Here we provide RD Sharma Class 12 Ex 3.5 Solutions Chapter 3 Binary Operations for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 12 Ex 3.5 Solutions Chapter 3 Binary Operations book pdf download. Now you will get step-by-step solutions to each question.

Textbook | NCERT |

Class | Class 12th |

Subject | Maths |

Chapter | 3 |

Exercise | 3.5 |

Category | RD Sharma Solutions |

Table of Contents

### Question 1. Construct the composition table for ×_{4} on set S = {0,1,2,3}.

**Solution:**

*Given that *×_{4 } *on set S = {0, 1, 2, 3}*

*We have to find the composition table for *×_{4} *by using multiplication and modulo operations.*

*As 0 x _{4} 0 = 0 (0 × 0 = 0 and 0 modulo 4 = 0 % 4 = 0)*

*As 1 x _{4} 1 = 1 (1 × 1 = 1 and 1 modulo 4 = 1 % 4 =1)*

*As 1 x _{4} 2 = 2 (1 × 2 = 2 and 2 modulo 4 = 2 % 4 = 2)*

*As 3 x _{4 }1 = 3 (3 × 1 = 3 and 3 modulo 4 = 3 % 4 = 3)*

*As 4 x _{4} 4 = 0 (2 × 2 = 4 and 4 modulo 4 = 4 % 4 =* 0)

.

*So on. . .*

*Here % – (modulo operator) means it gives the remainder of the number when divided by the other number (i.e., 4)*

*Therefore, the composition table is:*

x_{4} | 0 | 1 | 2 | 3 |

0 | 0 | 0 | 0 | 0 |

1 | 0 | 1 | 2 | 3 |

2 | 0 | 2 | 0 | 2 |

3 | 0 | 3 | 2 | 1 |

### Question 2. Construct the composition table for +_{5} on set S = {0, 1, 2, 3, 4}.

**Solution:**

*Given that + _{5} on set S = {0,1,2,3,4}*

*We have to find the composition table for + _{5} by using addition and modulo operations.*

*As 0 + _{5} 0 = 0 (0 + 0 = 0 and 0 modulo 5 = 0 % 5 = 0)*

*As 1 + _{5} 0 = 1 (1 + 0 = 1 and 1 modulo 5 = 1 % 5 = 1)*

*As 1 + _{5} 1 = 2 (1 + 1 = 2 and 2 modulo 5 = 2 % 5 =2)*

*As 1 + _{5} 2 = 3 (1 + 2 = 3 and 3 modulo 5 = 3 % 5 = 3)*

*As 3 + _{5} 1 = 4 (3 + 1 = 4 and 4 modulo 5 = 4 % 5 = 4)*

*As 2 + _{5} 2 = 4 (2 + 2 = 4 and 4 modulo 5 = 4 % 5 = 4)*

*.*

*So on . . .*

*Here % – (modulo operator) means it gives the remainder of the number when divided by the other number (i,e., 5)*

*Therefore, the composition table is:*

+_{5} | 0 | 1 | 2 | 3 | 4 |

0 | 0 | 1 | 2 | 3 | 4 |

1 | 1 | 2 | 3 | 4 | 0 |

2 | 2 | 3 | 4 | 0 | 1 |

3 | 3 | 4 | 0 | 1 | 2 |

4 | 4 | 0 | 1 | 2 | 3 |

### Question 3. Construct the composition table for ×_{6} on set S = {0, 1, 2, 3, 4, 5}.

**Solution:**

*Given that × _{6} on set S = {0, 1, 2, 3, 4, 5}*

*We have to find the composition table for × _{6} by using multiplication and modulo operations.*

