# Class 11 Maths Ncert Solutions Ex 1.6

## Class 11 Maths Ncert Solutions Chapter 1 Ex 1.6

Q.1: If A and B are two sets such that n(A) = 16, n(B) = 24 and n(AB)$n\;(A \cup B)$ = 39. Find n(AB)$n\;(A \cap B)$.

Given:

n(A) = 16, n(B) = 23 and n(AB)$n\;(A \cup B)$ = 39

As we know that n(AB)=n(A)+n(B)n(AB)$n\;(A \cup B) \; = \; n(A) \; + \; n(B) \; – \; n(A \cap B)$

39 = 16 + 24 – n(AB)$n\;(A \cap B)$

n(AB)$n\;(A \cap B)$ = 40 – 39 = 1

Therefore, n(AB)$n\;(A \cap B)$ = 1

Q.2: If A and B are two sets such that AB$A \cup B$ has 17 elements, A has 9 elements and B has 14 elements. How many elements does AB$A \cap B$ have?

Given:

n(A) = 9, n(B) = 14 and, n(AB)$n(A \cup B)$ = 17

As we know that n(AB)=n(A)+n(B)n(AB)$n\;(A \cup B) \; = \; n(A) \; + \; n(B) \; – \; n\;(A \cap B)$

17 = 9 + 14 – n(AB)$n\;(A \cap B)$

n(AB)$n\;(A \cap B)$ = 23 – 17 = 6

Therefore, AB$A \cap B$ = 6

Q.3: In a group there are 450 people, 200 speaks Hindi and 270 can speak English. How many people can speak both English and Hindi?

Let us assume that:

H = the set of people who speaks Hindi

And, E = the set of people who speaks English

Given:

n(HE)$n\;(H \cup E)$ = 450, n(H) = 200 and n(E) = 270

n(HE)$n\;(H \cap E)$ = ?

As we know that, n(HE)=n(H)+n(E)n(HE)$n\;(H \cup E) \; = \; n(H) \; + \; n(E) \; – \; n\;(H \cap E)$

450 = 200 + 270 – n(HE)$n\;(H \cap E)$

450 = 470 – n(HE)$n\;(H \cap E)$

n(HE)$n\;(H \cap E)$ = 470 – 450

n(HE)$n\;(H \cap E)$ = 20

Therefore, 20 people can speak both English and Hindi.

Q.4: If X and Y are two sets such that X has 22 elements, Y has 34 elements, and n(XY)$n\;(X \cap Y)$ has 10 elements, how many elements does n(XY)$n\;(X \cup Y)$ have?

Given:

n(XY)$n\;(X \cap Y)$ = 10, n(X) = 22 and, n(Y) = 34

n(XY)$n\;(X \cup Y)$ = ?

As we know that, n(XY)=n(X)+n(Y)n(XY)$n\;(X \cup Y) \; = \; n(X) \; + \; n(Y) \; – \; n\;(X \cap Y)$

n(XY)$n\;(X \cup Y)$ = 22 + 34 – 10

n(XY)$n\;(X \cup Y)$ = 56 – 10

n(XY)$n\;(X \cup Y)$ = 46

Therefore, the set n(XY)$n\;(X \cup Y)$ has 46 elements.

Q.5: If A and B are two sets such that A has 45 elements, n (A ∩ B) has 15 elements and n (A ∪ B) has 70 elements, how many elements does B have?

Given:

n(A) = 45, n(AB)$n\;(A \cap B)$ = 15 and, n(AB)$n\;(A \cup B)$ = 70

n(B)= ?

As we know that, n(AB)=n(A)+n(B)n(AB)$n\;(A \cup B) \; = \; n(A) \; + \; n(B) \; – \; n(A \cap B)$

70 = 45 + n(B) – 15

70 = 30 + n(B)

n(B) = 40

Therefore, the set n(B) has 40 elements.

Q.6:There are 70 people, out of which 35 like tea, 55 like coffee, and each person likes at least one of the two beverages. How many people like both tea and coffee?

Let us assume that:

T = the set of people who like tea

And, C = the set of people who like coffee

Given:

n(T) = 35, n(C) = 55 and n(CT)$n\;(C \cup T)$ = 70

n(CT)$n\;(C \cap T)$ = ?

As we know that n(CT)=n(C)+n(T)n(CT)$n\;(C \cup T) \; = \; n(C) \; + \; n(T) \; – \; n\;(C \cap T)$

70 = 55 + 35 – n(CT)$n\;(C \cap T)$

70 = 90 – n(CT)$n\;(C \cap T)$

n(CT)$n\;(C \cap T)$ = 20

Therefore, there are 20 people who like both tea and coffee.

Q.7: There are 70 students in a group, 35 like cricket, 15 like both tennis and cricket. How many like tennis only and not cricket? How many like tennis?

Let us assume that:

C = the set of students who like cricket

And, T = the set of students who like tennis

Given:

n(C) = 35, n(CT)$n\;(C \cup T)$ = 70, n(CT)$n\;(C \cap T)$ = 15

n(T) = ?

As we know that n(CT)=n(C)+n(T)n(CT)$n\;(C \cup T) \; = \; n\;(C) \; + \; n\;(T) \; – \; n\;(C \cap T)$

70 = 35 + n(T) – 15

70 = 20 + n(T)

n(T) = 50

Therefore, there are 50 students who like to play tennis.

Now, (TC)(TC)=T$(T – C) \cup (T \cap C) \; = \; T$

Also, (TC)(TC)$(T – C) \cap (T \cap C)$ = Ø

Therefore, n(T) = n(T – C) + n(TC)$n\;(T \cap C)$

50 = n(T – C) + 15

n(T – C) = 50 – 15

n(T – C) = 35

Therefore, there are 35 students who play only tennis.

Q.8: In a committee, 60 people speak French, 30 speak Spanish and 20 speak both Spanish and French. How many speak at least one of these two languages?

Let us assume that:

F = the set of people who speaks French

And, S = the set of people who speaks Spanish

Given:

n(F) = 60, n(S) = 30, n(FS)$n\;(F \cap S)$ = 20

n(FS)$n\;(F \cup S)$ = ?

As we know that n(SF)=n(S)+n(F)n(SF)$n\;(S \cup F) \; = \; n(S) \; + \; n(F) \; – \; n\;(S \cap F)$

n(FS)$n(F \cup S)$ = 30 + 60 – 20

n(FS)$n(F \cup S)$ = 70

Therefore, there are 70 people who can speak at least one of the two languages.