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NCERT Solutions for Class 8 Maths Chapter 6 We Distribute, Yet Things Multiply - Exercise 6.3

Mastering the "Distribute" part of your chapter title is simple with our NCERT Solutions for Class 8 Maths Ch 6: We Distribute, Yet Things Multiply – Exercise 6.3. This exercise focuses on the Distributive Law, where a single term (monomial) outside a bracket is multiplied by every term inside the bracket.

Class 8 Math Lessons (Ch 6): Learn to Expand Algebraic Expressions with Ease! The solutions to Exercise 6.3 will be presented in a step-by-step approach to help students correctly "distribute" terms and simplify complex expressions. These solutions follow CBSE Guidelines to ensure you handle signs and variables accurately during expansion.

1.0Download NCERT Solutions for Class 8 Maths Chapter 6 We Distribute, Yet Things Multiply Ex 6.3 : Free PDF

Our easy-to-follow NCERT Solutions for Class 8 Maths Chapter 6 Exercise 6.3 are available for download as a PDF. Perfect for offline study and quick reference.

NCERT Solutions Class 8 Maths Chapter 6 Ex 6.3

Download PDF

2.0Detailed NCERT Class 8 Maths Chapter 6 Solutions of Exercise 6.1

1.Compute these products using the suggested identity. (i) $4 6^{2}$ using Identity 1 A for $(a + b)^{2}$ (ii) $397 \times 403$ using Identity 1C for $(a + b) (a - b)$ (iii) $9 1^{2}$ using Identity 1 B for $(a - b)^{2}$ (iv) $43 \times 45$ using Identity 1 C for $(a + b) (a - b)$

Sol. (i) $4 6^{2} = (40 + 6)^{2}$

$= 4 0^{2} + 2 \times 40 \times 6 + 6^{2} [∵ (a + b)^{2} = a^{2} + 2 ab + b^{2}] = 1600 + 480 + 36 = 2116$

(ii) $397 \times 403 = (400 - 3) (400 + 3)$

$[∵ (a + b) \times (a - b) = a^{2} - b^{2}]$

$= 40 0^{2} - 3^{2}$ $= 160000 - 9$ $= 159991$ (iii) $9 1^{2} = (100 - 9)^{2}$

$= 10 0^{2} - 2 \times 100 \times 9 + 9^{2} = 10000 - 1800 + 81 = 8281 [∵ (a - b)^{2} = a^{2} + b^{2} - 2 ab]$

(iv) $43 \times 45 = (44 - 1) (44 + 1)$

$[∵ a^{2} - b^{2} = (a + b) \times (a - b)]$

2.Use either a suitable identity or the distributive property to find each of the following products. (i) $(p - 1) (p + 11)$ (ii) $(3 a - 9 b) (3 a + 9 b)$ (iii) $- (2 y + 5) (3 y + 4)$ (iv) $(6 x + 5 y)^{2}$ (v) $(2 x - \frac{1}{2})^{2}$ (vi) $(7 p) \times (3 r) \times (p + 2)$

Sol. (i) $(p - 1) (p + 11) = p (p + 11) - 1 (p + 11) = p^{2} + 11 p - p - 11 = p^{2} + 10 p - 11$ (ii) $(3 a - 9 b) (3 a + 9 b)$

$= (3 a)^{2} - (9 b)^{2} = 9 a^{2} - 81 b^{2} [∵ (a + b) \times (a - b) = a^{2} - b^{2}]$

(iii) $- (2 y + 5) (3 y + 4) = (- 2 y - 5) (3 y + 4) = - 2 y (3 y + 4) - 5 (3 y + 4)$

$= - 6 y^{2} - 8 y - 15 y - 20 = - 6 y^{2} - 23 y - 20$

(iv) $(6 x + 5 y)^{2}$

$= (6 x)^{2} + 2 (6 x) (5 y) + (5 y)^{2} [∵ (a + b)^{2} = a^{2} + 2 ab + b^{2}] = 36 x^{2} + 60 x y + 25 y^{2}$

(v) $(2 x - \frac{1}{2})^{2} = (2 x)^{2} - 2 \times 2 x \times \frac{1}{2} + (\frac{1}{2})^{2} [∵ (a - b)^{2} = a^{2} + b^{2} - 2 ab]$

$= 4 x^{2} - 2 x + \frac{1}{4}$

(vi) $(7 p) \times (3 r) \times (p + 2) = 7 p \times 3 r \times (p + 2)$

3.For each statement identify the appropriate algebraic expression(s). (i) Two more than a square number.

