Sets in daily life (Class- 9, Experience- 1)

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• MCQ Test
• Exercise Solution

[Exercise- Page 27]

E.1.1. Express in tabular method: 

a) A = {x ∈ N: –3 < x ≤ 5} 

b) B = {x ∈ Z: x is a prime number and x² ≤ 50}

c) C = {x ∈ Z: x⁴ < 264}

Solution:

Representing each set using tabular method: 

a) A = {x ∈ N: −3 < x ≤ 5} 

[Natural numbers, often denoted by the symbol N, are a set of numbers used for counting and ordering objects. They start from 1 and continue indefinitely, including all positive integers greater than zero.] 

So, the elements of set A are: {1, 2, 3, 4, 5}. 

b) B = {x ∈ Z: x is a prime number and x² ≤ 50} 

[A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. In simpler terms, a prime number is a whole number greater than 1 that cannot be formed by multiplying two smaller natural numbers.]

 

So, the elements of set B are: {2, 3, 5, 7}.

c) C = {x ∈ Z: x⁴ < 264} 

So, the elements of set C are: {-4, -3, -2, -1, 0, 1, 2, 3, 4} 

Note: In the tables, Yes indicates that the element satisfies the condition specified for the set, and No indicates it does not. For set B, the Prime Number column indicates whether the number is prime or not, and other column indicates whether the square of the number is less than or equal to 50.

E.1.2. Express in set builder method: 

a) A = {1, 3, 5,…,101} 

b) B = {4, 9, 16, 25, 36, 49, 64, 81, 100}

Solution:

Expressing each set in set-builder notation:

a) A = {x ∈ N: x is odd and 1 ≤ x ≤ 101}

Explanation:

• x ∈ N denotes that x is a natural number.

• x is odd specifies that the elements of the set are odd numbers.

• 1 ≤ x ≤ 101 specifies the range of numbers from 1 to 101.

b) B = {x ∈ N: x = n² where n is a natural number and 2 ≤ n ≤ 10}

Explanation:

• x ∈ N denotes that x is a natural number.

• x = n² specifies that the elements of the set are perfect squares.

• n is a natural number" specifies that n is a natural number.

• 2 ≤ n ≤ 10 restricts the value of n to be between 2 and 10, inclusive. 

This ensures that we're considering the squares of natural numbers from 2 to 10.

E.1.3. If A = {1, 2, 3, 4, 5}, B = {0, 1, 3, 5, 6} and C = {1, 5, 6}, then find the following sets. 

a) A ∪ B 

b) A ∩ C 

c) B \ C 

d) A ∪ (B ∩ C) 

e) A ∩ (B ∪ C)

Solution:

Let's find the specified sets using the given sets A, B, and C: 

Given sets: 

A= {1,2,3,4,5} 

B= {0,1,3,5,6} 

C= {1,5,6} 

a) A ∪ B 

(Union of A and B): 

A ∪ B= {1,2,3,4,5} ∪ {0,1,3,5,6}

A ∪ B= {0,1,2,3,4,5,6} 

b) A ∩ C 

(Intersection of A and C): 

A ∩ C= {1,2,3,4,5} ∩ {1,5,6}

A ∩ C= {1,5} 

c) B ∖ C 

(Set difference of B and C): 

B ∖ C= {0,1,3,5,6} \ {1,5,6}

B ∖ C= {0,3} 

d) A ∪ (B ∩ C) 

(Union of A and the intersection of B and C): 

(B ∩ C)= {0,1,3,5,6} ∩ {1,5,6} = {1,5,6}

A ∪ (B ∩ C)= A∪{1,5,6}

A ∪ (B ∩ C)= {1,2,3,4,5} ∪ {1,5,6}

A ∪ (B ∩ C)= {1,2,3,4,5,6} 

e) A ∩ (B ∪ C) 

(Intersection of A and the union of B and C): 

B ∪ C={0,1,3,5,6} ∪ {1,5,6} ={0,1,3,5,6}

A ∩ (B ∪ C)= A ∩ {0,1,3,5,6}

A ∩ (B ∪ C)={1,2,3,4,5} ∩ {0,1,3,5,6}

A ∩ (B ∪ C)={1,3,5}

E.1.4. If U = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {1, 3, 5, 7}, B = {0, 2, 4, 6} and C = {3, 4, 5, 6, 7}, then verify the following relations: 

a) (A ∪ B)′ = A′ ∩ B′

b) (B ∩ C)′ = B′ ∪ C′

c) (A ∪ B) ∩ C = (A ∩ C) ∩ (B ∩ C) 

d) (A ∩ B) ∪ C = (A ∪ C) ∩ (B ∪ C)

