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Topic 1: Problems Including Two Sets | Class 10 Mathematics Solution

Class 10 Mathematics

Topic 1: Problems Including Two Sets

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1. a) In the given Venn-diagram, F and B are the sets of students who play football and basketball respectively. Let's find the number of students in these sets operations.
(a) n(F ∩ B)
(b) n(F ∪ B)
(c) n( F ∪ B )
(d) nₒ(F)
(e) n(B – F)
(f) n(U)

Solution

Given

From Venn-diagram, we get

(a) n(F ∩ B) = 18

(b) n(F ∪ B) = 12 + 18 + 10 = 40

(d) nₒ(F) = 12

(e) n(B – F) = 10

(f) n(U) = 12 + 18 + 10 + 8 = 48

b) In the adjoining Venn-diagram, N and H are the sets of people who like Nepali and Hindi movies respectively. If n(U) = 135 and n(N ∪ H) = 120, find:
(a) n(N ∩ H)
(b) n( N ∪ H )
(c) nₒ(N)
(d) nₒ(H)

Solution

Given

If n(U) = 135 and n(N ∪ H) = 120

(a) From Venn-diagram, n (N ∪ H) = nₒ(N) + nₒ(H) + n (N ∩ H) or, 120 = (85 – x) + (55 – x) + x or, 120 = 140 – x or, x = 20 ∴ n(N ∩ H) = x = 20

(b) n( N ∪ H ) = n (U) – n (N ∪ H) = 135 – 120 = 15

(d) nₒ(H) = 55 – x = 55 – 20 = 35

2. a) If n(U) = 50, n(A) = 25, n(B) = 27 and n(A ∩ B) = 10, find:
(a) n(A ∪ B)
(b) n( A ∪ B )
(c) n(A – B)
(d) nₒ(B)

Solution

Given

Here, n(U) = 50, n(A) = 25, n(B) = 27 and n(A ∩ B) = 10

(a) We know, n (A ∪ B) = n(A) + n(B) – n (A ∩ B) = 25 + 27 – 10 = 42

(b) We have, n( A ∪ B ) = n (U) – n (A ∪ B) = 50 – 42 =8

(d) We have, nₒ(B) = n (B) – n (A ∩ B) = 27 – 10 = 17

b) If n(U) = 60, n(A) = 30, n(B) = 40 and n(A ∪ B) = 55, find:
(a) n(A ∩ B)
(b) n( A ∪ B )
(c) nₒ(A)
(d) n(only B)

Solution

Given

Here, n(U) = 60, n(A) = 30, n(B) = 40 and n(A ∪ B) = 55

(a) We know, n (A ∩ B) = n(A) + n(B) – n (A ∪ B) = 30 + 40 – 55 = 15

(b) We have, n( A ∪ B ) = n (U) – n (A ∪ B) = 60 – 55 =5

(d) We have, n(only B) = n (B) – n (A ∩ B) = 40 – 15 = 25

c) P and Q are the subsets of a universal set U. If n(U) = 90, n(P) = 45, n(Q) = 35 and n( P ∪ Q ) = 15, illustrate this information in a Venn-diagram and find:
(a) n (P ∪ Q)
(b) n(P ∩ Q)
(c) nₒ(P)
(d) n(Q – P)

Solution

Given

Here, n (U) = 90, n(P) = 45, n(Q) = 35 and n( P ∪ Q ) = 15 Let, n (P ∩ Q) = x. Illustration of the given information in a Venn-diagram:

(a) We know, n (P ∪ Q) = n (U) – n( P ∪ Q ) = 90 – 15 = 75

(b) From Venn-diagram, n (U) = nₒ(P) + nₒ(Q) + n (P ∩ Q) + n( P ∪ Q ) or, 90 = (45 – x) + (35 – x) + x + 15 or, 90 = 95 – x or, x =5 ∴ n(P ∩ Q) = x = 5

(b) nₒ(P) = 45 – x = 45 – 5 = 40

d) If n(A) = 65, n(B) = 80 and A ⊂ B, find:
(a) n(A ∪ B)
(b) n(A ∩ B)
(c) n(A – B)
(d) n(B – A)

Solution

Given

Here, n(A) = 65, n(B) = 80 and A ⊂ B

(a) We know, n (A ∪ B) = n(B) = 80

(b) We have, n(A ∩ B) = n (A) = 65

(d) We have, n(B – A) = n (B) – n (A ∩ B) = 80 – 65 = 15

e) If nₒ(A) = 12, nₒ(B) = 15, n( A ∪ B ) = 11 and n(U) = 45, find:
(a) n(A ∪ B)
(b) n(A ∩ B)
(c) n(A)
(d) n(B)

Solution

Given

Here, nₒ(A) = 12, nₒ(B) = 15, n( A ∪ B ) = 11 and n(U) = 45

(a) We know, n (A ∪ B) = n (U) – n( A ∪ B ) = 45 – 11 = 34

(b) We know, n (A ∪ B)= nₒ(A) + nₒ(B) + n (A ∩ B) or, 34 = 12 + 15 + n (A ∩ B) or, 34 = 27 + n (A ∩ B) ∴ n(A ∩ B) = 7

(d) n(B) = nₒ(B) + n(A ∩ B) = 15+ 7 = 22

3. a) If n (A) = 40 and n (B) = 50, find:
(a) the possible maximum value of n (A ∩ B) and minimum value of n (A ∪ B).
(b) the possible minimum value of n (A ∩ B) and maximum value of n (A ∪ B).

