Question

At one party, 1/5 of the guests only wanted to have a soft drink. Of the remainder, half of the guests preferred to have coffee and 2/3 preferred to have tea. If 12 guests opted for both coffee and tea, then how many guests attended the party in total?

Answer 1

Step-by-step explanation:

let coffee be c, tea a, and soft drink s

let total guests =t

n(s)=t/5

remaining = t-t/5=4t/5 =n(c U a)

n(c) = 1/2*4t/5 =2t/5

n(a)= 2/3*4t/5 = 8t/15

n(c n a) = 12

n(cUa)= n(a)+n(c)- n(cna)

=8t/15 + 2t/5 - 12

= 14t/15-12

4t/5 =14t/15 -12

12t/15 =14t/15 -12

12 = 2t/15

t=12×15/2=15×6=90

total no of guests =90

here I tried to solve the problem by considering 1/5 of guests taking soft drink

(instead of only soft drink. when only soft drink is considered it should be treated in a different way for which I don't have enough info like guests who like both c and s, or both a and s)