**BODMAS Questions for Class 6 with all brackets**

In order to simplify expressions involving more than one bracket, use the following steps:

(i) If If a vinculum is present, then perform operations under it.

(ii) After solving vinculum, see the innermost bracket and perform operations within it.

(iii) See the next innermost bracket and perform operations within it.

(iv) Continue this process till all the brackets are removed.

(iv) while removing brackets , use BODMAS rule to simplify expression that lies with in the brackets.

Note: In the absence of any sign before a bracket , we take the sign as multiplication.

**BODMAS rule simplification questions for class 6 (Solved Examples )**

Different types of BODMAS Questions with all brackets are given here:

**(1)****5 x {3 + 2 x [6 ÷ (2 + 1)]}
**Solution: 5 x {3 + 2 x [6 ÷ (2 + 1)]} = 5 x {3 + 2 x [6 ÷ 3]} (Perform the addition within the inner brackets)

= 5 x {3 + 2 x 2} (Perform the division within the inner brackets)

= 5 x {3 + 4} ( Perform the multiplication within the inner brackets)

= 5 x 7

=35

Hence 5 x {3 + 2 x [6 ÷ (2 + 1)]} = 35

**(2) 3 x {4 + 2 x (7 – 5) – 1}
**Solution: 3 x {4 + 2 x (7 – 5) – 1} = 3 x {4 + 2 x 2 – 1}

= 3 x {4 + 4 – 1}

= 3 x {8 – 1}

= 3x 7

=21

Therefore 3 x {4 + 2 x (7 – 5) – 1} = 21

**(3) [(2 + 3) x 4 + 1] ÷ (6 – 3)
**Solution: [(2 + 3) x 4 + 1] ÷ (6 – 3) = [5 x 4 + 1] ÷ 3

= [20 + 1] ÷ 3

=21 ÷ 3

=7

Hence [(2 + 3) x 4 + 1] ÷ (6 – 3) = 7

**(4)****(4 + 3) x (8 – 2) ÷ (5 – 2)
**

**Solution:**(4 + 3) x (8 – 2) ÷ (5 – 2) = 7x 6 ÷ 3

=7 x 2

=14

Therefore (4 + 3) x (8 – 2) ÷ (5 – 2) =14

**(5) (3 + 4) x 2 – 5 ÷ 5
**Solution : (3 + 4) x 2 – 5 ÷ 5 = 7 x 2 – 5 ÷ 5

= 7 x 2 – 1

=14-1

=13

Therefore (3 + 4) x 2 – 5 ÷ 5 = 13

**(6) 2 x (5 + 3) ÷ 4 – 1
Solution: **2 x (5 + 3) ÷ 4 – 1 =2 x 8 ÷ 4 – 1

= 2 x 2 – 1

=4-1

=3

Therefore 2 x (5 + 3) ÷ 4 – 1 = 3

**(7) 3 x (4 + 2) – 8 ÷ 2 + 5
**Solution : 3 x (4 + 2) – 8 ÷ 2 + 5 =3 x 6 – 8 ÷ 2 + 5

=3 x 6 – 4 + 5

=18 – 4 + 5

= 23-4

=19

Therefore 3 x (4 + 2) – 8 ÷ 2 + 5 =19

**(8)(2 + 3) × (4 + 5) – 6 ÷ 2**

Solution: (2 + 3) × (4 + 5) – 6 ÷ 2 = 5 × 9 – 6 ÷ 2

= 5 × 9 – 3

= 45 – 3

=42

Hence (2 + 3) × (4 + 5) – 6 ÷ 2 =42