Solution

Given

Here, n (A) = 40 and n (B) = 50

(a) The value of n (A ∩ B) is maximum when A ⊂ B. So, the possible maximum value of n (A ∩ B) = n (A) = 40 Again, the value of n (A ∪ B) is minimum when A ⊂ B. So, the possible minimum value of n (A ∪ B) = n (B) = 50

(b) The value of n (A ∩ B) is minimum when A and B are disjoint sets. So, the possible minimum value of n (A ∩ B) = 0 Again, the value of n (A ∪ B) is maximum when A and B are disjoint sets. So, the possible maximum value of n (A ∪ B) = n (A) + n (B) = 40 + 50 = 90

b) If M and N are two sets containing 15 and 10 elements, find:
(a) the greatest value of n (M ∩ N) and least value of n (M ∪ N).
(b) the least value of n (M ∩ N) and greatest value of n (M ∪ N).

Solution

Given

Here, n (M) = 15 and n (N) = 10

(a) The value of n (M ∩ N) is greatest when N ⊂ M. So, the greatest value of n (M ∩ N) = n (N) = 10 Again, the value of n (M ∪ N) is least when N ⊂ M. So, the least value of n (M ∪ N) = n (M) = 15

(b) The value of n (M ∩ N) is least when M and N are disjoint sets. So, the least value of n (M ∩ N) = 0 Again, the value of n (M ∪ N) is greatest when M and N are disjoint sets. So, the greatest value of n (M ∪ N) = n (M) + n (N) = 15 + 10 = 25

c) P and Q are the subsets of universal set U. If n (U) = 100, n (P) = 75 and n (Q) = 55, find:
(a) the greatest value of n (P ∪ Q)
(b) the least value of n (P ∩ Q).

Solution

Given

Here, P and Q are the subsets of universal set U. If n (U) = 100, n (P) = 75 and n (Q) = 55

(a) The value of n (P ∪ Q) cannot be greater than n (U). So, the greatest value of n (P ∪ Q) = n (U) = 100

(b) The value of n (P ∩ Q) is least when n (P ∪ Q) the greatest value. We have, n (P ∩ Q) = n (P) + n (Q) – n (P ∪ Q) = 75 + 55 – 100 = 30

4. a) A survey was conducted among 125 students of class 10 studying in Gyan Mandir Secondary School regarding the places for educational tour and it was found that 65 students preferred to visit Lumbini (L), 75 preferred to visit Pokhara (P), and 25 preferred both the places.
(a) Write the cardinality of set of students who preferred to visit both Pokhara and Lumbini in set notational form.
(b) Show the above information in a Venn-diagram.
(c) Find the number of students who preferred neither of two places.
(d) Find the ratio of the number of students who preferred Lumbini only and Pokhara only.

Solution

Given

Here, n (U) = 125, n (L) = 65, n (P) = 75 and n (P ∩ L) = 25

(a) The cardinality of set of students who preferred to visit both Pokhara and Lumbini is n (P ∩ L) = 25

(b) The given information is shown in a Venn-diagram:

(c) We have, n (P ∪ L) = n(P) + n(L) – n (P ∩ L) = 75 + 65 – 25 = 115 Again, n( P ∪ L ) = n (U) – n (P ∪ L) = 125 – 115 = 10 Thus, 10 students preferred neither of two places.

(d) No. of students who preferred Lumbini only, nₒ(L) = n (L) – n (P ∩ L) = 65 – 25 = 40 No. of students who preferred Pokhara only, nₒ(P) = n (P) – n (P ∩ L) = 75 – 25 = 50 So, the ratio of nₒ(L) and nₒ(P) = 4050 = 4 : 5

b) A survey was carried out in a village of Chandrapur municipality of Rautahat district regarding the local languages. Among 1,400 people participated in the survey, it was found that 600 people can speak Bhojpuri, 900 can speak Maithili and 350 people can speak Bhojpuri as well as Maithili languages.
(a) If B and M represent the sets of people who can speak Bhojpuri and Maithili respectively, write the cardinality of n(B ∩ M).
(b) Draw a Venn-diagram to show the above data.
(c) How many people cannot speak both the languages?
(d) How many more or less people can speak Maithili only than Bhojpuri only?

Solution

Given

Here, n (U) = 1400, n (B) = 600, n (M) = 900 and n (B ∩ M) = 350

(a) The cardinality of n (B ∩ M) = 350

(b) The given information is shown in a Venn-diagram:

(c) We have, n (B ∪ M)= n(B) + n(M) – n (B ∩ M) = 600 + 900 – 350 = 1150 Again, n( B ∪ M ) = n (U) – n (B ∪ M) = 1400 – 1150 = 250 Thus, 250 people can speak neither language.

(d) No. of people who can speak Bhojpuri only, nₒ(B) = n (B) – n (B ∩ M) = 600 – 350 = 250 No. of people who can speak Maithili only, nₒ(M) = n (M) – n (B ∩ M) = 900 – 350 = 550 Difference of nₒ(M) and nₒ(B) = 550 – 250 = 300 So, the number of people who can speak Maithili only is 300 more than the number of people who can speak Bhojpuri only.

c) During ‘Visit Nepal 2020’, among 7,500 Chinese tourists who visited Nepal, 60% of them had already visited Bhutan, 50 % had visited Sri Lanka, and 30 % have visited both the countries.
(a) If B and S denote the sets of tourists who visited Bhutan and Sri Lanka respectively, write the number of tourists who visited Bhutan in set notation.
(b) Illustrate this information in a Venn-diagram.
(c) How many tourists have visited neither Bhutan nor Srilanka?
(d) How many tourists have already visited only one of these countries?

