Selina Solutions Concise Maths Class 7 Chapter 11 Fundamental Concepts (Including Fundamental Operations) Exercise 11C has answers curated by highly experienced faculty at BYJUâ€™S. This exercise deals with multiplication operations on monomial and polynomial, as per the latest syllabus of ICSE board. Students are advised to solve the exercise problems using the solutions PDF as a major reference guide. Selina Solutions Concise Maths Class 7 Chapter 11 Fundamental Concepts (Including Fundamental Operations) Exercise 11C, PDF links are available here with a free download option.

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#### Exercise 11C page: 129

**1. Multiply:**

**(i) 3x, 5x ^{2}y and 2y**

**(ii) 5, 3a and 2ab ^{2}**

**(iii) 5x + 2y and 3xy**

**(iv) 6a â€“ 5b and â€“ 2a**

**(v) 4a + 5b and 4a â€“ 5b**

**Solution:**

(i) 3x, 5x^{2}y and 2y

Product = 3x Ã— 5x^{2}y Ã— 2y

We can write it as

= 3 Ã— 5 Ã— 2 Ã— x Ã— x^{2} Ã— y Ã— y

So we get

= 30x^{3}y^{2}

(ii) 5, 3a and 2ab^{2}

Product = 5 Ã— 3a Ã— 2ab^{2}

We can write it as

= 5 Ã— 3 Ã— 2 Ã— a Ã— ab^{2}

So we get

= 30a^{2}b^{2}

(iii) 5x + 2y and 3xy

Product = 3xy (5x + 2y)

We can write it as

= 3xy Ã— 5x + 3xy Ã— 2y

So we get

= 15x^{2}y + 6xy^{2}

(iv) 6a â€“ 5b and â€“ 2a

Product = – 2a (6a â€“ 5b)

We can write it as

= -2a Ã— 6a + 2a Ã— 5b

So we get

= -12a^{2} + 10ab

(v) 4a + 5b and 4a â€“ 5b

Product = (4a + 5b) (4a â€“ 5b)

So we get

= 16a^{2} â€“ 25b^{2}

**2. Copy and complete the following multiplications:**

**Solution:**

**3. Evaluate:**

**(i) (c + 5) (c â€“ 3)**

**(ii) (3c â€“ 5d) (4c â€“ 6d)**

**(iii) (1/2a + 1/2b) (1/2a â€“ 1/2b)**

**(iv) (a ^{2} + 2ab + b^{2}) (a + b)**

**(v) (3x â€“ 1) (4x ^{3} â€“ 2x^{2} + 6x â€“ 3)**

**Solution:**

(i) (c + 5) (c â€“ 3)

It can be written as

= c (c â€“ 3) + 5 (c â€“ 3)

By further calculation

= c^{2} â€“ 3c + 5c â€“ 15

= c^{2} + 2c â€“ 15

(ii) (3c â€“ 5d) (4c â€“ 6d)

It can be written as

= 3c (4c â€“ 6d) â€“ 5d (4c â€“ 6d)

By further calculation

= 12c^{2} â€“ 18cd â€“ 20cd + 30d^{2}

= 12c^{2} â€“ 38cd + 30d^{2}

(iii) (1/2a + 1/2b) (1/2a â€“ 1/2b)

It can be written as

= 1/2a (1/2a â€“ 1/2b) + 1/2b (1/2a â€“ 1/2b)

By further calculation

= 1/4a^{2} â€“ 1/4ab + 1/4ab â€“ 1/4b^{2}

= 1/4a^{2} â€“ 1/4b^{2}

(iv) (a^{2} + 2ab + b^{2}) (a + b)

It can be written as

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

By further calculation

= a^{3} + 2a^{2}b + ab^{2} + a^{2}b + 2ab^{2} + b^{3}

= a^{3} + b^{3} + 3a^{2}b + 3ab^{2}

(v) (3x â€“ 1) (4x^{3} â€“ 2x^{2} + 6x â€“ 3)

It can be written as

= 3x (4x^{3} â€“ 2x^{2} + 6x â€“ 3) â€“ 1 (4x^{3} â€“ 2x^{2} + 6x â€“ 3)

By further calculation

= 12x^{4} â€“ 6x^{3} + 18x^{2} â€“ 9x â€“ 4x^{3} + 2x^{2} â€“ 6x + 3

= 12x^{4} â€“ 6x^{3} â€“ 4x^{3} + 18x^{2} + 2x^{2} â€“ 9x â€“ 6x + 3

So we get

= 12x^{4} â€“ 10x^{3} + 20x^{2} â€“ 15x + 3

**4. Evaluate:**

**(i) (a + b) (a â€“ b).**

**(ii) (a ^{2} + b^{2}) (a + b) (a â€“ b), using the result of (i).**

**(iii) (a ^{4} + b^{4}) (a^{2} + b^{2}) (a + b) (a â€“ b), using the result of (ii).**

**Solution:**

(i) (a + b) (a â€“ b).

