Selina Solutions Concise Maths Class 7 Chapter 11 Fundamental Concepts (Including Fundamental Operations) Exercise 11B has answers prepared by a set of expert teachers at BYJUâ€™S. The addition and subtraction of like and unlike terms are the important concepts discussed under this exercise. The solutions contain explanations in an interactive manner to improve interest among students. For further hold on the concepts, students can download Selina Solutions Concise Maths Class 7 Chapter 11 Fundamental Concepts (Including Fundamental Operations) Exercise 11B PDF, from the links which are available below.

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

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#### Exercise 11B page: 125

**1. Fill in the blanks:**

**(i) 8x + 5x = â€¦â€¦.**

**(ii) 8x â€“ 5x = â€¦â€¦**

**(iii) 6xy ^{2} + 9xy^{2} = â€¦â€¦.**

**(iv) 6xy ^{2} â€“ 9xy^{2} = â€¦â€¦.**

**(v) The sum of 8a, 6a and 5b = â€¦â€¦.**

**(vi) The addition of 5, 7xy, 6 and 3xy = â€¦â€¦..**

**(vii) 4a + 3b â€“ 7a + 4b = â€¦â€¦â€¦.**

**(viii) â€“ 15x + 13x + 8 = â€¦â€¦..**

**(ix) 6x ^{2}y + 13xy^{2} â€“ 4x^{2}y + 2xy^{2} = â€¦â€¦â€¦**

**(x) 16x ^{2} â€“ 9x^{2} = â€¦â€¦. and 25xy^{2} â€“ 17xy^{2} = â€¦â€¦â€¦**

**Solution: **

(i) 8x + 5x = 13x

(ii) 8x â€“ 5x = 3x

(iii) 6xy^{2} + 9xy^{2} = 15xy^{2}

(iv) 6xy^{2} â€“ 9xy^{2} = -3xy^{2}

(v) The sum of 8a, 6a and 5b = 14a + 5b

It can be written as

8a + 6a + 5b = 14a + 5b

(vi) The addition of 5, 7xy, 6 and 3xy = 11 + 10xy

It can be written as

5 + 7xy + 6 + 3xy = 11 + 10xy

(vii) 4a + 3b â€“ 7a + 4b = 7b â€“ 3a

It can be written as

4a + 3b â€“ 7a + 4b = (4 â€“ 7)a + (3 + 4)b

= -3a + 7b

(viii) â€“ 15x + 13x + 8 = 8 â€“ 2x

It can be written as

-15x + 13x + 8 = (-15 + 13) x + 8 = -2x + 8

(ix) 6x^{2}y + 13xy^{2} â€“ 4x^{2}y + 2xy^{2} = 2x^{2}y + 15xy^{2}

It can be written as

6x^{2}y + 13xy^{2} â€“ 4x^{2}y + 2xy^{2} = (6 â€“ 4) x^{2}y + (13 + 2) xy^{2}

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

(x) 16x^{2} â€“ 9x^{2} = 7x^{2} and 25xy^{2} â€“ 17xy^{2} = 8xy^{2}

**2. Add:**

**(i) -9x, 3x and 4x**

**(ii) 23y ^{2}, 8y^{2} and â€“ 12y^{2}**

**(iii) 18pq, -15pq and 3pq**

**Solution:**

(i) -9x, 3x and 4x

It can be written as

= -9x + 3x + 4x

So we get

= -9x + 7x

= -2x

(ii) 23y^{2}, 8y^{2} and â€“ 12y^{2}

It can be written as

= 23y^{2} + 8y^{2} â€“ 12y^{2}

So we get

= 31y^{2} â€“ 12y^{2}

= 19y^{2}

(iii) 18pq, -15pq and 3pq

It can be written as

= 18pq â€“ 15pq + 3pq

So we get

= 3pq + 3pq

= 6pq

**3. Simplify:**

**(i) 3m + 12m â€“ 5m**

**(ii) 7n ^{2} â€“ 9n^{2} + 3n^{2}**

**(iii) 25zy â€“ 8zy â€“ 6zy**

**(iv) -5ax ^{2} + 7ax^{2} â€“ 12ax^{2}**

**(v) â€“ 16am + 4mx + 4am â€“ 15mx + 5am**

**Solution:**

(i) 3m + 12m â€“ 5m

It can be written as

= 15m â€“ 5m

So we get

= 10m

(ii) 7n^{2} â€“ 9n^{2} + 3n^{2}

It can be written as

= (7 + 3) n^{2} â€“ 9n^{2}