Solution

Given

Here, n (U) = 7500, n (B) = 60% of 7500 = 4500, n (S) = 50% of 7500 = 3750 and n (B ∩ S) = 30% of 7500 = 2250

(a) The number of tourists who visited Bhutan, n (B) = 4500

(b) The given information is shown in a Venn-diagram:

(c) We have, n (B ∪ S)= n(B) + n(S) – n (B ∩ S) = 4500 + 3750 – 2250 = 6000 Again, n( B ∪ S ) = n (U) – n (B ∪ S) = 7500 – 6000 = 1500 Thus, 1500 tourists have visited neither Bhutan nor Sri Lanka.

(d) No. of tourists have already visited Bhutan only, nₒ(B) = n (B) – n (B ∩ S) = 4500 – 2250 = 2250 No. of tourists have already visited Sri Lanka only, nₒ(S) = n (S) – n (B ∩ S) = 3750 – 2250 = 1500 Thus, the number of tourists who have already visited only one of these countries = nₒ(B) + nₒ(S) = 2250 + 1500 = 3750

5. a) In a survey conducted in a community of Gorkha district regarding cultural dances, it was found that out of 500 people, 180 people can perform Sorathi dance (S), 250 can perform Ghatu dance (G) and 120 can perform neither of these two dances.
(a) Write the set notation that represents the number of people who can perform neither Sorathi nor Ghatu dance.
(b) Represent the above information in a Venn-diagram.
(c) How many people can perform both the dances?
(d) Compare the number of people who can perform Sorathi only and Ghatu only.

Solution

Given

Here, n (U) = 500, n (S) = 180, n (G) = 250 and n( S ∪ G ) = 120

(a) The set notation that represents the number of people who can perform neither Sorathi nor Ghatu dance is n( S ∪ G ) = 120.

(b) Let, n (S ∩ G) = x. The given information is shown in a Venn-diagram:

(c) From Venn-diagram, n (U) = nₒ(S) + nₒ(G) + n (S ∩ G) + n( S ∪ G ) or, 500 = (180 – x) + (250 – x) + x + 120 or, 500 = 550 – x or, x = 50 So, 50 people can perform both the dances.

(d) No. of people who can perform Sorathi only, nₒ(S) = 180 – 50 = 130 No. of people who can perform Ghatu only, nₒ(G) = 250 – 50 = 200 Also, difference between nₒ(G) and nₒ(S) = 200 – 130 = 70 So, the number of people who can perform Sorathi only is 70 less than the number of people who can perform Ghatu only. OR, The ratio of nₒ(S) and nₒ(G) = 130200 = 13 : 20

b) Last Wednesday, a School successfully accomplished its school’s day. For this program, every student participated in at least one of the activities, athletics or music. In a class of 45 students, 21 participated in athletics (A) and 29 participated in music (M).
(a) Write the cardinality of n( A ∪ M )?
(b) Draw a Venn-diagram to represent the above information.
(c) How many students participated in both the activities?
(d) Compare the number of students who participated in athletics only and music only.

Solution

Given

Here, n (U) = 45, n (A) = 21, n (M) = 29 and n( A ∪ M ) = 0

(a) The cardinality of n( A ∪ M ) is 0.

(b) Let, n (A ∩ M) = x. The given information is shown in a Venn-diagram:

(c) From Venn-diagram, n (U) = nₒ(A) + nₒ(M) + n (A ∩ M) + n( A ∪ M ) or, 45 = (21 – x) + (29 – x) + x + 0 or, 45 = 50 – x or, x =5 So, 5 students participated in both the activities.

(d) No. of students who participated in athletics only, nₒ(A) = 21 – 5 = 16 No. of students who participated in music only, nₒ(M) = 29 – 5 = 24 Also, difference between nₒ(A) and nₒ(M) = 24 – 16 = 8 So, the number of students who participated in music only is 8 more than the number of students who participated in athletics only. OR, The ratio of nₒ(A) and nₒ(M) = 1624 = 2 : 3

c) In a survey conducted among 150 students of a school, it was found that 70 students liked cricket, 62 liked basketball and 30 students did not like any of these two games.
(a) If C and B denote the set of the students who liked cricket and basketball respectively, write the cardinality of n( B ∪ C ).
(b) Present the information in a Venn-diagram.
(c) Find the number of students who liked cricket only.
(d) Compare the number of students who liked both games and who liked except these two games.

Solution

Given

Here, n (U) = 150, n (C) = 70, n (B) = 62 and n( B ∪ C ) = 30

(a) The cardinality of n( B ∪ C ) is 30.

(b) Let, n (B ∩ C) = x. The given information is shown in a Venn-diagram:

(c) From Venn-diagram, n (U) = nₒ(B) + nₒ(C) + n (B ∩ C) + n( B ∪ C ) or, 150 = (62 – x) + (70 – x) + x + 30 or, 150 = 162 – x or, x = 12 Also, the number of students who liked cricket only = nₒ(C) = 70 – 12 = 58

(d) Difference between the number of students who liked both games and those who liked neither game = n( B ∪ C ) – n (B ∩ C) = 30 – 12 = 18 So, the number of students who liked both games is 18 less than the number of students who liked neither game. OR, The ratio of n (B ∩ C) and n( B ∪ C ) = 1230 = 2 : 5

6. a) Among 54 SEE appeared students from a school, 18 students got ‘A+’ grade in Mathematics only, 25 got ‘ A+’ grade in English only and 7 students did not get ‘A+’ grade in these two subjects.
(a) Write the cardinalities of the given sets in set notational forms.
(b) Show the above information in Venn-diagram.
(c) How many students got ‘ A+’ grade in Mathematics?
(d) How many more or less students got ‘A+’ grade not in English than not in Mathematics?

