May 21, 2024

BODMAS Simplification Questions for Class 5 Solved Examples & Practice PDF

BODMAS simplification questions for class 5

BODMAS simplification questions for class 5 contain mathematical expressions that may involve addition, subtraction, multiplication, and division of whole numbers or fractions, and may also include the use of brackets. To solve these types of problems student must know the rule of BODMAS .

BODMAS is an acronym that stands for the order of operations in math: Brackets, Of, Division, Multiplication, Addition, and Subtraction. It is a mathematical rule that helps to simplify arithmetic expressions and equations.  Understanding the BODMAS rule is a must for Class 5 students.

In some countries PEMDAS  “Parentheses, Exponents, Multiplication/Division, Addition/Subtraction,”  is used to solve  mathematical expressions. Despite the differences in their acronyms, BODMAS and PEMDAS have the same order of operations.

Here is the order of operations using both BODMAS and PEMDAS:
BODMAS: Brackets, Order, Division/Multiplication, Addition/Subtraction
PEMDAS: Parentheses, Exponents, Multiplication/Division, Addition/Subtraction

In practice, both BODMAS and PEMDAS work in the same way, and either acronym can be used to correctly solve arithmetic expressions.

Students can exactly determine the order of operations required to solve a problem and arrive at the correct answer by applying this rule. Here are some examples of BODMAS simplification questions for class 5 with answers:

BODMAS simplification questions for class 5 (Solved Examples) 

Example 1:  7 + 4 x 2 +3 -1
Solution
:  According to the BODMAS , we will first  perform the multiplication operation: 4 x 2 = 8
Then, perform the addition operation: 7 + 8 = 15
Therefore, 7 + 4 x 2 = 15

Example 2 :  20 ÷ 5 + 3 x 2
Solution:  First,  we will perform the division  20 ÷ 5 = 4
Then, perform the multiplication operation: 3 x 2 = 6
After that perform the addition  4 + 6 = 10
Therefore, 20 ÷ 5 + 3 x 2 = 10

Example 3:  (9 + 3) x 2 – 6
Solution:  (9 + 3) x 2 – 6 = 12 x 2 – 6        ( Performing the operation within the brackets  9 + 3 = 12 )
= 24-6                                                                (Perform the multiplication operation  12 x 2 = 24 )
=18                                                                     (Perform the subtraction  24 – 6 = 18  )
Therefore, (9 + 3) x 2 – 6 = 18

Example 4:  6 ÷ (2 + 1) x (7 – 4)
Solution: 
6 ÷ (2 + 1) x (7 – 4)  = 6 ÷ 3 x 3      (Performing the operation within the brackets) 
 = 2 x 3
= 6

Example 5:  3 x 4 + 7 – 5 ÷ 5
Solution:
3 x 4 + 7 – 5 ÷ 5 =3 x 4 + 7 – 1
=12 + 7 – 1
= 19-1
=18

Example 6: \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 ]
=\left [ 40 \div 10 \right ]
=4

Example 7 :  ( 11 + ( 24 – 6 ÷ 6 ) )
Solution:
( 11 + ( 24 – 6 ÷ 6 ) )  =( 11 + ( 24 – 1 ) )
= 11+23
= 34

Example 8:  12-[6÷3+{8÷2(8-6)}]
Solution : 
12-[6÷3+{8÷2(8-6)}] =12-[6÷3+{8÷2 x 2}]
=12-[6÷3+{4 x 2}]
=12-[6÷3+ 8]
=12-[2 + 8]
= 12- 10
= 2

Example 9:  10 \times 10+ [400\div \{100-(50-\overline{3\times 10})\}]
Solution: 10 \times 10+ [400\div \{100-(50-\overline{3\times 10})\}] = 10 \times 10+ [400\div \{100-(50-30)\}]
=10 \times 10+ [400\div \{100-20\}]
=10 \times 10+ [400\div 80]
= 10 \times 10+ 5
=100+ 5
= 105

Example 10 : 8 ÷ 4 × (6 + 2 of 4) + 32 – 2
Solution: 8 ÷ 4 × (6 + 2 of 4) + 32 – 2 =8 ÷ 4 × (6 + 2 x 4) + 32 – 2
=8 ÷ 4 × (6 + 8) + 32 – 2
=8 ÷ 4 × 14 + 32 – 2
=2 × 14 + 32 – 2
= 28  + 32 – 2
=62-2
= 60

Example 11 : 10+4\div 2-3\times 2+4\div 2 \times 2-4
Solution: 10+4\div 2-3\times 2+4\div 2 \times 2-4 =10+2-3\times 2+2 \times 2-4
=10+2-6+4-4
= 16-6-4
=10-4
=6

Example 12 : 6\div 6+6\times 6 -6
Solution: 6\div 6+6\times 6 -6 =1+6\times 6 -6
= 1+36-6
=37-6
=31

Example 13 :  24+[6-{5-2(4-3)}]
Solution:
24+[6-{5-2(4-3)}]=24+[6-{5-2x 1}]
=24+[6-{5-2}]
=24+[6-{5-2}]
=24+[6-3]
=24+3
=27

Example 14: \{0.50+0.15\div 0.05\} \times \frac{2}{7}
Solution : 
\{0.50+0.15\div 0.05\} \times \frac{2}{7} =\{0.50+3\} \times \frac{2}{7}
=3.50\times \frac{2}{7}
=\frac{7}{7}
= 1

Example15:  \large \frac{\frac{7}{3}\times \frac{2}{3}\div \frac{3}{5}}{2+1\frac{2}{3}}

Solution :  \large \frac{\frac{7}{3}\times \frac{2}{3}\div \frac{3}{5}}{2+1\frac{2}{3}} =\large \frac{\frac{7}{3}\times \frac{2}{3}\times \frac{5}{3}}{2+\frac{5}{3}}

=\large \frac{ \frac{70}{27}}{\frac{6+5}{3}}

=\large \frac{ \frac{70}{27}}{\frac{11}{3}}

=\frac{70}{27}\times \frac{3}{11}

=\frac{70}{99}

 

To practice more questions on BODMAS  simplification class 5, Download the PDF given below:

BODMAS Simplification Questions For Class 5 [PDF Download]

PDF Name BODMAS Simplification Questions For Class 5
Language  English
No. of Pages 6
PDF Size 253 KB
Category worksheets
Quality Excellent
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