So we get

= 10n^{2} â€“ 9n^{2}

= n^{2}

(iii) 25zy â€“ 8zy â€“ 6zy

It can be written as

= 25zy â€“ 14zy

So we get

= 11zy

(iv) -5ax^{2} + 7ax^{2} â€“ 12ax^{2}

It can be written as

= (-5 â€“ 12) ax^{2} + 7ax^{2}

So we get

= -17ax^{2} + 7ax^{2}

= -10ax^{2}

(v) â€“ 16am + 4mx + 4am â€“ 15mx + 5am

It can be written as

= (-16 + 4 + 5) am + (4 â€“ 15) mx

So we get

= – 7am â€“ 11mx

**4. Add:**

**(i) a + b and 2a + 3b**

**(ii) 2x + y and 3x â€“ 4y**

**(iii) -3a + 2b and 3a + b**

**(iv) 4 + x, 5 â€“ 2x and 6x**

**Solution:**

(i) a + b and 2a + 3b

It can be written as

= a + b + 2a + 3b

So we get

= a + 2a + b + 3b

= 3a + 4b

(ii) 2x + y and 3x â€“ 4y

It can be written as

= 2x + y + 3x â€“ 4y

So we get

= 2x + 3x + y â€“ 4y

= 5x â€“ 3y

(iii) -3a + 2b and 3a + b

It can be written as

= -3a + 2b + 3a + b

So we get

= -3a + 3a + 2b + b

= 3b

(iv) 4 + x, 5 â€“ 2x and 6x

It can be written as

= 4 + x + 5 â€“ 2x + 6x

So we get

= x â€“ 2x + 6x + 4 + 5

= 5x + 9

**5. Find the sum of:**

**(i) 3x + 8y + 7z, 6y + 4z â€“ 2x and 3y â€“ 4x + 6z**

**(ii) 3a + 5b + 2c, 2a + 3b â€“ c and a + b + c**

**(iii) 4x ^{2} + 8xy â€“ 2y^{2} and 8xy â€“ 5y^{2} + x^{2}**

**(iv) 9x ^{2} â€“ 6x + 7, 5 â€“ 4x and 6 â€“ 3x^{2}**

**(v) 5x ^{2} â€“ 2xy + 3y^{2}, -2x^{2} + 5xy + 9y^{2} and 3x^{2} â€“ xy â€“ 4y^{2}**

**Solution:**

(i) 3x + 8y + 7z, 6y + 4z â€“ 2x and 3y â€“ 4x + 6z

It can be written as

= 3x + 8y + 7z + 6y + 4z â€“ 2x + 3y â€“ 4x + 6z

By further calculation

= 3x â€“ 2x â€“ 4x + 8y + 6y + 3y + 7z + 4z + 6z

So we get

= 3x â€“ 6x + 17y + 17z

= -3x + 17y + 17z

(ii) 3a + 5b + 2c, 2a + 3b â€“ c and a + b + c

It can be written as

= 3a + 5b + 2c + 2a + 3b â€“ c + a + b + c

By further calculation

= 3a + 2a + a + 5b + 3b + b + 2c â€“ c + c

So we get

= 6a + 9b + 3c â€“ c

= 6a + 9b + 2c

(iii) 4x^{2} + 8xy â€“ 2y^{2} and 8xy â€“ 5y^{2} + x^{2}

It can be written as

= 4x^{2} + 8xy â€“ 2y^{2} + 8xy â€“ 5y^{2} + x^{2}

By further calculation

= 4x^{2} + x^{2} + 8xy + 8xy â€“ 2y^{2} â€“ 5y^{2}

So we get

= 5x^{2} + 16xy â€“ 7y^{2}

(iv) 9x^{2} â€“ 6x + 7, 5 â€“ 4x and 6 â€“ 3x^{2}

It can be written as

= 9x^{2} â€“ 6x + 7 + 5 â€“ 4x + 6 â€“ 3x^{2}

By further calculation

= 9x^{2} â€“ 3x^{2} â€“ 6x â€“ 4x + 7 + 5 + 6

So we get

= 6x^{2} â€“ 10x + 18

(v) 5x^{2} â€“ 2xy + 3y^{2}, -2x^{2} + 5xy + 9y^{2} and 3x^{2} â€“ xy â€“ 4y^{2}

It can be written as

= 5x^{2} â€“ 2xy + 3y^{2} – 2x^{2} + 5xy + 9y^{2} + 3x^{2} â€“ xy â€“ 4y^{2}

By further calculation

= 5x^{2} â€“ 2x^{2} + 3x^{2} â€“ 2xy + 5xy â€“ xy + 3y^{2} + 9y^{2} â€“ 4y^{2}

So we get

= 6x^{2} + 2xy + 8y^{2}

**6. Find the sum of:**

**(i) x and 3y**

**(ii) -2a and +5**

**(iii) -4x ^{2} and + 7x**

**(iv) +4a and -7b**

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

**(vi) 11 and â€“by**

**Solution:**

(i) x and 3y

The sum of x and 3y is x + 3y.

