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Tuesday, July 1, 2025

Class 9 Mathematics - Chapter 01 - Set Theory - Exercise 1.7 Solutions - Book Back Solutions - Samacheer Kalvi Tamilnadu State Board

 



1.     Which of the following is correct?

(1)   {7} {1,2,3,4,5,6,7,8,9,10} (2)     7 {1,2,3,4,5,6,7,8,9,10}

(3)   7 {1,2,3,4,5,6,7,8,9,10}      (4)     {7} Ë {1,2,3,4,5,6,7,8,9,10}

Answer:

(2) 7 {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

First option is Set, 7, which not an element, third option given is 7 not an element, but 7 is an element of the given set, and, Fourth option given is, Set, 7 is not a subset of the given set.  But, 7 is an element of the given set, then Set, 7 is a subset of the given set.

 

2.     The set P = {x | x Z, –1< x < 1} is a

(1)   Singleton set                            (2)     Power set

(3)   Null set                                     (4)     Subset

Answer:

(1) Singleton set

When we write the set in Roster form, we get, Set, P = {0}.  Hence, there are only one element in Set, P.  Therefore, it is a singleton set.


 


3.     If U ={x | x N, x < 10} and A = {x | x N, 2 ≤ x < 6} then (A′)′ is

(1)   {1, 6, 7, 8, 9}                           (2)     {1, 2, 3, 4}

(3)   {2, 3, 4, 5}                                (4)     { }

Answer:

(3) {2, 3, 4, 5}

When we write the given sets in Roster form, we get,

Set, U = {1, 2, 3, 4, 5, 6, 7, 8, 9} and, Set, A = {2, 3, 4, 5}

Now A’ = {1, 6, 7, 8, 9}

Hence, (A’)’ = {2, 3, 4, 5}.

4.     If B A then n(A ∩ B) is

(1)   n(A–B)                                     (2)     n(B)

(3)   n(B – A)                                    (4)     n(A)

Answer:

(2) n(B)

We know that, When B A, then A ∩ B = B.

5.     If A = {x, y, z} then the number of non- empty subsets of A is

(1)   8                                                (2)     5     

(3)   6                                                (4)     7

Answer:

(4) 7

Total number of subsets of a set = 2n = 23 = 8.  But, there includes null set also.  When we exclude null set from the number of subset, we get, 7.


 

6.     Which of the following is correct?

(1)   {a, b}                                 (2)     {a, b}

(3)   {a} {a, b}                              (4)     a {a, b}

Answer:

Ø {a, b}

Null set is a subset of every set.

7.     If AB = A∩B, then

(1)   A≠B                                          (2)     A = B

(3)   A B                                        (4)     B A

Answer:

(2) A = B

8.     If B – A is B, then A∩B is

(1)   A                                                (2)     B

(3)   U                                               (4)    

Answer:

(4) Ø

Given B – A = B, then we know that, A and B are disjoint sets.

9.     From the adjacent diagram n[P(AΔB)] is

(1)   8                     (2)     16

(3)   32                  (4)     64

Answer:

(3) 32

A ∆ B = { 60, 85, 75, 90, 70}, n(A ∆ B) = 5, n(P(A ∆ B)) = 25 = 32

10.   If n(A) = 10 and n(B) = 15, then the minimum and maximum number of elements in A ∩ B is

(1)   10,15                                         (2)     15,10

(3)   10,0                                           (4)     0,10

Answer:

(4) (0, 10)

If the two sets A and B are disjoint, then, A ∩ B has no elements.  In the given numbers, there should be maximum of 10 elements in A ∩ B.

11.   Let A = {} and B = P(A), then A∩B is

(1)   { , {} }                                 (2)     {}

(3)                                                  (4)     {0}

Answer:

(2) {Ø}

Given, Set A = Null Set.  Then Power Set P(A) = {, {}}, That is, Set, B = {, {}}.   Now A ∩ B = {Ø}.

12.   In a class of 50 boys, 35 boys play Carrom and 20 boys play Chess then the number of boys play both games is

(1)   5                                                (2)     30

(3)   15                                              (4)     10.

Answer:

(1) 5

Here, Let A be the boys play carrom and Set, B be the boys play Chess.  Total Strength of the class is 50.  That is n(A B) = 50.  n(A) = 35 and n(B) = 20.  We know that, (A B) = n(A) + n(B) – n(A ∩ B) 50 = 35 + 20 – n(A ∩ B) n(A ∩ B) = 5.

13.   If U = {x : x Î N and x <10},  A = {1,2, 3,5, 8} and
B = {2,5,6,7,9}, then n[(A
È B)’] is

(1)   1                                                (2)     2

(3)   4                                                (4)     8

Answer:

(1) 1

Converting the given sets, in Roster Form, We get, U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {1, 2, 3, 5, 8}, and, B = {2, 5, 6, 7, 9}.

A B = {1, 2, 3, 5, 6, 7, 8, 9}.

