Let number of people read newspaper H is `n(H)`, newspaper T is `n(T)` and newspaper `I` is `n(I)`
`:. n(H) = 25,n(T)=26,n(I)=26`
`n(HcapT)=11,n(TcapI)=8,(HcapI)=9 and n(HcapTcapI)=3`
(i) Number of people who read at least one newspaper will be `n(H cup T cup I)`
then `n(HcupTcupI)=n(H)+n(T)+n(I)-n(HcapT)`
`-n(TcapI)-n(HcapI)+n(HcapT capI)`
`=25+26+26-11-8-9+3=52`
Therefore, 52 people read at least one newspaper.
(ii) `:.` Number of people who read newspaper H and T only
`= n(H cap T)-n(H cap T cap I)`
`= 11-3=8`
Similarly, number of people who read newspaper T and I only
`= n(T cap I)-n(H cap T cap I)`
`= 8-3=5`
and number of people read newspaper H and I only
`=n(H cap I)-n(H cap T cap I)`
`= 9-3 =6`
Therefore, number of people who read only two newspaper
`= 8+5 +6 =19`
These are not included the people who read three newspaper
Therefore, number of people who read at least two newspaper = number of people who read two + three newspaper
`= 19 +3 = 22`
`:.` Number of people who read only one newspaper `= 52 -22 = 30`
Therefore, 30 people read only one newspaper.