Solution

Given

(a) Let, M and E denote the sets of students who got ‘A+’ grade in Mathematics and English respectively. Then, n (U) = 54, nₒ(M) = 18, nₒ(E) = 25 and n( M ∪ E ) = 7

(b) Let, n (M ∩ E) = x. The given information is shown in a Venn-diagram:

(c) From Venn-diagram, n (U) = nₒ(M) + nₒ(E) + n (M ∩ E) + n( M ∪ E ) or, 54 = 18 + 25 + x + 7 or, 54 = 50 + x or, x =4 Also, the no. of students who got ‘A+’ grade in Mathematics = n(M) = 18 + 4 = 22.

(d) Again, the no. of students who got ‘A+’ grade in English = n(E) = 25 + 4 = 29. The number of students got ‘A+’ grade not in English = n( E )= 54 – 29 = 25 The number of students got ‘A+’ grade not in Mathematics= n( M )= 54 – 22 = 32 Difference between n( M ) and n( E ) = 32 – 25 = 7 So, the students who got ‘A+’ grade not in English is 7 less than the number of students who got ‘A+’ grade not in Mathematics.

b) In a survey of a group of people, 20 % are using cellular data but not Wi-Fi, 65% are using Wi-Fi but not cellular data and 5 % of them use neither cellular data nor Wi-Fi,
(a) If C and W denote the sets of people who are using cellular data and Wi-Fi respectively, write the cardinality of n( C ∪ W ).
(b) Represent the above information in a Venn-diagram.
(c) Find the percent of people who are using Wi-Fi.
(d) By what percent of the people who are not using cellular data is more or less than the people who are not using Wi-Fi?

Solution

Given

Let, n (U) = 100. Then, nₒ(C) = 20, nₒ(W) = 65 and n( C ∪ W ) = 5

(a) The cardinality of n( C ∪ W ) is 5

(b) Let, n (C ∩ W) = x. The given information is shown in a Venn-diagram:

(c) From Venn-diagram, n (U) = nₒ(C) + nₒ(W) + n (C ∩ W) + n( C ∪ W ) or, 100 = 20 + 65 + x + 5 or, 100 = 90 + x or, x = 10 Also, the percent of people who are using Wi-Fi, n (W) = (65 + 10) % = 75%

(d) Again, the percent of the people who are using cellular data=n(C)=(20 +10)% = 30% The percent of the people who are not using cellular data =n( C )=(100 – 30)%= 70% The percent of the people who are not using Wi-Fi =n( W )=(100 – 75)%= 25% The percent of the people who are not using cellular data is 45% more than the people who are not using Wi-Fi.

7. a) In a survey of 15,000 students of different schools, 6,000 of them were found to have tuition classes before the SEE examination. Among them, 3,000 studied only Mathematics, 1,800 only Science and 600 studied other subjects but not these two subjects.
(a) If M and S denote the sets of students who studied Mathematics and Science respectively, write the cardinality of nₒ(M), nₒ(S) and n( M ∪ S ).
(b) Represent the above information in a Venn-diagram.
(c) Find the number of students who studied Science.
(d) What percent of the students of the survey studied either Mathematics or Science?

Solution

Given

Here, n (U) = 15000 and n(U1) = 6000

(a) nₒ(M) = 3000, nₒ(S) = 1800 and n( M ∪ S ) = 600

(b) Let, n (M ∩ S) = x. The given information is shown in a Venn-diagram:

(c) From Venn-diagram, n (U1) = nₒ(M) + nₒ(S) + n (M ∩ S) + n( M ∪ S ) or, 6000 = 3000 + 1800 + x + 600 or, 6000 = 5400 + x or, x = 600 Also, the number of students who studied Science, n (S) = 1800 + 600 = 2400

(d) The number of students who studied either Mathematics or Science, n (M ∪ S) = nₒ(M) + nₒ(S) + n (M ∩ S) = 3000 + 1800 + 600 = 5400 Again, the percent of the students of the survey who studied either Mathematics or Science = n (M ∪ S)n(U) × 100% = 540015000 × 100% = 36%

b) Of 200 students who appeared the SEE from Dhaulagiri Secondary School, 25% students got ‘A+’ grade in different subjects. Out of the students who got ‘A+’ grade, 40% students got in English only, 20% got in Mathematics only and 30% in other subjects.
(a) Write the set notation representing the number of students who got A+ grades.
(b) Show the above information in a Venn-diagram.
(c) How many students got ‘A+’ grade in English?
(d) Find the ratio of number of students who got ‘A+’ grade in English to those students who got in Mathematics.

Solution

Given

Let, E and M denote the sets of students who who got ‘A+’ grade in English and Mathematics respectively. Then, n (U) = 200, n(U1) = 25% of 200 = 50, nₒ(E) = 40% of 50 = 20, nₒ(M) = 20% of 50 = 10 and n( E ∪ M ) = 30% of 50 = 15

(a) The set notation representing the number of students who got A+ grades, n(U 1)= 50

(b) Let, n (E ∩ M) = x. The given information is shown in a Venn-diagram:

(c) From Venn-diagram, n (U1) = nₒ(E) + nₒ(M) + n (E ∩ M) + n( E ∪ M ) or, 50 = 20 + 10 + x + 15 or, 50 = 45 + x or, x =5 Also, the number of students who got ‘A+’ grade in English, n (E) = 20 + 5 = 25

(d) The number of students who got ‘A+’ grade in Mathematics, n (M) = 10 + 5 = 15 Again, the ratio of number of students who got ‘A+’ grade in English to those students who got in Mathematics = 2515 = 5 : 3

8. a) In a survey of a group of people, it was found that 65% of them liked comedy movies, 63% liked action movies, 33% liked both types of movies, and 150 people did not like both types of movies.
(a) Draw a Venn-diagram to illustrate the above information.
(b) Find the number of people participated in the survey.
(d) Find the number of people who liked both types of movies.
(e) Amrita calculated that the number of people who did not like comedy movie is 150 more than the number of people who liked action movie only. Justify her calculation.

