by BODMAS rule simplification questions for class 6
The aim of simplification questions for class 6 is normally to evaluate a student’s ability with addition, subtraction, multiplication, and division. In these inquiries, complex expressions or equations must be made simpler by applying one or more operations to them.
The questions may involve the use of brackets or parentheses, exponents or indices, fractions, and decimals. Students may be asked to simplify expressions by combining like terms, factorising, or expanding.
Generally, simplification questions for class 6 are designed to assess a student’s learning of fundamental mathematical ideas . These questions play an important part in the growth of math ability and in building a foundation for future study of more difficult topics in the subject.
For simplifying an expression , the order of performing the various operations should be strictly in the following order :
(i) Brackets (ii) of (iii) Division (iv) Multiplication (v) Addition (vi) Subtraction
Remember the word ‘BODMAS ‘ , where the letters B, O, D, M, A, S stand for Brackets , of, Division, Multiplication, Addition and Subtraction respectively.
Simplification questions for class 6 with answers:
(1) Simplify : 12-[20÷{8-2(9-5-2)}]
Sol. 12-[20÷{8-2(9-5-2)}]= 12-[20÷{8-2 x 2}] (removing small brackets)
=12-[20÷{8-4}]
=12-[20÷4] (removing braces)
= 12-5 (removing big brackets)
=7
(2) 4+4+4+4÷4
4+4+4+4÷4 =4+4+4+1 (solving division)
=13
(3) 25− [4 + {16 − (12 − 2)}]
25− [4 + {16 − (12 − 2)}] = 25− [4 + {16 − 10}] (removing small brackets)
= 25− [4 + 6] (removing curly braces)
=25-10 (removing big brackets)
=15
(4) 36 − [18 − {14 − (15 − 4 ÷2 ×2)}]
36 − [18 − {14 − (15 − 4 ÷2 ×2)}]= 36 − [18 − {14 − (15 − 2 ×2)}] (Performing division in small brackets)
=36 − [18 − {14 − (15 − 4)}] (performing multiplication)
= 36 − [18 − {14 − 11}] (removing small brackets)
=36- [18-3] (removing curly braces)
=36-15 (removing big brackets)
=21
(5) 10×10+[400÷{100-(50-
10×10+[400÷{100-(50-)}] =10×10+[400÷{100-(50-30)}] (removing bar)
=10×10+[400÷{100-20}] (removing small brackets)
=10×10+[400÷80] (removing braces)
=10×10+5 (removing big brackets)
= 100+5 (performing multiplication)
= 105
(6) 20−[4+{16−(12− 3)}]
Sol. 20 – [4 + {16 – (12 – 3)}] = 20 – [4 + {16 – 9}] (Removing small brackets)
= 20 – [4 + 7] (Removing curly brackets)
= 20 – 11 (Removing square brackets)
= 9
(7 ) 13-[20 ÷{8-2(9-5-2)}]
Sol. 13-[20 ÷{8-2(9-5-2)}]= 13-[20 ÷{8-2 x 2}] (removing small brackets)
=13-[20 ÷{8-4}]
=13-[20 ÷4] (removing braces)
= 13-5 (removing big brackets)
=8
(8) [30 ÷ {10- (8- )}]
Sol. [30 ÷ {10- (8- )}] = [30 ÷ {10- (8- 4)}] (removing bar)
=[30 ÷ {10- 4}] (removing small brackets)
=[30 ÷ 6] (removing braces)
=5 (removing big brackets)
9) 10 + {8 + {5 + (2 + 3) – 6}}
Solution: 10 + {8 + {5 + (2 + 3) – 6}}=10 + {8 + {5 + 5 – 6}} (Solving expression inside small bracket)
=10+{8+4} (solving expression inside curly braces)
=10+12
= 22
Therefore 10 + {8 + {5 + (2 + 3) – 6}}= 22.
