In a group of 50 persons, 14 drink tea but not coffee and 30 drink tea

1.	In a group of 50 persons, 14 drink tea but not coffee and 30 drink tea. Find: i. how may drink tea and coffee both. ii. how many drink coffee but not tea.
Answer» Let total number of people n(P) = 50 A number of people who drink Tea n(T) = 30. A number of people who drink coffee n(C). n(T–C) = 14 i. how may drink tea and coffee both. We can see that T is disjoint union of n(T–C) and n (T ∩ C). (If A and B are disjoint then n (A ∪ B) = n(A) + n(B)) ∴ T = n(T–C) ∪ n (T ∩ C). ⇒ n(T) = n(T–C) + n (T ∩ C). ⇒ 30 = 14 + n (T ∩ C). ⇒ n(T ∩ C) = 16. 16 People drink both coffee and tea. ii. how many drink coffee but not tea. We know n (P) = n(T) + n(C) – n (T ∩ C) Substituting the values we get 50 = 30+n(C) – 16 n(C) = 36. We can see that T is disjoint union of n(C–T) and n (T ∩ C). (If A and B are disjoint then n (A ∪ B) = n(A) + n(B)) ∴ C = n(C–T) ∪ n (T ∩ C). ⇒ n(C) = n(C–T) + n (T ∩ C). ⇒ 36 = n(C–T) + 16. ⇒ n(C–T) = 20. 20 People drink coffee but not tea.

In a group of 50 persons, 14 drink tea but not coffee and 30 drink tea. Find: i. how may drink tea and coffee both. ii. how many drink coffee but not tea.

Answer»

Let total number of people n(P) = 50

A number of people who drink Tea n(T) = 30.

A number of people who drink coffee n(C).

n(T–C) = 14

i. how may drink tea and coffee both.

We can see that T is disjoint union of n(T–C) and n (T ∩ C).

(If A and B are disjoint then n (A ∪ B) = n(A) + n(B))

∴ T = n(T–C) ∪ n (T ∩ C).

⇒ n(T) = n(T–C) + n (T ∩ C).

⇒ 30 = 14 + n (T ∩ C).

⇒ n(T ∩ C) = 16.

16 People drink both coffee and tea.

ii. how many drink coffee but not tea.

We know

n (P) = n(T) + n(C) – n (T ∩ C)

Substituting the values we get

50 = 30+n(C) – 16

n(C) = 36.

We can see that T is disjoint union of n(C–T) and n (T ∩ C).

(If A and B are disjoint then n (A ∪ B) = n(A) + n(B))

∴ C = n(C–T) ∪ n (T ∩ C).

⇒ n(C) = n(C–T) + n (T ∩ C).

⇒ 36 = n(C–T) + 16.

⇒ n(C–T) = 20.

20 People drink coffee but not tea.