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

Mastering the expansion and multiplication of algebraic terms becomes easier with our NCERT Solutions for Class 8 Maths Chapter 6: We Distribute, Yet Things Multiply. This chapter helps students understand the basics of algebraic multiplication, including combining numerical coefficients and applying the laws of exponents to variables.

The step-by-step solutions are designed to help students solve problems related to algebraic expressions with clarity and confidence. These NCERT Solutions follow CBSE guidelines and provide accurate explanations to strengthen students’ understanding of fundamental algebraic operations and their applications in real-life mathematical problems.

1.0Download NCERT Solutions for Class 8 Maths Chapter 6 We Distribute, Yet Things Multiply Ex 6.1: Free PDF

Our easy-to-follow NCERT Solutions for Class 8 Maths Chapter 6 Exercise 6.2 are available for download as a PDF. Perfect for offline study and quick reference.

NCERT Solutions Class 8 Maths Chapter 6 Ex 6.1

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

1.Observe the multiplication grid below. Each number inside the grid is formed by multiplying two numbers. If the middle number of a $3 \times 3$ frame is given by the expression pq, as shown in the figure, write the expressions for the other numbers in the grid.

x	1	2	3	4	5	6	7	8	9	10
1	1	2	3	4	5	6	7	8	9	10	$3 \times 5$	$3 \times 6$	$3 \times 7$
2	2	4	6	8	10	12	14	16	-18	20	$4 \times 5$	$4 \times 6$	$4 \times 7$
3	3	6	9	12	15	18	21	24	27	30
4	4	8	12	16	20	24	28	32	36	40	$5 \times 5$	$5 \times 6$	$5 \times 7$
5	5	10	15	20	25	30	35	40	45	-50-
6	6	12	18	24	30	36	42	48	54	60
7	7	14	21	28	35	42	49	56	63	70
8	8	16	24	32	40	48	56	64	72	80		pq
9	9	18	27	36	45	54	63	72	81	90
10	10	20	30	40	50	60	70	80	90	100

Sol.

2. Expand the following products. (i)

(3 + u) (v - 3)

(ii)

\frac{2}{3} (15 + 6 a)

(iii)

(10 a + b) (10 c + d)

(iv)

(3 - x) (x - 6)

(v)

(- 5 a + b) (c + d)

(vi)

(5 + z) (y + 9)

Sol. (i) we have, $(3 + u) (v - 3)$

$= 3 (v - 3) + u (v - 3) = 3 v - 9 + uv - 3 u = 3 v - 3 u + uv - 9$

(ii) Here, $\frac{2}{3} (15 + 6 a)$

$= \frac{2}{3} \times 15 + \frac{2}{3} \times 6 a = 10 + 4 a$

(iii) Here, $(10 a + b) (10 c + d)$

$= 10 a \times 10 c + 10 a \times d + b \times 10 c + b \times d = 100 a c + 10 a d + 10 b c + b d$

(iv) Here, $(3 - x) (x - 6)$

$= 3 (x - 6) - x (x - 6) = 3 x - 18 - x^{2} + 6 x = - x^{2} + 9 x - 18$

(v) we have, $(- 5 a + b) (c + d)$

$= - 5 a (c + d) + b (c + d) = - 5 a c - 5 a d + b c + b d$

(vi) we have, $(5 + z) (y + 9)$

3.Find 3 examples where the product of two numbers remains unchanged when one of them is increased by 2 and the other is decreased by 4 .

Sol. Let the number be a and b Then, $ab = (a + 2) (b - 4)$ $\Rightarrow ab = ab - 4 a + 2 b - 8$ $\Rightarrow ab - ab + 4 a + 8 = 2 b$ $\Rightarrow 4 a + 8 = 2 b$ (divide throughout by 2) $\Rightarrow 2 a + 4 = b$ $\Rightarrow b = 2 a + 4$ For $a = 1, b = 2 \times 1 + 4 = 6$ $ab = 1 \times 6 = 6$ and $(a + 2) (b - 4) = 3 \times 2 = 6$ Hence, $ab = (a + 2) (b - 4)$ Let $a = 2$ , then $b = 2 \times 2 + 4 = 8$ Let $a = 3$ , then $b = 2 \times 3 + 4 = 10$ Three such pairs are 1 and 6; 2 and 8; 3 and 10 .

