PEMDAS RULE

PEMDAS is the rule that can be used to simplify or evaluate complicated numerical expressions with more than one binary operation.

Very simply way to remember PEMDAS rule!

----> Parentheses

----> Exponent/Radical 

M ----> Multiplication

----> Division

----> Addition

----> Subtraction

pemdas-rule.png

Important Notes :

1. In the simplification of a particular numerical expression, if both multiplication and division are there, do the operations one by one in the order from left to right.

2. Always multiplication can not be expected before division. Do one by one in the order from left to right.

3. In the simplification of a particular numerical expression, if both addition and subtraction are there, do the operations one by one in the order from left to right.

Examples :

18 ÷ 9 x 5 = 2 x 5 = 10

20 - 5 + 7 = 15 + 7 = 22 

In the simplification of the two numerical expressions above, we have both division and multiplication. From left to right, division comes first and multiplication comes next. So, do division first and multiplication next.

Video Lesson

Note :

Inside the parentheses, if there are two or more operations, follow PEMDAS Rule inside the parentheses.

Problem 1 :

Evaluate :

12 + 4 x 3.25

Solution :

Multiply ----> = 12 + 4 x 3.25

Add ----> = 12 + 13

Answer ----> = 25

Problem 2 :

Evaluate :

35 ÷ 7 + 22 

Solution :

Exponent ----> = 35 ÷ 7 + 22

Divide ----> = 35 ÷ 7 + 4

Add ----> = 5 + 4

Answer ----> = 9

Problem 3 :

Evaluate :

2.5 x (13 - 5)

Solution :

Parentheses ----> = 2.5 x (13 - 5)

Multiply ----> = 2.5 x 8

Answer ----> = 20

Problem 4 :

Evaluate :

45 ÷ (6 + 3) x 3 - 6 ÷ 2

Solution :

Parentheses ----> = 45 ÷ (6 + 3) x 3 - 6 ÷ 2

Divide ----> = 45 ÷ 9 x 3 - 6 ÷ 2

Multiply ----> = 5 x 3 - 6 ÷ 2

Divide ----> = 15 - 6 ÷ 2

Subtract ----> = 15 - 3

Answer ----> = 12

Problem 5 :

Evaluate :

48 - 3(16 + 15 ÷ 5 x 3 - 2 x 3) + 8

Solution :

Parentheses ----> = 48 - 3(16 + 15 ÷ 5 x 3 - 2 x 3) + 8

Divide ----> = 48 - 3(16 + 15 ÷ 5 x 3 - 2 x 3) + 8

Multiply ----> = 48 - 3(16 + 3 x 3 - 2 x 3) + 8

Multiply ----> = 48 - 3(16 + 9 - 2 x 3) + 8

Add ----> = 48 - 3(16 + 9 - 6) + 8

Parentheses ----> = 48 - 3(25 - 6) + 8

Multiply ----> = 48 - 3(19) + 8

Subtract ----> = 48 - 57 + 8

Subtract ----> = -9 + 8

Answer ----> = -1

Problem 6 :

Evaluate :

7 + [(25 - 1) ÷ (32 - 3)] - 3

Solution :

Brackets ----> = 7 + [(25 - 1) ÷ (32 - 3)] - 3

Parentheses ----> = 7 + [(25 - 1) ÷ (32 - 3)] - 3

Exponent ----> = 7 + [24 ÷ (32 - 3)] - 3

Brackets ----> = 7 + [24 ÷ 6] - 3

Add ----> = 7 + 4 - 3

Subtract ---> = 11 - 3

Answer ----> 8

Problem 7 :

Evaluate :

(63 ÷ 7) + 22 x (9 + 1) ÷ 11

Solution :

Parentheses = (63 ÷ 7) + 22 x (9 + 1) ÷ 11

Parentheses ----> = 9 + 22 x (9 + 1) ÷ 11

Multiply ----> = 9 + 22 x 10 ÷ 11

Divide ----> = 9 + 220 ÷ 11

Add ----> = 9 + 20

Answer ----> = 29

Problem 8 :

Evaluate :

(146 - 2) ÷ (3 x 4) - 15 + 9

Solution :

Parentheses ----> = (146 - 2) ÷ (3 x 4) - 15 + 9

Parentheses ----> = 144 ÷ (3 x 4) - 15 + 9

Divide ----> = 144 ÷ 12 - 15 + 9

Subtract ----> =  12 - 15 + 9

Subtract ----> = -3 + 9

Answer ----> = 6

Problem 9 :

Evaluate :

78 ÷ 13 + (24 - 12) x 5

Solution :

Parentheses ----> = 78 ÷ 13 + (24 - 12) x 5

Divide ----> = 78 ÷ 13 + 12 x 5

Multiply ----> = 6 + 12 x 5

Add ----> = 6 + 60

Answer ----> = 66

Problem 10 :

Evaluate :

(24 + 1) x 2 ÷ (13 - 3) - 19

Solution :

Parentheses ----> = (24 + 1) x 2 ÷ (13 - 3) - 19

Parentheses ----> = 25 x 2 ÷ (13 - 3) - 19

Multiply ----> = 25 x 2 ÷ 10 - 19

Divide ----> = 50 ÷ 10 - 19

Subtract ----> = 5 - 19

Answer ----> = -14

Problem 11 :