$2 + s$
$(s + 2)^{2}$
$s^{2} + 2$
$s^{2} + 4$
$2 s^{2}$
$2^{2} s$ (ii) The sum of the squares of two consecutive numbers
$m^{2} + n^{2}$
$(m + n)^{2}$
$m^{2} + 1$
$m^{2} + (m + 1)^{2}$
$m^{2} + (m - 1)^{2}$
$(m + (m + 1))^{2}$
$(2 m)^{2} + (2 m + 1)^{2}$

Sol. (i) Two more than a square is $s^{2} + 2$ (ii) Sum of the squares of two consecutive numbers is $m^{2} + (m + 1)^{2}$ 4. Consider any 2 by 2 square of numbers in a calendar, as shown in the figure.

Find products of numbers lying along each diagonal $- 4 \times 12 = 48, 5 \times 11 = 55$ . Do this for the other 2 by 2 squares. What do you observe about the diagonal products? Explain why this happens.

Hint: Label the numbers in each 2 by 2 square as

$a$	$(a + 1)$
$a + 7$	$(a + 8)$

Sol. Case -I

6	7
13	14

Here, $6 \times 14 = 84$ $13 \times 7 = 91$ Difference $= 91 - 84 = 7$

Case II

9	10
16	17

Here, $9 \times 17 = 153$ $16 \times 10 = 160$ Difference $= 160 - 153 = 7$ We observe that the difference of the diagonal products in both cases is always 7. 5. Verify which of the following statements are true. (i) $(k + 1) (k + 2) - (k + 3)$ is always a multiple of 2 . (ii) $(2 q + 1) (2 q - 3)$ is a multiple of 4 . (iii) Squares of even numbers are multiples of 4 , and squares of odd numbers are 1 more than multiples of 8 . (iv) $(6 n + 2)^{2} - (4 n + 3)^{2}$ is 5 less than a square number.

Sol.: $(k + 1) (k + 2) - (k - 3)$ is always a multiple of 2 . $(k + 1) (k + 2)$ is product of 2 consecutive numbers. ∴ It is always a multiple of 2 . Case-I: If k is even, then ( $k + 3$ ) is an odd number. So, complete given calculation will be in the form of even-odd that will result into an odd number. Hence, ( $k + 1$ ) ( $k + 2$ ) - ( $k + 3$ ) will not be a multiple of 2 because it will give odd number. For $k =$ even number. Case-II: If k is odd, then ( $k + 3$ ) is an even number. So, complete given calculation will be in the form of even-even that will result into an even number.

Hence, $(k + 1) (k + 2) - (k + 3)$ will not be a multiple of 2 because it will give even number. For $k =$ odd number. Therefore, the given statement is not always true. 6. A number leaves a remainder of 3 when divided by 7 , and another number leaves a remainder of 5 when divided by 7 . What is the remainder when their sum, difference, and product are divided by 7 ?

Sol. Let the numbers be x and y . $x = 7 a + 3$ , $y = 7 b + 5$ Sum $= x + y = 7 a + 3 + 7 b + 5$ $= 7 (a + b) + 8 = 7 (a + b) + 7 + 1 = 7 (a + b + 1) + 1$ ∴ The remainder on division by 7 is 1 . Difference $= x - y$ $= (7 a + 3) - (7 b + 5)$ $= 7 a + 3 - 7 b - 5 = 7 (a - b) - 2$ $= 7 (a - b) - 7 + 5$ $(∵ - 2 = - 7 + 5)$ $= 7 (a - b - 1) + 5$ ∵ The remainder on division by 7 is 5 . Product = xy $= (7 a + 3) (7 b + 5)$ $= 49 ab + 35 a + 21 b + 15$ $= (49 ab + 35 a + 21 b + 14) + 1$ $= 7 (7 ab + 5 a + 3 b + 2) + 1$ ∴ The remainder on division by 7 is 1 . 7. Choose three consecutive numbers, square the middle one, and subtract the product of the other two. Repeat the same with other sets of numbers. What pattern do you notice? How do we write this as an algebraic equation? Expand both sides of the equation to check that it is a true identity.