Solution

Given

Let, C and A denote the sets of people who liked comedy movies and action movies respectively and n (U) = x. Then, n (C) = 65% of x = 0.65x, n(A) = 63% of x = 0.63x, n (C ∩ A) = 33% of x = 0.33x and n( C ∪ A )= 150

(a) Representing the given data in a Venn-diagram:

(b) From Venn-diagram, n(U) = nₒ(C) + nₒ(A) + n (C ∩ A) + n( C ∪ A ) or, x = 0.32x + 0.3x + 0.33x + 150 or, x = 0.95x + 150 or, x = 3000 Thus, 3000 people participated in the survey.

(d) The number of people who did not like comedy movie, n( C ) = n (U) – n (C) = 3000 – 0.65 × 3000 = 1050 The number of people who liked action movie only, nₒ(A) = 0.3 × 3000 = 900 Also, difference = 1050 – 900 = 150 Thus, Amrita’s calculation is correct.

b) In an examination, 80 % examinees passed in English, 70 % in Mathematics, 60 % passed in both the subjects, and 45 examinees failed in both subjects.
(a) Draw a Venn-diagram to represent the above information.
(b) Find the number of examinees participated in the survey.
(f) Find the number of examinees who passed in both subjects. (g) Ram calculated that the number of examinees who failed in English is double the number of examinees who passed mathematics only. State whether he is right or wrong by testing his calculation.

Solution

Given

Let, E and M denote the sets of examinees who passed in English and Mathematics respectively and n (U) = x. Then, n (E) = 80% of x = 0.8x, n(M) = 70% of x = 0.7x, n (E ∩ M) = 60% of x = 0.6x and n( E ∪ M )= 45

(a) Representing the given data in a Venn-diagram:

(b) From Venn-diagram, n(U) = nₒ(E) + nₒ(M) + n (E ∩ M) + n( E ∪ M ) or, x = 0.2x + 0.1x + 0.6x + 45 or, x = 0.9x + 45 or, x = 450 Thus, 450 examinees participated in the survey.

(d) The number of examinees who failed in English, n( E ) = n (U) – n (E) = 450 – 0.8 × 450 = 90 The number of examinees who passed mathematics only, nₒ(M) = 0.1 × 450 = 45 n( E ) 90 Also, n (M) = 45= 2 ∴ n( E ) = 2 × nₒ(M) o Thus, Ram’s calculation is correct.

9. a) In a survey of a community, it was found that 65% of people liked folk songs, 55% liked modern songs and 10% of people did not like both types of songs.
(a) Illustrate the above information in a Venn-diagram.
(b) What percentage of people liked either folk songs or modern songs?
(c) If 360 people liked both types of songs, how many people were surveyed?
(d) How many times is the number of people who liked both type of songs of the number of people who did not like both type of songs?

Solution

Given

Let, F and M denote the sets of people who liked folk songs and modern songs respectively and n (U) = 100. Then, n (F) = 65, n(M) = 55, n( F ∪ M )= 10

(a) Let, n (E ∩ M) = x. Representing the given data in a Venn-diagram:

(b) We have, n (F ∪ M) = n (U) – n( F ∪ M )= 100 – 10 = 90 So, 90% people liked either folk songs or modern song.

(c) From Venn-diagram, n(U) = nₒ(F) + nₒ(M) + n (F ∩ M) + n( F ∪ M ) or, 100 = (65 – x) + (55- x) + x + 10 or, 100 = 130 – x or, x = 30 So, 30% people liked both types of songs. According to question, n (F ∩ M) = 360 Suppose, n (U) = y. Then, 30% of y = 360 or, y = 1200 Thus, 1200 people participated in the survey.

(d) No. of people who liked both type of songs, n (F ∩ M) = 360 No. of people who did not like both type of songs, n( F ∪ M )=10% of 1200 = 120 n (F ∩ M) 360 Also, = 120= 3 i.e., n (F ∩ M) = 3 × n( F ∪ M ) n( F ∪ M ) Thus, the number of people who liked both type of songs is 3 times the number of people who did not like both type of songs.

b) In a survey of some farmers in a community, 70% of them are found cultivating rice, 60% cultivating wheat, 20% are not cultivating both the crops, and 450 farmers are found cultivating both the crops.
(a) Draw a Venn-diagram to illustrate the above information.
(b) Find the percentage of farmers who are cultivating either rice or wheat.
(c) Find the total number of farmers participated in the survey.
(d) By how many times is the number of farmers who are cultivating both the crops more than the number of farmers who are cultivating none of these crops?

Solution

Given

Let, R and W denote the sets of farmers who are cultivating rice and wheat respectively and n (U) = 100. Then, n (R) = 70, n(W) = 60, n( F ∪ M )= 20

(a) Let, n (R ∩ W) = x. Representing the given data in a Venn-diagram:

(b) We have, n (R ∪ W) = n (U) – n( R ∪ W )= 100 – 20 = 80 So, 80% farmers are cultivating either rice or wheat.

