10th maths Exercise 1.1 Solution

10th Maths Guide Exercise 1.1  English Medium 

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10th maths Guide English Medium 

10th maths Exercise 1.1 Solution (Answers)

 10th maths Example 1.1 Solutions

Question: 

If A = {1, 3, 5} and B = {2, 3} then 

(i) find A × B and B × A.

 (ii) Is A × B = B × A? If not why ? 

(iii) Show that n(A × B) = n(B × A) = n(A) × n(B).

Solution : 

Given that A = {1, 3, 5} and B = {2, 3}

(i) A × B = {1, 3, 5} × {2, 3}  

= {(1, 2), (1, 3), (3, 2), (3, 3), (5, 2), (5, 3)}

B × A = {2, 3} × {1, 3, 5}  

= {(2, 1), (2, 3), (2, 5), (3, 1), (3, 3), (3, 5)} .      

(ii)  From (1) and (2) we conclude that  

 A × B ≠ B × A as (1, 2) ≠ (2, 1) 

and   (1, 3) ≠) (3, 1), etc.

(iii)  n(A) = 3; n(B) = 2. 

From (1) and (2) we observe that,  

 n(A × B) = n(B × A) = 6;

we see that, n(A) × n(B) = 3 × 2 = 6 and  

n(B) × n(A) = 2 × 3 = 6Hence, 

n(A × B) = n(B × A) = n(A) × n(B) = 6.

Thus, n(A × B) = n(B × A) = n(A) × n(B).

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10th Maths Example 1.2 solutions

 If A × B = {(3, 2), (3, 4), (5, 2), (5, 4)} then find A and B.

Solution :

 A × B = {(3, 2), (3, 4), (5, 2), (5, 4)} 

We have  

 A = {set of all first coordinates of elements of A × B}. Therefore A = {3, 5} 

B = {set of all second coordinates of ele-ments of A × B}. ThereforeB = {2, 4} 

Thus A = {3, 5} and B = {2, 4}.

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Samacheer kalvi 10th maths Example 1.3 solution

Let A = {x ∈ N | 1 < x < 4}, B = {x ∈ W |  0 ≤ x < 2} and C = {x ∈ N | x < 3}.Then verify that (i) A×(B∪C)= (A×B) ∪ (A×C)  (ii) A × (B ∩ C) = (A × B) ∩ (A × C)

Solution : 

A = {x ∈ N | 1 < x < 4} = {2, 3}, 

 B = {x ∈ W | 0 ≤ x < 2} = {0, 1}.

 C = {x ∈ N | x < 3} = {1, 2}

(i) A × (B ∪ C) = (A × B) ∪ (A × C) B ∪ C = {0, 1} ∪ {1, 2} = {0, 1, 2} A × (B ∪ C)  

= {2, 3} × {0, 1, 2} 

 = {(2, 0), (2, 1), (2, 2), (3, 0), (3, 1), (3, 2)}       

A × B = {2, 3} × {0, 1}           

  = {(2, 0), (2, 1), (3, 0), (3, 1)}

A × C  = {2, 3} × {1, 2}            

 = {(2, 1), (2, 2), (3, 1), (3, 2)} (A × B) ∪ (A × C) 

 = {(2, 0), (2, 1), (3, 0),   (3, 1)} ∪ {(2, 1), (2, 2), (3, 1), (3, 2)}  

= {(2, 0), (2, 1), (2, 2), (3, 0), (3, 1), (3, 2)}      ... (2) 

From (1) and (2), A × (B ∪ C)

 = (A × B) ∪ (A × C) is verified.

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(ii) A × (B ∩ C) = (A × B) ∩ (A × C) 

B ∩ C  = {0, 1} ∩ {1, 2}= {1}

 A × (B ∩ C)  = {2, 3} × {1}    = {(2, 1), (3, 1)} ... (3)

 A × B  = {2, 3} × {0, 1}         = {(2, 0), (2, 1), (3, 0), (3, 1)} 

A × C  = {2, 3} ×{1, 2}         = {(2, 1), (2, 2), (3, 1), (3, 2)}

 (A × B) ∩ (A × C)   

    = {(2, 0), (2, 1), (3, 0), (3, 1)} ∩     {(2, 1), (2, 2), (3, 1), (3, 2)}  

= {(2, 1), (3, 1)}   ... (4)

 From (3) and (4),  

A×(B ∩ C) = (A ×B) ∩ (A × C) is verified.

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10th maths EXERCIsE 1.1 solution 

10th maths Exercise 1.1 Question 1.

 Find A × B, A × A and B × A (i) A = {2, – 2, 3} and B = {1, – 4}  (ii) A = B = {p, q} (iii) A = {m, n} ; B = φ

Solution:

(i)  Given A = {2, – 2, 3}, B = {1, – 4}. 

