Selina Solutions Concise Maths Class 7 Chapter 11 Fundamental Concepts (Including Fundamental Operations) help students get a clear idea of operations and methods to be followed in solving problems effortlessly. The solutions contain explanations of basic concepts in an understandable language. Shortcut tricks and important formulas are covered in a stepwise manner for better conceptual knowledge among students. Selina Solutions Concise Maths Class 7 Chapter 11 Fundamental Concepts (Including Fundamental Operations) PDF can be downloaded from the links provided below.
Chapter 11 explains the elementary treatment and operations on various terms, as per the latest syllabus and ICSE exam. The main aim of preparing chapter-wise solutions is to provide a better hold on the subject among students.
Selina Solutions Concise Maths Class 7 Chapter 11: Fundamental Concepts (Including Fundamental Operations) Download PDF
Exercises of Selina Solutions Concise Maths Class 7 Chapter 11 – Fundamental Concepts (Including Fundamental Operations)
Access Selina Solutions Concise Maths Class 7 Chapter 11: Fundamental Concepts (Including Fundamental Operations)
Exercise 11A page: 121
1. Separate constant terms and variable terms from the following:
8, x, 6xy, 6 + x, – 5xy2, 15az2, 32z/ xy, y2/ 3x
Solution:
The constant term is 8.
The variable terms are x, 6xy, 6 + x, – 5xy2, 15az2, 32z/ xy, y2/ 3x.
2. For each expression, given below, state whether it is a monomial, binomial or trinomial:
(i) 2x ÷ 15
(ii) ax + 9
(iii) 3x2 × 5x
(iv) 5 + 2x – 3b
(v) 2y – 7z/3 ÷ x
(vi) 3p × q ÷ z
(vii) 12z ÷ 5x + 4
(viii) 12 – 5z – 4
(ix) a3 -3ab2 × c
Solution:
(i) 2x ÷ 15 = 2x/15
It has one term and hence it is a monomial.
(ii) ax + 9
It has two terms and hence it is a binomial.
(iii) 3x2 × 5x = 15x3
It has one term and hence it is a monomial.
(iv) 5 + 2x – 3b
It has three terms and hence it is a trinomial.
(v) 2y – 7z/3 ÷ x = 2y – 7z/3x
It has two terms and hence it is a binomial.
(vi) 3p × q ÷ z = 3pq/z
It has one term and hence it is a monomial.
(vii) 12z ÷ 5x + 4 = 12z/5x + 4
It has two terms and hence it is a binomial.
(viii) 12 – 5z – 4 = 8 – 5z
It has two terms and hence it is a binomial.
(ix) a3 -3ab2 × c = a3 – 3ab2c
It has two terms and hence it is a binomial.
3. Write the coefficient of:
(i) xy in -3axy
(ii) z2 in p2yz2
(iii) mn in – mn
(iv) 15 in -15p2
Solution:
(i) xy in -3axy
The coefficient of xy in -3axy = -3a
(ii) z2 in p2yz2
The coefficient of z2 in p2yz2 = p2y
(iii) mn in – mn
The coefficient of mn in – mn = -1
(iv) 15 in -15p2
The coefficient of 15 in -15p2 = -p2
4. For each of the following monomials, write its degree:
(i) 7y
(ii) –x2y
(iii) xy2z
(iv) -9y2z3
(v) 3m3n4
(vi) -2p2q3r4
Solution:
(i) The degree of 7y is 1.
(ii) The degree of –x2y = 2 + 1 = 3
(iii) The degree of xy2z = 1 + 2 + 1 = 4
(iv) The degree of -9y2z3 = 2 + 3 = 5
(v) The degree of 3m3n4 = 3 + 4 = 7
(vi) The degree of -2p2q3r4 = 2 + 3 + 4 = 9
5. Write the degree of each of the following polynomials:
(i) 3y3 – x2y2 + 4x
(ii) p3q2 – 6p2q5 + p4q4
(iii) -8mn6 + 5m3n
(iv) 7 – 3x2y + y2
(v) 3x – 15
(vi) 2y2z + 9yz3
Solution:
(i) The degree of 3y3 – x2y2 + 4x is 4
x2y2 is the term which has the highest degree.
(ii) The degree of p3q2 – 6p2q5 + p4q4 is 8
p4q4 is the term which has the highest degree.
(iii) The degree of -8mn6 + 5m3n is 7
-8mn6 is the term which has the highest degree.
(iv) The degree of 7 – 3x2y + y2 is 3
– 3x2y is the term which has the highest degree.
(v) The degree of 3x – 15 is 1
3x is the term which has the highest degree.
(vi) The degree of 2y2z + 9yz3 is 4
9yz3 is the term which has the highest degree.
