Selina Solutions Concise Mathematics Class 6 Chapter 19: Fundamental Operations Exercise 19(C)

Selina Solutions Concise Mathematics Class 6 Chapter 19 Fundamental Operations Exercise 19(C) has accurate answers prepared by experts at BYJU’S, with the intention to improve grasping abilities of students. Multiplication of algebraic expressions is the major concept discussed under this exercise. Students who aim to score well in the annual exam are suggested to solve the textbook questions, using these solutions. To speed up the problem solving abilities, students can make use of Selina Solutions Concise Mathematics Class 6 Chapter 19 Fundamental Operations Exercise 19(C) PDF, from the links which are provided, with a free download option.

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Access Selina Solutions Concise Mathematics Class 6 Chapter 19: Fundamental Operations Exercise 19(C)

Exercise 19(C)

1. Fill in the blanks:

(i) 6 × 3 = ………. and 6x × 3x = …………

(ii) 6 × 3 = ……… and 6x² × 3x³ = …………

(iii) 5 × 4 = ………. and 5x × 4y = …………

(iv) 4 × 7 = …………. and 4ax × 7x = ……….

(v) 6 × 2 = …………. and 6xy × 2xy = ……….

Solution:

(i) 6 × 3 = 18

Hence,

6x × 3x = 6 × 3 × x × x

We get,

= 18 × x²

= 18x²

Therefore, 6 × 3 = 18 and 6x × 3x = 18x²

(ii) 6 × 3 = 18

Hence,

6x² × 3x³ = 6 × 3 × x^{2 + 3}

= 18 × x⁵

= 18x⁵

Therefore, 6 × 3 = 18 and 6x² × 3x³ = 18x⁵

(iii) 5 × 4 = 20 and 5x × 4y = 5 × 4 × x × y

= 20xy

Therefore, 5 × 4 = 20 and 5x × 4y = 20xy

(iv) 4 × 7 = 28

Hence,

4ax × 7x = 4 × 7 × a × x × x

= 28 × a × x²

= 28ax²

Therefore, 4 × 7 = 28 and 4ax × 7x = 28ax²

(v) 6 × 2 = 12

Hence,

6xy × 2xy = 6 × 2 × x^{1 + 1} × y^{1 + 1}

= 12 × x² × y²

= 12x²y²

Therefore, 6 × 2 = 12 and 6xy × 2xy = 12x²y²

2. Fill in the blanks:

(i) 4x × 6x × 2 = ………

(ii) 3ab × 6ax = …………

(iii) x × 2x² × 3x³ = ………

(iv) 5 × 5a³ = …………

(v) 6 × 6x² × 6x²y² = ………

Solution:

(i) 4x × 6x × 2 = 4 × 6 × 2 × x × x

= 48 × x²

We get,

= 48x²

Hence, 4x × 6x × 2 = 48x²

(ii) 3ab × 6ax = 3 × 6 × a × a × b × x

= 18 × a² × b × x

We get,

= 18a²bx

Hence, 3ab × 6ax = 18a²bx

(iii) x × 2x² × 3x³ = 2 × 3 × x × x² × x³

= 6 × x^{1 + 2 + 3}

= 6 × x⁶

= 6x⁶

Hence, x × 2x² × 3x³ = 6x⁶

(iv) 5 × 5a³ = 5 × 5 × a³

= 25 × a³

We get,

= 25a³

Hence, 5 × 5a³ = 25a³

(v) 6 × 6x² × 6x²y² = 6 × 6 × 6 × x² × x² × y²

= 216 × x^{2+ 2} × y²

= 216 × x⁴ × y²

We get,

= 216x⁴y²

Hence, 6 × 6x² × 6x²y² = 216x⁴y²

3. Find the value of:

(i) 3x³ × 5x⁴

(ii) 5a² × 7a⁷

(iii) 3abc × 6ac³

(iv) a²b² × 5a³b⁴

(v) 2x²y³ × 5x³y⁴

Solution:

(i) 3x³ × 5x⁴

3x³ × 5x⁴ = 3 × 5 × x³ × x⁴

= 15 × x^{3 + 4}

We get,

= 15 × x⁷

= 15x⁷

Hence, the value of 3x³ × 5x⁴ is 15x⁷

(ii) 5a² × 7a⁷

5a² × 7a⁷ = 5 × 7 × a² × a⁷

= 35 × a^{2 + 7}

= 35 × a⁹

We get,

= 35a⁹

Hence, the value of 5a² × 7a⁷ is 35a⁹

(iii) 3abc × 6ac³

3abc × 6ac³ = 3 × 6 × a × a × b × c × c³

=18 × a^{1+ 1} × b × c¹⁺³

= 18 × a² × b × c⁴

We get,

= 18a²bc⁴

Hence, the value of 3abc × 6ac³ is 18a²bc⁴

(iv) a²b² × 5a³b⁴

a²b² × 5a³b⁴ = 5 × a² × a³ × b² × b⁴

= 5 × a²⁺³ × b²⁺⁴

= 5 × a⁵ × b⁶

We get,

= 5a⁵b⁶

Hence, the value of a²b² × 5a³b⁴ is 5a⁵b⁶

(v) 2x²y³ × 5x³y⁴

2x²y³ × 5x³y⁴ = 2 × 5 × x² × x³ × y³ × y⁴

= 10 × x²⁺³ × y³⁺⁴

We get,

= 10 × x⁵ × y⁷

= 10x⁵y⁷

Hence, the value of 2x²y³ × 5x³y⁴ is 10x⁵y⁷

4. Multiply:

(i) a + b by ab

(ii) 3ab – 4b by 3ab

(iii) 2xy – 5by by 4bx

(iv) 4x + 2y by 3xy

(v) 1 + 4x by x

Solution:

(i) a + b by ab

The multiplication of a + b by ab is calculated as,

(a + b) × ab = a × ab + b × ab

= a¹⁺¹b + ab¹⁺¹

We get,

= a²b + ab²

Hence, (a + b) by ab = a²b + ab²

(ii) 3ab – 4b by 3ab

The multiplication of 3ab – 4b by 3ab is calculated as,

(3ab – 4b) × 3ab = 3ab × 3ab – 4b × 3ab

= 9a¹⁺¹b¹⁺¹ – 12ab¹⁺¹

We get,

= 9a²b² – 12ab²

Therefore, (3ab – 4b) by 3ab = 9a²b² – 12ab²

(iii) 2xy – 5by by 4bx

The multiplication of 2xy – 5by by 4bx is calculated as,

(2xy – 5by) × 4bx = 2xy × 4bx – 5by × 4bx

= 8bx¹⁺¹y – 20b¹⁺¹xy

We get,

= 8bx²y – 20b²xy

Therefore, (2xy – 5by) by 4bx = 8bx²y – 20b²xy

(iv) 4x + 2y by 3xy

The multiplication of 4x + 2y by 3xy is calculated as,

(4x + 2y) × 3xy = 4x × 3xy + 2y × 3xy

On simplification, we get

= 12x¹⁺¹y + 6xy¹⁺¹

= 12x²y + 6xy²

Therefore, (4x + 2y) by 3xy = 12x²y + 6xy²

(v) 1 + 4x by x

The multiplication of (1 + 4x) by x is calculated as,

(1 + 4x) × x = 1 × x + 4x × x

On simplification, we get

= x + 4x¹⁺¹

= x + 4x²

Therefore, (1 + 4x) by x = x + 4x²

5. Multiply:

(i) – x + y – z and – 2x

(ii) xy – yz and x²yz²

(iii) 2xyz + 3xy and – 2y²z

(iv) – 3xy² + 4x²y and – xy

(v) 4xy and – x²y – 3x² y²

Solution:

(i) – x + y – z and – 2x

The multiplication of the given expression is calculated as,

(- x + y – z) × – 2x = – x × – 2x + y × – 2x – z × – 2x

On further calculation, we get

= 2x¹⁺¹ – 2xy + 2xz

= 2x² – 2xy + 2xz

Hence, the multiplication of (- x + y – z) and – 2x is 2x² – 2xy + 2xz

(ii) xy – yz and x²yz²

The multiplication of the given expression is calculated as,

(xy – yz) × (x²yz²) = xy × x²yz² – yz × x²yz²

We get,

= x¹⁺²y¹⁺¹z² – x²y¹⁺¹z¹⁺²

= x³y²z² – x²y²z³

Hence, the multiplication of (xy – yz) and x²yz² = x³y²z² – x²y²z³

(iii) 2xyz + 3xy and – 2y²z

The multiplication of the given expression is calculated as,

(2xyz + 3xy) × – 2y²z = 2xyz × – 2y²z + 3xy × – 2y²z

On further calculation, we get

= – 4xy¹⁺²z¹⁺¹ – 6xy¹⁺²z

= – 4xy³z² – 6xy³z

Hence, the multiplication of 2xyz + 3xy and – 2y²z = – 4xy³z² – 6xy³z

(iv) – 3xy² + 4x²y and – xy

The multiplication of the given expression is calculated as,

(- 3xy² + 4x²y) × – xy = 3x¹⁺¹y²⁺¹ – 4x²⁺¹y¹⁺¹

On calculation, we get

= 3x²y³ – 4x³y²

Hence, the multiplication of – 3xy² + 4x²y and – xy = 3x²y³ – 4x³y²

(v) 4xy and – x²y – 3x² y²

The multiplication of the given expression is calculated as,

(- x²y – 3x²y²) × 4xy = – x²y × 4xy – 3x²y² × 4xy

On further calculation, we get

= – 4x²⁺¹y¹⁺¹ – 12x²⁺¹y²⁺¹

= – 4x³y² – 12x³y³

Hence, the multiplication of 4xy and – x²y – 3x² y² = – 4x³y² – 12x³y³

6. Multiply:

(i) 3a + 4b – 5c and 3a

(ii) – 5xy and – xy² – 6x²y

Solution:

(i) 3a + 4b – 5c and 3a

The multiplication of the given expression is calculated as,

(3a + 4b – 5c) × 3a = 3a × 3a + 4b × 3a – 5c × 3a

On further calculation, we get

= 9a¹⁺¹ + 12ab – 15ac

= 9a² + 12ab – 15ac

Therefore, the multiplication of 3a + 4b – 5c and 3a = 9a² + 12ab – 15ac

(ii) – 5xy and – xy² – 6x²y

The multiplication of the given expression is calculated as,

– 5xy × (- xy² – 6x²y) = – 5xy × – xy² – 5xy × – 6x²y

On further calculation, we get

= 5x¹⁺¹y¹⁺² + 30x¹⁺²y¹⁺¹

= 5x²y³ + 30x³y²

Therefore, the multiplication of – 5xy and – xy² – 6x²y = 5x²y³ + 30x³y²

7. Multiply:

(i) x + 2 and x + 10

(ii) x + 5 and x – 3

(iii) x – 5 and x + 3

(iv) x – 5 and x – 3

(v) 2x + y and x + 3y

Solution:

(i) x + 2 and x + 10

The given expression is calculated as follows

(x + 2) × (x + 10) = x × (x + 10) + 2 × (x + 10)

We get,

= x² + 10x + 2x + 20

= x² + 12x + 20

Hence, the multiplication of (x + 2) and (x + 10) = x² + 12x + 20

(ii) x + 5 and x – 3

The given expression is calculated as follows

(x + 5) × (x – 3) = x × (x – 3) + 5 × (x – 3)

On simplification, we get

= x² – 3x + 5x – 15

= x² + 2x – 15

Hence, the multiplication of (x + 5) and (x – 3) = x² + 2x – 15

(iii) x – 5 and x + 3

The given expression is calculated as follows

(x – 5) × (x + 3) = x × (x + 3) – 5 × (x + 3)

On further calculation, we get

= x² + 3x – 5x – 15

= x² – 2x – 15

Hence, the multiplication of (x – 5) and (x + 3) = x² – 2x – 15

(iv) x – 5 and x – 3

The given expression is calculated as,

(x – 5) × (x – 3) = x × (x – 3) – 5 × (x – 3)

On further calculation, we get

= x² – 3x – 5x + 15

= x² – 8x + 15

Hence, the multiplication of (x – 5) and (x – 3) = x² – 8x + 15

(v) 2x + y and x + 3y

The given expression is calculated as,

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

On simplification, we get

= 2x² + 6xy + xy + 3y²

= 2x² + 7xy + 3y²

Hence, the multiplication of (2x + y) and (x + 3y) = 2x² + 7xy + 3y²

8. Multiply:

(i) 3abc and – 5a²b²c

(ii) x – y + z and -2x

(iii) 2x – 3y – 5z and -2y

(iv) – 8xyz + 10 x²yz³ and xyz

(v) xyz and – 13xy²z + 15x²yz – 6xyz²

Solution:

(i) 3abc and – 5a²b²c

The given expression is calculated as follows,

3abc × – 5a²b²c = 3 × – 5 × a × a² × b × b² × c × c

On further calculation, we get

= – 15 × a¹⁺² × b¹⁺² × c¹⁺¹

= – 15 × a³ × b³ × c²

= – 15a³b³c²

Therefore, the multiplication of 3abc and – 5a²b²c = – 15a³b³c²

(ii) x – y + z and -2x

The given expression is calculated as follows,

(x – y + z) × – 2x = x × – 2x – y × – 2x + z × – 2x

On simplification, we get

= – 2x¹⁺¹ + 2xy – 2xz

= – 2x² + 2xy – 2xz

Therefore, the multiplication of x – y + z and -2x = – 2x² + 2xy – 2xz

(iii) 2x – 3y – 5z and -2y

The given expression is calculated as follows,

(2x – 3y – 5z) × – 2y = 2x × – 2y – 3y × – 2y – 5z × – 2y

On further calculation, we get

= – 4xy + 6y¹⁺¹ + 10yz

= – 4xy + 6y² + 10yz

Therefore, the multiplication of 2x – 3y – 5z and -2y = – 4xy + 6y² + 10yz

(iv) – 8xyz + 10 x²yz³ and xyz

The given expression is calculated as follows,

(- 8xyz + 10x²yz³) × xyz = – 8xyz × xyz + 10x²yz³ × xyz

On further calculation, we get

= – 8x¹⁺¹y¹⁺¹z¹⁺¹ + 10x²⁺¹y¹⁺¹z³⁺¹

= – 8x²y²z² + 10x³y²z⁴

Therefore, the multiplication of – 8xyz + 10 x²yz³ and xyz = – 8x²y²z² + 10x³y²z⁴