*As 0 x _{6} 0 = 0 (0 × 0 = 0 and 0 modulo 6 = 0 % 6 = 0)*

*As 1 x _{6} 1 = 1 (1 × 1 = 1 and 1 modulo 6 = 1 % 6 =1)*

*As 1 x _{6} 2 = 2 (1 × 2 = 2 and 2 modulo 6 = 2 % 6 = 2)*

*As 3 x _{6} 1 = 3 (3 × 1 = 3 and 3 modulo 6 = 3 % 6 = 3)*

*As 2 x _{6} 2 = 4 (2 × 2 = 4 and 4 modulo 6 = 4 % 6 = 4)*

*As 5 x _{6} 1 = 5 (5 × 1 = 5 and 5 modulo 6 = 5 % 6 = 5)*

*.*

*so on . . .*

*Here % – (modulo operator) means it gives the remainder of the number when divided by the other number (i.e., 6)*

*Therefore, the composition table is :*

x_{6} | 0 | 1 | 2 | 3 | 4 | 5 |

0 | 0 | 0 | 0 | 0 | 0 | 0 |

1 | 0 | 1 | 2 | 3 | 4 | 5 |

2 | 0 | 2 | 4 | 0 | 2 | 4 |

3 | 0 | 3 | 0 | 3 | 0 | 3 |

4 | 0 | 4 | 2 | 0 | 4 | 2 |

5 | 0 | 5 | 4 | 3 | 2 | 1 |

### Question 4. Construct the composition table for ×_{5} on set S = {0, 1, 2, 3, 4}.

**Solution:**

*Given that × _{5} on set S = {0,1,2,3,4}*

*We have to find the composition table for × _{5} by using multiplication and modulo operations.*

*As 0 x _{5} 0 = 0 (0 × 0 = 0 and 0 modulo 5 = 0 % 5 = 0)*

*As 1 x _{5} 1 = 1 (1 × 1 = 1 and 1 modulo 5 = 1 % 5 =1)*

*As 1 x _{5} 2 = 2 (1 × 2 = 2 and 2 modulo 5 = 2 % 5 = 2)*

*As 2 x _{5} 2 = 4 (2 × 2 = 4 and 4 modulo 5 = 4 % 5 = 4)*

*As 3 x _{5} 3 = 2 (3 × 4 = 12 and 12 modulo 5 = 12 % 5 = 2)*

*.*

*So on . . .*

*Here % – (modulo operator) means it gives the remainder of the number when divided by the other number (i.e., 5)*

*Therefore, the composition table is:*

x_{5} | 0 | 1 | 2 | 3 | 4 |

0 | 0 | 0 | 0 | 0 | 0 |

1 | 0 | 1 | 2 | 3 | 4 |

2 | 0 | 2 | 4 | 1 | 3 |

3 | 0 | 3 | 1 | 4 | 2 |

4 | 0 | 4 | 3 | 2 | 1 |

### Question 5. Construct the composition table for ×_{10} on set S = {1, 3, 7, 9}.

**Solution:**

*Given that × _{10} on set S = {1,3,7,9}*

*We have to find the composition table for × _{10} by using multiplication and modulo operations.*

*As 1 x _{10} 1 = 1 (1 × 1 = 1 and 1 modulo 10 = 1 % 10 = 1)*

*As 1 x _{10} 7 = 7 (1 × 7 = 7 and 7 modulo 10 = 2 % 10 = 7)*

*As 3 x _{10} 1 = 3 (3 × 1 = 3 and 3 modulo 10 = 3 % 10 = 3)*

*As 9 x _{10} 7 = 3 (9 × 7 = 63 and 63 modulo 10 = 63 % 10 = 3)*

*.*

*So on . . .*

*Here % – (modulo operator) means it gives the remainder of the number when divided by the other number (i.e., 10)*

*Therefore, the composition table is:*

x_{10} | 1 | 3 | 7 | 9 |

1 | 1 | 3 | 7 | 9 |

3 | 3 | 9 | 1 | 7 |

7 | 7 | 1 | 9 | 3 |

9 | 9 | 7 | 3 | 1 |

*From the composition table, we can observe that elements 1 multiples (i.e., 1st row) are same as top most row.*

*Therefore, 1 *∈ S is the identity element for x_{10}.

*We have to find the inverse of 3:*

*As 3 x _{10} 7 = 1 (3 × 7 = 21 and 21 modulo 10 = 1) in the composition table.*

*So the inverse of 3 is 7*

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