6. Group the like terms together:
(i) 9x2, xy, -3x2, x2 and -2xy
(ii) ab, -a2b, -3ab, 5a2b and -8a2b.
(iii) 7p, 8pq, -5pq, -2p and 3p
Solution:
(i) 9x2, xy, -3x2, x2 and -2xy
9x2, -3x2 and x2 are like terms
xy and -2xy are like terms.
(ii) ab, -a2b, -3ab, 5a2b and -8a2b
-a2b, 5a2b and -8a2b are like terms
ab and – 3ab are like terms.
(iii) 7p, 8pq, -5pq, -2p and 3p
7p, -2p and 3p are like terms
8pq and -5pq are like terms.
7. Write the numerical coefficient of each of the following:
(i) y
(ii) – y
(iii) 2x2y
(iv) -8xy3
(v) 3py2
(vi) -9a2b3
Solution:
(i) The numerical coefficient of y is 1.
(ii) The numerical coefficient of – y is – 1.
(iii) The numerical coefficient of 2x2y is 2.
(iv) The numerical coefficient of -8xy3 is -8.
(v) The numerical coefficient of 3py2 is 3.
(vi) The numerical coefficient of -9a2b3 is -9.
8. In -5x3y2z4; write the coefficient of:
(i) z2
(ii) y2
(iii) yz2
(iv) x3y
(v) –xy2
(vi) -5xy2z
Also, write the degree of the given algebraic expression.
Solution:
(i) The coefficient of z2 is -5x3y2z2.
(ii) The coefficient of y2 is -5x3z4.
(iii) The coefficient of yz2 is -5x3yz2.
(iv) The coefficient of x3y is -5yz4.
(v) The coefficient of –xy2 is 5x2z4.
(vi) The coefficient of -5xy2z is x2z3.
So the degree of the given algebraic expression = 3 + 2 + 4 = 9.
Exercise 11B page: 125
1. Fill in the blanks:
(i) 8x + 5x = …….
(ii) 8x – 5x = ……
(iii) 6xy2 + 9xy2 = …….
(iv) 6xy2 – 9xy2 = …….
(v) The sum of 8a, 6a and 5b = …….
(vi) The addition of 5, 7xy, 6 and 3xy = ……..
(vii) 4a + 3b – 7a + 4b = ……….
(viii) – 15x + 13x + 8 = ……..
(ix) 6x2y + 13xy2 – 4x2y + 2xy2 = ………
(x) 16x2 – 9x2 = ……. and 25xy2 – 17xy2 = ………
Solution:
(i) 8x + 5x = 13x
(ii) 8x – 5x = 3x
(iii) 6xy2 + 9xy2 = 15xy2
(iv) 6xy2 – 9xy2 = -3xy2
(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 – 7a + 4b = 7b – 3a
It can be written as
4a + 3b – 7a + 4b = (4 – 7)a + (3 + 4)b
= -3a + 7b
(viii) – 15x + 13x + 8 = 8 – 2x
It can be written as
-15x + 13x + 8 = (-15 + 13) x + 8 = -2x + 8
(ix) 6x2y + 13xy2 – 4x2y + 2xy2 = 2x2y + 15xy2
It can be written as
6x2y + 13xy2 – 4x2y + 2xy2 = (6 – 4) x2y + (13 + 2) xy2
= 2x2y + 15xy2
(x) 16x2 – 9x2 = 7x2 and 25xy2 – 17xy2 = 8xy2
2. Add:
(i) -9x, 3x and 4x
(ii) 23y2, 8y2 and – 12y2
(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) 23y2, 8y2 and – 12y2
It can be written as
= 23y2 + 8y2 – 12y2
So we get
= 31y2 – 12y2
= 19y2
(iii) 18pq, -15pq and 3pq
It can be written as
= 18pq – 15pq + 3pq
So we get
= 3pq + 3pq
= 6pq
3. Simplify:
(i) 3m + 12m – 5m
(ii) 7n2 – 9n2 + 3n2
(iii) 25zy – 8zy – 6zy
(iv) -5ax2 + 7ax2 – 12ax2
(v) – 16am + 4mx + 4am – 15mx + 5am
