---
title: "NCERT Solutions Class 8 Maths We Distribute, yet Things Multiply"
url: https://www.swavid.com/maths/class/8/chapter/we-distribute-yet-things-multiply/ncert-solutions
dateModified: 2026-10-07T15:27:13+00:00
---

# NCERT Solutions Class 8 Maths We Distribute, yet Things Multiply

This chapter's questions cover the distributive property of multiplication, algebraic identities, and their application to patterns, fast multiplication, and geometric area calculations. The exercises require students to expand algebraic expressions, verify identities, and solve problems involving patterns and geometric figures.

Free PDF (5 pages): https://www.swavid.com/api/seo/pdf/ncert/maths/class-8/swavid-ncert-solutions-class-8-maths-chapter-6-we-distribute-yet-things-multiply-19a8057753.pdf

## 6.1 Some Properties of Multiplication

### Question 1

*3 marks · Short answer*

By how much does the product increase if the first number (23) is increased by 1?

**Solution**

1. Let the initial two numbers be $a$ and $b$, where $a = 23$ and $b = 27$.
2. If the first number $a$ is increased by $1$, the product becomes $(a + 1)b = ab + b$.
3. The increase in the product is equal to $b$, which is $27$.

**Answer:** The product increases by $27$.

> Common mistake: Students often mistakenly think the product increases by the first number (23) instead of the second number (27).

### Question 2

*3 marks · Short answer*

What if the second number (27) is increased by 1?

**Solution**

1. The initial product of the two numbers is $23 \times 27$.
2. When the second number is increased by 1, the multiplication becomes $23 \times (27 + 1)$.
3. Using the distributive property and commutativity, $23 \times (27 + 1) = 23 \times 27 + 23$.
4. Therefore, the product increases by 23.

**Answer:** The product increases by 23.

> Common mistake: Increasing the product by the second number instead of the first number.

### Question 3

*3 marks · Short answer*

How about when both numbers are increased by 1?

**Solution**

1. Let the initial two numbers be $a = 23$ and $b = 27$.
2. When both numbers are increased by 1, the new product is $(a + 1)(b + 1) = ab + a + b + 1$.
3. Substituting $a = 23$ and $b = 27$, the increase is $a + b + 1 = 23 + 27 + 1 = 51$.

**Answer:** The product increases by $51$.

> Common mistake: Students forget to add the final $+1$ term from the expansion $(a+1)(b+1)$ and write $50$ instead of $51$.

### Question 4

*3 marks · Short answer*

Do you see a pattern that could help generalise our observations to the product of any two numbers?

**Solution**

1. Let the two numbers be $a$ and $b$, so their initial product is $ab$.
2. If $a$ is increased by 1, the increase in product is $b$.
3. If $b$ is increased by 1, the increase in product is $a$.
4. If both $a$ and $b$ are increased by 1, the product becomes $(a + 1)(b + 1) = ab + a + b + 1$, so the increase is $a + b + 1$.

**Answer:** The general increase when both numbers are increased by 1 is $a + b + 1$.

> Common mistake: Not including the constant term 1 when expanding $(a + 1)(b + 1)$.

## Figure it Out

### Question 1

*3 marks · Short answer*

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

**Solution**

1. Let the middle number of the $3 \times 3$ frame in the multiplication grid be $pq$, where the row factor is $p$ and the column factor is $q$.
2. Moving up the frame decreases the row factor by $1$, and moving down increases it by $1$.
3. Moving left decreases the column factor by $1$, and moving right increases it by $1$, giving the expressions for the other numbers in the grid.

**Answer:** The expressions for the numbers in the $3 \times 3$ grid frame are: top row $(p-1)(q-1)$, $(p-1)q$, $(p-1)(q+1)$; middle row $p(q-1)$, $pq$, $p(q+1)$; bottom row $(p+1)(q-1)$, $(p+1)q$, $(p+1)(q+1)$.

> Common mistake: Confusing the row and column factors when shifting positions inside the $3 \times 3$ grid frame.

## Exercises

### Question 2

*3 marks · Short answer*

Expand the following products. (i) $(3 + u) (v – 3)$ (ii) $\frac{2}{3} (15 + 6a)$ (iii) $(10a + b) (10c + d)$ (iv) $(3 – x) (x – 6)$ (v) $(–5a + b) (c + d)$ (vi) $(5 + z) (y + 9)$

**Part (i)**

1. $(3 + u) (v – 3) = 3(v – 3) + u(v – 3)$
2. $= 3v – 9 + uv – 3u$

Answer (i): $3v – 9 + uv – 3u$

**Part (ii)**

1. $\frac{2}{3} (15 + 6a) = \frac{2}{3} \times 15 + \frac{2}{3} \times 6a$
2. $= 10 + 4a$

Answer (ii): $10 + 4a$

**Part (iii)**

1. $(10a + b) (10c + d) = 10a(10c + d) + b(10c + d)$
2. $= 100ac + 10ad + 10bc + bd$

