Question

in a group of 500 persons,300take tea,150 take coffee,250 take cold drink ,90 take tea and coffee,110 take tea and cold drink 80 take coffee and cold drink and 30 take none of the three drinks,.find the number of persons who take all the three drinks.also find the number of persons who drink exactly one drink ?

Answer 1

let the set of  people who take tea =A
set of people who take coffee =B
set of people who take cold drink= C

Acc to ques,

n(A)=300, n(B)=150, n(C)=250

also, n(A n B)=90, n(A n C)=110, n(B n C)=80

30 take none of 3 drinks

Hence, (A u B)=500-30=470
by formula,

n(A u B u C)= n(A)+ n(B) +n(C) - n(A n B ) - n(B n C)- n(A n C)+ n(A n B n C)
470 = 300 + 150 + 250 - 90 - 110 - 80 + n(A n B n C)

n(A n B n C)= 50

therefore, no of people who take all the 3 drinks=50

Answer 2

people who take tea = A

who take coffee =B

people who take cold drink= C

n(A)=300, n(B)=150, n(C)=250

also, n(A n B)=90, n(A n C)=110, n(B n C)=80

30 take none of 3 drinks

(A u B u C ) = 470

n(A u B u C)= n(A)+ n(B) +n(C) - n(A n B ) - n(B n C)- n(A n C)+ n(A n B n C)

470 = 300 + 150 + 250 - 90 - 110 - 80 + n(A n B n C)

n(A n B n C)= 50

no of people who take all the 3 drinks=50