Solution:
(i) 3m + 12m – 5m
It can be written as
= 15m – 5m
So we get
= 10m
(ii) 7n2 – 9n2 + 3n2
It can be written as
= (7 + 3) n2 – 9n2
So we get
= 10n2 – 9n2
= n2
(iii) 25zy – 8zy – 6zy
It can be written as
= 25zy – 14zy
So we get
= 11zy
(iv) -5ax2 + 7ax2 – 12ax2
It can be written as
= (-5 – 12) ax2 + 7ax2
So we get
= -17ax2 + 7ax2
= -10ax2
(v) – 16am + 4mx + 4am – 15mx + 5am
It can be written as
= (-16 + 4 + 5) am + (4 – 15) mx
So we get
= – 7am – 11mx
4. Add:
(i) a + b and 2a + 3b
(ii) 2x + y and 3x – 4y
(iii) -3a + 2b and 3a + b
(iv) 4 + x, 5 – 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 – 4y
It can be written as
= 2x + y + 3x – 4y
So we get
= 2x + 3x + y – 4y
= 5x – 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 – 2x and 6x
It can be written as
= 4 + x + 5 – 2x + 6x
So we get
= x – 2x + 6x + 4 + 5
= 5x + 9
5. Find the sum of:
(i) 3x + 8y + 7z, 6y + 4z – 2x and 3y – 4x + 6z
(ii) 3a + 5b + 2c, 2a + 3b – c and a + b + c
(iii) 4x2 + 8xy – 2y2 and 8xy – 5y2 + x2
(iv) 9x2 – 6x + 7, 5 – 4x and 6 – 3x2
(v) 5x2 – 2xy + 3y2, -2x2 + 5xy + 9y2 and 3x2 – xy – 4y2
Solution:
(i) 3x + 8y + 7z, 6y + 4z – 2x and 3y – 4x + 6z
It can be written as
= 3x + 8y + 7z + 6y + 4z – 2x + 3y – 4x + 6z
By further calculation
= 3x – 2x – 4x + 8y + 6y + 3y + 7z + 4z + 6z
So we get
= 3x – 6x + 17y + 17z
= -3x + 17y + 17z
(ii) 3a + 5b + 2c, 2a + 3b – c and a + b + c
It can be written as
= 3a + 5b + 2c + 2a + 3b – c + a + b + c
By further calculation
= 3a + 2a + a + 5b + 3b + b + 2c – c + c
So we get
= 6a + 9b + 3c – c
= 6a + 9b + 2c
(iii) 4x2 + 8xy – 2y2 and 8xy – 5y2 + x2
It can be written as
= 4x2 + 8xy – 2y2 + 8xy – 5y2 + x2
By further calculation
= 4x2 + x2 + 8xy + 8xy – 2y2 – 5y2
So we get
= 5x2 + 16xy – 7y2
(iv) 9x2 – 6x + 7, 5 – 4x and 6 – 3x2
It can be written as
= 9x2 – 6x + 7 + 5 – 4x + 6 – 3x2
By further calculation
= 9x2 – 3x2 – 6x – 4x + 7 + 5 + 6
So we get
= 6x2 – 10x + 18
(v) 5x2 – 2xy + 3y2, -2x2 + 5xy + 9y2 and 3x2 – xy – 4y2
It can be written as
= 5x2 – 2xy + 3y2 – 2x2 + 5xy + 9y2 + 3x2 – xy – 4y2
By further calculation
= 5x2 – 2x2 + 3x2 – 2xy + 5xy – xy + 3y2 + 9y2 – 4y2
So we get
= 6x2 + 2xy + 8y2
6. Find the sum of:
(i) x and 3y
(ii) -2a and +5
(iii) -4x2 and + 7x
(iv) +4a and -7b
(v) x3, 3x2y and 2y2
(vi) 11 and –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) -4x2 and + 7x
The sum of -4x2 and + 7x is -4x2 + 7x.
(iv) +4a and -7b
The sum of +4a and -7b is + 4a – 7b.
(v) x3, 3x2y and 2y2
The sum of x3, 3x2y and 2y2 is x3 + 3x2y + 2y2.
(vi) 11 and –by
The sum of 11 and -by is 11 – 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 – 2y
