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We will discuss here about the simplification of numerical
expressions. We know how to perform the four fundamental operations, that is
namely, addition, subtraction, multiplication and division involving whole
numbers, fractional numbers and decimals. We perform only one operation at a
time. Now we will learn how to perform two or more operations together.
Let us simplify the expression 96 ÷ 3 + 4 × 8 – 2.
We can solve it in many ways as:
Case I: 96 ÷ 3 + 4 × 8 – 2 = 32 + 4 × 6 = 32 + 24 = 56
Case II: 96 ÷ 3 + 4 × 8 – 2 = 32 + 32 – 2 = 32 + 30 = 62
Case III: 96 ÷ 3 + 4 × 8 – 2 = 32 + 4 × 6 = 36 × 6 = 216
In all the cases, the answers are different. Can you say which answer is wrong and which answer be called correct. Mathematics avoids such ambiguity.
Hence, to avoid such confusions, it is very important to follow some rules. The following rules are used to evaluate numerical expressions involving a combination of the fundamental operations +, -, × and ÷.
Rule I: First do division and multiplication in order from left to right.
Rule II: Then do addition and subtraction in order from left to right.
In short, it can be remembered as DMAS.
D → stands for division, M → for multiplication, A → for addition and S →for subtraction.
Some expressions involve the word ‘of’; for example, 6 of 8, one-fourth of 7, etc.
The word ‘of’ indicates the multiplication.
Thus, 6 of 8 means 6 × 8, one-fourth of 7 means \(\frac{1}{4}\) × 7, etc.
To simplify an expression involving ‘of’, first of all, i.e., before division and multiplication, we carried out the ‘of’ operation.
Solved Examples on Simplification of Numerical Expressions
Simplification of Numerical Operations
1. Simplify: 16 + 35 ÷ 7 × 6 – 1
Solution:
16 + 35 ÷ 7 × 6 – 1 = 16 + 5 × 6 – 1, [Performing “÷”]
= 16 + 30 – 1 [Performing “×”]
= 46 – 1 [Performing “+”]
= 45 [Performing “-“]
2. Simplify: 90 ÷ 3 of 6 × 8 – 5
Solution:
90 ÷ 3 of 6 × 8 – 5 = 90 ÷ 18 × 8 – 5, [Performing “of”]
= 5 × 8 – 5 [Performing “÷”]
= 40 – 5 [Performing “×”]
= 35 [Performing “-“]
3. Simplify the numerical Expression:
(i) 3 of 25 ÷ 5 + 56
(ii) 20 ÷ (2 of 3 + 8 – 4) + 14
Solution:
(i) 3 of 25 ÷ 5 + 56
= 3 × 25 ÷ 5 + 56
= 3 × 5 + 56
= 15 + 56
= 71.
(ii) 20 ÷ (2 of 3 + 8 – 4) + 14
= 20 ÷ (2 × 3 + 8 – 4) + 14 [Performing “of”]
= 20 ÷ (6 + 8 – 4) + 14 [Performing “×”]
= 20 ÷ (14 – 4) + 14 [Performing “+”]
= 20 ÷ 10 + 14 [Performing “-“]
= 2 + 14 [Performing “÷”]
= 16. [Performing “+”]
Simplification of Brackets:
To simplify a numerical expression having two or more operations, we perform operation like: Division first, followed by Multiplication, Addition and then Subtraction. A standard result called BODMAS is applied for simplification of these operations.
The word BODMAS stands for:
B → Brackets
O → of means (Multiplication ×)
D → Division
M → Multiplication
A → Addition
S → Subtraction
Some numerical expressions, involve grouping of symbols, are called brackets. Brackets tell us which operation is done first.
If the brackets are present in the problem, we first
simplify the brackets. There are four kind of brackets.
1. ( )
→ simple brackets or round brackets or or small brackets parenthesis.
2. { }
→ Braces or Curly brackets.
3. [ ]
→ Square brackets or big brackets.
4. ______
→ This is a line called bar, vinculum. If two or more types of brackets are
involved in the problem, then they are removed in this order ‘_________’, ( ),
{ }, [ ].
Solved Examples to Simplify the Numerical Expressions:
1. Simplify the following: (Simplification of Brackets)
(i) [12 + {7 – (8 ÷ 2)}] × 3
= [12 + {7 – 4}] × 3 (Round brackets removed)
= [12 + 3] × 3 (Curly brackets removed)
= 15 × 3 (Square brackets removed)
= 45
(ii) 14 + [22 – {8 + (6 ÷ 2)}]
= 14 + [22 – {8 + 3}] (Round brackets removed)
= 14 + [22 – 11] (Curly brackets removed)
= 14 + 11 (Square brackets removed)
= 25.
5th Grade Numbers
5th Grade Math Problems
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