\displaystyle \textbf{Question 1: }\text{If }A\text{ and }B\text{ are two sets such that }n(A\cup B)=50,\ n(A)=28
\displaystyle \text{and }n(B)=32,\text{ find }n(A\cap B).
\displaystyle \text{Answer:}
\displaystyle \text{Given }n(A\cup B)=50,\quad n(A)=28\quad\text{and}\quad n(B)=32.
\displaystyle \text{We know that}2019-04-30_8-54-04
\displaystyle n(A\cup B)=n(A)+n(B)-n(A\cap B).
\displaystyle \therefore n(A\cap B)=n(A)+n(B)-n(A\cup B).
\displaystyle =28+32-50
\displaystyle =10.
\displaystyle \therefore n(A\cap B)=10.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{If }P\text{ and }Q\text{ are two sets such that }P\text{ has }40\text{ elements,}
\displaystyle P\cup Q\text{ has }60\text{ elements and }P\cap Q\text{ has }10\text{ elements,}
\displaystyle \text{how many elements does }Q\text{ have?}
\displaystyle \text{Answer:}
\displaystyle \text{Given }n(P)=40,\quad n(P\cup Q)=60,\quad n(P\cap Q)=10.
\displaystyle \text{We know that}2019-04-30_8-55-06
\displaystyle n(P\cup Q)=n(P)+n(Q)-n(P\cap Q).
\displaystyle \therefore 60=40+n(Q)-10.
\displaystyle 60=30+n(Q).
\displaystyle \therefore n(Q)=30.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{In a school there are }20\text{ teachers who teach Mathematics or Physics.}
\displaystyle \text{Of these, }12\text{ teach Mathematics and }4\text{ teach both Mathematics and Physics.}
\displaystyle \text{How many teach Physics?}
\displaystyle \text{Answer:}
\displaystyle \text{Given }n(M\cup P)=20,\quad n(M)=12,\quad n(M\cap P)=4.
\displaystyle \text{We know that}2019-04-30_8-55-19
\displaystyle n(M\cup P)=n(M)+n(P)-n(M\cap P).
\displaystyle \therefore 20=12+n(P)-4.
\displaystyle 20=8+n(P).
\displaystyle \therefore n(P)=12.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{In a group of }70\text{ people, }37\text{ like coffee, }52\text{ like tea,}
\displaystyle \text{and each person likes at least one of the two drinks. How many like both}
\displaystyle \text{coffee and tea?}
\displaystyle \text{Answer:}
\displaystyle \text{Given }n(T\cup C)=70,\quad n(T)=52,\quad n(C)=37.
\displaystyle \text{We know that}2019-04-30_9-07-18
\displaystyle n(T\cup C)=n(T)+n(C)-n(T\cap C).
\displaystyle \therefore 70=52+37-n(T\cap C).
\displaystyle 70=89-n(T\cap C).
\displaystyle \therefore n(T\cap C)=19.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Let }A\text{ and }B\text{ be two sets such that }n(A)=20,\ n(A\cup B)=42
\displaystyle \text{and }n(A\cap B)=4.\text{ Find: (i) }n(B)\text{ (ii) }n(A-B)\text{ (iii) }n(B-A).
\displaystyle \text{Answer:}
\displaystyle \text{Given }n(A)=20,\quad n(A\cup B)=42,\quad n(A\cap B)=4.
\displaystyle \text{(i) We know that}2019-04-30_9-07-33
\displaystyle n(A\cup B)=n(A)+n(B)-n(A\cap B).
\displaystyle \therefore 42=20+n(B)-4.
\displaystyle 42=16+n(B).
\displaystyle \therefore n(B)=26.
\displaystyle \\

\displaystyle \text{(ii) }n(A-B)=n(A)-n(A\cap B).
\displaystyle \therefore n(A-B)=20-4=16.
\displaystyle \\

