I think the formula we need for this involves the union of 3 sets. You have probably seen the 2 set version, which states

N(A u B) = N(A) + N(B) - N(A n B), where N(A) is the number of elements in set A.

So basically to get the number of elements in 2 sets, you add the number of elements in each set and subtract out the number in both (since these are double counted). This extends to 3 sets as

N(AuBuC) = N(A)+N(B)+N(C)-N(AnB)-N(AnC)-N(BnC)+N(AnBnC)

So you are adding the number in each individual set, then subtracting out the number in any 2 sets (these are double counted), then adding back the number in all 3 (elements in A,B, and C are counted 3 times, then subtracted out 3 times, so they have to be added back at the end).

Applying this formula is helpful in this problem, since

N(CuTuS)=N(C)+N(T)+N(S)-N(CnT)-N(CnS)-N(SnT)+N(CnSnT)
94 = 48 + 54 + 64 - 28 - 30 - 32 + N(CnSnT)
18 = N(CnSnT)

Then, the number who only like coffee would be

N(C)-N(CnT)-N(CnS)+N(CnSnT) = 48-28-30+18 = 8

and the number who only like tea would be

N(T)-N(CnT)-N(TnS)+N(CnSnT) = 54-28-32+18 = 12

So, the ratio we are interested in is 8/12 = 2/3.

ankanas wrote:In a survey among students at all IIMs, it was found that 48% preferred coffee, 54% liked tea and 64% smoked. Of the total, 28% liked coffee and tea, 32% smoked and drank tea, 30% smoked and drank coffee. Only 6% did none of these. If the total number of students is 2000 then

Q)the ratio of the number of students who like only coffee to the number who like only tea is
a)5:3
b)8:9
c)2:3
d)3:2

The answer is C..

Anyone????

Where is this question from? It's odd that there's only 4 answer choices.