July 16, 2024
BODMAS Questions for Class 6 with all brackets

BODMAS Questions for Class 6 with all brackets| BODMAS Practice Questions PDF

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) \left [ \left \{ (30-\overline{9-6})\div 3 \right \}\times 6+6 \right ]


Solution : \left [ \left \{ (30-\overline{9-6})\div 3 \right \}\times 6+6 \right ]  =\left [ \left \{ (30-3)\div 3 \right \}\times 6+6 \right ]
=\left [ \left \{ 27\div 3 \right \}\times 6+6 \right ]
= \left [ 9\times 6+6 \right ]
=54+6
=60

(19)  12-\left [ 6\div 3+\left \{8\div 2(8-6)) \right \} \right ]
Solution : 12-\left [ 6\div 3+\left \{8\div 2(8-6)) \right \} \right ] = 12-\left [ 6\div 3+\left \{8\div 2\times 2) \right \} \right ]
=12-\left [ 6\div 3+\left \{4\times 2) \right \} \right ]
12-\left [ 6\div 3+8 \right ]
= 12-\left [ 2+8 \right ]
=12-10
=2
Therefore 12-\left [ 6\div 3+\left \{8\div 2(8-6)) \right \} \right ] =2

(20) \left [ 40\div \left \{ 19-3(6-\overline{4-1}) \right \} \right ]
Solution : \left [ 40\div \left \{ 19-3(6-\overline{4-1}) \right \} \right ]    = \left [ 40\div \left \{ 19-3(6-3) \right \} \right ]
=\left [ 40\div \left \{ 19-3\times 3 \right \} \right ]
= \left [ 40\div \left \{ 19-9\right \} \right ]
=40\div 10
=4
Therefore \left [ 40\div \left \{ 19-3(6-\overline{4-1}) \right \} \right ]    =4

(21)  \left [ 105\div \left \{ 23+2(9-\overline{5-2}) \right \} \right ]
Solution : \left [ 105\div \left \{ 23+2(9-\overline{5-2}) \right \} \right ]  = \left [ 105\div \left \{ 23+2(9-3) \right \} \right ]
=\left [ 105\div \left \{ 23+2\times 6 \right \} \right ]
=\left [ 105\div \left \{ 23+12 \right \} \right ]
=105\div 35
= 3
Therefore \left [ 105\div \left \{ 23+2(9-\overline{5-2}) \right \} \right ]  =3

(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) 4(10+15\div 5 \times 4 -2\times 2)
Solution: 4(10+15\div 5 \times 4 -2\times 2)4(10+3 \times 4 -2\times 2)
=4(10+12 -2\times 2)
=4(10+12 -4)
= 4(22-4)
=4 x 18
=72
Therefore  4(10+15\div 5 \times 4 -2\times 2) =  72

(36) 37[-5+\left \{ 28-(19-7)) \right \}]
Solution:  37[-5+\left \{ 28-(19-7)) \right \}]= 37[-5+\left \{ 28-(19-7)) \right \}]
=37[-5+\left \{ 28-12 \right \}]
=37[-5+16]
=37 x 11
= 407
Therefore  37[-5+\left \{ 28-(19-7)) \right \}]=407

(37) 18-[8\div 2\left \{ 8\div \overline{4\times 4} \right \}]
Solution :18-[8\div 2\left \{ 8\div \overline{4\times 4} \right \}]   = 18-[8\div 2\left \{ 8\div 16 \right \}]
=18-[8\div 2\times \frac{1}{2}]
= 18-[4\times \frac{1}{2}]
=18-2
=16

Hence 18-[8\div 2\left \{ 8\div \overline{4\times 4} \right \}]= 16

(38)  ((18+4)+(18\div 2))\times 7
Solution: ((18+4)+(18\div 2))\times 7 =(22+9)\times 7
=31 x 7
=217
Hence ((18+4)+(18\div 2))\times 7=217

(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 =64-3(13+2 \times \frac{12}{8} -3 \times 3) +11
=64-3(13+3 -3 \times 3) +11
=64-3(13+3 -9) +11
=64-3 \times 7 +11
=64-21 +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)3-\left \{ 10-[6+7\div (16-2))]+2 \right \}
Solution : 3-\left \{ 10-[6+7\div (16-2))]+2 \right \}=3-\left \{ 10-[6+7\div 14]+2 \right \}
=3-\left \{ 10-[6+0.5]+2 \right \}
=3-\left \{ 10-6.5+2 \right \}
=3-\left \{ 12-6.5 \right \}
=3-5.5
=-2.5

(48)  5+5\div 5+5\times 5-5
Solution : 5+5\div 5+5\times 5-5=5+1+5\times 5-5
=5+1+25-5
= 31-5
=26
Hence  5+5\div 5+5\times 5-5 = 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)    (16\div 4)+((15+2)+6)
Solution: (16\div 4)+((15+2)+6) = 4+((15+2)+6)
=4+ (17+6)
= 4+23
=27
So (16\div 4)+((15+2)+6)= 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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