Sol. Let us take the numbers 7, 8, 9 Now, $8^{2} - 7 \times 9 = 64 - 63 = 1$ Let us take the numbers $10, 11, 12$ Then $1 1^{2} - 10 \times 12 = 121 - 120 = 1$ Generalizing: Let the numbers be $a - 1, a, a + 1$ Then $a^{2} - (a + 1) (a - 1) = 1$ LHS $= a^{2} - (a + 1) (a - 1)$ $= a^{2} - (a^{2} - 1)$ $= a^{2} - a^{2} + 1$ $= 1$ $∵ {$ LHS $=$ RHS $}$

Hence, the identity is correct. 8. What is the algebraic expression describing the following steps: add any two numbers?

Multiply this by half of the sum of the two numbers? Prove that this result will be half of the square of the sum of the two numbers.

Sol. Let the two numbers be a and b . Step 1: $a + b$ Step 2: $(a + b) \times \frac{1}{2} (a + b)$ $∴ (a + b) \times \frac{1}{2} (a + b) = \frac{1}{2} (a + b)^{2}$ 9. Which is larger? Find out without fully computing the product. (i) $14 \times 26$ or $16 \times 24$ (ii) $25 \times 75$ or $26 \times 74$

Sol. (i) Let p $= 14 \times 26$

$p^{'} = 16 \times 24 = (14 + 2) (26 - 2) = 14 \times 26 + 2 \times 26 - 14 \times 2 - 2 \times 2 = 14 \times 26 + 2 (26 - 14 - 2) = 14 \times 26 + 2 \times 10 p^{'} = p + 2 \times 10 ∵ p^{'} > p or 16 \times 24 > 14 \times 26$

(ii) Let

10.A tiny park is coming up in Dhauli. The plan is shown in the figure. The two square plots, each of area $g^{2} sq$ . ft ., will have a green cover. All the remaining area is a walking path w ft . wide that needs to be tiled. Write an expression for the area that needs to be tiled.

Sol. Length $= w + g + 2 w + g + w = 4 w + 2 g$ Breadth $= w + g + w = 2 w + g$ Area of park $= (4 w + 2 g) (2 w + g)$ $= 8 w^{2} + 4 wg + 4 wg + 2 g^{2}$ $= 8 w^{2} + 8 wg + 2 g^{2}$ Area of path = Area of park - Area of green cover $= 8 w^{2} + 8 wg + 2 g^{2} - 2 g^{2}$ $= 8 w^{2} + 8 wg$ $∴ (8 w^{2} + 8 w g)$ sq. feet area needs to be tiled. 11. For each pattern shown below, (i) Draw the next figure in the sequence. (ii) How many basic units are there in Step 10? (iii) Write an expression to describe the number of basic units in Step y.

Sol. (a)

(i) Step 1:2 vertical strips of 3 units each +1 horizontal strip of 3 units

= 3

strips of 3 units each

= 9

units squares

= (1 + 2)^{2}

units squares Step 2: 4 strips of 4 units each

= 16

units squares

= (2 + 2)^{2}

units squares

Step 3: 5 strips of 5 units each = 25 units squares $= (3 + 2)^{2}$ units squares

Step 4: 6 strips of 6 units each $= 2$ are vertical and 4 are horizontal. (ii) Number of unit squares in step $10 = (10 + 2) 2 = 144$ (iii) Number of unit squares in step $y = (y + 2) 2$ (b) (i)

Step 3

$□$	$□$	$□$	$□$
$□$	$□$	$□$	$□$
$□$	$□$	$□$	$□$
$□$	$□$	$□$	$□$

□ □ □ Number of unit squares in step $1 = 5 = 2^{2} + 1$ Number of unit squares in step 2 $= 11 = 3^{2} + 2$ Number of unit squares in step $3 = 19 = 4^{2} + 3$ Number of unit squares in step $4 = 29 = 5^{2} + 4$