Solution:

Let's verify the given relations using the provided sets U, A, B, and C:

a) (A ∪ B)^c = A^c ∩ B^c 

Given set: 

U = {0,1,2,3,4,5,6,7,8,9} 

A = {1,3,5,7} 

B = {0,2,4,6} 

C = {3,4,5,6,7}

Here,

A ∪ B = {1,3,5,7} ∪ {0,2,4,6} 

A ∪ B = {0,1,2,3,4,5,6,7}

(A ∪ B)^c = U - (A ∪ B) 

(A ∪ B)^c = U - {0,1,2,3,4,5,6,7}

(A ∪ B)^c = {0,1,2,3,4,5,6,7,8,9} - {0,1,2,3,4,5,6,7}

(A ∪ B)^c = {8,9} (left-side)

Again,

A^c = U - A 

A^c = {0,1,2,3,4,5,6,7,8,9} - {1,3,5,7} 

A^c = {0,2,4,6,8,9} 

B^c = U - B

B^c = {0,1,2,3,4,5,6,7,8,9} - {0,2,4,6} 

B^c = {1,3,5,7,8,9} 

A^c ∩ B^c = {0,2,4,6,8,9} ∩ {1,3,5,7,8,9} 

A^c ∩ B^c = {8,9} (Right-side)

Therefore {8,9} = {8,9}, So, (A ∪ B)^c = A^c ∩ B^c This relation is true.  (Answer)

b) (B ∩ C)^c = B^c ∪ C^c 

Given set: 

U = {0,1,2,3,4,5,6,7,8,9} 

A = {1,3,5,7} 

B = {0,2,4,6} 

C = {3,4,5,6,7}

Here,

(B ∩ C) = {0,2,4,6} ∩ {3,4,5,6,7}

(B ∩ C) = {4,6}

(B ∩ C)^c = U - (B ∩ C)

(B ∩ C)^c = {0,1,2,3,4,5,6,7,8,9} - {4,6}

(B ∩ C)^c = {0,1,2,3,5,7,8,9}  (left-side)

Again,

B^c = U - B 

B^c = {0,1,2,3,4,5,6,7,8,9} - {0,2,4,6} 

B^c = {1,3,5,7,8,9} 

C^c = U - C 

C^c = {0,1,2,3,4,5,6,7,8,9} - {3,4,5,6,7} 

C^c = {0,1,2,8,9} 

B^c ∪ C^c= {1,3,5,7,8,9} ∪ {0,1,2,8,9}

B^c ∪ C^c= {0,1,2,3,5,7,8,9}  (Right-side)

Therefore {0,1,2,3,5,7,8,9} = {0,1,2,3,5,7,8,9}, So, (B ∩ C)^c = B^c ∪ C^c This relation is true.  (Answer)

c) (A ∪ B) ∩ C = (A ∩ C) ∩ (B ∩ C) 

Given set: 

U = {0,1,2,3,4,5,6,7,8,9} 

A = {1,3,5,7} 

B = {0,2,4,6} 

C = {3,4,5,6,7}

Here,

A ∪ B = {1,3,5,7} ∪ {0,2,4,6}

A ∪ B = {0,1,2,3,4,5,6,7}

 

(A ∪ B) ∩ C = {0,1,2,3,4,5,6,7} ∩ C

(A ∪ B) ∩ C = {0,1,2,3,4,5,6,7} ∩ {3,4,5,6,7}

(A ∪ B) ∩ C = {3,4,5,6,7}  (left-side)

Again,

A ∩ C = {1,3,5,7} ∩ {3,4,5,6,7}

A ∩ C = {3,5,7}

B ∩ C = {0,2,4,6} ∩ {3,4,5,6,7}

B ∩ C = {4,6} 

(A ∩ C) ∩ (B ∩ C)= {3,5,7} ∩ {4,6} 

(A ∩ C) ∩ (B ∩ C)= { }  (Right-side)

Therefore {3,4,5,6,7} ≠ { },  So, (A ∪ B) ∩ C = (A ∩ C) ∩ (B ∩ C) this relation is not true   (Answer)

d) (A ∩ B) ∪ C = (A ∪ C) ∩ (B ∪ C) 