(c) From Venn-diagram, n(U) = nₒ(R) + nₒ(W) + n (R ∩ W) + n( R ∪ W ) or, 100 = (70 – x) + (60- x) + x + 20 or, 100 = 150 – x or, x = 50 So, 50% farmers are cultivating both types of crops. According to question, n (R ∩ W) = 450 Suppose, n (U) = y. Then, 50% of y = 450 or, y = 900 Thus, 900 farmers participated in the survey.

(d) No. of farmers who are cultivating both the crops, n (R ∩ W) = 450 No. of farmers cultivating none of these crops, n( R ∪ W ) =20% of 900 =180 n (R ∩ W) 450 Also, = 180= 2.5 i.e., n (R ∩ W) = 2.5 × n( R ∪ W ) n( R ∪ W ) Thus, the number of who are cultivating both the crops is 2.5 times the number of farmers cultivating none of these crops.

10. a) 75 students in a class like picnic (P) or hiking (H) or both. Out of them, 10 like both the activities. The ratio of the number of students who like picnic to those who like hiking is 2 : 3.
(a) If n(P) = 2x, what is the value of n(H)?
(b) Represent the above information in a Venn-diagram.
(c) Find the number of students who like picnic.
(d) Find the percentage of students who like picnic only.

Solution

Given

Here, n (U) = 75, n (P ∩ H) = 10, n( P ∪ H ) = 0

(a) Given, n(P) = 2x then n(H) = 3x

(b) The given information is shown in a Venn-diagram:

(c) From Venn-diagram, n (U) = nₒ(P) + nₒ(H) + n (P ∩ H) + n( P ∪ H ) or, 75 = (2x – 10) + (3x – 10) + 10 + 0 or, 75 = 5x – 10 or, x = 17 So, the number of students who like picnic, n(P) = 2x = 2 × 17 = 34

(d) No. of students who like picnic only, nₒ(P) = 34 – 10 = 24 Thus, the percentage of students who like picnic only = 2475 × 100% = 32%

b) A survey is conducted in a community regarding the activities that are beneficial for healthy life. Out of 100 people participating in the survey, it is found that the ratio of number of people who are doing yoga (Y) and going for jogging (J) regularly in the early morning is 4 : 5. If 25 people are following both the activities and 35 people are not following both of these activities, answer the following questions.
(a) If n(Y) = 4x, what is the value of n(J)?
(b) Draw a Venn-diagram to represent the above information.
(c) Find the number of people who are going for jogging.
(d) Find the percent of people who are going for jugging only.

Solution

Given

Here, n (U) = 100, n (Y ∩ J) = 25, n( Y ∪ J ) = 35

(a) Given, n(Y) = 4x then n(J) = 5x

(b) The given information is shown in a Venn-diagram:

(c) From Venn-diagram, n (U) = nₒ(Y) + nₒ(J) + n (Y ∩ J) + n( Y ∪ J ) or, 100 = (4x – 25) + (5x – 25) + 25 + 35 or, 100 = 9x + 10 or, x = 10 So, the number of people who are going for jogging, n(J) = 5x = 5 × 10 = 50

(d) No. of people who are going for jogging only, nₒ(J) = 50 – 25 = 25 Thus, the percentage of people who are going for jogging only = 25100 × 100% = 25%

c) In a group of students, the ratio of the number of students who liked music (M) and sports (S) is 9 : 7. Out of which 25 liked both the activities, 20 liked music only, and 15 liked none of the activities.
(a) If n(M) = 9x, what is the value of n(S)?
(b) Represent the above information in a Venn-diagram.
(c) Find the total number of students in the group.
(d) Find the number of students who liked at most one of these activities.

Solution

Given

Here, n (M ∩ S) = 25, nₒ(M) = 20 and n( M ∪ S ) = 15

(a) Given, n(M) = 9x then n(S) = 7x

(b) The given information is shown in a Venn-diagram:

(c) According to question, nₒ(M) = 20 or, 9x – 25 = 20 or, x =5 We have, n (U) = nₒ(M) + nₒ(S) + n (M ∩ S) + n( M ∪ S ) = 20 + (7 × 5 – 25) + 25 + 15 = 70 So, the total number of students in the group is 70.

(d) No. of students who liked at most one of these activities, n( M ∩ S ) = 70 – 25 = 45

d) Out of 120 students appeared in an examination, the number of students who passed in Mathematics only is twice the number of students who passed in Science only. Also, 50 students passed in both subjects and one-third students failed in both subjects.
(a) If M and S denote the sets of students who passed in Mathematics and Science respectively, write the cardinality of n( M ∪ S ).
(b) Present the given information in a Venn-diagram.
(c) Find the number of students who passed in Mathematics.
(d) If the number of students who failed in both subjects were successful in Science, what is the ratio of the number of students who passed in only Mathematics and only Science?

