# Class 11 RD Sharma Solutions – Chapter 1 Sets – Exercise 1.8 | Set 2

**Question 11. In a survey of 60 people, it was found that 25 people read newspaper H, 26 people read newspaper T, 26 read newspaper I, 9 read both H and i, 11 read both H and T and I, 3 read all three newspapers. Find:**

**(i) the number of people who read atleast one of the newspaper:**

**(ii)The number of the people who read exactly one newspaper.**

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**Solution:**

(i)Let n(P) denote the total number of people.n(H) denote the total number who read newspaper H

n(T) denote the total number who read newspaper T

n(I) denote number of people who read newspaper I

According to formula:

n(P)=60, n(H)= 25, n(T)=26, n(I)=26

n(H∩I)= 9, n(H∩T)=11, n(T∩I)=8, n(H∩T∩I)=3

Here we need to find, number of people who read at least one of the newspaper:

i.e. (H∪T∪I)

n(H∪T∪I) = n(A) + n(B) + n(C) – n(A∩B) – n(B∩C) – n(A∩C) + n(H∩T∩I)

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

=25+52-28+3

=52

(ii)The number of people who read newspaper H only = 25 – (8+3+6) = 8The number of people who read newspaper T only = 26 – (8+3+5) = 10

The number of people who read newspaper I only = 26 – (6+3+5) = 12

The number of people who read exactly one newspaper = 8+10+12 =30

**Question 12. Of the members of the three athletic teams in a certain school, 21 are in the basketball team, 26 in hockey team and 29 in the football team. 14 play hockey and **basketball**, 15 paly hockey and football, 12 play football and basketball and 8 play all three games. How many members are there in all?**

**Solution:**

Let assume that

n(P) is the number of members in the basketball team.

n(B) is the number of the people in the basketball team.

n(H) is the number of the people in the hockey team.

n(F) is the number of the people in the Football team.

n(B) = 21 n(H) = 26 n(F) = 29

n(H ∩ B) = 14 n(H ∩ F) = 15 n(F ∩ B) = 12, n(H ∩ B ∩ F) = 8

P = B ∪ H ∪ F

n(P) = n(B ∪ H ∪ F)

= n(B) + n(H) + n(F) – n(B ∩ H) – n(H ∩ F) – n(B ∩ F) + n(B ∩ H ∩ F)

21 + 26 + 29 – 14 – 15 – 12 + 8 = 43

**Question 13. In a group of 1000 people, there are 750 who can speak Hindi and 400 who can speak Bengali. How many can speak only Hindi/How many can speak Bengali? How many can speak both Hindi and Bengali?**

**Solution:**

Let assume that

n(P) the number of people

n(H) the number of people who can speak Hindi

n(B) the number of people who can speak Bengali

n(P) = 1000 n(H) = 750 n(B) = 400

P = (H ∪ B) = n(H) + n(B) – n(H ∩ B)

1000 = 750 + 400 – n(H ∩ B)

n(H ∩ B) = 150

So we can say that 150 can speak both Hindi and Bengali

H = (H – B) ∪ (H ∪ B)

750 = n(H-B) + 150

n(H-B) = 600

Similarly, B = (B-H) ∪ (H ∩ B)

400 = n(B-H) + 150

n(B-H) = 400 – 150 = 250

**Question 14. A survey of 500 television viewers produced the following information: 285 watch football,195 watch hockey, 115 watch basketball, 50 do not watch any of the the three games. How many watch all the three games? How many watch exactly of the three games?**

**Solution:**

Let assume that

n(P) the number of people

n(F) the number of people who watch football

n(H) the number of people who watch hockey

n(B) the number of people who watch basketball

n(P) = 500 n(F)=285 n(H) = 195 n(B) = 115 n(F ∩ B) = 45 n(F ∩ H) = 70

n(H ∩ B) = 50 and n(F∪H ∪ B) = 50

n(F ∪ H ∪ B’) = n(P) – n(F ∪ H ∪ B)

50 = 500 – (285 + 195 +115 -70 -50 – 45)

n(F∩ H ∩ B) = 20

Number of people who watch only football = 285 – (50+20+25)

= 285 – 195 = 190

Number of people who watch only hockey = 195 – (50 +20 + 30)

= 195 – 100 = 95

Number of people who watch only basketball = 115 – (25 + 20 + 30)

= 40

Number of people who watch exactly one of the three games = Number of people who watches

either football only or basketball only.

190 + 95 + 40 = 325

**Question 15. In a survey of 100 persons**,** it was found that 28 read magazine A, 30 read magazine B, 42 read magazine C, B read 10 read 10 read magazines A and C, 5 read magazine B the read three magazine. Find:**

**(i) How many read none of three magazines?**

**(ii) How many read magazine C only?**

**Solution:**

Let assume that

n(P) denote the total number of person

n(A) denotes the number of the people who read magazine A

n(B) denotes the number of the people who read magazine B

n(C) denote the number of people who read magazine

n(P) = 100 n(A) = 28 n(B) = 30 n(C) = 42 n(A ∩ B) = 8

n(A ∩ C) =10, n(B ∩ C) = 5, n(A ∩ B ∩ C) = 3

According to the formula

n(A ∪ B ∪) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(A ∩ C) + n(A∩ B ∩ C)

= 28 + 30 + 42 – 8 -10 – 5 + 3

=100 – 20 = 80

Number of people who read none of the three magazine:

= n (A ∪ B ∪ C)’

= n(P) – n(A ∪ B ∪ C)

100 – 80 =20

(ii)Number people who read magazine C only:= 42 – (& + 3 + 2)

= 30

**Question 16. 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 how many students were studying English and Hindi?**

**Solution:**

n(U)=100 (Total number of students)

n(E)=26 (Number of student studying English)

n(S)=48 (Number of student studying Sanskrit)

n(E∩S)=8

n(S∩H)=8

n(E∩H∩S)=3

The number of students who study English only =18

Number of students who study no language =24

Number of students who study Hindi only =[100−(18+5+3+5+35)]−24

=100−66−24

=100−90

=10

Number of students who study Hindi =10+3+5

=18

And Number of students who study English and Hindi = 3

**Question 17. In a survey it was found that 21 liked product P1, 26 liked product P2, and 29 liked product P3, 14 liked products P1 and P2, 12 liked products P3 and P1, 14 liked products P2 and P1 and 8 liked all the three products, find how many liked products P3 only?**

**Solution:**

Let assume that A, B, and C be the set of people who like product P1, P2, and P3 respectively.

n(A) = 21, n(B) = 26, n(C) = 29, n(A ∩ B) = 14, n(C ∩ A) = 12, n(B ∩ C) = 14, n(A ∩ B ∩ C) = 8

People who many liked product C only

= n(C) – n(C ∩ A) – n(B ∩ C) + n(A ∩ B ∩ C)

= 29 -12 – 14 + 8

= 11

Hence, 11 liked product P3 only.