Given set: 

U = {0,1,2,3,4,5,6,7,8,9} 

A = {1,3,5,7} 

B = {0,2,4,6} 

C = {3,4,5,6,7}

Here,

A ∩ B = {1,3,5,7} ∩ {0,2,4,6} 

A ∩ B = ∅ 

(A ∩ B) ∪ C = ∅ ∪ C 

(A ∩ B) ∪ C = C 

(A ∩ B) ∪ C = {3,4,5,6,7}  (left-side)

Again,

A ∪ C = {1,3,5,7} ∪ {3,4,5,6,7} 

A ∪ C = {1,3,4,5,6,7} 

B ∪ C = {0,2,4,6} ∪ {3,4,5,6,7} 

B ∪ C = {0,2,3,4,5,6,7} 

(A ∪ C) ∩ (B ∪ C)=  {1,3,4,5,6,7} ∩ {0,2,3,4,5,6,7}

(A ∪ C) ∩ (B ∪ C)= {3,4,5,6,7} (Right-side)

Therefore {3,4,5,6,7} = {3,4,5,6,7},  So, (A ∩ B) ∪ C = (A ∪ C) ∩ (B ∪ C) This relation is true.   (Answer)

E.1.5. Find the following: 

a) N ∩ 2N 

b) N ∩ A 

c) 2N ∩ P 

Where, N is the set of all natural numbers, A is the set of all odd numbers, P is the set of all prime numbers.

Solution:

Let's break down each part:

a) N ∩ 2N 

This represents the intersection of the set of all natural numbers (N) and the set of all numbers that are twice a natural number (2N) 

N consists of all natural numbers: {1, 2, 3, 4, 5, ...}

2N consists of all numbers that are twice a natural number: {2, 4, 6, 8, 10, ...} 

The intersection of N and 2N will include only the numbers that are present in both sets, which are the even numbers. 

So, N ∩ 2N = {2, 4, 6, 8, 10, ...} 

b) N ∩ A 

This represents the intersection of the set of all natural numbers (N) and the set of all odd numbers (A). 

N consists of all natural numbers: {1, 2, 3, 4, 5, ...} 

A consists of all odd numbers: {1, 3, 5, 7, 9, ...} 

The intersection of N and A will include only the numbers that are present in both sets, which are the odd natural numbers. 

So, N ∩ A = {1, 3, 5, 7, 9, ...}. 

c) 2N ∩ P

 

This represents the intersection of the set of all numbers that are twice a natural number (2N) and the set of all prime numbers (P). 

2N consists of all numbers that are twice a natural number: {2, 4, 6, 8, 10, ...} 

P consists of all prime numbers: {2, 3, 5, 7, 11, 13, ...} 

The intersection of 2N and P will include only the numbers that are present in both sets, which are the prime numbers that are twice a natural number. 

So, 2N ∩ P = {2}

E.1.6. Let U be the set of all triangles and A be the set of all right triangles. Then describe the set A^c.

Solution:

U represents the set of all triangles.

A represents the set of all right triangles.

The complement of a set A, denoted as A' or sometimes A^c, represents all elements that are in the universal set U but not in A.

So, A' or A^c represents all triangles that are not right triangles.

In other words, A' consists of all triangles in U that are not right triangles. This includes triangles that are obtuse or acute. 

Therefore, the set A^c can be described as - the set of all triangles in U that are not right triangles.

E.1.7. Show the followings with Venn diagrams. For any sets A, B, C - 

a) (A ∪ B)′ = A′ ∩ B′

b) (B ∩ C)′ = B′ ∪ C′

c) (A ∪ B) ∩ C = (A ∩ C) ∪ (B ∩ C) 

d) (A ∩ B) ∪ C = (A ∪ C) ∩ (B ∪ C)

Solution:

Let's represent each of these set identities using Venn diagrams:

a) (A ∪ B)′ = A′ ∩ B′

For any sets of A and B, Venn diagrams are drawn below-

So, It is proved by Venn diagrams, (A ∪ B)′ = A′ ∩ B′

b) (B ∩ C)′ = B′ ∪ C′

For any sets of A and B, Venn diagrams are drawn below-

So, it is shown that (B ∩ C)′ = B′ ∪ C′ which is proved by Venn diagrams.

c) (A ∪ B) ∩ C = (A ∩ C) ∪ (B ∩ C)

For any sets of A, B and C, Venn diagrams are drawn below-

Let's break it down:

1. Draw a rectangle to represent the universal set.

2. Draw three circles, one for each set A, B, and C, with overlapping regions.

3. Take the union of sets A and B, then take the intersection with set C.

4. For (A ∩ C) ∪ (B ∩ C), first, take the intersection of sets A and C, and the intersection of sets B and C, then take the union of these intersections.