(ii) Step 1 has

(1 + 1)^{2} + 1

or 5 squares

Step 2 has $(2 + 1)^{2} + 2$ or 11 squares Step 3 has $(3 + 1)^{2} + 3$ or 19 squares Hence step 10 has $(10 + 1)^{2} + 10$ or 131 squares (iii) Step y has $[(y + 1)^{2} + y]$ squares

3.0 4.0Key Concepts of Chapter 6 We Distribute, Yet Things Multiply Exercise 6.3

The Distributive Law: For any three terms a, b, and c, the property is defined as: a(b + c) = ab + ac.
Multiplying a Monomial by a Binomial: Multiply the monomial with the first term of the binomial, then multiply it with the second term, and add the results.
Multiplying a Monomial by a Trinomial: Similar to a binomial, the monomial is distributed to all three terms inside the bracket.
Simplification of Expressions: Often, you will need to expand brackets first and then collect like terms to reach the final answer.

5.0NCERT Solutions for Class 8 Maths Chapter 6 We Distribute, Yet Things Multiply : All Exercises

NCERT Solutions for Class 8 Maths Chapter 6 We Distribute, Yet Things Multiply - Exercise 6.1

NCERT Solutions for Class 8 Maths Chapter 6 We Distribute, Yet Things Multiply - Exercise 6.2

NCERT Solutions for Class 8 Maths Chapter 6 We Distribute, Yet Things Multiply - Exercise 6.3

6.0Benefits of NCERT Solutions for Class 8 Maths Chapter 6 Exercise 6.3

1. Conceptual Clarity: Clearly shows how the monomial "visits" every term inside the parentheses.

2. Efficiency: Teaches students to simplify algebraic expressions before substituting numerical values to save time.

3. Sign Precision: Helps students avoid common mistakes when a negative monomial is distributed.

NCERT Class 8 Maths Ch. 6 We Distribute, Yet Things Multiply Other Exercises:-

Exercise 6.1

Exercise 6.2

Exercise 6.3

NCERT Solutions Class 8 Maths All Chapters:-

Chapter 1 - A Square and a Cube

Chapter 2 - Power play

Chapter 3 - A story of Numbers

Chapter 4 - Quadrilaterals

Chapter 5 - Number Play

Chapter 6 - We distribute, yet things multiply

Chapter 7 - Proportional Reasoning

Frequently Asked Questions

Join ALLEN!

(Session 2026 - 27)

Name

Mobile Number

Class

Choose class

Goal

Choose your goal

Preferred Programs

Preferred Mode

State

Choose State

I agree to

I authorise ALLEN Career Institute Pvt Ltd to send me regular updates via Phone calls, Whatsapp, SMS, Robocalls (Automated Calls), Emails, or on postal address.

NCERT Solutions for Class 8 Maths Chapter 6 We Distribute, Yet Things Multiply - Exercise 6.3

Mastering the "Distribute" part of your chapter title is simple with our NCERT Solutions for Class 8 Maths Ch 6: We Distribute, Yet Things Multiply – Exercise 6.3. This exercise focuses on the Distributive Law, where a single term (monomial) outside a bracket is multiplied by every term inside the bracket.

Class 8 Math Lessons (Ch 6): Learn to Expand Algebraic Expressions with Ease! The solutions to Exercise 6.3 will be presented in a step-by-step approach to help students correctly "distribute" terms and simplify complex expressions. These solutions follow CBSE Guidelines to ensure you handle signs and variables accurately during expansion.

1.0Download NCERT Solutions for Class 8 Maths Chapter 6 We Distribute, Yet Things Multiply Ex 6.3 : Free PDF

Our easy-to-follow NCERT Solutions for Class 8 Maths Chapter 6 Exercise 6.3 are available for download as a PDF. Perfect for offline study and quick reference.

