BODMAS PRACTICE QUESTIONS
The simple rules to remember the BODMAS is given below.
- First, simplify operations within Brackets. ( ), { }, [ ]
- Second, evaluate the exponential form. 22
- Third, do the division and multiplication one by one from left to right. ÷ and ×
- Finally, do the addition and subtraction one by one from left to right. + and -
Note :
- If an expression has more than one set of brackets, do the innermost bracket first.
- If there is a division sign between two fractions, multiply by the reciprocal of the divisor fraction.
Question 1 :
Simplify 25 - [20 - {10 - (7-5-3)}]
Solution :
|
= 25 - [20 - {10 - (7-5-3)}] = 25 - [20 - {10 + 1}] = 25 - [20 - 11] = 25 - 9 = 16 |
Inner Bracket ( ) Inner Bracket { } Bracket [ ] Subtraction |
Question 2 :
Find out the answer for 100 - 3[20 + {50 - 40}]
Solution :
|
= 100 - 3[20 + {50 - 40}] = 100 - 3[20 + 10] = 100 - 3 × 30 = 100 - 90 = 10 |
Inner Bracket { } Bracket [ ] Multiplication Subtraction |
Question 3 :
7 + (8 - 3 × 2)
Solution :
|
= 7 + (8 - 3 × 2) = 7 + (8 - 6) = 7 + 2 = 9 |
Bracket, Multiplication Bracket, Subtraction Addition |
Question 4 :
What would be the answer for 50 - [20 + {30 - (20 - 5)}]
Solution :
|
= 50 - [20 + {30 - (20 - 5)}] = 50 - [20 + {30 - 15}] = 50 - [20 + 15] = 50 - 35 = 15 |
Inner Bracket ( ) Inner Bracket { } Bracket [ ] Addition |
Question 5 :
Find the value of 150 - [10 + {3 - (20 - 5)}]
Solution :
|
= 150 - [10 + {3 - (20 - 5)}] = 150 - [10 + {3 - 15}] = 150 - [10 - 12] = 150 + 2 = 152 |
Inner Bracket ( ) Inner Bracket { } Bracket [ ] Addition |
Question 6 :
Simplify 1 ÷ 3/7 × (6 + 8 × 3 - 2) + [1/5 ÷ 7/25 - {3/7 + 8/14}]
Solution :
Step 1 :
Find the value of 1 ÷ 3/7 × (6 + 8 × 3 - 2).
|
= 1 ÷ 3/7 × (6 + 8 × 3 - 2) = 1 ÷ 3/7 × (6 + 24 - 2) = 1 ÷ 3/7 × (30 - 2) = 1 × 7/3 × 28 = 196/3 |
Bracket, Multiplication Bracket, Addition Bracket, Subtraction Change ÷ into ×
|
The value of 1 ÷ 3/7 × (6 + 8 × 3 - 1) = 196/3 ----(1)
Step 2 :
Find the value of [1/5 ÷ 7/25 - {3/7 + 8/14}]
= [1/5 ÷ 7/25 - {3/7 + 8/14}]
The LCM of 7, 14 is 14
= [1/5 ÷ 7/25 - {3/7 × 2/2 + 8/14}]
= [1/5 ÷ 7/25 - {6/14 + 8/14}
= [1/5 ÷ 7/25 - {14/14}]
= [1/5 ÷ 7/25 - 1]
= [1/5 × 25/7 - 1]
= [5/7 - 1]
= -2/7
The value of [1/5 ÷ 7/25 - {3/7 + 8/14}] = -2/7 ----(2)
Add (1) + (2), we get
= (196/3) + (-2/7)
= (196/3) - (2/7)
By using cross multiplication,
= (1372 - 6)/21
= (1366)/21
= 65.04
Question 7 :
Using the rule of BODMAS, determine the answer of
18 ÷ 10 - 4 + 32 ÷ (4 + 10 ÷ 2 - 1)
Solution :
|
= 18 ÷ 10 - 4 + 32 ÷ (4 + 10 ÷ 2 - 1) = 18 ÷ 10 - 4 + 32 ÷ (4 + 5 - 1) = 18 ÷ 10 - 4 + 32 ÷ (9 - 1) = 18 ÷ 10 - 4 + 32 ÷ 8 = 1.8 - 4 + 4 = 1.8 |
Bracket, Division Bracket, Addition ( ), Subtraction Division Addition |
Question 8 :
10 - [6 - {7 - (6 - 8 - 5)}]
Solution :
|
= 10 - [6 - {7 - (6 - 8 - 5)}] = 10 - [6 - {7 + 7}] = 10 - [6 - 14] = 10 + 8 18 |
Inner Bracket ( ) Inner Bracket { } Inner Bracket [ ] Addition |
Question 9 :
What will the answer of this question
5 × 1/4 ÷ 3/7 + [45/24 - 2/3 + 5/6 × 2/5]
Solution :
= 5 × 1/4 ÷ 3/7 + [45/24 - 2/3 + 5/6 × 2/5]
= 5 × 1/4 ÷ 3/7 + [45/24 - 2/3 + 1/3]
= 5 × 1/4 ÷ 3/7 + [45/24 - 1/3]
The LCM of 24, 3 is 24
= 5 × 1/4 ÷ 3/7 + [45/24 - 1/3 × 8/8]
= 5 × 1/4 ÷ 3/7 + [45/24 - 8/24]
= 5 × 1/4 ÷ 3/7 + 37/24
Changing ÷ into ×.
= 5 × 1/4 × 7/3 + 37/24
= 35/12 + 37/24
The LCM of 12, 24 is 24
= 35/12 × 2/2 + 37/24
= 70/24 + 37/24
= 107/24
= 4.46
Question 10 :
1800 ÷ 10 {(12 - 6) + (24 - 12)}
Solution :
|
= 1800 ÷ 10{(12 - 6) + (24 - 12)} = 1800 ÷ 10 {6 + 12} = 1800 ÷ 10 × 18 = 180 × 18 = 3240 |
Inner Brackets ( ) Inner Bracket { } Division Multiplication |
Recent Articles
-
Finding Range of Values Inequality Problems
May 21, 24 08:51 PM
Finding Range of Values Inequality Problems -
Solving Two Step Inequality Word Problems
May 21, 24 08:51 AM
Solving Two Step Inequality Word Problems -
Exponential Function Context and Data Modeling
May 20, 24 10:45 PM
Exponential Function Context and Data Modeling