By further calculation
= 2x + x + 7x + 3y + 5y – 2y
So we get
= 10x + 8y – 2y
= 10x + 6y
8. The two adjacent sides of a rectangle are 6a + 9b and 8a – 4b. Find its perimeter.
Solution:
It is given that
Sides of a rectangle are 6a + 9b and 8a – 4b
So length = 6a + 9b and breadth = 8a – 4b
We know that
Perimeter = 2 (length + breadth)
Substituting the values
= 2 (6a + 9b + 8a – 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) a3 – 1 + a, 3a – 2a2
(v) p + 2, 1
Solution:
(i) 2a + b, a + b
It can be written as
= (2a + b) – (a + b)
So we get
= 2a + b – a – b
= 2a – a + b – b
= a
(ii) -2b + 2c, b + 3c
It can be written as
= (-2b + 2c) – (b + 3c)
So we get
= -2b + 2c – b – 3c
= -2b – b + 2c – 3c
= -3b – c
(iii) 5a + b, -6b + 2a
It can be written as
= (5a + b) – (-6b + 2a)
So we get
= 5a + b + 6b – 2a
= 5a – 2a + b + 6b
= 3a + 7b
(iv) a3 – 1 + a, 3a – 2a2
It can be written as
= (a3 – 1 + a) – (3a – 2a2)
So we get
= a3 – 1 + a – 3a + 2a2
= a3 + 2a2 + a – 3a – 1
= a3 + 2a2 – 2a – 1
(v) p + 2, 1
It can be written as
= p + 2 – 1
So we get
= p + 1
10. Subtract:
(i) 4x from 8 – x
(ii) -8c from c + 3d
(iii) – 5a – 2b from b + 6c
(iv) 4p + p2 from 3p2 – 8p
(v) 5a – 3b + 2c from 4a – b – 2c
Solution:
(i) 4x from 8 – x
It can be written as
= (8 – x) – 4x
By further calculation
= 8 – x – 4x
= 8 – 5x
(ii) -8c from c + 3d
It can be written as
= (c + 3d) – (-8c)
By further calculation
= c + 3d + 8c
= 9c + 3d
(iii) – 5a – 2b from b + 6c
It can be written as
= (b + 6c) – (-5a – 2b)
By further calculation
= b + 6c + 5a + 2b
= 5a + 3b + 6c
(iv) 4p + p2 from 3p2 – 8p
It can be written as
= (3p2 – 8p) – (4p + p2)
By further calculation
= 3p2 – 8p – 4p – p2
= 2p2 – 12p
(v) 5a – 3b + 2c from 4a – b – 2c
It can be written as
= (4a – b – 2c) – (5a – 3b + 2c)
By further calculation
= 4a – b – 2c – 5a + 3b – 2c
= -a + 2b – 4c
11. Subtract -5a2 – 3a + 1 from the sum of 4a2 + 3 – 8a and 9a – 7.
Solution:
We know that
Sum of 4a2 + 3 – 8a and 9a – 7 can be written as
= 4a2 + 3 – 8a + 9a – 7
By further calculation
= 4a2 + a – 4
Here
(4a2 + a – 4) – (-5a2 – 3a + 1) = 4a2 + a – 4 + 5a2 + 3a – 1
By further calculation
= 4a2 + 5a2 + a + 3a – 4 – 1
So we get
= 9a2 + 4a – 5
12. By how much does 8x3 – 6x2 + 9x – 10 exceed 4x3 + 2x2 + 7x – 3?
Solution:
We know that
8x3 – 6x2 + 9x – 10 exceed 4x3 + 2x2 + 7x – 3
It can be written as
= (8x3 – 6x2 + 9x – 10) – (4x3 + 2x2 + 7x – 3)
By further calculation
= 8x3 – 6x2 + 9x – 10 – 4x3 – 2x2 -7x + 3
So we get
= 8x3 – 4x3 – 6x2 – 2x2 + 9x – 7x – 10 + 3
= 4x3 – 8x2 + 2x – 7
13. What must be added to 2a3 + 5a – a2 – 6 to get a2 – a – a3 + 1?
Solution:
The answer can be obtained by subtracting 2a3 + 5a – a2 – 6 from a2 – a – a3 + 1
= (-a3 + a2 – a + 1) – (2a3 + 5a – a2 – 6)
It can be written as
= -a3 + a2 – a + 1 – 2a3 – 5a + a2 + 6
By further calculation
= – a3 – 2a3 + a2 + a2 – a – 5a + 1 + 6
= -3a3 + 2a2 – 6a + 7
14. What must be subtracted from a2 + b2 + 2ab to get – 4ab + 2b2?
Solution:
The answer can be obtained by subtracting – 4ab + 2b2 from a2 + b2 + 2ab
= a2 + b2 + 2ab – (– 4ab + 2b2)