NCERT Solutions Class 8 Maths Chapter 6 Ex 6.3

Download PDF

2.0Detailed NCERT Class 8 Maths Chapter 6 Solutions of Exercise 6.1

1.Compute these products using the suggested identity. (i) $4 6^{2}$ using Identity 1 A for $(a + b)^{2}$ (ii) $397 \times 403$ using Identity 1C for $(a + b) (a - b)$ (iii) $9 1^{2}$ using Identity 1 B for $(a - b)^{2}$ (iv) $43 \times 45$ using Identity 1 C for $(a + b) (a - b)$

Sol. (i) $4 6^{2} = (40 + 6)^{2}$

$= 4 0^{2} + 2 \times 40 \times 6 + 6^{2} [∵ (a + b)^{2} = a^{2} + 2 ab + b^{2}] = 1600 + 480 + 36 = 2116$

(ii) $397 \times 403 = (400 - 3) (400 + 3)$

$[∵ (a + b) \times (a - b) = a^{2} - b^{2}]$

$= 40 0^{2} - 3^{2}$ $= 160000 - 9$ $= 159991$ (iii) $9 1^{2} = (100 - 9)^{2}$

$= 10 0^{2} - 2 \times 100 \times 9 + 9^{2} = 10000 - 1800 + 81 = 8281 [∵ (a - b)^{2} = a^{2} + b^{2} - 2 ab]$

(iv) $43 \times 45 = (44 - 1) (44 + 1)$

$[∵ a^{2} - b^{2} = (a + b) \times (a - b)]$

2.Use either a suitable identity or the distributive property to find each of the following products. (i) $(p - 1) (p + 11)$ (ii) $(3 a - 9 b) (3 a + 9 b)$ (iii) $- (2 y + 5) (3 y + 4)$ (iv) $(6 x + 5 y)^{2}$ (v) $(2 x - \frac{1}{2})^{2}$ (vi) $(7 p) \times (3 r) \times (p + 2)$

Sol. (i) $(p - 1) (p + 11) = p (p + 11) - 1 (p + 11) = p^{2} + 11 p - p - 11 = p^{2} + 10 p - 11$ (ii) $(3 a - 9 b) (3 a + 9 b)$

$= (3 a)^{2} - (9 b)^{2} = 9 a^{2} - 81 b^{2} [∵ (a + b) \times (a - b) = a^{2} - b^{2}]$

(iii) $- (2 y + 5) (3 y + 4) = (- 2 y - 5) (3 y + 4) = - 2 y (3 y + 4) - 5 (3 y + 4)$

$= - 6 y^{2} - 8 y - 15 y - 20 = - 6 y^{2} - 23 y - 20$

(iv) $(6 x + 5 y)^{2}$

$= (6 x)^{2} + 2 (6 x) (5 y) + (5 y)^{2} [∵ (a + b)^{2} = a^{2} + 2 ab + b^{2}] = 36 x^{2} + 60 x y + 25 y^{2}$

(v) $(2 x - \frac{1}{2})^{2} = (2 x)^{2} - 2 \times 2 x \times \frac{1}{2} + (\frac{1}{2})^{2} [∵ (a - b)^{2} = a^{2} + b^{2} - 2 ab]$

$= 4 x^{2} - 2 x + \frac{1}{4}$

(vi) $(7 p) \times (3 r) \times (p + 2) = 7 p \times 3 r \times (p + 2)$

3.For each statement identify the appropriate algebraic expression(s). (i) Two more than a square number.

$2 + s$
$(s + 2)^{2}$
$s^{2} + 2$
$s^{2} + 4$
$2 s^{2}$
$2^{2} s$ (ii) The sum of the squares of two consecutive numbers
$m^{2} + n^{2}$
$(m + n)^{2}$
$m^{2} + 1$
$m^{2} + (m + 1)^{2}$
$m^{2} + (m - 1)^{2}$
$(m + (m + 1))^{2}$
$(2 m)^{2} + (2 m + 1)^{2}$

Sol. (i) Two more than a square is $s^{2} + 2$ (ii) Sum of the squares of two consecutive numbers is $m^{2} + (m + 1)^{2}$ 4. Consider any 2 by 2 square of numbers in a calendar, as shown in the figure.

Find products of numbers lying along each diagonal $- 4 \times 12 = 48, 5 \times 11 = 55$ . Do this for the other 2 by 2 squares. What do you observe about the diagonal products? Explain why this happens.

Hint: Label the numbers in each 2 by 2 square as

$a$	$(a + 1)$
$a + 7$	$(a + 8)$

Sol. Case -I

6	7
13	14

Here, $6 \times 14 = 84$ $13 \times 7 = 91$ Difference $= 91 - 84 = 7$

Case II

9	10
16	17