5. The shaded regions in both cases should be identical.

So, it is shown that (A ∪ B) ∩ C = (A ∩ C) ∪ (B ∩ C) which is proved by Venn diagrams.

d) (A ∩ B) ∪ C = (A ∪ C) ∩ (B ∪ C)

For any sets of A, B and C, Venn diagrams are drawn below-

Let's break it down:

1. Draw a rectangle to represent the universal set.

2. Draw three circles, one for each set A, B, and C, with overlapping regions.

3. Take the intersection of sets A and B, then take the union with set C.

4. For (A ∪ C) ∩ (B ∪ C), first, take the union of sets A and C, and the union of sets B and C, then take the intersection of these unions.

5. The shaded regions in both cases should be identical.

So, it is shown that (A ∩ B) ∪ C = (A ∪ C) ∩ (B ∪ C) which is proved by Venn diagrams.

E.1.8. Out of 40 students in a class, 25 like birds and 15 like cats. There are 10 students who like both birds and cats. Determine by a Venn diagram how many students like neither bird nor cat.

Solution:

To solve this problem using a Venn diagram, let's denote:

• A as the set of students who like birds,

• B as the set of students who like cats.

• U as the set of Total students

Given that there are 10 students who like both birds and cats, we'll represent this in the overlapping region of the two circles. Now, let's fill in the information we have:

• A (Birds) = 25

• B (Cats) = 15

• A ∩ B (Both Birds and Cats) = 10

Now, we can calculate the number of students who like neither birds nor cats.

Total number of students U = 40 (given)

Using the principle of inclusion-exclusion:

    Total = A + B - (A ∩ B) + Neither

        40 = 25 + 15 - 10 + Neither   [Substituting the value we have]

Neither = 40 - 25 - 15 + 10     [Now, solving for -Neither ]

Neither = 10

So, 10 students like neither birds nor cats.

E.1.9. If P = {a, b}, Q = {0, 1, 2} and R = {0, 1, a}, then find the values of expressions below. 

a) P × Q, P × P, Q × Q, Q × P and P × ∅ 

b) (P × Q) ∩ (P × R) 

c) P × (Q ∩ R) 

d) (P × Q) ∩ R

e) n(P × Q), n(Q × Q) 

f) Give your logic on the equality of (c) and (d). 

Solution:

Let's evaluate each expression using the given sets: 

a) P × Q,  P × P,  Q × Q,  Q × P  and  P × ∅ 

P×Q represents the Cartesian product of sets P and Q.

P×P represents the Cartesian product of set P with itself.

Q×Q represents the Cartesian product of set Q with itself.

Q×P represents the Cartesian product of set Q with set P.

P×∅ represents the Cartesian product of set P with the empty set.

Given: 

P= {a,b}

Q= {0,1,2}

R= {0,1,a}

Let's evaluate each expression:

P×Q= {(a,0), (a,1), (a,2), (b,0), (b,1), (b,2)}

P×P= {(a,a), (a,b), (b,a), (b,b)}

Q×Q= {(0,0), (0,1), (0,2), (1,0), (1,1), (1,2), (2,0), (2,1), (2,2)}

Q×P= {(0,a), (0,b), (1,a), (1,b), (2,a), (2,b)}

P×∅= ∅

b) (P × Q) ∩ (P × R)

For this expression, we need to find the common elements between P × Q and P × R.

P × Q= {(a,0), (a,1), (a,2), (b,0), (b,1), (b,2)} 

P × R= {(a,0), (a,1), (a,a), (b,0), (b,1), (b,a)}

Their intersection is {(a,0), (a,1), (b,0), (b,1)}.

c) P × (Q ∩ R)

We first evaluate Q ∩ R, which is the intersection of sets Q and R. Q ∩ R= {0,1}.

Then, P × (Q ∩ R) is the Cartesian product of set P with Q ∩ R.