Answer (iii): $100ac + 10ad + 10bc + bd$

**Part (iv)**

1. $(3 – x) (x – 6) = 3(x – 6) – x(x – 6)$
2. $= 3x – 18 – x^2 + 6x$
3. $= –x^2 + 9x – 18$

Answer (iv): $-x^2 + 9x – 18$

**Part (v)**

1. $(-5a + b) (c + d) = -5a(c + d) + b(c + d)$
2. $= -5ac – 5ad + bc + bd$

Answer (v): $-5ac – 5ad + bc + bd$

**Part (vi)**

1. $(5 + z) (y + 9) = 5(y + 9) + z(y + 9)$
2. $= 5y + 45 + yz + 9z$

Answer (vi): $5y + 45 + yz + 9z$

**Answer:** Expanded forms of the given products.

> Common mistake: Forgetting to multiply signs properly when dealing with negative terms.

### Question 3

*3 marks · Short answer*

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.

**Solution**

1. Let the two numbers be $a$ and $b$.
2. According to the question, $(a + 2)(b - 4) = ab$.
3. Expanding the left side gives $ab - 4a + 2b - 8 = ab$, which simplifies to $2b - 4a = 8$ or $b = 2a + 4$.
4. Choosing different values for $a$ gives three pairs of numbers: for $a = 1, b = 6$; for $a = 2, b = 8$; and for $a = 3, b = 10$.

**Answer:** Three examples are $(1, 6)$, $(2, 8)$, and $(3, 10)$.

> Common mistake: Forgetting to expand the product correctly or equating it incorrectly.

### Question 4

*3 marks · Short answer*

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

**Part (i)**

1. $(a + ab – 3b^2) (4 + b) = 4(a + ab – 3b^2) + b(a + ab – 3b^2)$
2. $= 4a + 4ab – 12b^2 + ab + ab^2 – 3b^3$
3. $= 4a + 5ab + ab^2 – 12b^2 – 3b^3$

Answer (i): $4a + 5ab + ab^2 – 12b^2 – 3b^3$

**Part (ii)**

1. $(4y + 7) (y + 11z – 3) = 4y(y + 11z – 3) + 7(y + 11z – 3)$
2. $= 4y^2 + 44yz – 12y + 7y + 77z – 21$
3. $= 4y^2 + 44yz – 5y + 77z – 21$

Answer (ii): $4y^2 + 44yz – 5y + 77z – 21$

**Answer:** Expanded algebraic expressions.

> Common mistake: Missing out on combining like terms after multiplication.

### Question 5

*3 marks · Short answer*

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

**Solution**

1. Expanding the first expression gives $(a – b)(a + b) = a^2 – b^2$.
2. Expanding the second expression gives $(a – b)(a^2 + ab + b^2) = a^3 – b^3$.
3. Expanding the third expression gives $(a – b)(a^3 + a^2b + ab^2 + b^3) = a^4 – b^4$.
4. A clear pattern is visible where each expansion results in the difference of two powers, so the next identity in the pattern is $(a – b)(a^4 + a^3b + a^2b^2 + ab^3 + b^4) = a^5 – b^5$.

**Answer:** $a^2 – b^2$, $a^3 – b^3$, $a^4 – b^4$; next identity is $(a – b)(a^4 + a^3b + a^2b^2 + ab^3 + b^4) = a^5 – b^5$.

> Common mistake: Not recognizing the cancellation of intermediate terms during expansion.

## Frequently asked questions

### How many questions are there in NCERT Solutions for Class 8 Maths Chapter 6 We Distribute, yet Things Multiply?

This chapter follows the new NCERT book for the 2026-27 session and contains a total of 9 short answer questions across different sections. You can find all these questions along with SwaVid's free PDF and step-by-step solutions on this page only.

### Which topics are covered in the questions of Chapter 6?

The questions cover concepts like generalising product increments, multiplication grid patterns using algebra, and the distributive property in equations. They also focus on algebraic patterns, identities, and the expansion of algebraic expressions.

### What are the hardest question types in this chapter and how should I approach them?

The hardest questions involve the expansion of algebraic expressions and applying the distributive property of multiplication in equations. To approach them, carefully break down the terms using algebraic identities before simplifying the final product.

### How can I write answers to get full marks in Class 8 Maths Chapter 6?

To secure full marks, write down each step clearly, state the algebraic property or identity being used, and show the intermediate expansion steps. Refer to SwaVid's step-by-step solutions available on this page only to understand the proper presentation format.

### Is a free PDF available for Class 8 Maths Chapter 6 We Distribute, yet Things Multiply?

Yes, SwaVid provides a complete free PDF of the chapter solutions for the 2026-27 session. You can easily access and download the step-by-step solutions directly from this page only.

## Related pages

- [Class 8 Maths chapters](https://www.swavid.com/maths/class/8)

Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student