Question

In one small school with 97 students, a questionnaire about favorite flavors of ice cream was done, and here are the results:
25 students like chocolate and vanilla flavor
16 students like vanilla and strawberry flavor
17 students like chocolate and strawberry flavor
9 students like all three flavor
5 students do not like icecream
How many students like only vanilla flavor if the number is equal to the number of students who like only strawberry flavor and that number is two times less than the number of students who like only chocolate flavor?

Answer 1

Answer:

41 students like vanilla flavour

33 like strawberry flavour

Answer 2

Given:

No. of students in class = 97

No. of students who like chocolate and vanilla flavours = 25

No. of students who like strawberry and vanilla flavours  = 16

No. of students who like chocolate and strawberry flavours = 17

No. of students who like all three flavours = 9

No. of students who do not like ice cream = 5

To find:

No. of students who like only vanilla flavour.

Solution:

Let C, S, V represent chocolate, strawberry and vanilla flavours respectively. Then, according to the given information,

n(U) = 97, n(C∩V) = 25, n(V∩S) = 16, n(C∩S) = 17, n(C∩S∩V) = 9

Also given that, n(V) = n(S) = n(C) - 2

Now, no. of students who like ice cream = Total no. of students - No. of students who do not like ice cream

No. of students who like ice cream = $97-5=92$

i.e., n(C∪S∪V) = 92

We also have the formula,

n(C∪S∪V) = n(C) + n(S) + n(V) - n(C∩V) - n(V∩S) - n(C∩S) + n(C∩V∩S)

$92=n(C)+n(S)+n(S) -25-17-16+9$

$92+25+17+16-9=n(C)+2n(S)$

$141=n(C)+2[n(C)-2]$

$141=n(C)+2n(C)-4$

$145=3n(C)$

$n(C)=\frac{145}{3}=48$

$n(S)=n(C)-2$

$n(S)=48-2=46=n(V)$

Hence, 46 students like only vanilla flavoured ice cream, 46 students like only strawberry flavoured ice cream and 48 students like only chocolate flavoured ice cream.

46 students like only vanilla flavoured ice cream.