It can be written as
= a2 + b2 + 2ab + 4ab – 2b2
By further calculation
= a2 + b2 – 2b2 + 2ab + 4ab
= a2 – b2 + 6ab
15. Find the excess of 4m2 + 4n2 + 4p2 over m2 + 3n2 – 5p2.
Solution:
The answer can be obtained by subtracting m2 + 3n2 – 5p2 from 4m2 + 4n2 + 4p2
= (4m2 + 4n2 + 4p2) – (m2 + 3n2 – 5p2)
It can be written as
= 4m2 + 4n2 + 4p2 – m2 – 3n2 + 5p2
By further calculation
= 4m2 – m2 + 4n2 – 3n2 + 4p2 + 5p2
= 3m2 + n2 + 9p2
Exercise 11C page: 129
1. Multiply:
(i) 3x, 5x2y and 2y
(ii) 5, 3a and 2ab2
(iii) 5x + 2y and 3xy
(iv) 6a – 5b and – 2a
(v) 4a + 5b and 4a – 5b
Solution:
(i) 3x, 5x2y and 2y
Product = 3x × 5x2y × 2y
We can write it as
= 3 × 5 × 2 × x × x2 × y × y
So we get
= 30x3y2
(ii) 5, 3a and 2ab2
Product = 5 × 3a × 2ab2
We can write it as
= 5 × 3 × 2 × a × ab2
So we get
= 30a2b2
(iii) 5x + 2y and 3xy
Product = 3xy (5x + 2y)
We can write it as
= 3xy × 5x + 3xy × 2y
So we get
= 15x2y + 6xy2
(iv) 6a – 5b and – 2a
Product = – 2a (6a – 5b)
We can write it as
= -2a × 6a + 2a × 5b
So we get
= -12a2 + 10ab
(v) 4a + 5b and 4a – 5b
Product = (4a + 5b) (4a – 5b)
So we get
= 16a2 – 25b2
2. Copy and complete the following multiplications:
Solution:
3. Evaluate:
(i) (c + 5) (c – 3)
(ii) (3c – 5d) (4c – 6d)
(iii) (1/2a + 1/2b) (1/2a – 1/2b)
(iv) (a2 + 2ab + b2) (a + b)
(v) (3x – 1) (4x3 – 2x2 + 6x – 3)
Solution:
(i) (c + 5) (c – 3)
It can be written as
= c (c – 3) + 5 (c – 3)
By further calculation
= c2 – 3c + 5c – 15
= c2 + 2c – 15
(ii) (3c – 5d) (4c – 6d)
It can be written as
= 3c (4c – 6d) – 5d (4c – 6d)
By further calculation
= 12c2 – 18cd – 20cd + 30d2
= 12c2 – 38cd + 30d2
(iii) (1/2a + 1/2b) (1/2a – 1/2b)
It can be written as
= 1/2a (1/2a – 1/2b) + 1/2b (1/2a – 1/2b)
By further calculation
= 1/4a2 – 1/4ab + 1/4ab – 1/4b2
= 1/4a2 – 1/4b2
(iv) (a2 + 2ab + b2) (a + b)
It can be written as
= a (a2 + 2ab + b2) + b (a2 + 2ab + b2)
By further calculation
= a3 + 2a2b + ab2 + a2b + 2ab2 + b3
= a3 + b3 + 3a2b + 3ab2
(v) (3x – 1) (4x3 – 2x2 + 6x – 3)
It can be written as
= 3x (4x3 – 2x2 + 6x – 3) – 1 (4x3 – 2x2 + 6x – 3)
By further calculation
= 12x4 – 6x3 + 18x2 – 9x – 4x3 + 2x2 – 6x + 3
= 12x4 – 6x3 – 4x3 + 18x2 + 2x2 – 9x – 6x + 3
So we get
= 12x4 – 10x3 + 20x2 – 15x + 3
4. Evaluate:
(i) (a + b) (a – b).
(ii) (a2 + b2) (a + b) (a – b), using the result of (i).
(iii) (a4 + b4) (a2 + b2) (a + b) (a – b), using the result of (ii).
Solution:
(i) (a + b) (a – b).