P × (Q ∩ R) = {(a,0), (a,1), (b,0), (b,1)}

d) (P × Q) ∩ R

For this expression, we need to find the common elements between 

P × Q and set R.

P × Q= {(a,0), (a,1), (a,2), (b,0), (b,1), (b,2)} 

R= {0,1,a}

Their intersection is {(a,0), (a,1)}.

These are the values of the given expressions.

e) n(P × Q), n(Q × Q) 

Given, 

P= {a,b}

Q= {0,1,2}

R= {0,1,a}

here,

P × Q = {a,b} × {0,1,2}

P × Q = {(a,0), (a,1), (a,2), (b,0), (b,1), (b,2)}

The number of elements in a set n(P × Q) is  6  

∴ n(P × Q) = 6   (Answer)

f) Give your logic on the equality of (c) and (d).

That means, Give your logic on the equality of P × (Q ∩ R)  and (P × Q) ∩ R

Given, 

P= {a,b}

Q= {0,1,2}

R= {0,1,a}

We have got the answer from (c) of P × (Q ∩ R) are {(a,0), (a,1), (b,0), (b,1)}  

We have got the answer from (d) of (P × Q) ∩ R are {(a,0), (a,1)}

Now, let's compare the results:

P × (Q ∩ R) = {(a,0), (a,1), (b,0), (b,1)}

(P × Q) ∩ R = {(a,0), (a,1)}

Both expressions yield the same result, showing that P × (Q ∩ R) and (P × Q) ∩ R are indeed equal for this example. The logic behind this equality lies in the distributive property of Cartesian products over intersections, which essentially allows us to distribute the Cartesian product operation over the intersection operation without changing the result.

E.1.10. If P = {0, 1, 2, 3}, Q = { 1, 3, 4} and R = P ∩ Q,

 i) Determine P × R and R × Q . 

ii) Find the values of n(P × R) and n(R × Q).

Solution:

і) Determine P × R and R × Q

Given, 

P = {0, 1, 2, 3}, 

Q = {1, 3, 4} and 

R = P ∩ Q 

First, let's find the intersection of sets P and Q to obtain R 

R = P ∩ Q R 

    = {0, 1, 2, 3} ∩ {1, 3, 4} 

    = {1,3} 

Now, let's compute the Cartesian product of sets P and R,

P ✕ R = {0, 1, 2, 3} ✕ {1,3}

P ✕ R = {(0,1),(0,3),(1,1),(1,3),(2,1),(2,3),(3,1),(3,3)} 

So, the values of the expression P ✕ R are (0,1),(0,3),(1,1),(1,3),(2,1),(2,3),(3,1) and (3,3)   (Answer)

Again, let's compute the Cartesian product of sets R and Q,

R ✕ Q = {1,3} ✕ {1, 3, 4}

R ✕ Q = {(1,1),(1,3),(1,4),(3,1),(3,3),(3,4)}

So, So, the values of the expression P ✕ R are (1,1),(1,3),(1,4),(3,1),(3,3) and (3,4)       (Answer) 

ii) Find the values of n(P × R) and n(R × Q).

Here, 

P ✕ R = {(0,1),(0,3),(1,1),(1,3),(2,1),(2,3),(3,1),(3,3)}

To find the number of elements in P ✕ R, we need to find n(P ✕ R). 

Counting the elements in the set P ✕ R, we find that there are 8 ordered pairs.

So, the value of  n(P ✕ R) = 8    (Answer)

Again, we found that the Cartesian product of sets R and Q is given by, 

R ✕ Q = {(1,1),(1,3),(1,4),(3,1),(3,3),(3,4)}

Here, To find the number of elements in R ✕ Q, we need to find n(R ✕ Q).

Counting the elements in the set R ✕ Q, we find that there are 8 ordered pairs.

So, the value of  n(R ✕ Q) = 6    (Answer)

E.1.11. If P × Q = {(0, a ), (1, c), (2, b)} then determine P and Q.

Solution:

Given that 

(P × Q) = {(0, a), (1, c), (2, b)} 

let's find sets P and Q. 

The set (P × Q) contains ordered pairs where the first element is from set P and the second element is from set Q. 

From (P × Q), we can extract the first elements of the ordered pairs to find set P: 

P = {0, 1, 2} 

Similarly, we can extract the second elements of the ordered pairs to find set Q: 

Q = {a, b, c} 

So, the sets p = {0, 1, 2} and Q = {a, b, c}    (Answer)