Evaluate :

[12 - 36 ÷ (42 - 4) ÷ 3 + 2] x 5

Solution :

Brackets ----> = [12 - 36 ÷ (42 - 4) ÷ 3 + 2] x 5

Parentheses ----> = [12 - 36 ÷ (42 - 4) ÷ 3 + 2] x 5

Exponent ----> = [12 - 36 ÷ (42 - 4) ÷ 3 + 2] x 5

Parentheses ----> = [12 - 36 ÷ (16 - 4) ÷ 3 + 2] x 5

Divide ----> = [12 - 36 ÷ 12 ÷ 3 + 2] x 5

Divide ----> = [12 - 3 ÷ 3 + 2] x 5

Subtract ----> = [12 - 1 + 2] x 5

Add ----> = [11 + 2] x 5

Multiply ----> = 13 x 5

Answer ----> = 65

Problem 12 :

Evaluate :

a- (b2 + c)2 ÷ a + (bc + a)

if a = 4, b = -3 and c = 7.

Solution :

a- (b2 + c)2 ÷ a3 + (bc + a)

Substitute a = 4, b = -3 and c = 7.

4- ([(-3)2 + 7)2 ÷ 43 + [(-4)7 +7]

Evaluation :

Brackets ----> = 4- [(-3)2 + 7]2 ÷ 43 + [(-4)7 + 7]

Exponent ----> = 4- [(-3)2 + 7]2 ÷ 43 + [(-4)7 + 7]

Brackets ----> = 4- [9 + 7]2 ÷ 43 + [(-4)7 + 7]3

Brackets ----> = 4- 162 ÷ 43 + [(-4)7 + 7]

Multiply ----> = 4- 162 ÷ 43 + [(-4)7 + 7]

Brackets ----> = 4- 162 ÷ 43 + [-28 + 7]

Exponent ----> = 4- 162 ÷ 43 + [-21]

Exponent ----> = 64 - 162 ÷ 43 + [-21]

Exponent ----> = 64 - 256 ÷ 43 + [-21]

Divide ----> = 64 - 256 ÷ 64 + [-21]

Multiply ----> = 64 - 4 + [-21]

Subtract ----> = 64 - 4 - 21

Subtract ----> = 60 - 21

Answer ----> = 39

Problem 13 :

Evaluate the following expression for x = -1 and y = 2 :

x+ 3y3

Solution :

= x+ 3y3

Substitute x = -1 and y = 2.

= (-1)+ 3(2)2

Evaluation :

Exponent ----> = (-1)+ 3(2)2

Exponent ----> = 1 + 3(2)2

Multiply ----> = 1 + 3(4)

Add ----> = 1 + 12

Answer ----> 13

Problem 14 :

Evaluate the following expression for x = 1 and y = 1.

(y3 + x) ÷ 2 + x

Solution :

= (y3 + x) ÷ 2 + x

Substitute x = 1 and y = 1.

= (13 + 1) ÷ 2 + 1

Evaluation :

Parentheses ----> = (13 + 1) ÷ 2 + 1

Exponent ----> = (13 + 1) ÷ 2 + 1

Parentheses ----> = (1 + 1) ÷ 2 + 1

Divide ----> = 2 ÷ 2 + 1

Add ----> = 1 + 1

Answer ----> = 2

Problem 15 :

Evaluate the following expression for y = 3 and z = 7.

z3 - (y ÷ 3 - 1)

Solution :

= z3 - (y ÷ 3 - 1)

Substitute y = 3 and z = 7.

= 73 - (3 ÷ 3 - 1)

Evaluation :

Parentheses ----> = 73 - (3 ÷ 3 - 1)

Divide ----> = 73 - (3 ÷ 3 - 1)

Parentheses = 73 - (1 - 1)

Exponent ----> = 73 - 0

Answer ----> = 243

Problem 16 :

Evaluate :

pemdasrule1.png

Solution :

pemdasrule2.png

Problem 17 :

What is the value of

pemdasrule3.png

if a = -1/2, b = 3/2 and c = 5/2?

Solution :

pemdasrule4.png

Problem 18 :

What is the value of

pemdasrule5.png

if p = 4, q = 1/2 and r = 2?

Solution :

pemdasrule6.png
pemdas-rule-1

Click here to get step by step answers for the above questions.

Kindly mail your feedback to [email protected]

We always appreciate your feedback.

©All rights reserved. onlinemath4all.com

Recent Articles

  1. Using Partial Sums to Find Convergence or Diveregence

    Dec 29, 24 02:21 AM

    Using Partial Sums to Find Convergence or Diveregence of an Infinite Series

    Read More

  2. Digital SAT Math Problems and Solutions (part - 92)

    Dec 27, 24 10:53 PM

    digitalsatmath80.png
    Digital SAT Math Problems and Solutions (part - 92)

    Read More

  3. SAT Math Resources (Videos, Concepts, Worksheets and More)

    Dec 27, 24 10:48 PM

    SAT Math Resources (Videos, Concepts, Worksheets and More)

    Read More