# 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

----> Subtract

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. Division does not always come before multiplication. 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 above simplification, 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.

### Solved Problems

Problem 1 :

Evaluate :

6 + 7 x 8

Solution :

 Evaluation= 6 + 7 x 8= 6 + 56= 62 OperationMultiplyAddResult

Problem 2 :

Evaluate :

102 - 16 ÷ 8

Solution :

 Evaluation= 102 - 16 ÷ 8= 100 - 16 ÷ 8= 100 - 2= 98 OperationExponentDivideSubtractResult

Problem 3 :

Evaluate :

(25 + 11) x 2

Solution :

 Evaluation= (25 + 11) x 2= 36 x 2= 72 OperationParenthesesMultiplyResult

Problem 4 :

Evaluate :

3 + 6 x (5 + 4) ÷ 3 -7

Solution :

 Evaluation= 3 + 6 x (5 + 4) ÷ 3 -7= 3 + 6 x 9 ÷ 3 -7= 3 + 54 ÷ 3 -7= 3 + 18 -7= 21 - 7= 14 OperationParenthesesMultiplyDivideAddSubtractResult

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 OperationParenthesesDivideMultiplyAddSubtractMultiplySubtractAddResult

Problem 6 :

Evaluate :

6 + [(16 - 4) ÷ (22 + 2)] - 2

Solution :

 Evaluation= 6 + [(16 - 4) ÷ (22 + 2)] - 2= 6 + [12 ÷ (22 + 2)] - 2= 6 + [12 ÷ (4 + 2)] - 2= 6 + [12 ÷ 6] - 2= 6 + 2 - 2= 8 - 2= 6 OperationSquare BracketExponentParenthesesSquare BracketAddSubtractResult

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 OperationParenthesesMultiplyDivideAddResult

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 OperationParenthesesDivideSubtractSubtractResult

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 OperationParenthesesDivideMultiplyAddResult

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 OperationParenthesesMultiplyDivideSubtractResult

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 OperationBracket2Parentheses2Exponent2ParenthesesDivisionMultiplicationSubtractionBracketMultiplicationResult

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= 43 - ((-3)2 + 7) ÷ 4 + (4(-3) + 7)= 43 - ((-3)2 + 7) ÷ 4 + (4(-3) + 7)= 43 - (9 + 7) ÷ 4 + (4(-3) + 7)= 43 - 16 ÷ 4 + (-12 + 7)= 43 - 16 ÷ 4 - 5= 64 - 16 ÷ 4 - 5= 64 - 4 - 5= 60 - 5= 55 OperationParentheses2Exponent2Parentheses2Parentheses2Exponent2DivisionSubtractionSubtractionResult

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)2 + 3(2)2= 1 + 3(4)= 1 + 12= 13 ExponentMultiplicationAdditionResult

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 Parentheses1ExponentParenthesesDivisionAdditionResult

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 1Parentheses1Division1Subtraction1ExponentResult

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