\displaystyle \text{(iii) }n(B-A)=n(B)-n(A\cap B).
\displaystyle \therefore n(B-A)=26-4=22.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{A survey shows that }76\%\text{ of the Indians like oranges, whereas }62\%
\displaystyle \text{like bananas. Assuming every Indian likes at least one of the two fruits, what}
\displaystyle \text{percentage of the Indians like both oranges and bananas?}
\displaystyle \text{Answer:}
\displaystyle \text{Let the total population be }100.
\displaystyle \therefore n(O\cup B)=100,\quad n(O)=76,\quad n(B)=62.
\displaystyle \text{We know that}2019-05-01_8-16-14.png
\displaystyle n(O\cup B)=n(O)+n(B)-n(O\cap B).
\displaystyle \therefore 100=76+62-n(O\cap B).
\displaystyle 100=138-n(O\cap B).
\displaystyle \therefore n(O\cap B)=38.
\displaystyle \therefore 38\%\text{ of the Indians like both oranges and bananas.}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{In a group of }950\text{ persons, }750\text{ can speak Hindi and }460\text{ can}
\displaystyle \text{speak English. Find: (i) how many can speak both Hindi and English}
\displaystyle \text{(ii) how many can speak Hindi only (iii) how many can speak English only.}
\displaystyle \text{Answer:}
\displaystyle \text{Given }n(H)=750,\quad n(E)=460,\quad n(H\cup E)=950.
\displaystyle \text{(i) We know that}
\displaystyle n(H\cup E)=n(H)+n(E)-n(H\cap E).
\displaystyle \therefore 950=750+460-n(H\cap E).
\displaystyle 950=1210-n(H\cap E).
\displaystyle \therefore n(H\cap E)=260.
\displaystyle \therefore 260\text{ persons can speak both Hindi and English.}
\displaystyle \\ 2019-05-01_8-30-28

\displaystyle \text{(ii) }n(H)=n(H-E)+n(H\cap E).
\displaystyle \therefore 750=n(H-E)+260.
\displaystyle \therefore n(H-E)=490.
\displaystyle \therefore 490\text{ persons can speak Hindi only.}
\displaystyle \\

\displaystyle \text{(iii) }n(E)=n(E-H)+n(H\cap E).
\displaystyle \therefore 460=n(E-H)+260.
\displaystyle \therefore n(E-H)=200.
\displaystyle \therefore 200\text{ persons can speak English only.}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{In a group of }50\text{ persons, }14\text{ drink tea but not coffee and }30
\displaystyle \text{drink tea. Assuming every person drinks at least one of tea or coffee, find:}
\displaystyle \text{(i) how many drink both tea and coffee (ii) how many drink coffee but not tea.}
\displaystyle \text{Answer:}
\displaystyle \text{Given }n(T)=30,\quad n(T-C)=14,\quad n(T\cup C)=50.
\displaystyle \text{(i) We know that}2019-05-01_8-31-07
\displaystyle n(T)=n(T-C)+n(T\cap C).
\displaystyle \therefore 30=14+n(T\cap C).
\displaystyle \therefore n(T\cap C)=16.
\displaystyle \therefore 16\text{ persons drink both tea and coffee.}
\displaystyle \\

\displaystyle \text{(ii) Since }T\cup C=(T-C)\cup(T\cap C)\cup(C-T),
\displaystyle n(T\cup C)=n(T-C)+n(T\cap C)+n(C-T).
\displaystyle \therefore 50=14+16+n(C-T).
\displaystyle \therefore n(C-T)=20.
\displaystyle \therefore 20\text{ persons drink coffee but not tea.}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{In a survey of }60\text{ people, it was found that }25\text{ read newspaper }H,
\displaystyle 26\text{ read newspaper }T,\ 26\text{ read newspaper }I,\ 9\text{ read both }H\text{ and }I,
\displaystyle 11\text{ read both }H\text{ and }T,\ 8\text{ read both }T\text{ and }I,\text{ and }3\text{ read all}
\displaystyle \text{three newspapers. Find: (i) the number of people who read at least one newspaper}
\displaystyle \text{(ii) the number of people who read exactly one newspaper.}
\displaystyle \text{Answer:}
\displaystyle \text{Given }n(H)=25,\quad n(T)=26,\quad n(I)=26,
\displaystyle n(H\cap I)=9,\quad n(H\cap T)=11,\quad n(T\cap I)=8
\displaystyle \text{and }n(H\cap T\cap I)=3.
\displaystyle \text{(i) By the inclusion--exclusion principle,}
\displaystyle n(H\cup T\cup I)=n(H)+n(T)+n(I)-n(H\cap T)
\displaystyle \hspace{2.4cm}-n(T\cap I)-n(I\cap H)+n(H\cap T\cap I).
\displaystyle =25+26+26-11-8-9+3
\displaystyle =52.
\displaystyle \therefore 52\text{ people read at least one of the newspapers.}
\displaystyle \text{Also, the number of people who read none of the newspapers}=60-52=8.
\displaystyle \\

