Experts have designed these Class 7 Maths Notes and Chapter 2 Arithmetic Expressions Class 7 Notes for effective learning.
Class 7 Maths Chapter 2 Notes Arithmetic Expressions
Class 7 Maths Notes Chapter 2 – Class 7 Arithmetic Expressions Notes
Simple Expressions Class 7 Notes
Arithmetical phrases such as 15 + 10, 20 – 5, 12 × 15, 36 ÷ 3 are called arithmetic expressions. A simple arithmetic expression consists of two numbers connected by a symbol (+ or – or ×, or ÷).
For example, 15 + 10, 36 – 12, 10 × 8, 50 ÷ 5, etc.
Every arithmetical expression has a value. This value can be found by evaluating it. We use the equality sign ‘=’ to denote the relationship between an arithmetic expression and its value. 15 + 10 = 25. Hence, 25 is the value of the given simple expression.
Different simple expressions may have the same value.
12 + 10 and 11 × 2 have the same value, 22.
Example 1.
Fill in the blanks to make the expressions equal on both sides of the ‘=’ sign:
(a) 16 + 4 = ____________ + 5
(b) 20 + ____________ = 5 × 10
Solution:
(a) LHS = 16 + 4 = 20
RHS = LHS, so RHS = 20
Now, ____________ + 5 = 20
(A number increased by 5 is equal to 20)
The required number = 20 – 5 = 15
(b) RHS = 5 × 10 = 50
LHS = RHS, so LHS = 50
Now, 20 + ____ = 50
(20 increased by a number is equal to 50)
The required number = 50 – 20 = 30
Note: LHS = Left Hand Side, RHS = Right Hand Side
Comparing Expressions
Any two expressions can be compared. To compare them, we first need to evaluate them.
Example 2.
Compare
(a) 20 + 30 and 150 – 90
(b) 125 ÷ 5 and 10 × 2
Solution:
(a) 20 + 30 = 50; 150 – 90 = 60
50 < 60
So, 20 + 30 < 150 = 90 (b) 125 ÷ 5 = 25, 10 × 2 = 20 25 > 20
So, 125 ÷ 5 > 10 × 2
Note:
(i) = means both sides are equal.
(ii) > means the left side is greater than the right side.
(iii) < means the left side is less than the right side.
Reading and Evaluating Complex Expressions Class 7 Notes
Arithmetical expressions involving at least three numbers are called complex arithmetical expressions.
For example: 2 × 4 + 6, 8 × 10 – (36 ÷ 2), etc.
Expressions may or may not have brackets. If brackets are present, they are evaluated first.
Example 1.
Evaluate 80 – (36 ÷ 2).
Solution:
80 – (36 ÷ 2) = 80 – 18 = 62
However, brackets may not always be present.
For example: 2 × 4 + 6, 8 × 10 – 36 ÷ 2, etc.
To decide the order of two or more operations in mathematical expressions, we use the rule of DMAS.
DMAS (Divide – Multiply – Add – Subtract)
We first perform division, next multiplication, then addition, and finally subtraction.
If an operation is not present, we go to the next one.
We put brackets using the above sequence.
Example 2.
Simplify the following:
(a) 4 × 8 + 3
(b) 2 × 12 ÷ 3 – 6
(c) 72 × 9 ÷ 3 – 150
(d) 2 + 3 × 8 ÷ 4 – 5
Solution:
(a) 4 × 8 + 3
Divide is not present, so we multiply first = 32 + 3
Next, we add = 35
(b) 2 × 12 ÷ 3 – 6
We first divide = 2 × 4 – 6
Next, we multiply = 8 – 6
Finally, we subtract = 2
(c) 72 × 9 ÷ 3 – 150
We first divide = 72 × 3 – 150
Next, we multiply = 216 – 150
We have no ‘addition’, so we skip it.
Finally, we subtract = 66.
(d) 2 + 3 × 8 ÷ 4 – 5
Here, 2 + 3 × 8 ÷ 4 – 5
We first divide = 2 + 3 × 2 – 5
Now, we multiply = 2 + 6 – 5
Now we add 8 – 5
Finally, we subtract = 3
Removing Brackets and Tinkering with the Terms Class 7 Notes
When the brackets preceded by a positive sign are removed, the signs of the terms inside the brackets remain the same.
For example, 70 + (20 + 30) = 70 + 20 + 30; 100 + (20 – 10) = 100 + 20 – 10 etc.
When the brackets preceded by a negative sign are removed, the signs of the terms inside the brackets change.
For example, 70 – (20 + 30) = 70 – 20 – 30; 100 – (20 – 10) = 100 – 20 + 10.
Example 1.
Remove the brackets for each of the following:
(a) 200 + (56 + 44)
(b) 200 – (60 – 20)
(c) 521 – (231 + 107)
Solution:
(a) 200 + (56 + 44) = 200 + 56 + 44
(b) 200 – (60 – 20) = 200 – 60 + 20
(c) 521 – (231 + 107) = 521 – 231 – 107
Tinkering with the Terms
When one of the terms is increased or decreased, the value of the expression changes. This is called tinkering.
For example, 99 – 20 = 79.
If we add 1 to both 99 and 79, the expression becomes 100 – 20 = 80.
We have tinkered the terms of the expression. Such tinkering may help find the value of the expression.
More on Removing Brackets and Tinkering Terms Class 7 Notes
Consider the numbers 5, 6 and 4.
5 × (2 + 4) = 5 × 6 = 30 and 5 × 2 + 5 × 4 = 10 + 20 = 30.
This means that 5 × (2 + 4) = 5 × 2 + 5 × 4.
If a, b and c are any three whole numbers, then a × (b + c) = a × b + a × c.
We say multiplication is distributive over addition.
Also, 5 × (6 – 4) = 5 × 2 = 10 and 5 × 6 – 5 × 4 = 30 – 20 = 10.
This means that 5 × (6 – 4) = 5 × 6 – 5 × 4.
If a, b and c are any three whole numbers and b > c, then a × (b – c) = a × b – a × c.
We say multiplication is distributive over subtraction.
More on Tinkering
Tinkering can be done using the above property.
Example 1.
Given that 75 × 48 = 3600, find 75 × 50.
Solution:
75 × 50 = 75 × (48 + 2)
= 75 × 48 + 75 × 2
= 3600 + 150
= 3750
Example 2.
Find 132 × 98.
Solution:
132 × 98 = 132 × (100 – 2)
= 13200 – 264
= 12936
We multiply by 100 and subtract twice the number from the product.