4.Expand (i) $(a + ab - 3 b^{2}) (4 + b)$ , and (ii) $(4 y + 7) (y + 11 z - 3)$ .

Sol. (i) Here, $(a + ab - 3 b^{2}) (4 + b)$

$= (4 + b) (a + ab - 3 b^{2}) = 4 (a + ab - 3 b^{2}) + b (a + ab - 3 b^{2}) = 4 a + 4 ab - 12 b^{2} + ab + a b^{2} - 3 b^{3} = 4 a + 5 ab - 12 b^{2} + a b^{2} - 3 b^{3}$

(ii) Here, $(4 y + 7) (y + 11 z - 3)$

5.Expand (i) $(a - b) (a + b)$ , (ii) $(a - b) (a^{2} + ab + b^{2})$ and (iii) $(a - b) (a^{3} + a^{2} b + a b^{2} + b^{3})$ , Do you see a pattern? What would be the next identity in the pattern that you see? Can you check it by expanding?

Sol. (i) Here, $(a - b) (a + b)$

$= a (a + b) - b (a + b) = a^{2} + ab - ab - b^{2} = a^{2} - b^{2}$

(ii) Here, $(a - b) (a^{2} + ab + b^{2})$

$= a (a^{2} + ab + b^{2}) - b (a^{2} + ab + b^{2}) = a^{3} + a^{2} b + a b^{2} - a^{2} b - a b^{2} - b^{3} = a^{3} - b^{3}$

(iii) we have, $(a - b) (a^{3} + a^{2} b + a b^{2} + b^{3})$

$= a (a^{3} + a^{2} b + a b^{2} + b^{3}) - b (a^{3} + a^{2} b + a b^{2} + b^{3}) = a^{4} + a^{3} b + a^{2} b^{2} + a b^{3} - a^{3} b - a^{2} b^{2} - a b^{3} - b^{4} = a^{4} - b^{4}$

We observe the following pattern $(a - b) (a^{n} + a^{n - 1} b + \dots .. + b^{n}) = a^{n + 1} - b^{n + 1}$ Next identity in the pattern would be $(a - b) (a^{4} + a^{3} b + a^{2} b^{2} + a b^{3} + b^{4}) = a^{5} - b^{5}$

Multiple choice questions

1.If one factor of a product is increased by 1, the increase in the product is always equal to: (1) The other factor (2) The square of the other factor (3) Half of the other factor (4) Double the other factor

2. The expression $(x + 1) (y + 2) - x (y + 2)$ simplifies to: (1) $x + 1$ (2) $y + 2$ (3) $2 x + 1$ (4) 1

3. A shopkeeper sells 120 pens at ₹15 each. If the price of each pen increases by ₹1, then by how much will his total revenue increase? (1) ₹120 (2) ₹15 (3) ₹121 (4) ₹135

4. If $a = 2 x$ and $b = 3 x$ the increase in product of $a & b$ when both are increased by 1 is: (1) $5 x$ (2) $5 x + 1$ (3) $6 x + 1$ (4) $5 x + 2$

5. Two consecutive integers are multiplied. If both are increased by 1 , the increase in product is: (1) Equal to the larger integer (2) Equal to the smaller integer (3) Equal to the sum of both integers +1 (4) Always 2

6. If the sides of a rectangle are 18 cm and 24 cm , then when both sides are increased by 1 cm , the area increases by: (1) $42 cm^{2}$ (2) $43 cm^{2}$ (3) $44 cm^{2}$ (4) $45 cm^{2}$

7. A laptop costs ₹4,850 and a printer cost $₹1, 275$ . If the laptop price increases by ₹50 and the printer by ₹25, by how much does the product (laptop price × printer price) increase? (1) ₹172,500 (2) ₹180,250 (3) ₹186,250 (4) ₹192,000

8. Two numbers are 128 and 175. If the first is increased by 12 and the second by 15 , what is the increase in their product? (1) 3,880 (2) 4,000 (3) 4,200 (4) 4,300