(v) xyz and – 13xy²z + 15x²yz – 6xyz²

The given expression is calculated as follows,

xyz × (- 13xy²z + 15x²yz – 6xyz²) = xyz × – 13xy²z + xyz ×15x²yz – xyz × 6xyz²

On simplification, we get

= – 13x¹⁺¹y¹⁺²z¹⁺¹ + 15x¹⁺²y¹⁺¹z¹⁺¹ – 6x¹⁺¹y¹⁺¹z¹⁺²

We get,

= – 13x²y³z² + 15x³y²z² – 6x²y²z³

Therefore, the multiplication of xyz and – 13xy²z + 15x²yz – 6xyz² = – 13x²y³z² + 15x³y²z² – 6x²y²z³

9. Find the product of:

(i) xy – ab and xy + ab

(ii) 2abc – 3xy and 2abc + 3xy

(iii) a + b – c and 2a – 3b

(iv) 5x – 6y – 7z and 2x + 3y

(v) 5x – 6y – 7z and 2x + 3y + z

Solution:

(i) xy – ab and xy + ab

The product of the given expression is calculated as,

(xy – ab) × (xy + ab) = xy × (xy + ab) – ab × (xy + ab)

On simplification, we get

= xy × xy + xy × ab – ab × xy – ab × ab

= x²y² + abxy – abxy – a²b²

= x²y² – a²b²

Hence, the product of (xy – ab) and (xy + ab) = x²y² – a²b²

(ii) 2abc – 3xy and 2abc + 3xy

The product of the given expression is calculated as,

(2abc – 3xy) × (2abc + 3xy)

= 2abc × (2abc + 3xy) – 3xy × (2abc + 3xy)

We get,

= 2abc × 2abc + 2abc × 3xy – 3xy × 2abc – 3xy × 3xy

= 4a²b²c² + 6abcxy – 6abcxy – 9x²y²

= 4a²b²c² – 9x²y²

Hence, the product of 2abc – 3xy and 2abc + 3xy = 4a²b²c² – 9x²y²

(iii) a + b – c and 2a – 3b

The product of the given expression is calculated as,

(a + b – c) × (2a – 3b)

= a × (2a – 3b) + b × (2a – 3b) – c × (2a – 3b)

= a × 2a – a × 3b + b × 2a – b × 3b – c × 2a + c × 3b

= 2a¹⁺¹ – 3ab + 2ab – 3b¹⁺¹ – 2ac + 3bc

We get,

= 2a² – ab – 3b² – 2ac + 3bc

Hence, the product of a + b – c and 2a – 3b = 2a² – ab – 3b² – 2ac + 3bc

(iv) 5x – 6y – 7z and 2x + 3y

The product of the given expression is calculated as,

(5x – 6y – 7z) × (2x + 3y)

= (5x – 6y – 7z) × 2x + (5x – 6y – 7z) × 3y

= 5x × 2x – 6y × 2x – 7z × 2x + 5x × 3y – 6y × 3y – 7z × 3y

We get,

= 10x² – 12xy – 14xz + 15xy – 18y² – 21yz

= 10x² + 3xy – 14xz – 18y² – 21yz

Hence, the product of 5x – 6y – 7z and 2x + 3y = 10x² + 3xy – 14xz – 18y² – 21yz

(v) 5x – 6y – 7z and 2x + 3y + z

The product of the given expression is calculated as,

(5x – 6y – 7z) × (2x + 3y + z)

= (5x – 6y – 7z) × 2x + (5x – 6y – 7z) × 3y + (5x – 6y – 7z) × z

= 5x × 2x – 6y × 2x – 7z × 2x + 5x × 3y – 6y × 3y – 7z × 3y + 5x × z – 6y × z – 7z × z

We get,

= 10x² – 12xy – 14xz + 15xy – 18y² – 21yz + 5xz – 6yz – 7z²

= 10x² – 12xy + 15xy – 14xz + 5xz – 18y² – 21yz – 6yz – 7z²

= 10x² + 3xy – 9xz – 18y² – 27yz – 7z²

Hence, the product of 5x – 6y – 7z and 2x + 3y + z = 10x² + 3xy – 9xz – 18y² – 27yz – 7z²