Referring to the AP Inter 1st Year Maths Study Material Chapter 1 Sets Exercise 1g Solutions makes it easier to understand complex problems.
AP Inter 1st Year Maths Sets Solutions Exercise 1g
Question 1.
If X and Y are two sets such that n(X) & 17, n(Y) = 23 and n(X ∪ Y) = 38, find n(X ∩ Y),
Solution:
Given that, n(X) = 17; n(Y) = 23; n(X ∪ Y) = 38
We know that, n(X ∪ Y) = n(X) + n(Y) – n(X ∩ Y)
⇒ 38 = 17 + 23 – n(X ∩ Y)
⇒ 38 = 40 – n(X ∩ Y)
⇒ n(X ∩ Y) = 40 – 38 = 2
∴ n(X ∩ Y) = 2.
Question 2.
If X and Y are two sets such that X ∪ Y has 18 elements, X has 8 elements and Y has 15 elements, how many elements does X ∩ Y have ?
Solution:
n(X ∪ Y) = n(X) + n(Y) – n(X ∩ Y)
⇒ 18 = 8 + 15 – n(X ∩ Y)
⇒ 18 = 23 – n(X ∩ Y)
⇒ n(X ∩ Y) = 23 – 18 = 5
∴ n(X ∩ Y) = 5 elements.
Question 3.
In a group of 400 people, 250 can speak Hindi and 200 can speak English. How many people can speak both Hindi and English ?
Solution:
Let No. of people in a group be = 400 = n(H ∪ E)
No. of people who can speak Hindi = 250 = n(H)
No. of people who can speak English = 200 = n(E)
No. of people who can speak both = x = n(H ∩ E)
According to n(H ∪ E) = n(H) + n(E) – n(H ∩ E)
400 = 250 + 200 – x
400 = 450 – x
x = 450 – 400 = 50
∴ n(H ∩ E) = 50
∴ 50 people can speak both Hindi and English.
Question 4.
If S and T are two sets such that S has 21 elements. T has 32 elements and S ∩ T has 11 elements how many elements does S ∪ T have ?
Solution:
Given that n(S) = 21, n(T) = 32 and n(S ∩ T) = 11
We know that principle of inclusion & exclusion
n(S ∪ T) = n(S) + n(T) – n(S ∩ T)
⇒ n(S ∪ T) = 21 + 32 – 11 = 42
∴ 42 elements are in S ∪ T.
Question 5.
IfX and Yare two sets such that Xhas 40 elements, XuYhas 60elements and X ∩ Y has 10 elements, how many elements does Y have ?
Solution:
Given that, X has 40 elements = n(X) = 40
(X ∪ Y) has 60 elements = n(X ∪ Y) = 60
(X ∩ Y) has 10 elements = n(X ∩ Y)= 10
We know that, n(X ∪ Y) = n(X) + n(Y) – n(X ∩ Y)
⇒ 60 = 40 + n(Y) – 10
⇒ 60 = 30 + n(Y)
⇒ n(Y) = 60 – 30 = 30
∴ Y has 30 elements.
Question 6.
In a group of 70 people, 37 like coffee, 52 like tea and each person likes at least one of the two drinks. How many people like both coffee and tea ?
Solution:
Let No. of people in group n(C ∪ T) = 70
No. of people who likes coffee n(C) = 37
No. of people who likes tea n(T) = 52
Let No. of people who likes both n(C ∩ T) = x
Since everyone likes at least one, we know that
n(C ∪ T) = n(C) + n(T) – n(C ∩ T)
⇒ 70 = 37 + 52 – x
⇒ x = 89 – 70
⇒ x = 19
∴ ‘19’ people like both coffee and tea.
Question 7.
In a group of 65 people, 40 like cricket, 10 like both cricket and tennis. How many like tennis only and not cricket ? How many like tennis ?
Solution:
Total no. of people in a group n(C ∪ T) = 65
No. of people who likes only cricket n(C) = 40
No. of people who likes both cricket and tennis
n(C ∩ T) = 10
We know that,
n(C ∪ T) = n(C) + n(T) – n(C ∩ T)
⇒ 65 = 40 + n (T) – 10
⇒ n(T) = 65 – 30
⇒ n(T) = 35
So, 35 people like tennis (include both cricket)
Now, n(T) = n(T – C) + n(C ∩ T)
35 = n(T – C) + 10
⇒ n(T – C) = 35 – 10 = 25
∴ So, 25 people like only tennis.
Question 8.
In a committee, 50 people speak French. 20 speak Spanish and 10 speak both Spanish and French. How many speak at least one of these two languages?
Solution:
Given that,
No. of people who speaks French n(F) = 50,
No. of people who speaks Spanish n (S) = 20
No. of people who speaks both Spanish and French n(S ∩ F) = 10
We know that
n(S ∪ F) = n(F) + n(S) – n(S ∩ F)
n(S ∪ F) = 50 + 20 – 10 = 60
∴ 60 people speak at least one of these two languages.