G)=10
n(E)=4
n(F)=11
Therefore, n(G)=4,n(D)=6,n(C)=17,n(B)=21
Number of student who buy cake,buns,ice cream
=n(A)+n(B)+n(C)+n(D)+n(E)+n(F)+n(G)
=10+21+17+6+4+11+4=73
Therefore the students who buy nothing at all are=n(U)-73=123-73=50
Number of students who buy at least two items=n(D)+n(E)+n(F)+n(G)
=6+4+4+11=25
Number of students who buy 3 items=n(G)=4
A:student who ate cake only.
B:Student who ate ice-cream only.
C:students who ate buns only.
D:Students who ate ice cream & cake but not buns.
E:Students who ate cake & buns but not ice-cream.
F:Students who ate ice cream & buns but not cake.
G:students who ate cake,buns& ice cream.
Then according to the given information we have:
n(D)+n(B)+n(G)+n(F)=42
n(E)+n(G)+n(F)+n(C)=36
n(A)=10
n(G)+n(F)=15
n(D)+n(G)=10
n(E)=4
n(F)=11
Therefore, n(G)=4,n(D)=6,n(C)=17,n(B)=21
Number of student who buy cake,buns,ice cream
=n(A)+n(B)+n(C)+n(D)+n(E)+n(F)+n(G)
=10+21+17+6+4+11+4=73
Therefore the students who buy nothing at all are=n(U)-73=123-73=50
Number of students who buy at least two items=n(D)+n(E)+n(F)+n(G)
=6+4+4+11=25
Number of students who buy 3 items=n(G)=4
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