Now, (A B)’ = {4},

n(A B)’ = 1

14.   For any three sets P, Q and R, P −(Q  ∩  R) is

(1)   P −(Q È R)                               (2)     (P Ç ­Q)−R

(3)   (P −Q) È (P −R)                       (4)     (P −Q) Ç (P −R)

Answer:

(3) (P – Q) (P – R)

We know that, P −(Q  ∩  R) = (P – Q) (P – R)

15.   Which of the following is true?

(1)   A−B = A Ç B                           (2)     A−B = B −A

(3)   (A ­ È B)’ = A’ È­ B’                  (4)     (A  Ç­ B)’ = A’ È B’

Answer:

(4) (A ∩ B)’ = A’ B’

 

16.   If n(A ­È  BÈ  ­C) = 100, n(A) = 4x, n(B) = 6x, n(C) = 5x, n(A ­Ç B) = 20, n(B Ç ­C) = 15, n(A Ç ­C) = 25 and n(A Ç­ B Ç ­C) = 10 , then the value of x is

(1)   10                                              (2)     15

(3)   25                                              (4)     30

Answer:

(1) 10

We know that, n(A B C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(C ∩ A) + n(A ∩ B ∩ C)

Substituting the values, we get,

100 = 4x + 6x + 5x – 20 – 15 – 25 + 10

100 = 15x -60 + 10

100 = 15x – 50

15x = 100 + 50 = 150

x = 10

17.   For any three sets A, B and C, (A−B) Ç  (B −C) is equal to

(1)   A only                                       (2)     B only

(3)   C only                                       (4)     Æ

Answer:

(4) ϕ

As per the derivation, (A−B) Ç  (B −C) = Null Set.


 

18.   If J = Set of three sided shapes, K = Set of shapes with two equal sides and L = Set of shapes with right angle, then J Ç K Ç L is

(1)   Set of isosceles triangles         (2)     Set of equilateral triangles

(3)   Set of isosceles right triangles (4)    Set of right angled triangles

Answer:

(3) Set of isosceles right triangles

19.   The shaded region in the Venn diagram is

(1)   Z −(X ÈY)                               (2)     (X È ­Y) Ç Z

(3)   Z −(X ÇY)                               (4)     Z ­È (X ÇY)

Answer:

(3) Z – (X ∩ Y)

20.   In a city, 40% people like only one fruit, 35% people like only two fruits, 20% people like all the three fruits. How many percentage of people do not like any one of the above three fruits?

(1)   5                                                (2)     8

(3)   10                                              (4)     15

Answer:

(1) 5

 

The Venn Diagram gives you a clear idea of this question. 

40 + 35 + 20 + x = 100%

95% + x = 100%

x = 5%

 

Thursday, December 19, 2024

Radio towers are used for transmitting a range of communication services including radio and television. The tower will either act as an antenna itself or support one or more antennas on its structure.

 

Case Study Questions

Radio towers are used for transmitting a range of communication services including radio and television. The tower will either act as an antenna itself or support one or more antennas on its structure.

On a similar concept, a radio station tower was built in two stations A and B (B vertically below A). The tower is supported by wires AO and BO from a point O on the ground. Distance between the base C of the tower and the point O is 36 m. From O, the angles of  elevation of the tops of station B and station A are 30º and 45º respectively.

Based on the above, answer the following questions :

(i) Find the height of station B.

(ii) Find the height of station A.

(iii) Find the length of the wire OA.

(iv) Find the length of the wire OB. (CBSE 2022 Set 2)



Solution:

Look at the figure, which we construct now

(i)  Find the Height of  station B

So, we need to find the length of BC

We can take tan O

Tan 300 =

We know the value of tan 30 is 1/Ö3.  Substitute the values

        Cross multiply, we get

        BC =  =  = 12Ö3 m

Hence the length of tower B is 12Ö3 m.

(ii) Find the height of the station A

Here we want to find the length of AC.



CO is known and the angle is 450.  So we take tan 45

Tan 450    =

tan 450     =

We know that tan450 = 1, so substitute the values

        1       =

        AC = 36 m

Hence the height of tower A is 36 m.

(iii)  Find the length of OA

OA is the hypotenuse of the triangle AOC.



Here we know that the length of CO is 36 m and length of AC is also 36 m.

Here we take sin 450.

i.e.,

We know that sin450 = 1/Ö2.  Substitute the values

       

Cross multiply, we get

        AO = 36Ö2 m

Hence the length of the wire AO is 36Ö2 m.

(iv)  Find the length of the wire OB

Consider the triangle BCO.  Here BC and

OC are known and we want to find OB.  In this case we take cos 300 .



        Cos 300 =

We know that cost 30 = .  Substitute the values, we get

       

Cross multiply we get,

        BO =  =  = 12 x 2 x Ö3

        BO = 24 Ö3 m

Hence the length of the wire BO is 24Ö3 m

Summary of Answers:

(i) Height of station B = 12Ö3 m

(ii) Height of station A  = 36 m

(iii) Length of the wire OA = 36Ö2 m

(iv) Length of the wire OB = 24Ö3 m

 

Class 9 Mathematics - Chapter 01 - Set Theory - Exercise 1.7 Solutions - Book Back Solutions - Samacheer Kalvi Tamilnadu State Board

  1.     Which of the following is correct? (1)   {7} ∈ {1,2,3,4,5,6,7,8,9,10} (2)     7 ∈ {1,2,3,4,5,6,7,8,9,10} (3)   7 ∉ {1,2...