**(9) [(6 + 3) × 2 + 4] ÷ (9 – 2) **

Solution: [(6 + 3) × 2 + 4] ÷ (9 – 2) = [9 × 2 + 4] ÷ 7

=[18 + 4] ÷ 7

= 22 ÷ 7

= 3.14

Hence [(6 + 3) × 2 + 4] ÷ (9 – 2) = 3.14

**(10) {[2 + (4 × 3)] + 5} ÷ 7 **

Solution: {[2 + (4 × 3)] + 5} ÷ 7 = {[2 + 12] + 5} ÷ 7

= {14+ 5} ÷ 7

= 19 ÷ 7

= 2.71

So {[2 + (4 × 3)] + 5} ÷ 7 = 2.71

**(11) (8 – 2) ÷ {(3 + 2) × (7 – 5)}
Solution: **(8 – 2) ÷ {(3 + 2) × (7 – 5)} =6 ÷ {5 × 2}

=6 ÷ 10

=0.6

Hence (8 – 2) ÷ {(3 + 2) × (7 – 5)} =0.6

**(12) (10 – 5) × {[6 + (9 – 7)] ÷ 2} **

Solution : (10 – 5) × {[6 + (9 – 7)] ÷ 2} =5 × {[6 + 2] ÷ 2}

=5 × {8 ÷ 2}

=5 × 4

=20

Therefore (10 – 5) × {[6 + (9 – 7)] ÷ 2} =20

**(13){[4 × (6 + 3)] + 7} ÷ (5 – 1)**

Solution : {[4 × (6 + 3)] + 7} ÷ (5 – 1) ={[4 × 9] + 7} ÷ 4

= {36 + 7} ÷ 4

= 43÷ 4

=10.75

Therefore {[4 × (6 + 3)] + 7} ÷ (5 – 1) = 10.75

**(14) {[7 + (3 × 5)] × 2} – 4 **

Solution : {[7 + (3 × 5)] × 2} – 4 = {[7 + 15] × 2} – 4

={22 × 2} – 4

=44 -4

=40

Therefore {[7 + (3 × 5)] × 2} – 4 = 40

**(15) (12 + 3) ÷ {[2 + (5 – 3)] × 4}
Solution :** (12 + 3) ÷ {[2 + (5 – 3)] × 4} = 15 ÷ {[2 + 2] × 4}

= 15 ÷ {4 × 4}

= 15 ÷ 16

= 0.9375

Hence (12 + 3) ÷ {[2 + (5 – 3)] × 4} =0.9375

**(16) {[8 + (5 × 2)] × 3} + 2
**Solution : {[8 + (5 × 2)] × 3} + 2 = {[8 + 10] × 3} + 2

= {18 × 3} + 2

= 54 + 2

=56

Hence {[8 + (5 × 2)] × 3} + 2 = 56

**(17) (15 – 5) ÷ {[4 + (9 – 7)] × 2}
**Solution : (15 – 5) ÷ {[4 + (9 – 7)] × 2} =10 ÷ {[4 + 2] × 2}

=10 ÷ {6 × 2}

=10 ÷ 12

= 0.833

Therefore (15 – 5) ÷ {[4 + (9 – 7)] × 2} = 0.833

**(18)
**Solution :

=

=

=54+6

=60

**(19) **

**Solution : =**

=

=

=

=12-10

=2

Therefore

**(20) **

Solution :

=

=

=

=4

Therefore =4

**(21) **

**Solution : =**

=

=

=

= 3

Therefore

**(22) {[2 + (5 × 3)] ÷ 7} + {[(9 – 4) ÷ 2] – 1}
**Solution : {[2 + (5 × 3)] ÷ 7} + {[(9 – 4) ÷ 2] – 1} = {[2 + 15] ÷ 7} + {[5 ÷ 2] – 1}

= {17 ÷ 7} + {2.5- 1}

= {17 ÷ 7} + {2.5- 1}

=2.43+1.0

=3.43

Therefore {[2 + (5 × 3)] ÷ 7} + {[(9 – 4) ÷ 2] – 1} =3.43

**(23) {[8 + (2 × 3)] – (4 × 2)} ÷ {[(3 + 5) × 2] – 1}**

Solution : {[8 + (2 × 3)] – (4 × 2)} ÷ {[(3 + 5) × 2] – 1} = {[8 + 6] – 8} ÷ {[8 × 2] – 1}

= [14 – 8] ÷ [16 – 1]

