﻿ Class 11 NCERT Math Solution
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Exercise - 1.6

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 and n(X ∪ Y) = 38. n(X ∩ Y) = ?
n(X ∩ Y) = n(X) + n(Y) - n(X ∪ Y)
n(X ∩ Y) = 17 + 23 - 38
n(X ∩ Y) = 40 - 38
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 :-
```  Given that  n(X) = 8, n(Y) = 15 and n(X ∪ Y) = 18. n(X ∩ Y) = ?
n(X ∩ Y) = n(X) + n(Y) - n(X ∪ Y)
n(X ∩ Y) = 8 + 15 - 18
n(X ∩ Y) = 23 - 18
n(X ∩ Y) = 5
```

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 H be the set of people who speak Hindi.
Let E be the set of people who speak English.
n(H ∪ E) = 400, n(H) = 250, n(E) = 200, n(H ∩ E) = ?
n(H ∩ E) = n(H) + n(E) - n(H ∪ E)
n(H ∩ E) = 250 + 200 - 400
n(H ∩ E) = 450 - 400
n(H ∩ E) = 50
Thus, 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, n(S ∩ T) = 11, n(S ∪ T) = ?
n(S ∪ T) = n(S) + n(T) - n(S ∩ T)
n(S ∪ T) = 21 + 32 - 11
n(S ∪ T) = 53 - 11
n(S ∪ T) = 42
```

Question-5 :- If X and Y are two sets such that X has 40 elements, X ∪ Y has 60 elements and X ∩ Y has 10 elements, how many elements does Y have?

Solution :-
```  Given that  n(X) = 40, n(X ∪ Y) = 60 and n(X ∩ Y) = 10, n(Y) = ?
n(X ∩ Y) = n(X) + n(Y) - n(X ∪ Y)
10 = 40 + n(Y) - 60
n(Y) = 10 + 20
n(Y) = 30
```

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 C be the set of people who like Coffee.
Let T be the set of people who like Tea.
n(C ∪ T) = 70, n(C) = 37, n(T) = 52, n(C ∩ T) = ?
n(C ∩ T) = n(C) + n(T) - n(C ∪ T)
n(C ∩ T) = 37 + 52 - 70
n(C ∩ T) = 89 - 70
n(C ∩ T) = 19
Thus, 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 :-
```  Let C be the set of people who like Cricket.
Let T be the set of people who like Tennis.

(i) n(C ∪ T) = 65, n(C) = 40, n(C ∩ T) = 10, n(T) = ?
n(C ∩ T) = n(C) + n(T) - n(C ∪ T)
10 = 40 + n(T) - 65
n(T) = 10 + 25
n(T) = 35
Thus, 35 people like tennis.

(ii) n(C ∪ T) = 65, n(C) = 40, n(C ∩ T) = 10, n(T) = 35, n(T - C) = ?
n(T - C) = n(T) - n(C ∩ T)
n(T - C) = 35 - 10
n(T - C) = 25
Thus, 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 :-
```  Let F be the set of people who speak french.
Let S be the set of people who speak spanish.
n(F) = 50, n(S) = 20, n(F ∩ S) = 10, n(F ∪ S) = ?
n(F ∪ S) = n(F) + n(S) - n(F ∩ S)
n(F ∪ S) = 50 + 20 - 10
n(F ∪ S) = 70 - 10
n(F ∪ S) = 60
Thus, 60 people in the committee speak at least one of the two languages.
```
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