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

Selina Solutions Concise Mathematics Class 6 Chapter 19 Fundamental Operations Exercise 19(D) contains answers designed by expert faculty, in accordance with the ICSE exam. The solutions are in a stepwise manner, for better understanding of the concepts among students. Subject matter experts prepare 100% accurate solutions, with the aim to improve the academic performance of students. These solutions also enable them to increase their analytical and logical thinking. Selina Solutions Concise Mathematics Class 6 Chapter 19 Fundamental Operations Exercise 19(D), free PDF links are given below.

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

Exercise 19(D)

1. Divide:

(i) 3a by a

(ii) 15x by 3x

(iii) 16m by 4

(iv) 20x2 by 5x

(v) 30p2 by 10p2

Solution:

(i) 3a by a

3a Ã· a

This can be written as,

3a / a = (3 Ã— a) / a

= 3

Hence, 3a Ã· a = 3

(ii) 15x by 3x

15x Ã· 3x

15x / 3x = (15 Ã— x) / (3 Ã— x)

This can be written as,

= (3 Ã— 5 Ã— x) / (3 Ã— x)

We get,

= 5

Hence, 15x Ã· 3x = 5

(iii) 16m by 4

16m Ã· 4

16m / 4 = (16 Ã— m) / 4

This can be written as,

= (4 Ã— 4 Ã— m) / 4

We get,

= 4m

Hence, 16m Ã· 4 = 4m

(iv) 20x2 by 5x

20x2 Ã· 5x

20x2 / 5x = (20 Ã— x2) / (5 Ã— x)

This can be written as,

= (4 Ã— 5 Ã— x2-1) / 5

= 4 Ã— x

= 4x

Hence, 20x2 Ã· 5x = 4x

(v) 30p2 by 10p2

30p2 Ã· 10p2 = (30 Ã— p2) / (10 Ã— p2)

This can be written as,

= (3 Ã— 10 Ã— p2-2) / 10

= 3 Ã— p0

= 3 Ã— 1

= 3

Hence, 30p2 Ã· 10p2 = 3

2. Simplify:

(i) 2x5 Ã· x2

(ii) 6a8 Ã· 3a3

(iii) 20xy Ã· – 5xy

(iv) â€“ 24a2b2c2 Ã· 6ab

(v) â€“ 5x2y Ã· xy2

Solution:

(i) 2x5 Ã· x2

= (2 Ã— x5) / x2

= 2 Ã— x5-2

= 2 Ã— x3

We get,

= 2x3

Hence, 2x5 Ã· x2 = 2x3

(ii) 6a8 Ã· 3a3

= (6 Ã— a8) / (3 Ã— a3)

This can be written as,

= (2 Ã— 3 Ã— a8-3) / 3

We get,

= 2 Ã— a5

= 2a5

Hence, 6a8 Ã· 3a3 = 2a5

(iii) 20xy Ã· – 5xy

= (20 Ã— x Ã— y) / (- 5 Ã— x Ã— y)

This can be written as,

= (4 Ã— 5) / – 5

We get,

= – 4

Hence, 20xy Ã· – 5xy = – 4

(iv) â€“ 24a2b2c2 Ã· 6ab

= (- 24 Ã— a2 Ã— b2 Ã— c2) / (6 Ã— a Ã— b)

This can be written as,

= (-4 Ã— 6 Ã— a2-1 Ã— b2-1 Ã— c2) / 6

We get,

= – 4 Ã— a Ã— b Ã— c2

= – 4abc2

Hence, â€“ 24a2b2c2 Ã· 6ab = – 4abc2

(v) â€“ 5x2y Ã· xy2

= (- 5 Ã— x2 Ã— y) / (x Ã— y2)

