Class 11: Sets – Exercise 1.8 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: If $A$ and $B$ are two sets such that $n(A \cup B) =50, n(A) =28$ and $n(B) = 32$ , find $n(A \cap B)$ .

Answer:

Given $A ( \cup A ) = 50, \ n( A ) = 28, \ n( B ) = 32$

We know: $n ( A \cup B ) = n(A) + n ( B ) - n ( A \cap B)$

$\Rightarrow n ( A \cap B) = n( A) + n ( B ) - n ( A \cup B )$

$\Rightarrow n ( A \cap B ) = 28 + 32 - 50$

$\Rightarrow n ( A \cap B ) = 10$

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Question 2: If $P$ and $Q$ are two sets such that $P$ has $40$ elements, $P \cup Q$ has $50$ elements and $P \cap Q$ has $10$ elements, how many elements does $Q$ have?

Answer:

Given: $n(P ) = 40, \ n ( P \cup Q) = 60 , \ n ( P \cap Q) = 10$

We know: $n ( P \cup Q ) = n(P) + n ( Q ) - n ( P \cap Q)$

$\Rightarrow 60 = 40 + n(Q) -10$

$\Rightarrow n(Q) = 70 - 40 = 30$

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Question 3: In a school there are $20$ teachers who teach mathematics or physics. Of these, $12$ teach mathematics and $4$ teach physics and mathematics. How many teach physics?

Answer:

Given: $n (M \cup P) = 20, \ n(M) = 12, \ n(M \cap P) = 4$

We know: $n ( M \cup P ) = n(M) + n ( P ) - n ( M \cap P)$

$\Rightarrow 20 = 12 + n(P) - 4$

$\Rightarrow n(P) = 12$

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Question 4: In a group of $70$ people, $37$ like coffee, $52$ like tea and each person likes at east one of the two drinks. How many like both coffee and tea?

Answer:

Given: $n ( T \cup C ) = 70, \ n (T) = 52, \ n (C ) = 37$

We know: $n ( T \cup C ) = n(T) + n ( C ) - n ( T \cap C)$

$\Rightarrow 70 = 52 + 37 - n (T \cap C)$

$\Rightarrow T ( T \cap C) = 19$

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Question 5: Let $A$ and $B$ be two sets such that: $n(A) =20, n(A \cup B) = 42$ and $n(A \cap B) = 4$ . Find : (i) $n(A)$ (ii) $n(A-B)$ (iii) $n(B-A)$

Answer:

Given: $n ( A ) = 20, \ n (A \cup B) = 42, \ n(A \cap B) = 4$

i) $n ( A \cup B ) = n ( A ) + n ( B ) - n ( A \cap B)$

$\Rightarrow 42 = 20 + n ( B ) - 4$

$\Rightarrow n ( B ) = 26$

ii) $n ( A - B ) = n ( A ) - n ( A \cap B )$

$\Rightarrow n ( A - B )= 20 - 4 = 16$

iii) $n ( B - A ) = n ( B ) - n ( B \cap A )$

$\Rightarrow n ( B - A ) = 26 - 4 = 22$

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Question 6: A survey shows that $76\%$ of the Indians like oranges, whereas $62\%$ like bananas. What percentage of the Indians like both oranges and bananas?

Answer:

Let the population be $100$

Therefore $n ( O \cup B ) = 100$

Also $n ( O ) = 76, \ n ( B ) = 62$

We know: $n ( O \cup B ) = n(O) + n ( B ) - n ( O \cap B )$

$\Rightarrow 100 = 76 + 62 - n ( O \cap B )$

$\Rightarrow n ( O \cap B ) = 38$

Therefore $38\%$ of population likes both Banana and Oranges.

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Question 7: In a group of $950$ persons, $750$ can speak Hindi and $460$ can speak English. Find: (i) how many can speak both Hindi and English (ii) how many can speak Hindi only (iii) how many can speak English only

Answer:

Given: $n ( H ) = 750, \ n ( E ) = 460, \ n ( H \cup E ) = 950$

i) We know: $n ( H \cup E ) = n(H) + n ( E ) - n ( H \cap E )$

$\Rightarrow 950 = 750 + 460 - n ( H \cap E )$

$\Rightarrow n ( H \cap E ) = 260$

Therefore $260$ people can speak both Hindi and English

ii) $n( H ) = n ( H - E ) + n ( H \cap E)$

$\Rightarrow 750 = n ( H - E ) + 260$

$\Rightarrow n ( H - E ) = 750 -260 = 490$

Therefore $490$ people can speak Hindi only

iii) $n ( E ) = n ( E - H ) + n ( E \cap H)$

$\Rightarrow 460 = n ( E - H ) + 260$

$\Rightarrow n ( E - H ) = 200$

Therefore $200$ people speak English only.

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Question 8: In a group of $50$ persons, $14$ drink tea but not coffee and $30$ drink tea. Find: (i) how many drink tea and coffee both (ii) how many drink coffee but not tea.

