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?
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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)
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