A × B = {(2, 1), (2, – 4), (– 2, 1), (–2, – 4), (3, 1), (3, – 4)}i

A × A = {(2, 2), (2, – 2), (2, 3), (– 2, 2),  (– 2, – 2), (– 2, 3), (3, 2), (3, – 2), (3, 3)}

 B × A = {(1, 2), (1, – 2), (1, 3), (– 4, 2),  (– 4, – 2), (– 4, 3)}

(ii) Given 

A = B = {p, q} A × B = {(p, p), (p, q), (q, p), (q, q)}

 A × A = {(p, p), (p, q), (q, p), (q, q)} 

B × A = {(p, p), (p, q), (q, p), (q, q)}

(iii) A = {m, n}, B = φ

 If A = φ (or) B = φ,  then A × B = φ.  

and B × A =  φ 

A × B = φ and B × A = φ 

A × A = {(m, m), (m, n), (n, m), (n, n)}

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10th maths Exercise 1.1 Question 2. 

Let A = {1, 2, 3} and B = {x | x is a prime number less than 10}. Find A × B and  B × A.

Solution : 

Given A = {1, 2, 3}, B = {x | 

x is a prime number less than 10}. 

B = {2, 3, 5, 7} 

A × B = {(1, 2), (1, 3), (1, 5), (1, 7), (2, 2), (2, 3), (2, 5), (2, 7), (3, 2), (3, 3), (3, 5), (3, 7)} 

B × A = {(2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3), (5, 1), (5, 2), (5, 3), (7, 1), (7, 2), (7, 3)}

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 Exercise 1.1 Question 3. 

If B × A = {(– 2, 3), (– 2, 4), (0, 3), (0, 4), (3, 3), (3, 4)} find A and B.

Solution : 

Given  B × A  = {(–2, 3), (– 2, 4), (0, 3),     (0, 4), (3, 3), (3, 4)}  

∴ B = {– 2, 0, 3}, A = {3, 4}

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Exercise 1.1 Question 4. . 

If A = {5, 6}, B = {4, 5, 6}, C = {5, 6, 7}. Show that A × A = (B × B) ∩ (C × C).

Solution :

Given A = {5, 6}, B = {4, 5, 6}, C = {5, 6, 7}

LHS : A × A  = {5, 6} × {5, 6}        

 = {(5, 5), (5, 6), (6, 5), (6, 6)}  ...(1)

RHS : B ×B  = {4, 5, 6} ×{4, 5, 6}      

= {(4, 4), (4, 5), (4, 6), (5, 4), (5, 5),    (5, 6), (6, 4), (6, 5), (6, 6)} 

C × C = {5, 6, 7} × {5, 6, 7}     

= {(5, 5), (5, 6), (5, 7), (6, 5), (6, 6),    (6, 7), (7, 5), (7, 6), (7, 7)}

∴ (B × B) ∩ (C  × C)  = {(5, 5), (5, 6), (6, 5), (6, 6)}     ...(2) 

∴ From (1) and (2).  LHS  = RHS

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10th maths Exercise 1.1 Question 5

Given A = {1, 2, 3}, B = {2, 3, 5},  C = {3, 4} and D = {1, 3, 5}, check if  (A ∩ C) × (B ∩ D) = (A × B) ∩ (C × D) is true ?

Solution : Given A = {1, 2, 3}, B = {2, 3, 5},       

  C = {3, 4}, D = {1, 3, 5}    

 A ∩ C = {3},     B ∩ D = {3, 5}

∴ (A ∩ C) × (B ∩ D) = {(3, 3), (3, 5)}  ... (1) 

A × B = {(1, 2), (1, 3), (1, 5), (2, 2), (2, 3),            (2, 5), (3, 2), (3, 3), (3, 5)} 

C × D = {(3, 1), (3, 3), (3, 5), (4, 1), (4, 3), (4, 5)}

∴ (A × B) ∩ (C × D) = {(3, 3), (3, 5)}  ...(2) 

∴ From (1) and (2) LHS = RHS.

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10th maths Exercise 1.1 Question 6. 

Let A = {x ∈ W | x < 2}, B = {x ∈ N | 1 < x ≤ 4} and C = {3, 5}. 

Verify that 

 (i) A × (B ∪ C) = (A × B) ∪ (A × C) 

(ii) A × (B ∩ C) = (A × B) ∩ (A × C) 

(iii) (A ∪ B) × C = (A × C) ∪ (B × C)

Solution : 

Given  A = {x ∈ W | x < 2} ⇒ A = {0, 1}       

  B = {x ∈ N | 1 < x ≤ 4}   ⇒ B = {2, 3, 4}       

  C = {3, 5}

(i) To verify :  

A × (B ∪ C) = (A × B) ∪ (A × C)  

B ∪ C = {2, 3, 4, 5}

∴ A × (B ∪ C) = {(0, 2), (0, 3), (0, 4), (0, 5),    (1, 2), (1, 3), (1, 4), (1, 5)} ...(1)  

A × B = {(0, 2), (0, 3), (0, 4), (1, 2),       (1, 3), (1, 4)}  

A × C = {(0, 3), (0, 5), (1, 3), (1, 5)}

∴ (A × B) ∪ (A × C) = {(0, 2), (0, 3), (0, 4),   (0, 5), (1, 2), (1, 3), (1, 4), (1, 5)} ...(2) 

∴ From (1) and (2) LHS = RHS.