10) 15 – {10 – {6 – (2 + 1 ÷ 1)}}
Solution: 15 – {10 – {6 – (2 + 1 ÷ 1)}} = 15 – {10 – {6 – (2 + 1 )}} (Solving division inside small bracket)
=15-{10-{6-3}}
=15- {10-3} (solving expression inside curly braces)
= 15-7
=8
Therefore 15 – {10 – {6 – (2 + 1 ÷ 1)}}=8
11) 12 + {20 – {2 × (4 + 1)}}
Solution: 12 + {20 – {2 × (4 + 1)}} =12 + {20 – {2 × 5}}
=12 + {20 – 10}
=12+10
= 22
Therefore 12 + {20 – {2 × (4 + 1)}} =22
12) 45 – {24 + {3 × (4 – 2)}}
Solution: 45 – {24 + {3 × (4 – 2)}} = 45 – {24 + {3 × 2}}
=45 – {24 + 6}
=45-30
= 15
Therefore 45 – {24 + {3 × (4 – 2)}} =15
13) 21 + {9 – {2 × (5 – 2) + 1}}
Solution: 21 + {9 – {2 × (5 – 2) + 1}} =21 + {9 – {2 × 3 + 1}}
=21+{9-{6+1}}
=21+{9-7}
=21 + 2
= 23
Therefore 21 + {9 – {2 × (5 – 2) + 1}} =23
14) 30 – {20 – {5 × (3 – 1) + 1}}
Solution: 30 – {20 – {5 × (3 – 1) + 1}} = 30 – {20 – {5 × 2 + 1}}
= 30 – {20 – {10 + 1}}
= 30 – {20 – 11}
= 30-9
= 21
Therefore 30 – {20 – {5 × (3 – 1) + 1}} =21
15) 50 + {10 – {6 × (5 – 2) – 1}}
Solution: 50 + {10 – {6 × (5 – 2) – 1}} =50 + {10 – {6 × 3 – 1}}
=50 + {10 – {18-1}}
=50 + {10-17}
= 50-7
= 43
Therefore 50 + {10 – {6 × (5 – 2) – 1}} = 43.
16) 72 – {15 + {4 × (3 – 2) + 2}}
Solution: 72 – {15 + {4 × (3 – 2) + 2}} = 72 – {15 + {4 × 1 + 2}}
=72 – {15 + {4+2}}
= 72-{15+6}
=72-21
= 51
Therefore 72 – {15 + {4 × (3 – 2) + 2}} =51
17)40 + {5 – {2 × (5 – 3) + 1}}
Solution: 40 + {5 – {2 × (5 – 3) + 1}} =40 + {5 – {2 × 2 + 1}}
= 40 + {5 – {4 + 1}}
=40 + {5 – 5}
= 40
Therefore 40 + {5 – {2 × (5 – 3) + 1}} = 40.
18) 60 – {16 + {2 × (7 – 2)}}
Solution: 60 – {16 + {2 × (7 – 2)}} =60 – {16 + {2 × 5}}
=60 – {16 + 10}
=60- 26
= 34
Therefore 60 – {16 + {2 × (7 – 2)}} = 34.
19) 21 + 5 × 6 ÷ 3 – 4
Solution: 21 + 5 × 6 ÷ 3 – 4 =21 + 5 × 2 – 4
=21 + 10 – 4
=31-4
= 27
Therefor 21 + 5 × 6 ÷ 3 – 4 = 27
20) 12 – 3 × 2 + 6 ÷ 2 + 5
Solution:12 – 3 × 2 + 6 ÷ 2 + 5 =12 – 3 × 2 + 3+ 5
=12 – 6 + 3 + 5
= 20-6
= 14
Therefore 12 – 3 × 2 + 6 ÷ 2 + 5 = 14
21) 48 ÷ 4 + 7 × 3 – 5
Solution: 48 ÷ 4 + 7 × 3 – 5 = 12 + 7 × 3 – 5
=12 + 21 – 5
=33-5
= 28
Therefore 48 ÷ 4 + 7 × 3 – 5 = 28
22)16 – 4 × 3 + 12 ÷ 6 + 7
Solution: 16 – 4 × 3 + 12 ÷ 6 + 7 =16 – 4 × 3 + 2 + 7
=16 – 12 + 2 + 7
=25 -12
= 13
Therefore 16 – 4 × 3 + 12 ÷ 6 + 7 =13
23) 10 + 8 ÷ 2 × 5 – 6
Solution: 10 + 8 ÷ 2 × 5 – 6 =10 + 4 × 5 – 6
=10 + 20 – 6
=30-6
= 24
Therefore 10 + 8 ÷ 2 × 5 -6= 24
24) 30 ÷ 3 + 4 × 5 – 6
Solution: 30 ÷ 3 + 4 × 5 – 6 =30 ÷ 3 + 4 × 5 – 6
=10+ 4 x 5 -6
=10 + 20 – 6
=30-6
= 24
Therefore 30 ÷ 3 + 4 × 5 – 6 = 24
25) 24 – 6 × 2 + 12 ÷ 4 + 5
Solution: 24 – 6 × 2 + 12 ÷ 4 + 5 =24 – 6 × 2 + 3 + 5