9. Which of the following is the correct expansion of $(p - q) (p + q) (p + r)$ ? (1) $p^{3} + p r p - p q^{2} - q^{2} r$ (2) $p^{3} + p r^{2} - p q^{2} + p^{2} r - q^{2} r$ (3) $p^{3} - p q^{2} + p^{2} r - q^{2} r$ (4) $p^{3} + p^{2} r - p q^{2} + q^{2} r$

10. If: $(p + q + r + s) (p^{2} + pq + q^{2})$

Contains exactly 9 terms, how many distinct are in the simplified expression? (1) 9 (2) 8 (3) 7 (4) 10

11. If $(a + b)^{2} + (a - b)^{2} = 2 x$ and $(a + b)^{2} - (a - b)^{2} = 4 y$ , then what is the value of $x - y$ ? (1) ab (2) $a^{2} + b^{2} - 2 ab$ (3) $a^{2} + b^{2} - ab$ (4) $a^{2} + b^{2}$

12. If $2 (a^{2} + b^{2}) = 50$ and $(a + b)^{2} = 36$ Find the value of $(a - b)^{2}$ (1) 14 (2) 18 (3) 25 (4) 16

13. Let $x = 12345$ Evaluate $(2 x + 1) (2 x - 1)$ (1) 609,596,099 (2) 607,399,024 (3) $609, 596, 101$ (4) $608, 000, 000$

14. Compute $(0.999 + 0.001) (0.999 - 0.001)$ (1) 0.998001 (2) 0.998 (3) 0.997999 (4) 0.999

15. Given $(a + b)^{2} = 2025$ and $a - b = 5$ Find the ordered pair(s) (a, b). (1) $(25, 20)$ only (2) $(- 25, - 20)$ only (3) $(25, 20)$ ог $(- 20, - 25)$ (4) $(25, 20)$ or $(- 20, - 25)$ and $(- 25, - 20)$

16. If $(a + b)^{2} = 100$ and $(a - b)^{2} = 36$ find $ab$ and $a^{2} + b^{2}$ . (1) $ab = 16, a^{2} + b^{2} = 68$ (2) $ab = 8, a^{2} + b^{2} = 84$ (3) $ab = 16, a^{2} + b^{2} = 84$ (4) $ab = 32, a^{2} + b^{2} = 36$

17. The pattern is:

Step 1: 1 Step 2: 4 Step 3: 9 Step 4: 16 Step 5: ? The student claims: "Since the numbers are perfect squares, the next must be 25 ."

What critical flaw exists in this logic? (1) The numbers are not all squares (2) The pattern skips steps (3) The pattern is actually based on doubling (4) It's correct - no flaw

18. A rectangle is 20 cm long and 10 cm wide. A path of uniform width of ( 2 x ) in cm is built inside along all four sides. The area of the path is $176 cm^{2}$ . What is the value of x ? (1) 1 (2) 2 (3) 3 (4) 4

19. A square grid is formed at each step:

Step 1: $1 \times 1$ , Step 2: $2 \times 2$ , Step 3: $3 \times 3$ , etc. Each square holds exactly that many prime numbers. For example:

Step 1-1 $^{2} = 1$ prime - [2] Step $2 - 2^{2} = 4$ primes [3, 5, 7, 11] Step $3 - 3^{2} = 9$ primes $[13, 17, 19, 23, 29, 31$ , 37, 41, 43]

What is the largest prime number written in Step 4? (1) 101 (2) 107 (3) 113 (4) 127

20. A wall is in the shape of a rectangle with dimensions $(2 x + 5) m$ and $(x + 3) m$ . One square window of side ( $x + 1$ ) $m$ is to be excluded from painting. What is the area (in $m^{2}$ ) to be painted (in simplified form)? (1) $x^{2} + 8 x + 13$ (2) $x^{2} + 9 x + 14$ (3) $x^{2} + 10 x + 15$ (4) $x^{2} + 11 x + 15$

21. A rectangular board is ( $x + 6$ ) m long and $(x + 2) m$ wide. A square of side ( $x - 1$ ) m is cut from one corner. The remaining area (in $m^{2}$ ) of the board is: (1) $10 x + 1$ (2) $10 x + 11$ (3) $x^{2} + 10 x$ (4) $x^{2} + 11$