Answer:

Given: $n(T) = 30, \ n(T-C) = 14, \ n(T \cup C) = 50$

i) $n(T) = n(T-C) + n ( T \cap C)$

$\Rightarrow 30 = 14 + n ( T \cap C)$

$\Rightarrow n(T \cap C) = 16$

Therefore $16$ people drink both Tea and Coffee.

ii) $n( T \cup C) = n(T) + n(C) - n(T \cap C)$

$\Rightarrow 50 = 30 + n(C) - 16$

$\Rightarrow n(C) = 36$

Now we need to find $n(C-T)$

$n(C) = n(C - T) + n ( C \cap T)$

$\Rightarrow 36 = n ( C - T) + 16$

$\Rightarrow n(C-T) = 20$

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Question 9: In a survey of $60$ people, it was found that $25$ people read newspaper $H, \ 26$ read newspaper $T, \ 26$ read newspaper $I, \ 9$ read both $H$ and $I, \ 11$ read both $H$ and $T, \ 8$ read both $T$ and $I, \ 3$ read all three newspapers. Find: (i) the numbers of people who read at least one of the newspapers. (ii) the number of people who read exactly one newspaper.

Answer:

Given: $n(H) = 25, \ n(T) = 26, \ n(I) = 26$ $n( H \cap I) = 9, \ n( H \cap T) = 11, \ n( T \cap I) = 8, \$ $n ( H \cap T \cap I) = 3$

i) $n( H \cup T \cup I ) = n(H) + n(T) + n(I) - n( H \cap T) - n( T \cap I) - n( I \cap H) + n ( H \cap T \cap I)$

$= 25 + 26 + 26 - 9 - 11- 8 + 3$

$= 52$

Note: There are 8 people who do no read any news paper. We should not assume that all read newspaper.

ii)

$n( H \ or \ T \ or \ I) = n(H) + n(T) + n(I) - 2n(H \cap T) - 2n(H \cap I) - 2n(T \cap I)+ 3n(H \cap T \cap I)$

$= 25 + 26 + 26 - 22- 16- 18 + 9$

$= 30$

You can also do it by looking at the venn diagram. Venn diagram is more intutitive.

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Question 10: Of the members of three athletic teams in a certain school, $21$ are in the basketball team, $25$ in hockey team and $29$ in the football team. $14$ play hockey and basket ball, $15$ play hockey and football, $12$ play football and basketball and $8$ play all the three games. How many members are there in all?

Answer:

Given: $n(B) = 21, \ n(H) = 26, \ n(F) = 29$ $n(H \cap B) = 14$

$n(H \cap F) = 15$ $n(H \cap B \cap F) = 8$

We know: $n( H \cup B \cup F ) = n(H) + n(B) + n(F) - n( H \cap B) - n( B \cap F) - n( F \cap H) + n ( H \cap B \cap F)$

$\Rightarrow n ( H \cap B \cap F) = 26 + 21 + 29 -14 - 15 - 12 + 8 = 43$

Therefore total number of members in all the teams are $43$

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Question 11: In a group of $1000$ people, there are $750$ who can speak Hindi and $400$ who can speak Bengali. How many can speak Hindi only? How many can speak Bengali ? How many can speak both Hindi and Bengali?

Answer:

Given: $n ( H \cup B) = 1000, \ n(H) = 750, \ n(B) = 400$

We know: $n ( H \cup B ) = n(H) + n (B ) - n ( H \cap B )$

$\Rightarrow 1000 = 750+400 - n ( H \cap B)$

$\Rightarrow n ( H \cap B ) = 150$

Therefore the number of people who can speak both Hindi and Bengali is $150$ .

Number of people who can speak only Hindi

$n ( H - B ) = n ( H ) - n ( H \cap B)$ $= 750 - 150 = 600$

No of people who only speak Bengali

$n ( B - H ) = n(B) - n ( B \cap H )$ $= 400 - 150 = 250$

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Question 12: A survey of $500$ television viewers produced the following in-formation; $285$ watch football, $195$ watch hockey, $115$ watch basketball, $45$ watch football and basketball, $70$ watch football and hockey, $50$ watch hockey and basketball, $50$ do not watch any of the three games. How many watch all the three games? How many watch exactly one of the three games?

Answer:

Given: $n(F) = 285, \ n ( H ) = 195 , \ n(B) = 115$ $n ( F \cap B) = 45 , \ n( F \cap H) = 70, \ n( H \cap B) = 50$

$n ( F' \cap H' \cap B') = 50$

$\Rightarrow n ( F \cup H \cup B) ' = 50$

$\Rightarrow n ( U ) - n ( F \cup H \cup B) = 50$

$\Rightarrow 500 - n ( F \cup H \cup B) = 50$

$\Rightarrow n ( F \cup H \cup B) = 450$

Now, the number of students who watch all the games

$n ( F \cap H \cap B) = n ( F \cup H \cup B) - n(F) - n(H) - n(B) + n ( F \cap B) + n ( F \cap H) + n (H \cap B)$

$= 450 - 285 - 195 - 115 + 45 + 70 + 50 = 20$

No of students who watch exactly one game

$= n ( F) + n( H ) + n(B) - 2n( F \cap B) - 2n ( H \cap B ) - 2n ( H \cap F) + 3n ( H \cap B \cap F)$

$= 285 + 195 + 115 - 90 - 140 - 100 + 60 = 325$

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Question 13: In a survey of $100$ persons it was found that $28$ read magazine $A, \ 30$ read magazine $B, \ 42$ read magazine $C, \ S$ read magazines $A$ and $B, \ 10$ read magazines $A$ and $C, \ S$ read magazines $B$ and $C$ and $3$ read all the three magazines. Find: (i) How many read none of three magazines? (ii) How many read magazine C only?