\displaystyle \text{(ii) Number of people who read only }H2019-05-01_8-49-04
\displaystyle =n(H)-n(H\cap T)-n(H\cap I)+n(H\cap T\cap I)
\displaystyle =25-11-9+3=8.
\displaystyle \text{Number of people who read only }T
\displaystyle =n(T)-n(H\cap T)-n(T\cap I)+n(H\cap T\cap I)
\displaystyle =26-11-8+3=10.
\displaystyle \text{Number of people who read only }I
\displaystyle =n(I)-n(H\cap I)-n(T\cap I)+n(H\cap T\cap I)
\displaystyle =26-9-8+3=12.
\displaystyle \therefore \text{Number of people who read exactly one newspaper}=8+10+12=30.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Of the members of three athletic teams in a certain school, }21\text{ are in}
\displaystyle \text{the basketball team, }26\text{ are in the hockey team and }29\text{ are in the football}
\displaystyle \text{team. }14\text{ play hockey and basketball, }15\text{ play hockey and football, }12
\displaystyle \text{play football and basketball, and }8\text{ play all three games. How many members}
\displaystyle \text{are there in all?} 2019-05-01_8-55-34
\displaystyle \text{Answer:}
\displaystyle \text{Given }n(B)=21,\quad n(H)=26,\quad n(F)=29,
\displaystyle n(H\cap B)=14,\quad n(H\cap F)=15,\quad n(B\cap F)=12
\displaystyle \text{and }n(H\cap B\cap F)=8.
\displaystyle \text{By the inclusion--exclusion principle,}
\displaystyle n(H\cup B\cup F)=n(H)+n(B)+n(F)-n(H\cap B)
\displaystyle \hspace{2.5cm}-n(B\cap F)-n(F\cap H)+n(H\cap B\cap F).
\displaystyle =26+21+29-14-12-15+8
\displaystyle =43.
\displaystyle \therefore \text{The total number of members in all three teams is }43.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{In a group of }1000\text{ people, }750\text{ can speak Hindi and }400\text{ can}
\displaystyle \text{speak Bengali. Find: (i) how many can speak Hindi only (ii) how many can speak}
\displaystyle \text{Bengali only (iii) how many can speak both Hindi and Bengali.}
\displaystyle \text{Answer:}
\displaystyle \text{Given }n(H\cup B)=1000,\quad n(H)=750,\quad n(B)=400.
\displaystyle \text{We know that}2019-05-01_9-04-05.png
\displaystyle n(H\cup B)=n(H)+n(B)-n(H\cap B).
\displaystyle \therefore 1000=750+400-n(H\cap B).
\displaystyle 1000=1150-n(H\cap B).
\displaystyle \therefore n(H\cap B)=150.
\displaystyle \therefore 150\text{ people can speak both Hindi and Bengali.}
\displaystyle \\

\displaystyle \text{Number of people who can speak Hindi only}
\displaystyle =n(H)-n(H\cap B)=750-150=600.
\displaystyle \\

\displaystyle \text{Number of people who can speak Bengali only}
\displaystyle =n(B)-n(H\cap B)=400-150=250.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{A survey of }500\text{ television viewers produced the following information:}
\displaystyle 285\text{ watch football, }195\text{ watch hockey, }115\text{ watch basketball, }45\text{ watch}
\displaystyle \text{football and basketball, }70\text{ watch football and hockey, }50\text{ watch hockey}
\displaystyle \text{and basketball, and }50\text{ do not watch any of the three games. How many watch}
\displaystyle \text{(i) all three games (ii) exactly one of the three games?}
\displaystyle \text{Answer:}
\displaystyle \text{Given }n(F)=285,\quad n(H)=195,\quad n(B)=115,
\displaystyle n(F\cap B)=45,\quad n(F\cap H)=70,\quad n(H\cap B)=50.
\displaystyle \text{Also, }50\text{ viewers do not watch any of the three games.}
\displaystyle \therefore n(F\cup H\cup B)=500-50=450.
\displaystyle \\ 2019-05-01_9-22-25.png