(ii) -2a and +5

The sum of -2a and + 5 is -2a + 5.

(iii) -4x^{2} and + 7x

The sum of -4x^{2} and + 7x is -4x^{2} + 7x.

(iv) +4a and -7b

The sum of +4a and -7b is + 4a â€“ 7b.

(v) x^{3}, 3x^{2}y and 2y^{2}

The sum of x^{3}, 3x^{2}y and 2y^{2} is x^{3} + 3x^{2}y + 2y^{2}.

(vi) 11 and â€“by

The sum of 11 and -by is 11 â€“ by.

**7. The sides of a triangle are 2x + 3y, x + 5y and 7x -2y. Find its perimeter.**

**Solution:**

It is given that

Sides of a triangle are 2x + 3y, x + 5y and 7x -2y

We know that

Perimeter = Sum of all three sides of a triangle

Substituting the values

= 2x + 3y + x + 5y + 7x â€“ 2y

By further calculation

= 2x + x + 7x + 3y + 5y â€“ 2y

So we get

= 10x + 8y â€“ 2y

= 10x + 6y

**8. The two adjacent sides of a rectangle are 6a + 9b and 8a â€“ 4b. Find its perimeter.**

**Solution:**

It is given that

Sides of a rectangle are 6a + 9b and 8a â€“ 4b

So length = 6a + 9b and breadth = 8a â€“ 4b

We know that

Perimeter = 2 (length + breadth)

Substituting the values

= 2 (6a + 9b + 8a â€“ 4b)

By further calculation

= 2 (14a + 5b)

So we get

= 28a + 10b

**9. Subtract the second expression from the first:**

**(i) 2a + b, a + b**

**(ii) -2b + 2c, b + 3c**

**(iii) 5a + b, -6b + 2a**

**(iv) a ^{3} â€“ 1 + a, 3a â€“ 2a^{2}**

**(v) p + 2, 1**

**Solution:**

(i) 2a + b, a + b

It can be written as

= (2a + b) â€“ (a + b)

So we get

= 2a + b â€“ a â€“ b

= 2a â€“ a + b â€“ b

= a

(ii) -2b + 2c, b + 3c

It can be written as

= (-2b + 2c) â€“ (b + 3c)

So we get

= -2b + 2c â€“ b â€“ 3c

= -2b â€“ b + 2c – 3c

= -3b â€“ c

(iii) 5a + b, -6b + 2a

It can be written as

= (5a + b) â€“ (-6b + 2a)

So we get

= 5a + b + 6b â€“ 2a

= 5a â€“ 2a + b + 6b

= 3a + 7b

(iv) a^{3} â€“ 1 + a, 3a â€“ 2a^{2}

It can be written as

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

So we get

= a^{3} â€“ 1 + a â€“ 3a + 2a^{2}

= a^{3} + 2a^{2} + a â€“ 3a â€“ 1

= a^{3} + 2a^{2} â€“ 2a â€“ 1

(v) p + 2, 1

It can be written as

= p + 2 â€“ 1

So we get

= p + 1

**10. Subtract:**

**(i) 4x from 8 â€“ x**

**(ii) -8c from c + 3d**

**(iii) â€“ 5a â€“ 2b from b + 6c**

**(iv) 4p + p ^{2} from 3p^{2} â€“ 8p**

**(v) 5a â€“ 3b + 2c from 4a â€“ b â€“ 2c**

**Solution:**

(i) 4x from 8 â€“ x

It can be written as

= (8 â€“ x) â€“ 4x

By further calculation

= 8 â€“ x â€“ 4x

= 8 â€“ 5x

(ii) -8c from c + 3d

It can be written as

= (c + 3d) â€“ (-8c)

By further calculation

= c + 3d + 8c

= 9c + 3d

(iii) â€“ 5a â€“ 2b from b + 6c

It can be written as

= (b + 6c) â€“ (-5a â€“ 2b)