It can be written as
= a (a – b) + b (a – b)
By further calculation
= a2 – ab + ab – b2
= a2 – b2
(ii) (a2 + b2) (a + b) (a – b)
Substituting the result of (i)
= (a2 + b2) (a2 – b2)
It can be written as
= a2 (a2 – b2) + b2 (a2 – b2)
So we get
= a4 – a2b2 + a2b2 – b4
= a4 – b4
(iii) (a4 + b4) (a2 + b2) (a + b) (a – b)
Substituting the result of (ii)
= (a4 + b4) (a4 – b4)
It can be written as
= a4 (a4 – b4) + b4 (a4 – b4)
By further calculation
= a8 – a4b4 + a4b4 – b8
= a8 – b8
5. Evaluate:
(i) (3x – 2y) (4x + 3y)
(ii) (3x – 2y) (4x + 3y) (8x – 5y)
(iii) (a + 5) (3a – 2) (5a + 1)
(iv) (a + 1) (a2 – a + 1) and (a – 1) (a2 + a + 1); and then: (a + 1) (a2 – a + 1) + (a – 1) (a2 + a + 1)
(v) (5m – 2n) (5m + 2n) (25m2 + 4n2)
Solution:
(i) (3x – 2y) (4x + 3y)
It can be written as
= 3x (4x + 3y) – 2y (4x + 3y)
By further calculation
= 12x2 + 9xy – 8xy – 6y2
So we get
= 12x2 + xy – 6y2
(ii) (3x – 2y) (4x + 3y) (8x – 5y)
Substituting result of (i)
= (12x2 + xy – 6y2) (8x – 5y)
It can be written as
= 8x (12x2 + xy – 6y2) – 5y (12x2 + xy – 6y2)
By further calculation
= 96x3 + 8x2y – 48xy2 – 60x2y – 5xy2 + 30y3
So we get
= 96x3 + 8x2y – 60x2y – 48xy2 – 5xy2 + 30y3
= 96x3 – 52 x2y – 53xy2 + 30y3
(iii) (a + 5) (3a – 2) (5a + 1)
It can be written as
= a (3a – 2) + 5 (3a – 2) (5a + 1)
By further calculation
= (3a2 – 2a + 15a – 10) (5a + 1)
So we get
= (3a2 + 13a – 10) (5a + 1)
We can write it as
= 5a (3a2 + 13a – 10) + 1 (3a2 + 13a – 10)
By further calculation
= 15a3 + 65a2 – 50a + 3a2 + 13a – 10
= 15a3 + 68a2 – 37a – 10
(iv) (a + 1) (a2 – a + 1) and (a – 1) (a2 + a + 1); and then: (a + 1) (a2 – a + 1) + (a – 1) (a2 + a + 1)
Consider
(a + 1) (a2 – a + 1)
It can be written as
= a (a2 – a + 1) + 1 (a2 – a + 1)
By further calculation
= a3 – a2 + a + a2 – a + 1
So we get
= a3 + 1
(a – 1) (a2 + a + 1)
It can be written as
= a (a2 + a + 1) – 1 (a2 + a + 1)
By further calculation
= a3 + a2 + a – a2 – a – 1
So we get
= a3 – 1
Here
(a + 1) (a2 – a + 1) + (a – 1) (a2 + a + 1)
= a3 + 1 + a3 – 1
= 2a3
(v) (5m – 2n) (5m + 2n) (25m2 + 4n2)
It can be written as
= [5m (5m + 2n) – 2n (5m + 2n)] (25m2 + 4n2)
By further calculation
= (25m2 + 10mn – 10mn – 4n2) (25m2 + 4n2)
So we get
= (25m2 – 4n2) (25m2 + 4n2)
We can write it as
= 25m2 (25m2 + 4n2) – 4n2 (25m2 + 4n2)
By multiplying the terms
= 625m4 + 100m2n2 – 100m2n2 – 16n4
= 625m4 – 16n4
6. Multiply:
(i) mn4, m3n and 5m2n3
(ii) 2mnpq, 4mnpq and 5mnpq
(iii) pq – pm and p2m
(iv) x3 – 3y3 and 4x2y2
(v) a3 – 4ab and 2a2b
Solution:
(i) mn4, m3n and 5m2n3
It can be written as
= 5m2n3 × mn4 × m3n
By further calculation
= 5m(2 + 1 + 3) n(3 + 4 + 1)
= 5m6n8
(ii) 2mnpq, 4mnpq and 5mnpq
It can be written as
= 5mnpq × 2mnpq × 4mnpq
By further calculation
= 5 × 2 × 4 m(1 + 1 + 1) n(1 + 1 + 1) p(1 + 1 + 1) q(1 + 1 + 1)
= 40m3n3p3q3
(iii) pq – pm and p2m
It can be written as
= p2m × (pq – pm)
So we get
= p3qm – p3m2
(iv) x3 – 3y3 and 4x2y2
It can be written as
= 4x2y2 × (x3 – 3y3)
By further calculation
= 4x5y2 – 12x2y5