This can be written as,

= (- 5 Ã— x2-1) / y2-1

We get,

= (- 5 Ã— x) / y

= – 5x / y

Hence, â€“ 5x2y Ã· xy2 = – 5x / y

3. Divide:

(i) (- 3m / 4) by 2m

(ii) â€“ 15p6q8 by – 5p6q7

(iii) â€“ 21m5n7 by 14m2n2

(iv) 36a4x5y6 by 4x2a3y2

(v) 20x3a6 by 5xy

Solution:

(i) (- 3m / 4) by 2m

= – 3m / 4 Ã· 2m = – 3m / 4 Ã— 1 / 2m

= – (3 Ã— m) / (4 Ã— 2 Ã— m)

We get,

= – 3 / 8

Hence, (- 3m / 4) Ã· 2m = – 3 / 8

(ii) â€“ 15p6q8 by – 5p6q7

– 15p6q8 Ã· – 5p6q7 = (- 15 Ã— p6 Ã— q8) / (- 5 Ã— p6 Ã— q7)

This can be written as,

= (3 Ã— 5 Ã— q8-7) / 5

We get,

= 3 Ã— q

= 3q

Hence, – 15p6q8 Ã· – 5p6q7 = 3q

(iii) â€“ 21m5n7 by 14m2n2

– 21m5n7 Ã· 14m2n2 = (- 21 Ã— m5 Ã— n7) / (14 Ã— m2 Ã— n2)

This can be written as,

= (- 3 Ã— 7 Ã— m5-2 Ã— n7-2) / (2 Ã— 7)

= (- 3 Ã— m3 Ã— n5) / 2

We get,

= – 3m3n5 / 2

Hence, – 21m5n7 Ã· 14m2n2 = – 3m3n5 / 2

(iv) 36a4x5y6 by 4x2a3y2

36a4x5y6 Ã· 4x2a3y2 = (36 Ã— a4 Ã— x5 Ã— y6) / (4 Ã— x2 Ã— a3 Ã— y2)

This can be written as,

= (4 Ã— 9 Ã— a4-3 Ã— x5-2 Ã— y6-2) / 4

= 9 Ã— a1 Ã— x3 Ã— y4

We get,

= 9ax3y4

Hence, 36a4x5y6 Ã· 4x2a3y2 = 9ax3y4

(v) 20x3a6 by 5xy

20x3a6 Ã· 5xy = (20 Ã— x3 Ã— a6) / (5 Ã— x Ã— y)

This can be written as,

= (4 Ã— 5 Ã— x3-1 Ã— a6) / (5 Ã— y)

We get,

= (4 Ã— x2 Ã— a6) / y

= 4x2a6 / y

Hence, 20x3a6 Ã· 5xy = 4x2a6 / y

4. Simplify:

(i) (- 15m5n2) / (- 3m5)

(ii) 35x4y2 / – 15x2y2

(iii) (- 24x6y2) / (6x6y)

Solution:

(i) (-15m5n2) / (- 3m5) = (-15 Ã— m5 Ã— n2) / (- 3 Ã— m5)

This can be written as,

= (3 Ã— 5 Ã— m5-5 Ã— n2) / 3

= 5 Ã— m0 Ã— n2

= 5 Ã— 1 Ã— n2

= 5n2

Hence, (-15m5n2) / (- 3m5) = 5n2

(ii) 35x4y2 / – 15x2y2

35x4y2 / – 15x2y2 = (35 Ã— x4 Ã— y2) / (- 15 Ã— x2 Ã— y2)

This can be written as,

= – (5 Ã— 7 Ã— x4-2 Ã— y2-2) / (3 Ã— 5)

= – (7 Ã— x2 Ã— y0) / 3

We get,

= – 7x2y / 3

Hence, 35x4y2 / – 15x2y2 = – 7x2y / 3

(iii) (- 24x6y2) / (6x6y)

(- 24x6y2) / (6x6y) = (- 24 Ã— x6 Ã— y2) / (6 Ã— x6 Ã— y)