It can be written as

= a (a â€“ b) + b (a â€“ b)

By further calculation

= a^{2} â€“ ab + ab â€“ b^{2}

= a^{2} â€“ b^{2}

(ii) (a^{2} + b^{2}) (a + b) (a â€“ b)

Substituting the result of (i)

= (a^{2} + b^{2}) (a^{2} â€“ b^{2})

It can be written as

= a^{2} (a^{2} â€“ b^{2}) + b^{2} (a^{2} â€“ b^{2})

So we get

= a^{4} â€“ a^{2}b^{2} + a^{2}b^{2} â€“ b^{4}

= a^{4} â€“ b^{4}

(iii) (a^{4} + b^{4}) (a^{2} + b^{2}) (a + b) (a â€“ b)

Substituting the result of (ii)

= (a^{4} + b^{4}) (a^{4} â€“ b^{4})

It can be written as

= a^{4} (a^{4} â€“ b^{4}) + b^{4} (a^{4} â€“ b^{4})

By further calculation

= a^{8} â€“ a^{4}b^{4} + a^{4}b^{4} â€“ b^{8}

= a^{8} â€“ b^{8}

**5. Evaluate:**

**(i) (3x â€“ 2y) (4x + 3y)**

**(ii) (3x â€“ 2y) (4x + 3y) (8x â€“ 5y)**

**(iii) (a + 5) (3a â€“ 2) (5a + 1)**

**(iv) (a + 1) (a ^{2} â€“ a + 1) and (a â€“ 1) (a^{2} + a + 1); and then: (a + 1) (a^{2} â€“ a + 1) + (a â€“ 1) (a^{2} + a + 1)**

**(v) (5m â€“ 2n) (5m + 2n) (25m ^{2} + 4n^{2})**

**Solution:**

(i) (3x â€“ 2y) (4x + 3y)

It can be written as

= 3x (4x + 3y) â€“ 2y (4x + 3y)

By further calculation

= 12x^{2} + 9xy â€“ 8xy â€“ 6y^{2}

So we get

= 12x^{2} + xy â€“ 6y^{2}

(ii) (3x â€“ 2y) (4x + 3y) (8x â€“ 5y)

Substituting result of (i)

= (12x^{2} + xy â€“ 6y^{2}) (8x â€“ 5y)

It can be written as

= 8x (12x^{2} + xy â€“ 6y^{2}) â€“ 5y (12x^{2} + xy â€“ 6y^{2})

By further calculation

= 96x^{3} + 8x^{2}y â€“ 48xy^{2} â€“ 60x^{2}y â€“ 5xy^{2} + 30y^{3}

So we get

= 96x^{3} + 8x^{2}y â€“ 60x^{2}y â€“ 48xy^{2} â€“ 5xy^{2} + 30y^{3}

= 96x^{3} – 52 x^{2}y â€“ 53xy^{2} + 30y^{3}

(iii) (a + 5) (3a â€“ 2) (5a + 1)

It can be written as

= a (3a â€“ 2) + 5 (3a â€“ 2) (5a + 1)

By further calculation

= (3a^{2} â€“ 2a + 15a â€“ 10) (5a + 1)

So we get

= (3a^{2} + 13a â€“ 10) (5a + 1)

We can write it as

= 5a (3a^{2} + 13a â€“ 10) + 1 (3a^{2} + 13a â€“ 10)

By further calculation

= 15a^{3} + 65a^{2} â€“ 50a + 3a^{2} + 13a â€“ 10

= 15a^{3} + 68a^{2} â€“ 37a – 10

(iv) (a + 1) (a^{2} â€“ a + 1) and (a â€“ 1) (a^{2} + a + 1); and then: (a + 1) (a^{2} â€“ a + 1) + (a â€“ 1) (a^{2} + a + 1)

Consider

(a + 1) (a^{2} â€“ a + 1)

It can be written as

= a (a^{2} â€“ a + 1) + 1 (a^{2} â€“ a + 1)

By further calculation

= a^{3} â€“ a^{2} + a + a^{2} â€“ a + 1

So we get

= a^{3} + 1

(a â€“ 1) (a^{2} + a + 1)

It can be written as

= a (a^{2} + a + 1) â€“ 1 (a^{2} + a + 1)

By further calculation

= a^{3} + a^{2} + a â€“ a^{2} â€“ a â€“ 1

So we get

= a^{3} â€“ 1

Here

(a + 1) (a^{2} â€“ a + 1) + (a â€“ 1) (a^{2} + a + 1)