Solution

Given

1 Here, n(U) = 120, n (M ∩ S) = 50, n( M ∪ S ) = 3 of 120 = 40 Let, nₒ(S) = x then nₒ(M) = 2x.

(a) The cardinality of n( M ∪ S ) = 40

(b) The given information is shown in a Venn-diagram:

(c) From Venn-diagram, n (U) = nₒ(M) + nₒ(S) + n (M ∩ S) + n( M ∪ S ) or, 120 = 2x + x + 50 + 40 or, 120 = 3x + 90 or, x = 10 So, no. of students who passed in Mathematics = n (M) = (2x + 50) = 20 + 50 = 70

(d) Again, after the students who failed both subjects passed in Science, nₒ(S) = x + 40 = 50 and nₒ(M) = 20 So, the ratio of the number of students who passed in only Mathematics and only Science = 2050 = 2 : 5

11. a) In the local level election, Mr. Harka and Mrs. Sunita were two candidates for the post of the mayor in a municipality and 25,000 voters were in the voter list. Voters were supposed to cast the vote for a single candidate. 12,000 people cast vote for Harka, 10,000 people cast vote for Sunita and 1,000 people cast vote even for both the candidates.
(a) State the given sets in cardinality notations.
(b) Show the above information in a Venn-diagram.
(c) How many people didn’t cast the vote?
(d) Find the percentage of valid votes.

Solution

Given

(a) Let, H and S denote the sets of people who cast votes for Harka and Sunita respectively. Then, n(U) = 25,000, nₒ(H) = 12,000, nₒ(S) = 10,000 and n(H ∩ S) = 1,000.

(b) The given Venn-diagram illustrates the above information.

(c) The number of people who cast vote, n (H ∪ S) = nₒ(H) + nₒ(S) + n(H ∩ S) = 12,000 + 10,000 + 1,000 = 23,000 So, the number of people who did not cast vote, n( H ∪ S ) = n(U) – n (H ∪ S) = 25,000 – 23,000 = 2,000

(d) The number of valid votes = nₒ(H) + nₒ(S) = 12,000 + 10,000 = 22,000 So, percentage of valid votes = 2200023000 × 100% = 95.65%

b) There are 400 students in Sarada Secondary School. The students are allowed to cast vote either only for Ajay or for Binita as their school prefect. 200 students cast vote for Ajay, 175 cast vote for Binita and 15 of them cast vote even for both.
(a) What does n(A △ B ) represent?
(b) Represent the above data in a Venn-diagram.
(c) How many students did not cast vote?
(d) Calculate the percentage of valid votes.

Solution

Given

Let, A and B denote the sets of people who cast votes for Ajay and Binita respectively. Then, n(U) = 400, nₒ(A) = 200, nₒ(B) = 175 and n(A ∩ B) = 15.

(a) n(A △ B ) represents the number of valid votes.

(b) The given Venn-diagram illustrates the above information.

(c) The number of people who cast vote, n (A ∪ B) = nₒ(A) + nₒ(B) + n(A ∩ B) = 200 + 175 + 15 = 390 So, the number of people who did not cast vote, n( A ∪ B ) = n(U) – n (A ∪ B) = 400 – 390 = 10

(d) The number of valid votes = nₒ(A) + nₒ(B) = 200 + 175 = 375 So, percentage of valid votes = 375390 × 100% = 96.15%

12. a) In a survey conducted among the participants in a picnic programme, 60 people liked meat, 55 people didn’t like meat, 48 didn’t like fish and 25 liked meat but not fish.
(a) If M and F denote the sets of people who like meat and fish respectively and U is the universal set, write the relation among n(U), n(M) and n( M ).
(b) How many people were surveyed?
(c) How many people like fish but not meat?
(d) If the rest of the people who did not like meat and fish were vegetarians, find the ratio of the number of vegetarians and none-vegetarians.

Solution

Given

Let, M and F denote the sets of people who liked meat and fish respectively. Then, n (M) = 60, n( M ) = 55, n( F ) = 48 and nₒ(M) = 25.

(a) The relation among n(U), n(M) and n( M ) is n(U) =n(M) + n( M ).

(b) We know, n (U) = n (M) + n( M ) = 60 + 55 = 115

(c) We have, n (F) = n (U) – n( F ) = 115 – 48 = 67 Also, n (M ∩ F) = n (M) – nₒ(M) = 60 – 25 = 35 Again, nₒ(F) = n (F) – n (M ∩ F) = 67 – 35 = 32 ∴32 people like fish but not meat.

(d) No. of non-vegetarians, n (M ∪ F) = n (M) + n(F) – n(M ∩ F) = 60 + 67 – 35 = 92 No. of vegetarians, n( M ∪ F ) = n(U) – n (M ∪ F) = 115 – 92 = 23 ∴ The ratio of the number of vegetarians and non-vegetarians = 2392 = 1 : 4

b) In a group of 1,150 youths, each uses smart phones, either I-phone or Samsung or both or neither, 150 of them use only I-phone, 770 of them use Samsung and 900 use only one of these two smart phones.
(a) If I and S denote the sets of youths who use I-phone and Samsung respectively, write the cardinality of n(I △ S).
(b) Show the above information in a Venn-diagram.
(c) Find the number of youths who use both of these smart phones.
(d) Compare the number of youths who use either I-phone or Samsung, and none of these smart phones.

Solution

Given

Here, n (U) = 1150, nₒ(I) = 150, n(S) = 770 and nₒ(I) + nₒ(S) = 900.

(a) The cardinality, n(I △ S) = nₒ(I) + nₒ(S) = 900

(b) Let, n(I ∩ S) = x. The given Venn-diagram illustrates the above information.

(c) According to question, nₒ(I) + nₒ(S) = 900 or, 150 + (770 – x) = 900 or, 920 – x = 900 or, x = 20 ∴20 youths use both smart phones.