(v) a3 – 4ab and 2a2b
It can be written as
= 2a2b × (a3 – 4ab)
By further calculation
= 2a5b – 8a3b2
7. Multiply:
(i) (2x + 3y) (2x + 3y)
(ii) (2x – 3y) (2x + 3y)
(iii) (2x + 3y) (2x – 3y)
(iv) (2x – 3y) (2x – 3y)
(v) (-2x + 3y) (2x – 3y)
Solution:
(i) (2x + 3y) (2x + 3y)
It can be written as
= 2x (2x + 3y) + 3y (2x + 3y)
By further calculation
= 4x2 + 6xy + 6xy + 9y2
= 4x2 + 12xy + 9y2
(ii) (2x – 3y) (2x + 3y)
It can be written as
= 2x (2x + 3y) – 3y (2x + 3y)
By further calculation
= 4x2 + 6xy – 6xy – 9y2
= 4x2 – 9y2
(iii) (2x + 3y) (2x – 3y)
It can be written as
= 2x (2x – 3y) + 3y (2x – 3y)
By further calculation
= 4x2 – 6xy + 6xy – 9y2
= 4x2 – 9y2
(iv) (2x – 3y) (2x – 3y)
It can be written as
= 2x (2x – 3y) – 3y (2x – 3y)
By further calculation
= 4x2 – 6xy – 6xy + 9y2
= 4x2 – 12xy + 9y2
(v) (-2x + 3y) (2x – 3y)
It can be written as
= -2x (2x – 3y) + 3y (2x – 3y)
By further calculation
= – 4x2 + 6xy + 6xy – 9y2
= – 4x2 + 12xy – 9y2
Exercise 11D page: 132
1. Divide:
(i) – 16ab2c by 6abc
(ii) 25x2y by – 5y2
(iii) 8x + 24 by 4
(iv) 4a2 – a by – a
(v) 8m – 16 by – 8
Solution:
(i) – 16ab2c by 6abc
We can write it as
= – 16ab2c/ 6abc
= -8/3 b
(ii) 25x2y by – 5y2
We can write it as
= 25x2y/ -5y2
= -5 x2/y
(iii) 8x + 24 by 4
We can write it as
= (8x + 24)/4
Separating the terms
= 8x/4 + 24/4
= 2x + 6
(iv) 4a2 – a by – a
We can write it as
= (4a2 – a)/ -a
Separating the terms
= 4a2/-a – a/-a
= – 4a + 1
(v) 8m – 16 by – 8
We can write it as
= (8m – 16)/ -8
Separating the terms
= 8m/-8 – 16/-8
= – m + 2
2. Divide:
(i) n2 – 2n + 1 by n – 1
(ii) m2 – 2mn + n2 by m – n
(iii) 4a2 + 4a + 1 by 2a + 1
(iv) p2 + 4p + 4 by p + 2
(v) x2 + 4xy + 4y2 by x + 2y
Solution:
(i) n2 – 2n + 1 by n – 1
n2 – 2n + 1 by n – 1 = n – 1
(ii) m2 – 2mn + n2 by m – n
m2 – 2mn + n2 by m – n = m – n
(iii) 4a2 + 4a + 1 by 2a + 1
4a2 + 4a + 1 by 2a + 1 = 2a + 1
(iv) p2 + 4p + 4 by p + 2
p2 + 4p + 4 by p + 2 = p + 2
(v) x2 + 4xy + 4y2 by x + 2y
x2 + 4xy + 4y2 by x + 2y = x + 2y
3. The area of a rectangle is 6x2 – 4xy – 10y2 square unit and its length is 2x + 2y unit. Find its breadth.
Solution:
It is given that
Area of a rectangle = 6x2 – 4xy – 10y2 square unit
Length = 2x + 2y unit
We know that
Breadth = Area/ Length
So we get
= (6x2 – 4xy – 10y2)/ (2x + 2y)
= 3x – 5y units
4. The area of a rectangular field is 25x2 + 20xy + 3y2 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 = 25x2 + 20xy + 3y2 square unit
Length = 5x +3y unit
We know that
Breadth = Area/ Length
So we get
= (25x2 + 20xy + 3y2)/ (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) 2m3n5 by – mn
(ii) 5x2 – 3x by x
(iii) 10x3y – 9xy2 – 4x2y2 by xy
(iv) 3y3 – 9ay2 – 6ab2y by – 3y
(v) x5 – 15x4 – 10x2 by – 5x2
Solution:
(i) 2m3n5 by – mn
It can be written as
= 2m3n5/ -mn
= -2m2n4
(ii) 5x2 – 3x by x
It can be written as
= (5x2 – 3x)/ x
Separating the terms
= 5x2/x – 3x/x
= 5x – 3
(iii) 10x3y – 9xy2 – 4x2y2 by xy
It can be written as
= (10x3y – 9xy2 – 4x2y2)/ xy
Separating the terms