\displaystyle \text{(i) By the inclusion--exclusion principle,}
\displaystyle n(F\cup H\cup B)=n(F)+n(H)+n(B)-n(F\cap H)
\displaystyle \hspace{2.4cm}-n(F\cap B)-n(H\cap B)+n(F\cap H\cap B).
\displaystyle \therefore 450=285+195+115-70-45-50+n(F\cap H\cap B).
\displaystyle 450=430+n(F\cap H\cap B).
\displaystyle \therefore n(F\cap H\cap B)=20.
\displaystyle \therefore 20\text{ viewers watch all three games.}
\displaystyle \\

\displaystyle \text{(ii) Number of viewers who watch exactly one game}
\displaystyle =n(F)+n(H)+n(B)-2n(F\cap H)-2n(F\cap B)
\displaystyle \hspace{2.4cm}-2n(H\cap B)+3n(F\cap H\cap B).
\displaystyle =285+195+115-2(70)-2(45)-2(50)+3(20)
\displaystyle =595-140-90-100+60
\displaystyle =325.
\displaystyle \therefore 325\text{ viewers watch exactly one of the three games.}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{In a survey of }100\text{ persons, it was found that }28\text{ read magazine }A,
\displaystyle 30\text{ read magazine }B,\ 42\text{ read magazine }C,\ 8\text{ read magazines }A\text{ and }B,
\displaystyle 10\text{ read magazines }A\text{ and }C,\ 5\text{ read magazines }B\text{ and }C,\text{ and }3
\displaystyle \text{read all three magazines. Find: (i) how many read none of the three magazines}
\displaystyle \text{(ii) how many read magazine }C\text{ only.}
\displaystyle \text{Answer:}
\displaystyle \text{Given }n(A)=28,\quad n(B)=30,\quad n(C)=42,
\displaystyle n(A\cap B)=8,\quad n(A\cap C)=10,\quad n(B\cap C)=5
\displaystyle \text{and }n(A\cap B\cap C)=3.2019-05-01_9-34-12
\displaystyle \text{(i) By the inclusion--exclusion principle,}
\displaystyle n(A\cup B\cup C)=n(A)+n(B)+n(C)-n(A\cap B)
\displaystyle \hspace{2.4cm}-n(A\cap C)-n(B\cap C)+n(A\cap B\cap C).
\displaystyle =28+30+42-8-10-5+3
\displaystyle =80.
\displaystyle \therefore n\big((A\cup B\cup C)'\big)=n(U)-n(A\cup B\cup C).
\displaystyle =100-80=20.
\displaystyle \therefore 20\text{ persons read none of the three magazines.}
\displaystyle \\

\displaystyle \text{(ii) Number of persons who read magazine }C\text{ only}
\displaystyle =n(C)-n(A\cap C)-n(B\cap C)+n(A\cap B\cap C).
\displaystyle =42-10-5+3
\displaystyle =30.
\displaystyle \therefore 30\text{ persons read magazine }C\text{ only.}
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{In a survey of }100\text{ students, the number studying various languages}
\displaystyle \text{were: English only }18,\text{ English but not Hindi }23,\text{ English and Sanskrit }8,
\displaystyle \text{English }26,\text{ Sanskrit }48,\text{ Sanskrit and Hindi }8,\text{ and no language }24.
\displaystyle \text{Find: (i) how many students were studying Hindi}
\displaystyle \text{(ii) how many students were studying both English and Hindi.}
\displaystyle \text{Answer:}
\displaystyle \text{Given }n(E)=26,\quad n(S)=48,\quad n(E\cap S)=8,
\displaystyle n(S\cap H)=8,\quad n(E\cap H')=23
\displaystyle \text{and }n\big((E\cup H\cup S)'\big)=24.
\displaystyle \text{(ii) Since }E=(E\cap H')\cup(E\cap H),\text{ where the two sets are disjoint,}
\displaystyle n(E)=n(E\cap H')+n(E\cap H).
\displaystyle \therefore 26=23+n(E\cap H).2019-05-01_9-56-41.png
\displaystyle \therefore n(E\cap H)=3.
\displaystyle \therefore 3\text{ students study both English and Hindi.}
\displaystyle \\