(d) No. of youths who use either I-phone or Samsung, n (I ∪ S) = nₒ(I) + n(S) = 150 + 770 = 920 No. of youths who use none of these smart phones, n( I ∪ S ) = n(U) – n (I ∪ S) = 1150 – 920 = 230 ∴ The ratio of number of youths who use either I-phone or Samsung, and none of these smart phones = 920230 = 4 : 1

13. a) A marketing company found that, of 200 households surveyed, 80 used neither brand A nor B soaps, 60 used only brand A soap and for every household that used both brands of soap, 3 used only brand B soap.
(a) If the number of households that used both brands of soaps is x, write the number of households that used only brand B soap.
(b) Draw a Venn-diagram to show the above information.
(c) How many households used only one brand of soap?
(d) What percentage of the total number of households were found to use brand B soap?

Solution

Given

Let, A and B are sets of households surveyed. Here, n (U) = 200, n( A ∪ B ) = 80, nₒ(A) = 60 and n (A ∩ B) = x

(a) nₒ(B) = 3 × n (A ∩ B) = 3x

(b) The given Venn-diagram illustrates the above information.

(c) From Venn-diagram, n (U) = nₒ(A) + nₒ(B) + n(A ∩ B) + n( A ∪ B ) or, 200 = 60 + 3x + x + 80 or, 200 = 140 + 4x or, x = 15 Thus, the number of households that used only one brand of soap, nₒ(A) + nₒ(B) = 60 + 3x = 60 + 3 × 15 = 105

(d) Again, n (B) = nₒ(B) + n(A ∩ B) = 3x + x = 4 × 15 = 60 So, the percentage of the total number of households were found to use brand B soap = 60200 × 100% = 30%

b) 100 employees in an office were asked about their preference for tea and coffee. It was observed that for every person who preferred both the drinks, there were 2 people who preferred coffee and 3 people who preferred tea. The number of people who drink neither of the two drinks is same as those who drink both.
(a) How many people preferred both the drinks?
(b) How many people preferred only one drink?
(c) How many people preferred at most one drink?

Solution

Given

Let, T and C denote the sets of people who preferred tea and coffee respectively. Then, n (U) = 100 Let, n (T ∩ C) = x then n(C) = 2x, n(T) = 3x and n( T ∪ C ) = x

(a) We know, n (U) = n (T) + n (C) – n(T ∩ C) + n( T ∪ C ) or, 100 = 3x + 2x – x + x or, 100 = 5x or, x = 20 Thus, 20 people preferred both the drinks.

(b) No. of people who preferred only one drink , n(T △ C) = (2x – x) + (3x – x) = 3x = 3 × 20 = 60

c) Due to the heavy rainfall during a few monsoon days, 140 households were victimized throughout the country. In the first phase, Nepal government has decided to provide the support of either food, shelter or both to a few victimized households. 80 households got food support, 70 got shelter and 50 households got the support of food and shelter both. The government has managed the budget of Rs 10,000 per household for food, Rs 25,000 per household for shelter and Rs 35,000 per household for food and shelter.
(a) Calculate the amount of budget for food only.
(b) Calculate the amount of budget for shelter only.
(c) Calculate the total amount of budget allocated.
(d) How many households were remained to get support in the first phase?

Solution

Given

Let, F and S denote the sets of households who got food and shelter respectively. Then, n (U) = 140, n(F) = 80, n(S) = 70 and n (F ∩ S) =50 The budget for food per household = Rs. 10,000 The budget for shelter per household = Rs. 25,000 The budget for food and shelter both per household = Rs. 35,000

(a) No. of households for food, nₒ(F) = n(F) – n(F ∩ S) = 80 – 50 = 30 So, the amount of budget for food only = 30 × Rs. 10,000 = Rs. 3,00,000

(b) No. of households for shelter, nₒ(S) = n(S) – n(F ∩ S) = 70 – 50 = 20 So, the amount of budget for shelter only = 20 × Rs. 25,000 = Rs. 5,00,000

(c) The amount of budget for food and shelter both = 50 × Rs. 35,000 = Rs. 17,50,000 The total amount of budget allocated, n(F ∪ S) = nₒ(F) + nₒ(S) + n(F ∩ S) = Rs. 3,00,000 + Rs. 5,00,000 + Rs. 17,50,000 = Rs. 25,50,000

(d) No. of households that received at least one, n(F ∪ S) = nₒ(F) + nₒ(S) + n(F ∩ S) = 30 + 20 + 50 = 100 Also, no. of households that remained to get support in the first phase, n( F ∪ S ) = 140 – 100 = 40

d) In a survey, one-third of the number of children like only mango and 22 do not like mango at all. Also, two-fifth of the number of children like orange but 12 like none of them.
(a) Draw a Venn-diagram to show the above information.
(b) Find the total number of children in the survey.
(c) Find the number of children who like only one fruit.
(d) By how many more or less is the number of children who do not like orange at all than the number of children who do not like mango at all.

Solution

Given

Let, M and O denote the sets of children who like mango and orange respectively and x 2x n (U) = x then nₒ(M) = 3 , n( M ) = 22, n(O) = 5 and n( M ∪ O ) = 12

(a) The given Venn-diagram illustrates the above information.

(b) From Venn-diagram, n (U) = nₒ(M) + n(O) + n( M ∪ O ) x 2x or, x = 3 + 5 + 12 5x + 6x + 180 or, x = 15 or, 15x = 11x + 180 or, 4x = 180 or, x = 45 Thus, the total number of children in the survey is 45. x

(d) No. of children who do not like orange at all, n( O ) = 3 + 12 = 15 + 12 = 27 No. of children who do not like mango at all, n( M ) = 22 Difference = 27 – 22 = 5 Thus, the number of children who do not like orange at all is 5 more than the number of of children who do not like mango at all.