This can be written as,

= (- 4 Ã— 6 Ã— x6-6 Ã— y2-1) / 6

= – 4 Ã— x0 Ã— y1

= – 4y

Hence, (- 24x6y2) / (6x6y) = – 4y

5. Divide:

(i) 9x3 â€“ 6x2 by 3x

(ii) 6m2 â€“ 16m3 + 10m4 by â€“ 2m

(iii) 15x3y2 + 25x2y3 â€“ 36x4y4 by 5x2y2

(iv) 36a3x5 â€“ 24a4x4 + 18a5x3 by â€“ 6a3x3

Solution:

(i) 9x3 â€“ 6x2 by 3x

9x3 â€“ 6x2 Ã· 3x = (9 Ã— x3 â€“ 6 Ã— x2) / (3 Ã— x)

Separating the terms, we get

= (9 Ã— x3) / (3 Ã— x) â€“ (6 Ã— x2) / (3 Ã— x)

We get,

= 3 Ã— x3-1 â€“ 2 Ã— x2-1

= 3x2 â€“ 2x

Hence, 9x3 â€“ 6x2 Ã· 3x = 3x2 â€“ 2x

(ii) 6m2 â€“ 16m3 + 10m4 by â€“ 2m

6m2 â€“ 16m3 + 10m4 Ã· – 2m = (6 Ã— m2– 16 Ã— m3 + 10 Ã— m4) / – 2 Ã— m

Separating the terms, we get

= (6 Ã— m2 / – 2 Ã— m) â€“ (16 Ã— m3) / (- 2 Ã— m) + (10 Ã— m4) / (- 2 Ã— m)

= – 3 Ã— m2-1 + 8 Ã— m3-1 â€“ 5 Ã— m4-1

= – 3 Ã— m + 8 Ã— m2 â€“ 5 Ã— m3

We get,

= – 3m + 8m2 â€“ 5m3

Hence, 6m2 â€“ 16m3 + 10m4 Ã· – 2m = – 3m + 8m2 â€“ 5m3

(iii) 15x3y2 + 25x2y3 â€“ 36x4y4 by 5x2y2

15x3y2 + 25x2y3 â€“ 36x4y4 Ã· 5x2y2 = (15x3y2 + 25x2y3 â€“ 36x4y4) / (5x2y2)

= (15 Ã— x3 Ã— y2) / (5 Ã— x2 Ã— y2) + (25 Ã— x2 Ã— y3) / (5 Ã— x2 Ã— y2) â€“ (36 Ã— x4 Ã— y4) / (5 Ã— x2 Ã— y2)

On further calculation, we get

= 3 Ã— x3-2 Ã— y2-2 + 5 Ã— x2-2 Ã— y3-2 â€“ (36 Ã— x4-2 Ã— y4-2) / 5

We get,

= 3 Ã— x1 Ã— y0 + 5 Ã— x0 Ã— y1 â€“ (36 Ã— x2 Ã— y2) / 5

= 3x + 5y â€“ (36x2y2) / 5

Hence, 15x3y2 + 25x2y3 â€“ 36x4y4 Ã· 5x2y2 = 3x + 5y â€“ (36x2y2) / 5

(iv) 36a3x5 â€“ 24a4x4 + 18a5x3 by â€“ 6a3x3

36a3x5 â€“ 24a4x4 + 18a5x3 Ã· (â€“ 6a3x3) = (36a3x5 â€“ 24a4x4 + 18a5x3) / – 6a3x3

= (36.a3.x5) / (-6.a3.x3) â€“ (24.a4.x4) / (-6.a3.x3) + (18.a5.x3) / (-6.a3.x3)

We get,

= – 6.x5-3 + 4.a4-3.x4-3 â€“ 3.a5-3

= – 6x2 + 4ax â€“ 3a2

Hence, 36a3x5 â€“ 24a4x4 + 18a5x3 Ã· (â€“ 6a3x3) = – 6x2 + 4ax â€“ 3a2