\displaystyle \text{(i) Since }24\text{ students study no language,}
\displaystyle n(E\cup H\cup S)=100-24=76.
\displaystyle \text{By the inclusion--exclusion principle,}
\displaystyle n(E\cup H\cup S)=n(E)+n(H)+n(S)-n(E\cap H)
\displaystyle \hspace{2.4cm}-n(E\cap S)-n(H\cap S)+n(E\cap H\cap S).
\displaystyle \text{English only }=18\text{ and English but not Hindi }=23.
\displaystyle \therefore n(E\cap S)-n(E\cap H\cap S)=23-18=5.
\displaystyle \therefore n(E\cap H\cap S)=8-5=3.
\displaystyle \therefore 76=26+n(H)+48-3-8-8+3.
\displaystyle 76=n(H)+58.
\displaystyle \therefore n(H)=18.
\displaystyle \therefore 18\text{ students study Hindi.}
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{In a survey, it was found that }21\text{ persons liked product }P_1,\ 26
\displaystyle \text{liked product }P_2\text{ and }29\text{ liked product }P_3.\text{ If }14\text{ persons liked products }P_1
\displaystyle \text{and }P_2,\ 12\text{ liked products }P_3\text{ and }P_1,\ 14\text{ liked products }P_2\text{ and }P_3,
\displaystyle \text{and }8\text{ liked all three products, find how many liked product }P_3\text{ only.}
\displaystyle \text{Answer:}
\displaystyle \text{Given }n(P_1)=21,\quad n(P_2)=26,\quad n(P_3)=29,
\displaystyle n(P_1\cap P_2)=14,\quad n(P_3\cap P_1)=12,\quad n(P_2\cap P_3)=14
\displaystyle \text{and }n(P_1\cap P_2\cap P_3)=8.
\displaystyle \text{Number of persons who liked product }P_3\text{ only}
\displaystyle =n(P_3\cap P_1'\cap P_2')2019-05-01_10-20-09
\displaystyle =n\big(P_3\cap(P_1\cup P_2)'\big)
\displaystyle =n(P_3)-n\big(P_3\cap(P_1\cup P_2)\big)
\displaystyle =n(P_3)-n\big((P_3\cap P_1)\cup(P_3\cap P_2)\big)
\displaystyle =n(P_3)-\{n(P_3\cap P_1)+n(P_3\cap P_2)-n(P_1\cap P_2\cap P_3)\}
\displaystyle =29-(12+14-8)
\displaystyle =29-18=11.
\displaystyle \therefore 11\text{ persons liked product }P_3\text{ only.}
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{A market research group conducted a survey on }1000\text{ persons and}
\displaystyle \text{reported that }720\text{ persons liked product }A\text{ and }450\text{ persons liked}
\displaystyle \text{product }B.\text{ What is the least number of persons that must have liked both}
\displaystyle \text{products?}
\displaystyle \text{Answer:}
\displaystyle \text{Given }n(A)=720,\quad n(B)=450,\quad n(A\cup B)=1000.
\displaystyle \text{The least possible value of }n(A\cap B)\text{ occurs when }n(A\cup B)\text{ is maximum,}
\displaystyle \text{i.e., }n(A\cup B)=1000.2019-05-01_10-20-46
\displaystyle \text{We know that}
\displaystyle n(A\cup B)=n(A)+n(B)-n(A\cap B).
\displaystyle \therefore 1000=720+450-n(A\cap B).
\displaystyle 1000=1170-n(A\cap B).
\displaystyle \therefore n(A\cap B)=170.
\displaystyle \therefore \text{The least number of persons who must have liked both products is }170.
\displaystyle \\

Discover more from ICSE / ISC / CBSE Mathematics Portal for K12 Students

Subscribe to get the latest posts sent to your email.