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 

M ----> Multiply

----> Divide

----> Add

----> Subtract

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 division can not be expected before multiplication. Do 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.

Solved Problems

Problem 1 :

Evaluate :

5 + 2 x 4

Solution :

Evaluation

= 5 + 2 x 4

= 5 + 8

= 13

Operation

Multiply

Add

Result

Problem 2 :

Evaluate :

92 - 15 ÷ 3

Solution :

Evaluation

= 92 - 15 ÷ 3

= 81 - 15 ÷ 3

= 81 - 5

= 76

Operation

Exponent

Divide

Subtract

Result

Problem 3 :

Evaluate :

(16 + 8) x 5

Solution :

Evaluation

(16 + 8) x 5

= 24 x 5

= 120

Operation

Parentheses

Multiply

Result

Problem 4 :

Evaluate :

5 + 3 x (6 + 2) ÷ 8 - 6

Solution :

Evaluation

= 5 + 3 x (6 + 2) ÷ 8 - 6

= 5 + 3 x 8 ÷ 8 - 6

= 5 + 24 ÷ 8 - 6

5 + 3 - 6

= 8 - 6

= 2

Operation

Parentheses

Multiply

Divide

Add

Subtract

Result

Problem 5 :

Evaluate :

56 - 2(20 + 12 ÷ 4 x 3 - 2 x 2) + 10

Solution :

Evaluation

= 56 - 2(20 + 12 ÷ 4 x 3 - 2 x 2) + 10

= 56 - 2(20 + 12 ÷ 4 x 3 - 2 x 2) + 10

= 56 - 2(20 + 3 x 3 - 2 x 2) + 10

= 56 - 2(20 + 9 - 4) + 10

= 56 - 2(29 - 4) + 10

= 56 - 2(25) + 10

= 56 - 50 + 10

= 6 + 10

= 16

Operation

Parentheses

Divide

Multiply

Add

Subtract

Multiply

Subtract

Add

Result

Problem 6 :

Evaluate :

6 + [(16 - 4) ÷ (22 + 2)] - 2

Solution :

Evaluation

= 6 + [(16 - 4) ÷ (2+ 2)] - 2

= 6 + [12 ÷ (2+ 2)] - 2

= 6 + [12 ÷ (4 + 2)] - 2

= 6 + [12 ÷ 6] - 2

= 6 + 2 - 2

= 8 - 2

= 6

Operation

Square Bracket

Exponent

Parentheses

Square Bracket

Add

Subtract

Result

Problem 7 :

Evaluate :

(96 ÷ 12) + 14 x (12 + 8) ÷ 2

Solution :

Evaluation

= (96 ÷ 12) + 14 x (12 + 8) ÷ 2

= 8 + 14 x 20 ÷ 2

= 8 + 280 ÷ 2

= 8 + 140

= 148

Operation

Parentheses

Multiply

Divide

Add

Result

Problem 8 :

Evaluate :

(93 + 15) ÷ (3 x 4) - 24 + 8

Solution :

Evaluation

= (93 + 15) ÷ (3 x 4) - 24 + 8

= 108 ÷ 12 - 24 + 8

9 - 24 + 8

= -15 + 8

= -7

Operation

Parentheses

Divide

Subtract

Subtract

Result

Problem 9 :

Evaluate :

55 ÷ 11 + (18 - 6) x 9

Solution :

Evaluation

= 55 ÷ 11 + (18 - 6)  x 9

= 55 ÷ 11 + 12 x 9

= 5 + 12 x 9

= 5 + 108

= 113

Operation

Parentheses

Divide

Multiply

Add

Result

Problem 10 :

Evaluate :

(7 + 18) x 3 ÷ (2 + 13) - 28

Solution :

Evaluation

= (7 + 18) x 3 ÷ (2 + 13) - 28

= 25 x 3 ÷ 15 - 28

= 75 ÷ 15 - 28

= 5 - 28

= -23

Operation

Parentheses

Multiply

Divide

Subtract

Result

Problem 11 :

Evaluate :

[11 - 20 ÷ (52 - 13) ÷ 3 + 8] x 2

Solution :

Evaluation

[11 - 20 + (52 - 13) ÷ 3 x 8] x 2

[11 - 20 + (52 - 13) ÷ 3 x 8] x 2

[11 - 20 + (52 - 13) ÷ 3 x 8] x 2.

[11 - 20 + (25 - 13) ÷ 3 x 8] x 2

[11 - 20 + 12 ÷ 3 x 8] x 2

[11 - 20 + 4 x 8] x 2

[11 - 20 + 32] x 2

[-9 + 32] x 2

= 23 x 2

= 46

Operation

Bracket2

Parentheses2

Exponent2

Parentheses

Division

Multiplication

Subtraction

Bracket

Multiplication

Result

Problem 12 :

Evaluate :

a- (b2 + c) ÷ a + (ab + c)

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

Solution :

a- (b2 + c) ÷ a + (ab + c)

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

4- ((-3)2 + 7) ÷ 4 + (4(-3) + 7)

Evaluation

= 4((-3)2 + 7) ÷ 4 + (4(-3) + 7)

= 4- ((-3)2 + 7) ÷ 4 + (4(-3) + 7)

= 4(9 + 7) ÷ 4 + (4(-3) + 7)

= 4- 16 ÷ 4 + (-12 + 7)

4- 16 ÷ 4 - 5

= 64 - 16 ÷ 4 - 5

64 - 4 - 5

60 - 5

= 55

Operation

Parentheses2

Exponent2

Parentheses2

Parentheses2

Exponent2

Division

Subtraction

Subtraction

Result

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

(-1)+ 3(2)2

= 1 3(4)

= 1 + 12

= 13

Exponent

Multiplication

Addition

Result

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

(13 + 1) ÷ 2 + 1

= (13 + 1) ÷ 2 + 1

(1 + 1) ÷ 2 + 1

2 ÷ 2 + 1

1 + 1

= 2

Parentheses

1Exponent

Parentheses

Division

Addition

Result

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)

= 73 - (3 ÷ 3 - 1)

= 73 - (3 ÷ 3 - 1)

= 73 - (1 - 1)

73 - 0

= 243

1Parentheses

1Division

1Subtraction

1Exponent

Result

Problem 16 :

Evaluate :

Solution :

Problem 17 :

What is the value of

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

Solution :

Problem 18 :

What is the value of

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

Solution :

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.

©All rights reserved. onlinemath4all.com

Recent Articles

  1. Definition of nth Root

    Sep 29, 22 04:11 AM

    Definition of nth Root - Concept - Examples

    Read More

  2. Worksheet on nth Roots

    Sep 29, 22 04:08 AM

    Worksheet on nth Roots

    Read More

  3. Inverse Property of Multiplication Worksheet

    Sep 29, 22 12:02 AM

    Inverse Property of Multiplication Worksheet

    Read More