22. A rectangular region has an area of $4 x^{2} + 12 x + 9$ . This area is a perfect square. Its side length is (1) $x + 2$ (2) $x + 3$ (3) $2 x + 3$ (4) $2 x - 3$

23. If $(a x + b)^{2} = 16 x^{2} - 24 x + 9$ , then the product ab equals $____$ . (1) 10 (2) -12 (3) 12 (4) -10

24. In the expansion of $(2 x - 3) (x^{2} + 4 x + 5)$ , the coefficient of $x^{2}$ is $____$ . (1) 0 (2) 1 (3) 4 (4) 5

25. Statement-1: The expansion of $(3 x - 1)^{2}$ is equal to $(3 x)^{2} + (- 1)^{2}$ . Statement-2 : Multiplication of 5375 and 101 is equal to $(537500 + 5375)$ . (1) Statement-1 is true and statement-2 is false. (2) Statement-1 is false and statement-2 is true. (3) Both the statements are true. (4) Both the statements are false.

True or False

26.The difference of squares $a^{2} - b^{2}$ can never be factored if $a$ and $b$ are both negative.

27. The product $(x + 1) (x^{2} - x + 1)$ simplifies perfectly to $x^{3} + 1$ .

28. The expression $\frac{( x ^{2} - 4 )}{( x - 2 )}$ simplifies to $x + 2$ for all real x .

29. For any real $x$ , the expression $(x + 3) (x^{2} - 3 x + 9)$ always yields a cubic polynomial whose constant term is 27 .

Puzzle

30.The Garden Renovation Project

Mr. Verma owns a beautiful square garden with side length x meters. Recently, he decided to renovate the garden by adding a uniform paved walkway all around it, 5 meters wide. After construction, the total area of the garden plus the walkway became $(x + 5)^{2}$ square meters. Mr. Verma's friend suggested using the distributive property and algebraic identities to help find the area covered by just the walkway. They also discussed what happens if Mr. Verma decided to expand the walkway width to $2 \times 5 = 10$ meters instead, and how the total area and walkway area would change.

Finally, they wanted to compare the increase in walkway area when the width doubles. (i) Write an expression for the area of the walkway when the width is 5 meters using the square identity and distributive property. (ii) Simplify the expression found in (a) fully in terms of x . (iii) Find an expression for the area of the walkway if the walkway width is doubled to 10 meters, again using the square identity. (iv) Calculate the increase in the walkway area when the width doubles from 5 to 10 meters. (v) If the original garden side $x = 20$ meters, compute the walkway areas for both widths and the increase numerically.

ANSWER KEY

Multiple choice questions

Question	1	2	3	4	5	6	7	8	9	10	11	12	13	14	15
Answer	1	2	1	2	3	2	3	3	3	2	3	1	1	2	3
Question	16	17	18	19	20	21	22	23	24	25
Answer	1	4	2	3	2	2	3	2	4	2

3.0True of False

False
True
False
True

Puzzle 30. (i) $(x + 5)^{2} - x^{2}$ (ii) $10 x + 25$ (iii) $20 x + 100$ (iv) $10 x + 75$ (v) $275 m^{2}$

4.0Key Concepts of Chapter 6 We Distribute, Yet Things Multiply Exercise 6.1

Multiplying Two Monomials: To find the product, multiply the numerical coefficients together and the algebraic parts (variables) together.
Laws of Exponents: Remember that when multiplying variables with the same base, you add their powers: $x^{a} \times x^{b} = x^{a + b}$ .
Area as a Product: The area of a rectangle is Length x Breadth. If the dimensions are given as monomials, the area is their algebraic product.
Volume of a Rectangular Box: Volume is calculated as length x breadth x height. This involves finding the product of three monomials.

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 6.1

Conceptual Clarity: Helps students understand that the order of multiplication (commutative property) does not change the result.
Pattern Recognition: Reinforces the use of exponents when the same variable appears multiple times in a product.
Application-Based: Connects algebra to geometry by solving for area and volume.

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

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