=6 ÷ 15

0.4

Hence {[8 + (2 × 3)] – (4 × 2)} ÷ {[(3 + 5) × 2] – 1} =0.4

**(24) (7 – 3) × {[5 + (2 × 4)] ÷ 6}
**Solution : (7 – 3) × [5 + (2 × 4)] ÷ 6 =4 × {[5 + 8] ÷ 6}

=4 × 13 ÷ 6

=4x 2.167

= 8.667

Thus (7 – 3) × [5 + (2 × 4)] ÷ 6 = 8.667

**(25)30 ÷6+10-2 x 5**

Solution : 30 ÷6+10-2 x 5 = 5+10-2 x 5

= 5+10-10

=15-10

=5

Hence 30 ÷6+10-2 x 5 = 5

**(26) 48÷16 x 2+17-9
Solution: **48÷16 x 2+17-9 = 48÷16 x 2+17-9

= 4 x 2+17-9

= 8+17-9

=25-9

=16

So, 48÷16 x 2+17-9 = 16

**(27) (18+10)-(3×6) **

Solution : (18+10)-(3×6) = (18+10)-18

=28-18

=10

Hence (18+10)-(3×6) = 10

**(28) (10 x 8)÷(20÷5)**

Solution : (10 x 8)÷(20÷5) = (10 x 8)÷4

= 80÷4

=20

Therefore (10 x 8)÷(20÷5) =4

**(29) {20-(24-10)}+7 **

Solution: {20-(24-10)}+7 = {20-14}+7

=6+7

=13

Hence {20-(24-10)}+7 = 13

**(30)12 x 3÷3 x 4-2+6**

Solution : 12 x 3÷3 x 4-2+6 = 12 x 1 x 4-2+6

=48-2+6

=54-2

=52

Hence 12 x 3÷3 x 4-2+6 =52

**(31) 32-[(12-6)+8]
**Solution: 32-[(12-6)+8] =32-[6+8]

=32-14

=18

Therefore 32-[(12-6)+8] = 18

**(32) 100−{5-24÷6+(8-5)}
**Solution : 100−{5-24÷6+(8-5)} =100−{5-24÷6+3}

=100−{5-4+3}

=100−{8-4}

=100-4

=96

Hence 100−{5-24÷6+(8-5)} =96

**(33) 17+{[26-(15-8)+(8-4) ]-9}**

Solution : 17+{[26-(15-8)+(8-4) ]-9} = 17+{[26-7+4 ]-9}

= 17+{23-9}

=17+14

=31

Hence 17+{[26-(15-8)+(8-4) ]-9} =31

**(34) (10-8) x 3 +6 +2 x (9+7) **

Solution: (10-8) x 3 +6+2 x (9+7) =2 x 3 +6+2 x 16

=6 +6+32

=44

Therefore (10-8) x 3 +6+2 x (9+7) = 44

**(35) **

Solution:

=

=

= 4(22-4)

=4 x 18

=72

Therefore

**(36) **

Solution:

=

=

=37 x 11

= 407

Therefore

**(37) **

Solution :

=

=

=18-2

=16

Hence

**= 16**

**(38) **

Solution:

=31 x 7

=217

Hence

**(39) 25-4×(7+5)÷4 +3**

Solution : 25-4×(7+5)÷4 +3 =25-4×12÷4 +3

=25-4×3 +3

=25-12 +3

= 28-12

= 16

So 25-4×(7+5)÷4 +3 = 16

**(40) 64-3(13+2 × 12÷ 8 -3 x 3) +11**

Solution : 64-3(13+2 × 12÷ 8 -3 x 3) +11 =

=

=

=

=

=75-21

=54

Hence 64-3(13+2 × 12÷ 8 -3 x 3) +11 =54

**(41) ((12+6) ×5)-3-4**

Solution :((12+6) ×5)-3-4=(18 ×5)-3-4

=90-3-4

= 83

Hence ((12+6) ×5)-3-4 =83

**(42) (7+(8÷4+5))-7**

Solution : (7+(8÷4+5))-7 =(7+(2+5))-7

=(7+7)-7

= 14-7

=7

**(43) ((18-6)+(15÷3))+6**

Solution : ((18-6)+(15÷3))+6 =(12+5)+6

=17+6

=23

**(44) 15+{18-[4+(16-5)]}**

Solution : 15+{18-[4+(16-5)]}=15+{18-[4+11]}

=15+{18-15}

=15+{18-15}

= 15+3

=18

**(45) 18-[18-{18-(18-18)-18}]
**Solution: 18-[18-{18-(18-18)-18}] =18-[18-{18-(18-18)-18}]

=18-[18-{18-0-18}]

=18-[18-0]

=18-18

= 0

**(46) 20-3-[7-{2+(4-3)}] **

Solution: 20-3-[7-{2+(4-3)}] =20-3-[7-{2+1}]

=20-3-[7-3]

=20-3-4

= 20-12

=8

So 20-3-[7-{2+(4-3)}] = 8

**(47)**

Solution :

=

=

=

=3-5.5

=-2.5

**(48) **

Solution :

= 31-5

=26

Hence = 26

**(49) 3+3 x 3-3+3**

Solution : 3+3 x 3-3+3= 3+9-3+3

=15-3

=12

Therefore 3+3 x 3-3+3= 12

**(50) **

Solution: = 4+((15+2)+6)

=4+ (17+6)

= 4+23

=27

So

**= 27
**

In conclusion, for solving BODMAS questions with all brackets , it is necessary the understanding of order of operations . The BODMAS acronym stands for Brackets, Orders (exponents), Division, Multiplication, Addition, and Subtraction, and it represents the correct order in which mathematical operations should be performed.

Students must begin by resolving the operations contained within the innermost brackets in BODMAS questions that contain all brackets, then proceed to the outer brackets, and finally, apply the remaining operations in the BODMAS order. This procedure makes sure that the solution is accurate.

students can enhance their mathematical abilities and have a better knowledge of how to solve complex mathematical expressions by practising BODMAS questions with all brackets.

### BODMAS questions for class 6 with all brackets PDF DOWNLOAD

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