=24 – 12 + 3 + 5
= 32-12
= 20
Therefore 24 – 6 × 2 + 12 ÷ 4 + 5 = 20
26) 42 ÷ 6 – 3 × 7 + 10
Solution: 42 ÷ 6 – 3 × 7 + 10 =7 – 3 × 7 + 10
=7 – 21 + 10
=17-21
= -4
Therefore 42 ÷ 6 – 3 × 7 + 10 = -4
27) 60 – 12 ÷ 4 × 5 + 3
Solution: 60 – 12 ÷ 4 × 5 + 3 =60 – 3 × 5 + 3 (solving division)
=60-15+3 (solving multiplication)
= 63-15 (solving addition)
= 48
Therefore 60 – 12 ÷ 4 × 5 + 3 = 48
28) 25 + 4 × 7 – 18 ÷ 2
Solution: 25 + 4 × 7 – 18 ÷ 2 =25 + 4 × 7 – 9 (solving division)
=25+28-9 (solving multiplication)
= 53-9 (solving addition)
= 44 (solving subtraction)
Therefore 25 + 4 × 7 – 18 ÷ 2 = 44
29) (10 – 2 x 3) ÷ (2 + 1)
Solution: (10 – 2 x 3) ÷ (2 + 1) = (10 – 6) ÷ 3
= 4 ÷ 3
= 1.33
Therefore (10 – 2 x 3) ÷ (2 + 1) = 1.33
30) 5 + 2 x (7 – 3) ÷ 2
Solution: 5 + 2 x (7 – 3) ÷ 2 = 5 + 2 x 4 ÷ 2
= 5 + 2 x 2
=5+4
=9
Therefore 5 + 2 x (7 – 3) ÷ 2 = 9
31) (3 × 4 + 6) ÷ 3 – 2 × 5
Solution: (3 × 4 + 6) ÷ 3 – 2 × 5 =(12 + 6) ÷ 3 – 2 × 5
= 18 ÷ 3 – 2 × 5
=6 -2 x 5
= 6-10
=-4
32 ) -25+14 (5-3)
Solution : -25+14 (5-3) =-25+14
2
=-25+7
= -18
Therefore -25+14 (5-3) = -18
33 ) 7 – [6 – {5 – (4 – 3)}] ÷ 2
Solution: 7 – [6 – {5 – (4 – 3)}] ÷ 2 = 7 – [6 – {5 – 1}] ÷ 2
= 7 – [6 – 4] ÷ 2
= 7 – 2 ÷ 2
= 7-1
=6
Therefore 7 – [6 – {5 – (4 – 3)}] ÷ 2 = 6
34) 18 – [6 – {4 – (8 – 6 + 3 )}]
Solution : 18 – [6 – {4 – (8 – 6 + 3 )}] =18 – [6 – {4 – 5}]
=18 – [6 – (-1)]
= 18 – [6+1]
=18- 7
=11
35 ) (45 × 20÷10 + 21) ÷ 3
Solution: (45 × 20÷10 + 21) ÷ 3 =(45 × 2 + 21) ÷ 3
= (90 + 21) ÷ 3
= 111 ÷ 3
= 37
Therefore (45 × 20÷10 + 21) ÷ 3 =37
36) 5 × (2 + 3) – 7 ÷ 1
Solution: 5 × (2 + 3) – 7 ÷ 1 = 5 × 5 – 7 ÷ 1
= 5 x 5 -7
=25-7
=18
Therefore 5 × (2 + 3) – 7 ÷ 1 =18
37)
Solution : =
=
=
=4
38)
Solution : =
= 1000-100+20
= 1020-100
=920
Simplify questions for class 6 (unsolved with answers)
Simplify each of the following:
(i)(3 x 2) -(8-5)
(ii) 30÷ 6+10-2× 5
(iii) (18+10)-(3 x 6)
(iv) 48÷16×2+17-9
(v) (10×8) ÷(20÷5)
(vi) {20+(15+5)}-5
(vii) 121÷ 11+29-2 ×10
(viii) 36 – [18 – {14 – (15 – 4 ÷ 2 x 2)}]
(ix) 45 – [38 – {60 ÷ 3 – (6 – 9 ÷ 3) ÷ 3}]
(x) 60-48÷6×4+8
(xi) 39-15+18×9÷3
(xii) [29 – (-2) {6 – (7 – 3)}] ÷ [3 x {5 + (-3) x (-2)}]
(xiii) 100-72÷8+4×3
(xiv) 30+75÷15×4-18
(xv) 12-[20÷{8-2(9-5-2)}]
(xvi) 4+4+4+4÷4
(xvii) 25− [4 + {16 − (12 − 2)}]
(xviii) 36 − [18 − {14 − (15 − 4 ÷2 ×2)}]
(xix) 10×10+[400÷{100-(50-)}]
(xx) [40÷{19-3(6- )}]
Ans. (i) 2 (ii) 5 (iii) 10 (iv) 14 (v) 20 (vi) 35 (vii) 20 (viii) 21 (ix) 26 (x) 36 (xi) 78 (xii) 1
(xiii) 103 (xiv) 32 (xv) 7 (xvi) 13 (xvii) 15 (xviii) 21 (xix) 105 (xx) 4
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