In this section, you will learn PEMDAS rule that can be used to simplify or evaluate complicated numerical expressions with more than one binary operation easily.
Very simply way to remember PEMDAS rule :
P -----> Parentheses
E -----> Exponents
M -----> Multiplication
D -----> Division
A -----> Addition
S -----> Subtraction
Important Notes :
1. In a particular simplification, if you have both multiplication and division, do the operations one by one in the order from left to right.
2. Multiplication does not always come before division. We have to do one by one in the order from left to right.
3. In a particular simplification, if you have both addition and subtraction, do the operations one by one in the order from left to right.
Examples :
16 ÷ 4 x 3 = 4 x 3 = 12
18 - 3 + 6 = 15 + 6 = 21
In the first example above, we have both division and multiplication. From left to right, we have division first and multiplication next.
So we do division first and multiplication next.
Problem 1 :
Evaluate :
6 + 7 x 8
Solution :
Expression 6 + 7 x 8 |
Evaluation = 6 + 7 x 8 = 6 + 56 = 62 |
Operation Multiplication Addition Result |
Problem 2 :
Evaluate :
(25 + 11) x 2
Solution :
Expression (25 + 11) x 2 |
Evaluation = (25 + 11) x 2 = 36 x 2 = 72 |
Operation Parentheses Multiplication Result |
Problem 3 :
Evaluate :
10^{2} - 16 ÷ 8
Solution :
Expression 10^{2} - 16 ÷ 8 |
Evaluation = 10^{2} - 16 ÷ 8 = 100 - 16 ÷ 8 = 100 - 2 = 98 |
Operation Power Division Subtraction Result |
Problem 4 :
Evaluate :
3 + 6 x (5 + 4) ÷ 3 -7
Solution :
Expression 3 + 6 x (5+4) ÷ 3 -7 |
Evaluation = 3 + 6 x (5+4) ÷ 3 -7 = 3 + 6 x 9 ÷ 3 -7 = 3 + 54 ÷ 3 -7 = 3 + 18 -7 = 21 - 7 = 14 |
Operation Parentheses Multiplication Division Addition Subtraction Result |
Problem 5 :
Evaluate :
36 - 2(20 + 12 ÷ 4 x 3 - 2 x 2) + 10
Solution :
Problem 6 :
Evaluate :
6 + [(16 - 4) ÷ (2^{2} + 2)] - 2
Solution :
Expression 6+[(16-4)÷(2²+2)]-2 |
Evaluation = 6+[(16-4)÷(2^{2}+2)]-2 = 6+[12÷(2^{2}+2)]-2 = 6+[12÷(4+2)]-2 = 6+[12÷6]-2 = 6+2 - 2 = 8 - 2 = 6 |
Operation Parentheses Power Parentheses Parentheses Addition Subtraction Result |
Problem 7 :
Evaluate :
(96 ÷ 12) + 14 x (12 + 8) ÷ 2
Solution :
Expression (96÷12)+14x(12+8) ÷ 2 |
Evaluation =(96÷12)+14x(12+8) ÷ 2 = 8 + 14x20 ÷ 2 = 8 + 280 ÷ 2 = 8 + 140 = 148 |
Operation Parentheses Multiplication Division Addition Result |
Problem 8 :
Evaluate :
(93 + 15) ÷ (3 x 4) - 24 + 8
Solution :
Expression (93+15)÷(3x4)-24+8 |
Evaluation = (93+15)÷(3x4)-24+8 = 108 ÷ 12 - 24 + 8 = 9 - 24 + 8 = -15 + 8 = -7 |
Operation Parentheses Division Subtraction Subtraction Result |
Problem 9 :
Evaluate :
55 ÷ 11 + (18 - 6) x 9
Solution :
Expression 55÷11+(18-6)x9 |
Evaluation = 55÷11+(18-6)x9 = 55÷11 + 12x9 = 5 + 12x9 = 5 + 108 = 113 |
Operation Parentheses Division Multiplication Addition Result |
Problem 10 :
Evaluate :
(7 + 18) x 3 ÷ (2 + 13) - 28
Solution :
Expression (7+18)x3÷(2+13)- 28 |
Evaluation = (7+18)x3÷(2+13)-28 = 25 x 3 ÷ 15 - 28 = 75 ÷ 15 - 28 = 5 - 28 = -23 |
Operation Parentheses Multiplication Division Subtraction Result |
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