Selina Solutions Concise Maths Class 7 Chapter 11 Fundamental Concepts (Including Fundamental Operations) Exercise 11D provides students with an in-depth knowledge about the concepts. Exercise 11D has problems on division of monomial and polynomial functions, according to the textbook prescribed by the ICSE board. The solutions have steps which can be followed in solving the problems, with ease. To perform well in the annual exam, Selina Solutions Concise Maths Class 7 Chapter 11 Fundamental Concepts (Including Fundamental Operations) Exercise 11D PDF links, which are provided below for free download.

## Selina Solutions Concise Maths Class 7 Chapter 11: Fundamental Concepts (Including Fundamental Operations) Exercise 11D Download PDF

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#### Exercise 11D page: 132

**1. Divide:**

**(i) â€“ 16ab ^{2}c by 6abc**

**(ii) 25x ^{2}y by â€“ 5y^{2}**

**(iii) 8x + 24 by 4**

**(iv) 4a ^{2} â€“ a by â€“ a**

**(v) 8m â€“ 16 by â€“ 8**

**Solution:**

(i) â€“ 16ab^{2}c by 6abc

We can write it as

= â€“ 16ab^{2}c/ 6abc

= -8/3 b

(ii) 25x^{2}y by â€“ 5y^{2}

We can write it as

= 25x^{2}y/ -5y^{2}

= -5 x^{2}/y

(iii) 8x + 24 by 4

We can write it as

= (8x + 24)/4

Separating the terms

= 8x/4 + 24/4

= 2x + 6

(iv) 4a^{2} â€“ a by â€“ a

We can write it as

= (4a^{2} â€“ a)/ -a

Separating the terms

= 4a^{2}/-a â€“ a/-a

= – 4a + 1

(v) 8m â€“ 16 by â€“ 8

We can write it as

= (8m â€“ 16)/ -8

Separating the terms

= 8m/-8 â€“ 16/-8

= – m + 2

**2. Divide:**

**(i) n ^{2} â€“ 2n + 1 by n â€“ 1**

**(ii) m ^{2} â€“ 2mn + n^{2} by m â€“ n**

**(iii) 4a ^{2} + 4a + 1 by 2a + 1**

**(iv) p ^{2} + 4p + 4 by p + 2**

**(v) x ^{2} + 4xy + 4y^{2} by x + 2y**

**Solution:**

(i) n^{2} â€“ 2n + 1 by n â€“ 1

n^{2} â€“ 2n + 1 by n â€“ 1 = n â€“ 1

(ii) m^{2} â€“ 2mn + n^{2} by m â€“ n

m^{2} â€“ 2mn + n^{2} by m â€“ n = m – n

(iii) 4a^{2} + 4a + 1 by 2a + 1

4a^{2} + 4a + 1 by 2a + 1 = 2a + 1

(iv) p^{2} + 4p + 4 by p + 2

p^{2} + 4p + 4 by p + 2 = p + 2

(v) x^{2} + 4xy + 4y^{2} by x + 2y

x^{2} + 4xy + 4y^{2} by x + 2y = x + 2y

**3. The area of a rectangle is 6x ^{2} â€“ 4xy â€“ 10y^{2} square unit and its length is 2x + 2y unit. Find its breadth.**

**Solution:**

It is given that

Area of a rectangle = 6x^{2} â€“ 4xy â€“ 10y^{2} square unit

Length = 2x + 2y unit

We know that

Breadth = Area/ Length

So we get

= (6x^{2} â€“ 4xy â€“ 10y^{2})/ (2x + 2y)

= 3x â€“ 5y units

**4. The area of a rectangular field is 25x ^{2} + 20xy + 3y^{2} square unit. If its length is 5x + 3y unit, find its breadth. Hence, find its perimeter.**

**Solution:**

It is given that

Area of a rectangular field = 25x^{2} + 20xy + 3y^{2} square unit

Length = 5x +3y unit

We know that

Breadth = Area/ Length

So we get

= (25x^{2} + 20xy + 3y^{2})/ (5x + 3y)

= 5x + y units

Now the perimeter of the rectangular field = 2 (length + breadth)

Substituting the values

= 2 (5x + 3y + 5x + y)

So we get

= 2 (10x + 4y)

= 20x + 8y

**5. Divide:**

**(i) 2m ^{3}n^{5} by â€“ mn**

**(ii) 5x ^{2} â€“ 3x by x**

**(iii) 10x ^{3}y â€“ 9xy^{2} â€“ 4x^{2}y^{2} by xy**

**(iv) 3y ^{3} â€“ 9ay^{2} â€“ 6ab^{2}y by â€“ 3y**

**(v) x ^{5} â€“ 15x^{4} â€“ 10x^{2} by â€“ 5x^{2}**

**Solution:**

(i) 2m^{3}n^{5} by â€“ mn

It can be written as

= 2m^{3}n^{5}/ -mn

= -2m^{2}n^{4}

(ii) 5x^{2} â€“ 3x by x

It can be written as

= (5x^{2} â€“ 3x)/ x

Separating the terms

= 5x^{2}/x â€“ 3x/x

= 5x â€“ 3

(iii) 10x^{3}y â€“ 9xy^{2} â€“ 4x^{2}y^{2} by xy

It can be written as

= (10x^{3}y â€“ 9xy^{2} â€“ 4x^{2}y^{2})/ xy

Separating the terms

= 10x^{3}y/xy â€“ 9xy^{2}/xy â€“ 4x^{2}y^{2}/xy

= 10x^{2} â€“ 9y â€“ 4xy

(iv) 3y^{3} â€“ 9ay^{2} â€“ 6ab^{2}y by â€“ 3y

It can be written as

= (3y^{3} â€“ 9ay^{2} â€“ 6ab^{2}y)/ -3y

Separating the terms

= 3y^{3}/-3y â€“ 9ay^{2}/-3y â€“ 6ab^{2}y/ -3y

= -y^{2} + 3ay + 2ab^{2}

(v) x^{5} â€“ 15x^{4} â€“ 10x^{2} by â€“ 5x^{2}

It can be written as

= (x^{5} â€“ 15x^{4} â€“ 10x^{2})/ -5x^{2}

Separating the terms

= x^{5}/-5x^{2} â€“ 15x^{4}/-5x^{2} â€“ 10x^{2}/-5x^{2}

= -1/5x^{3} + 3x^{2} + 2