NCERT Solutions Class 7 Maths Chapter 2 Arithmetic Expressions Exercise 2.4

The NCERT Solutions Class 7 Maths Chapter 2: Arithmetic Expressions (Ganita Prakash), specifically Exercise 2.4, focuses on strengthening your understanding of the Distributive Property for integers and evaluating complex expressions through reasoning.

In this exercise, you will apply the Distributive Property for multiplication over both addition and subtraction, which is fundamental to all algebra that you will study in the future. And you will also start to think more deeply, not through calculation, about how to compare the value of two expressions based on specific alterations (such as adding brackets to change the order in which to evaluate the expressions).

1.0Download Class 7 Maths Chapter 2 Ex 2.4 NCERT Solutions PDF

Learn to master applying the Distributive Property and comparing the values of expressions with our comprehensive, step-by-step NCERT Solutions for Class 7 Maths Chapter 2 Exercise 2.4. To prepare for your exam or as a quick review for conceptual understanding, feel free to download the free PDF below.

2.0Key Concepts of Class 7 Exercise 2.4 Chapter 2 Arithmetic Expressions

• The Distributive Property- Completing blanks appropriately show that single multiplication over an addition or subtraction is the same as the addition or subtraction of the individual products:

a×(b+c)=(a×b)+(a×c)

a×(b-c)=(a×b)-(a×c)

• Comparing Logically- Determine if one expression is <, >, or = to another without calculating completely but by reasoning about the implications of regrouping (brackets) and changing operations. 
• Expression Creation- Given a set of numbers and operators, develop a variety of expressions that lead to the same goal, reinforcing the relationship between structure and value.
• Completing the Distributive Law: Fill-in-the-blanks style questions where you will be completing the Distributive Property on both addition and subtraction. 
• Expression Comparison: Questions where you will be comparing two expressions (often one has brackets and one does not) using the correct symbol (<,>,=). For example, comparing a+b×c vs. (a+b)×c. 
• Finding Equivalent Expressions: Questions where you will be asked to find alternative combinations of operations and groupings that equal a given target value.

3.0NCERT Solutions Class 7 Maths Chapter 2 Arithmetic Expressions : Other Exercises

4.0Detailed CBSE Class 7 Chapter 2 Exercise 2.4 Solutions

1. Fill in the blanks with numbers, and boxes by signs, so that the expressions on both sides are equal.

(a) 3×(6+7)=3×6+3×7

(b) (8+3)×4=8×4+3×4

(c) 3×(5+8)=3×5 □ 3× ____

(d) (9+2)×4=9×4 □ 2× ____

(e) 3×(+​+4)=3 ____ + ____

(f) (_+6)×4 =13×4+ ____

(g) 3× ____ + ____ =3×5+3×2

(h) (…+…)× ____ =2×4+3×4

(i) 5×(9−2)=5×9−5× ____

(j) (5−2)×7=5×7−2× ____

(k) 5×(8−3)=5×8 □ 5× ____

(l) (8−3)×7=8×7 □ 3×7

(m) 5×(12 ____ ) = ____ 5× ____

(n) (15− ____ ) ×7= ____ 6×7

(o) 5× ____ - ____ =5×9−5×4

(p) (__ ____ ) × ____ =17×7−9×7

Sol. (a) 3×(6+7)=3×6+3×7

(b) (8+3)×4=8×4+3×4

(c) 3×(5+8)=3×5+3×8

(d) (9+2)×4=9×4+2×4

(e) 3×(10+4)=30+12

(f) (13+6)×4=13×4+24

(g) 3×(5+2)=3×5+3×2

(h) (2+3)×4=2×4+3×4

(i) 5×(9−2)=5×9−5×2

(j) (5−2)×7=5×7−2×7

(k) 5×(8−3)=5×8−5×3

(l) (8−3)×7=8×7−3×7

(m) 5×(12−3)=60−5×3

(n) (15−6)×7=105−6×7

(o) 5×(9−4)=5×9−5×4

(p) (17−9)×7=17×7−9×7

2. In the boxes below, fill '<', '>' or '=' after analysing the expressions on the LHS and RHS. Use reasoning and understanding of terms and brackets to figure this out and not by evaluating the expressions.

(a) (8−3)×29 □ (3−8)×29

(b) 15+9×18 □ (15+9)×18

(c) 23×(17−9) □ 23×17+23×9

(d) (34−28)×42 □ 34×42−28×42

Sol. (a) (8−3)×29>(3−8)×29

Because, (3−8)×29=−(8−3)×29

⇒(8−3)×29>(3−8)×29

(b) 15+9×18<(15+9)×18

Because, (15+9)×18=15×18+9× 18 and 15×18>15

So, 15+9×18<(15+9)×18

(c) 23×(17−9)<23×17+23×9

Because, 23×(17−9)=23×17−23×9

Clearly, 23×17>23×17−23×9

⇒23×(17−9)<23×17+23×9

(d) (34−28)×42=34×42−28×42

3. Here is one way to make 14 :

2×(1+6)=14.

Are there other ways of getting 14? Fill them out below:

(a) ____ × ____ + ____ ) = 14

(b) ____ × ____ + ____ ) = 14

(c) ____ ×( ____ + ____ ) = 14

(d) ____ × ____ + ____ ) = 14

Sol. (a) 2×(5+2)=14

(b) 2×(3+4)=14

(c) 2×(4+3)=14

(d) 2×(6+1)=14

4. Find out the sum of the numbers given in each picture below in at least two different ways. Describe how you solved it through expressions.                            

(II)

Sol. For I: 5×4+4×8=20+32=52

or 4×(4+8)+4=4×12+4=52

For II: 8×(5+6)=8×11=88

or 8×5+8×6=40+48=88

5.0Key Features and Benefits: Class 7 Maths Chapter 2 Exercise 2.4

• Algebraic Preparation: This task is a transition from basic arithmetic to the Algebraic Expressions chapter since the emphasis is on structure and relationships.
• Precludes Order-of-Operations Errors: Practicing comparisons allows students to notice how parentheses/brackets and the Distributive Property change the value of an expression.
• Quick Calculation: The Distributive Property provides a readily accessible way to mentally calculate values such as 5×98=5×(100−2)=500−10=490.