Let U = set of students in a school.

A = set of students taking tea

B = set of students taking coffee

Then n(∪) = 600; n(A) = 150; n(B) = 225; n(A∩B) = 100 

∴ n(Au B) = n(A) + n(B)-n(A∩B) = 150 + 225-100 =275 

Now n(A’∩B’) = n(A∪B) 

= n(∪)-n(A∪5) = 600-275 = 325 

∴ 325 students were taking neither tea nor coffee.

Let U be the set of all students who took part in the survey.
Let T be the set of students taking tea.
Let C be the set of students taking coffee.
Accordingly, n(U) = 600, n(T) = 150, n(C) = 225, n(T ∩ C) = 100
To find: Number of student taking neither tea nor coffee i.e.,
we have to find n(T’ ∩ C’).
n(T’ ∩ C’) = n(T ∪ C)’
= n(U) – n(T ∪ C)
= n(U) – [n(T) + n(C) – n(T ∩ C)]
= 600 – [150 + 225 – 100]
= 600 – 275
= 325
Hence, 325 students were taking neither tea nor coffee.

given that, `uu= 600`
lets take tea taking student be T and coffee be C
`T nn C = 100 `
only tea `150-100 = 50`
only coffee`25-100= 125`
let the student taking neither of the 2 drinks be x
now, `x+125+50+100 = 600`
`x+275 = 600`
`x=325`
answer