Answer:

Given: $n ( A ) = 28, \ n ( B ) = 30 , \ n(C) = 42$ $n ( A \cap B)= 8 , \ n( A \cap C) = 10 , \ n ( B \cap C) = 5 \ \$ $n ( A \cap B \cap C) = 3$

i) Number of people who read none of the three magazines

$n ( A \cup B \cup C) = n ( U ) - n ( A \cup B \cup C)$

$= n ( U ) - \{ n(A) + n(B) + n(C) - n ( A \cap B) - n ( B \cap C) - n ( C \cap A) + n ( A \cap B \cap C) \}$

$= 100 - \{ 28 + 30 + 42 - 8 - 10 - 5 + 3 \}$

$= 100 - 80 = 20$

ii) Number of students who read $C$ only $= n ( C \cap A' \cap B')$

$= n \{ C \cap ( A \cup B)' \}$

$= n(C) - n \{ C \cap (A \cup B ) \}$

$= n(C) - n \{ (C \cap A) \cup ( C \cap B ) \}$

$= n(C) - n \{ (C \cap A) + ( C \cap B) - ( A \cap B \cap C) \}$

$= 42 - ( 10 + 5 - 3) = 42 - 12 = 30$

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Question 14: In a survey of $100$ students, the number of students studying the various languages were found to be: English only $18$ , English but not Hindi $23$ , English and Sanskrit $8$ , English $26$ , Sanskrit $48$ , Sanskrit and Hindi $8$ , no language $24$ . Find:
(i) How many students were studying Hindi?
(ii) How many students were studying English and Hindi?

Answer:

Given: $n ( E) = 26, \ n(S) = 48 , \ n(E \cap S) = 8$

$n ( E \cap H' \cap S') = 18, \ n(E \cap H') = 23, \ n( E' \cap H' \cap S') = 24$

ii) $n ( E \cap H') = n(E) - n ( E \cap H )$

$\Rightarrow 23 = 26 - n ( E \cap H)$

$\Rightarrow n ( E \cap H ) = 3$

i) $n( E \cap H' \cap S') = n( E) - n \{ E \cap ( H \cup S') \}$

$\Rightarrow 18 = 26 - n \{ ( E \cap H ) \cup ( E \cap S) \}$

$\Rightarrow 18 = 26 - \{ 3 + 8 - n (E \cap H \cap S ) \}$

$\Rightarrow n ( E \cap H \cap S) = 3$

$n ( E' \cap H' \cap S') = n(U) - n ( E \cup H \cup S)$

$24 = 100 - n ( E \cup H \cup S)$

Therefore number of students studying Hindi

$= n( E \cup H \cup S) - n ( E) - n ( S) + n ( E \cap H ) + n ( E \cap S) + n ( S \cap H) - n ( E \cap H \cap S)$

$= 76 -24 -48 + 3 + 8 + 8 - 3 = 18$

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Question 15: In a survey it was found that $21$ persons liked product $P_1, \ 26$ liked product $P_2$ and $29$ liked product $P_3$ . If $14$ persons liked products $P_1$ and $P_2; \ 12$ persons liked product $P_3$ and $P_1; \ 14$ persons liked products $P_2$ and $P_3$ and $8$ liked all the three products. Find how many liked product $P_3$ only.

Answer:

Given: $n(P_1) = 21, \ n ( P_2) = 26, \ n ( P_3) = 29$

$n ( P_1 \cap P_2) = 14, \ n ( P_3 \cap P_1 ) = 12, \ n ( P_2 \cap P_3) = 14$

$n ( P_1 \cap P_2 \cap P_3) = 8$

Number of people liking Product $P_3$ only

$= n ( P_3 \cap {P_1}' \cap {P_2}' )$

$= n \{ P_3 \cap ( P_1 \cup P_2)' \}$

$= n ( P_3) - n \{ P_3 \cap ( P_1 \cup P_2) \}$

$= n ( P_3) - n \{ ( P_3 \cap P_1) \cup ( P_3 \cap P_2 \}$

$= n ( P_3) - \{ n (P_3 \cap P_1) + n (P_3 \cap P_2) - n ( P_1 \cap P_2 \cap P_3 ) \}$

$= 29 - ( 12 + 14 - 8 ) = 11$

Therefore number of people liking $P_3$ is $11$

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Question 16: A market research group conducted survey on $1000$ persons and reported that $720$ persons liked product $A$ and $450$ persons liked product $B$ . What is the least number of persons that must have liked both products?