= a^{3} + 1 + a^{3} â€“ 1

= 2a^{3}

(v) (5m â€“ 2n) (5m + 2n) (25m^{2} + 4n^{2})

It can be written as

= [5m (5m + 2n) â€“ 2n (5m + 2n)] (25m^{2} + 4n^{2})

By further calculation

= (25m^{2} + 10mn â€“ 10mn â€“ 4n^{2}) (25m^{2} + 4n^{2})

So we get

= (25m^{2} – 4n^{2}) (25m^{2} + 4n^{2})

We can write it as

= 25m^{2} (25m^{2} + 4n^{2}) â€“ 4n^{2} (25m^{2} + 4n^{2})

By multiplying the terms

= 625m^{4} + 100m^{2}n^{2} â€“ 100m^{2}n^{2} â€“ 16n^{4}

= 625m^{4} – 16n^{4}

**6. Multiply:**

**(i) mn ^{4}, m^{3}n and 5m^{2}n^{3}**

**(ii) 2mnpq, 4mnpq and 5mnpq**

**(iii) pq â€“ pm and p ^{2}m**

**(iv) x ^{3} â€“ 3y^{3} and 4x^{2}y^{2}**

**(v) a ^{3} â€“ 4ab and 2a^{2}b**

**Solution:**

(i) mn^{4}, m^{3}n and 5m^{2}n^{3}

It can be written as

= 5m^{2}n^{3} Ã— mn^{4} Ã— m^{3}n

By further calculation

= 5m^{(2 + 1 + 3)} n^{(3 + 4 + 1)}

= 5m^{6}n^{8}

(ii) 2mnpq, 4mnpq and 5mnpq

It can be written as

= 5mnpq Ã— 2mnpq Ã— 4mnpq

By further calculation

= 5 Ã— 2 Ã— 4 m^{(1 + 1 + 1)} n^{(1 + 1 + 1)} p^{(1 + 1 + 1)} q^{(1 + 1 + 1)}

= 40m^{3}n^{3}p^{3}q^{3}

(iii) pq â€“ pm and p^{2}m

It can be written as

= p^{2}m Ã— (pq â€“ pm)

So we get

= p^{3}qm â€“ p^{3}m^{2}

(iv) x^{3} â€“ 3y^{3} and 4x^{2}y^{2}

It can be written as

= 4x^{2}y^{2} Ã— (x^{3} â€“ 3y^{3})

By further calculation

= 4x^{5}y^{2} â€“ 12x^{2}y^{5}

(v) a^{3} â€“ 4ab and 2a^{2}b

It can be written as

= 2a^{2}b Ã— (a^{3} â€“ 4ab)

By further calculation

= 2a^{5}b â€“ 8a^{3}b^{2}

**7. Multiply:**

**(i) (2x + 3y) (2x + 3y)**

**(ii) (2x â€“ 3y) (2x + 3y)**

**(iii) (2x + 3y) (2x â€“ 3y)**

**(iv) (2x â€“ 3y) (2x â€“ 3y)**

**(v) (-2x + 3y) (2x â€“ 3y)**

**Solution:**

(i) (2x + 3y) (2x + 3y)

It can be written as

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

By further calculation

= 4x^{2} + 6xy + 6xy + 9y^{2}

= 4x^{2} + 12xy + 9y^{2}

(ii) (2x â€“ 3y) (2x + 3y)

It can be written as

= 2x (2x + 3y) â€“ 3y (2x + 3y)

By further calculation

= 4x^{2} + 6xy â€“ 6xy â€“ 9y^{2}

= 4x^{2} â€“ 9y^{2}

(iii) (2x + 3y) (2x â€“ 3y)

It can be written as

= 2x (2x â€“ 3y) + 3y (2x â€“ 3y)

By further calculation

= 4x^{2} â€“ 6xy + 6xy â€“ 9y^{2}

= 4x^{2} â€“ 9y^{2}

(iv) (2x â€“ 3y) (2x â€“ 3y)

It can be written as

= 2x (2x â€“ 3y) â€“ 3y (2x â€“ 3y)

By further calculation

= 4x^{2} â€“ 6xy â€“ 6xy + 9y^{2}

= 4x^{2} â€“ 12xy + 9y^{2}

(v) (-2x + 3y) (2x â€“ 3y)

It can be written as

= -2x (2x â€“ 3y) + 3y (2x â€“ 3y)

By further calculation

= – 4x^{2} + 6xy + 6xy â€“ 9y^{2}

= – 4x^{2} + 12xy â€“ 9y^{2}