(ii) To verify : 

A × (B ∩ C) = (A × B) ∩ (A × C)  B ∩ C = {3} 

∴ A × (B ∩ C) = {(0, 3), (1, 3)} ...(1)      

 A × B = {(0, 2), (0, 3), (0, 4), (1, 2),       (1, 3), (1, 4)} 

 A × C = {(0, 3), (0, 5), (1, 3), (1, 5)}

∴ (A × B) ∩ (A × C) = {(0, 3), (1, 3)} ...(2) 

∴ From (1) and (2), LHS = RHS.

(iii)  (A ∪ B) × C = (A × C) ∪ (B × C)

  A ∪ B = {0, 1, 2, 3, 4}

∴ (A ∪ B) × C = {(0, 3), (0, 5), (1, 3), (1, 5),     (2, 3), (2, 5), (3, 3), (3, 5),     (4, 3), (4, 5)}  ...(1) 

 A × C = {(0, 3), (0, 5), (1, 3), (1, 5)}  

B × C = {(2, 3), (2, 5), (3, 3), (3, 5),      (4, 3), (4, 5)}

 ∴ (A × C) ∪ (B × C) = {(0, 3), (0, 5), (1, 3),      (1, 5), (2, 3), (2, 5)     (3, 3), (3, 5), (4, 3),      (4, 5)}  ...(2)

 ∴ From (1) and (2) LHS = RHS.

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10th maths Exercise 1.1 Question 7. Let A = The set of all natural numbers less than 8, 

B = The set of all prime num-bers less than 8, 

C = The set of even prime number. Verify that

 (i) (A ∩ B) × C = (A × C) ∩ (B × C)  

(ii) A × (B – C) = (A × B) – (A × C)

Solution : 

Given A = {1, 2, 3, 4, 5, 6, 7}      

 B = {1, 3, 5, 7}      

 C = {2}

(i) To verify : 

(A ∩ B) × C = (A × C) ∩ (B × C)  

A ∩ B = {1, 3, 5, 7}

∴ (A ∩ B) × C  = {(1, 2), (3, 2), (5, 2), (7, 2)}      ... (1)

 A × C  = {(1, 2), (2, 2),(3, 2),(4, 2),(5, 2),(6, 2), (7, 2)} 

B × C  = {(1, 2), (3, 2), (5, 2), (7, 2)}  ... (2) 

∴ From (1) and (2), LHS = RHS.

(ii) To verify : 

A × (B – C) = (A × B) – (A × C)  

B – C = {1, 3, 5, 7}

∴ A × (B – C) = {(1, 1), (1, 3), (1, 5), (1, 7), 

                             (2, 1), (2, 3), (2, 5), (2, 7),        

                             (3, 1), (3, 3), (3, 5), (3, 7),       

                                (4, 1), (4, 3), (4, 5), (4, 7),      

                                (5, 1), (5, 3), (5, 5), (5, 7),       

                                 (6, 1), (6, 3), (6, 5), (6, 7),        

                               (7, 1), (7, 3), (7, 5), (7, 7),}  

A × B = {(1, 1), (1, 3), (1, 5), (1, 7),        

             (2, 1), (2, 3), (2, 5), (2, 7),        

               (3, 1), (3, 3), (3, 5), (3, 7),        

                (4, 1), (4, 3), (4, 5), (4, 7),

               (5, 1), (5, 3), (5, 5), (5, 7),       

              (6, 1), (6, 3), (6, 5), (6, 7),       

               (7, 1), (7, 3), (7, 5), (7, 7),}  

A × C = {(1, 2), (2, 2), (3, 2), (4, 2), (5, 2), (6, 2), (7, 2)}

∴ (A × B) – (A × C)    = {(1, 1), (1, 3), (1, 5), (1, 7),                                   (2, 1), (2, 3), (2, 5), (2, 7),      

                             (3, 1), (3, 3), (3, 5), (3, 7),      

                             (4, 1), (4, 3), (4, 5), (4, 7),       

                              (5, 1), (5, 3), (5, 5), (5, 7),      

                                (6, 1), (6, 3), (6, 5), (6, 7),       

                            (7, 1), (7, 3), (7, 5), (7, 7),}        ...(2) 

∴ From (1) and (2), LHS = RHS.

10th maths Exercise 1.2 solutions Click here

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