By further calculation

= b + 6c + 5a + 2b

= 5a + 3b + 6c

(iv) 4p + p^{2} from 3p^{2} â€“ 8p

It can be written as

= (3p^{2} â€“ 8p) â€“ (4p + p^{2})

By further calculation

= 3p^{2} â€“ 8p â€“ 4p â€“ p^{2}

= 2p^{2} â€“ 12p

(v) 5a â€“ 3b + 2c from 4a â€“ b â€“ 2c

It can be written as

= (4a â€“ b â€“ 2c) â€“ (5a â€“ 3b + 2c)

By further calculation

= 4a â€“ b â€“ 2c â€“ 5a + 3b â€“ 2c

= -a + 2b â€“ 4c

**11. Subtract -5a ^{2} â€“ 3a + 1 from the sum of 4a^{2} + 3 â€“ 8a and 9a â€“ 7.**

**Solution:**

We know that

Sum of 4a^{2} + 3 â€“ 8a and 9a â€“ 7 can be written as

= 4a^{2} + 3 â€“ 8a + 9a â€“ 7

By further calculation

= 4a^{2} + a â€“ 4

Here

(4a^{2} + a â€“ 4) â€“ (-5a^{2} â€“ 3a + 1) = 4a^{2} + a â€“ 4 + 5a^{2} + 3a â€“ 1

By further calculation

= 4a^{2} + 5a^{2} + a + 3a â€“ 4 – 1

So we get

= 9a^{2} + 4a – 5

**12. By how much does 8x ^{3} â€“ 6x^{2} + 9x â€“ 10 exceed 4x^{3} + 2x^{2} + 7x â€“ 3?**

**Solution:**

We know that

8x^{3} â€“ 6x^{2} + 9x â€“ 10 exceed 4x^{3} + 2x^{2} + 7x â€“ 3

It can be written as

= (8x^{3} â€“ 6x^{2} + 9x â€“ 10) â€“ (4x^{3} + 2x^{2} + 7x â€“ 3)

By further calculation

= 8x^{3} â€“ 6x^{2} + 9x â€“ 10 â€“ 4x^{3} â€“ 2x^{2} -7x + 3

So we get

= 8x^{3} â€“ 4x^{3} â€“ 6x^{2} – 2x^{2} + 9x â€“ 7x â€“ 10 + 3

= 4x^{3} â€“ 8x^{2} + 2x – 7

**13. What must be added to 2a ^{3} + 5a â€“ a^{2} â€“ 6 to get a^{2} â€“ a â€“ a^{3} + 1?**

**Solution:**

The answer can be obtained by subtracting 2a^{3} + 5a â€“ a^{2} â€“ 6 from a^{2} â€“ a â€“ a^{3} + 1

= (-a^{3} + a^{2} â€“ a + 1) â€“ (2a^{3} + 5a â€“ a^{2} â€“ 6)

It can be written as

= -a^{3} + a^{2} â€“ a + 1 – 2a^{3} – 5a + a^{2} + 6

By further calculation

= – a^{3} – 2a^{3} + a^{2} + a^{2} â€“ a â€“ 5a + 1 + 6

= -3a^{3} + 2a^{2} â€“ 6a + 7

**14. What must be subtracted from a ^{2} + b^{2} + 2ab to get â€“ 4ab + 2b^{2}?**

**Solution:**

The answer can be obtained by subtracting â€“ 4ab + 2b^{2} from a^{2} + b^{2} + 2ab

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

It can be written as

= a^{2} + b^{2} + 2ab + 4ab – 2b^{2}

By further calculation

= a^{2} + b^{2} – 2b^{2} + 2ab + 4ab

= a^{2} – b^{2} + 6ab

**15. Find the excess of 4m ^{2} + 4n^{2} + 4p^{2} over m^{2} + 3n^{2} â€“ 5p^{2}.**

**Solution:**

The answer can be obtained by subtracting m^{2} + 3n^{2} â€“ 5p^{2} from 4m^{2} + 4n^{2} + 4p^{2}

= (4m^{2} + 4n^{2} + 4p^{2}) â€“ (m^{2} + 3n^{2} â€“ 5p^{2})

It can be written as

= 4m^{2} + 4n^{2} + 4p^{2} – m^{2} – 3n^{2} + 5p^{2}

By further calculation

= 4m^{2} – m^{2} + 4n^{2} – 3n^{2} + 4p^{2} + 5p^{2}

= 3m^{2} + n^{2} + 9p^{2}