In this post we will provide you with Simplification Questions for class 6 with answers which you can practice so as to improve your understanding of the topic.

We use the acronym **“BODMAS”** to make a mathematical equation with two or more operations simpler. The letters B, O, D, M, A, and S stand for brackets, orders (powers/indices or roots), division, multiplication, addition, and subtraction, respectively. The students can practice the various types of questions in the section below on simplifications to gain greater insight of the topic.

## Simplification Questions for class 6 PDF with Answers

Remember below table for correct implementation of BODMAS rule.

B | Brackets | ( ), { }, [ ] |

O | Order of | Square roots, indices, exponents and powers |

D | Division | ÷, / |

M | Multiplication | ×, * |

A | Addition | + |

S | Subtraction | – |

**1. 9 ÷ 3 + 2×7 – 16 **

Solution:

= 9/3+2**7-16 = 3+2**7-16 (As per BODMAS rule first division shall be performed)

= 3+14-16

= 17-16

= 1

**2. (3× 4 – 8) + (44 ÷11 + 6)**

Solution = (3*4-8)+(44/11+6)

= (12-8)+(44/11+6) (Performing Division)

= (4)+(44/11+6)

= 4+(44/11+6)

= 4+(4+6) (Performing Division)

= 4+(10)

= 4+10

= 14

**3. 3 × (2× 5 – 6) + 8 – 15 ÷3**

Solution: = 3*(2**5-6)+8-15/3

= 3*(10-6)+8-15/3 = 3*(4)+8-15/3

= 3*4+8-15/3

= 12+8-15/3

= 12+8-5

= 20-5

= 15

**4. (9 + 12) ÷ 7 + 36 ÷ 2 of 3**

Solution: (9+12) ÷ 7 + 36 ÷ 2 of 3 = 21 ÷ 7 + 36 ÷ 2 of 3

= 21 ÷ 7 + 36 ÷ 6

= 3 + 6 =9

**5.** **(2 of 4 + 6) ÷ 2 – 35 ÷ 7**

= (2 of 4 + 6) ÷ 2 – 35 ÷ 7 = (8 + 6) ÷ 2 – 35 ÷ 7 (Solving Of ) = 14÷2-5 ** ** (Performing division)

= 7-5 (Performing Subtraction)

= 2

**6. 16 + [7 + {(3 + 7) – 5̅̅̅−̅̅̅3̅}]**

**7. 13 + 6{10 – 6 ÷ 3}**

= 13+6*{10-6/3} = 13+6*{10-2}

= 13+6*{8} *

= 13+48

= 61

**8. 12 + [7 – (4 ÷ 2) –3̅̅̅−̅̅̅1̅]**

**9. 36 – [18 – {14 – (15 – 4 ÷ 2 × 2)}]**

= 36-[18-{14-(15-4/2**2)}] *

= 36-[18-{14-(15-4)}]

= 36-[18-{14-(11)}]

= 36-[18-{14-11}]

= 36-[18-{3}]

= 36-[18-3]

= 36-[15]

= 36-15

= 21

10. **85+ 6{10 + 27÷ 3} **

= 85+6*{10+27/3} = 85+6*{10+9}

= 85+6*{19} *

= 85+114

= 199

11. **8 + 4 ÷ 2 × 5 = ?**

(a) 30 (b) 50 (c) 18 (d) none of these

Ans: 18

Here are a few examples of simplification questions that could be appropriate for students in grade 8:

- Simplify: (12 + 18) x (15 – 9)
- Simplify: (8 x 7) + (6 x 5) + (4 x 3)
- Simplify: (12 x 8) – (15 + 9)
- Simplify: (20 – 15) x (12 + 8)
- Simplify: (18 x 9) + (6 x 4) – (12 + 8)

These questions involve more complex expressions and may require students to use their knowledge of order of operations, the properties of operations, and the rules of exponents to simplify the expressions. As with the previous questions, the specific level of difficulty and complexity of these questions may vary depending on the individual student and the curriculum being followed.

### What is PEDMAS?

The order or simplification of mathematical processes is known as BODMAS in India. The BODMAS regulation, however, is known as PEDMAS in various countries, such as the US (United States) or UK (United Kingdom), etc. We can reduce arithmetic problems with several mathematical operators by using the PEDMAS approach.

### What is BODMAS?

The order or simplification of mathematical processes is known as BODMAS in India.