= 10x3y/xy – 9xy2/xy – 4x2y2/xy
= 10x2 – 9y – 4xy
(iv) 3y3 – 9ay2 – 6ab2y by – 3y
It can be written as
= (3y3 – 9ay2 – 6ab2y)/ -3y
Separating the terms
= 3y3/-3y – 9ay2/-3y – 6ab2y/ -3y
= -y2 + 3ay + 2ab2
(v) x5 – 15x4 – 10x2 by – 5x2
It can be written as
= (x5 – 15x4 – 10x2)/ -5x2
Separating the terms
= x5/-5x2 – 15x4/-5x2 – 10x2/-5x2
= -1/5x3 + 3x2 + 2
Exercise 11E page: 133
Simplify:
1. x/2 + x/4
Solution:
x/2 + x/4
Taking LCM
= (2x + x)/4
= 3x/4
2. a/10 + 2a/5
Solution:
a/10 + 2a/5
Taking LCM
= (a + 4a)/ 10
= 5a/10
= a/2
3. y/4 + 3y/5
Solution:
y/4 + 3y/5
Taking LCM
= (5y + 12y)/ 20
= 17y/20
4. x/2 – x/8
Solution:
x/2 – x/8
Taking LCM
= (4x – x)/ 8
= 3x/8
5. 3y/4 – y/5
Solution:
3y/4 – y/5
Taking LCM
= (15y – 4y)/ 20
= 11y/20
6. 2p/3 – 3p/5
Solution:
2p/3 – 3p/5
Taking LCM
= (10p – 9p)/ 15
= p/15
7. k/2 + k/3 + 2k/5
Solution:
k/2 + k/3 + 2k/5
Here the LCM of 2, 3 and 5 is 30
= (15k + 10k + 12k)/ 30
= 37k/30
8. 2x/5 + 3x/4 – 3x/5
Solution:
2x/5 + 3x/4 – 3x/5
Here the LCM of 5 and 4 is 20
= (8x + 15x – 12x)/ 20
= 11x/20
9. 4a/7 – 2a/3 + a/7
Solution:
4a/7 – 2a/3 + a/7
Here the LCM of 3 and 7 is 21
= (12a – 14a + 3a)/ 21
= a/21
10. 2b/5 – 7b/15 + 13b/3
Solution:
2b/5 – 7b/15 + 13b/3
Here the LCM of 3, 5 and 15 is 15
= (6b – 7b + 65b)/ 15
= 64b/15
11. 6k/7 – (8k/9 – k/3)
Solution:
6k/7 – (8k/9 – k/3)
Here the LCM of 7, 9 and 3 is 63
= [54k – (56k – 21k)]/ 63
By further calculation
= (54k – 35k)/ 63
= 19k/63
12. 3a/8 + 4a/5 – (a/2 + 2a/5)
Solution:
3a/8 + 4a/5 – (a/2 + 2a/5)
Here the LCM of 8, 5 and 2 is 40
= [15a + 32a – (20a + 16a)]/ 40
By further calculation
= (47a – 36a)/ 40
= 11a/40
13. x + x/2 + x/3
Solution:
x + x/2 + x/3
Taking LCM
= (6x + 3x + 2x)/ 6
= 11x/6
14. y/5 + y – 19y/15
Solution:
y/5 + y – 19y/15
Here the LCM of 5 and 15 is 15
= (3y + 15y – 19y)/ 15
So we get
= – y/15
15. x/5 + (x + 1)/2
Solution:
x/5 + (x + 1)/2
Here the LCM of 5 and 2 is 10
= (2x + 5x + 5)/ 10
= (7x + 5)/ 10
Exercise 11F page: 136
Enclose the given terms in brackets as required:
1. x – y – z = x – (…………)
2. x2 – xy2 – 2xy – y2 = x2 – (………….)
3. 4a – 9 + 2b – 6 = 4a – (………..)
4. x2 – y2 + z2 + 3x – 2y = x2 – (…………)
5. -2a2 + 4ab – 6a2b2 + 8ab2 = -2a (…………….)
Solution:
1. x – y – z = x – (y + z)
2. x2 – xy2 – 2xy – y2 = x2 – (xy2 + 2xy + y2)
3. 4a – 9 + 2b – 6 = 4a – (9 – 2b + 6)
4. x2 – y2 + z2 + 3x – 2y = x2 – (y2 – z2 – 3x + 2y)
5. -2a2 + 4ab – 6a2b2 + 8ab2 = -2a (a – 2b + 3ab2 – 4b2)
Simplify:
6. 2x – (x + 2y – z)
Solution:
2x – (x + 2y – z)
We can write it as
= 2x – x – 2y + z
So we get
= x – 2y + z
7. p + q – (p – q) + (2p – 3q)
Solution:
p + q – (p – q) + (2p – 3q)
We can write it as
= p + q – p + q + 2p – 3q
So we get
= 2p – q
8. 9x – (-4x + 5)
Solution:
9x – (- 4x + 5)
We can write it as
= 9x + 4x – 5
So we get
= 13x – 5
9. 6a – (- 5a – 8b) + (3a + b)
Solution:
6a – (- 5a – 8b) + (3a + b)
We can write it as
= 6a + 5a + 8b + 3a + b
So we get
= 6a + 5a + 3a + 8b + b
= 14a + 9b
10. (p – 2q) – (3q – r)
Solution:
(p – 2q) – (3q – r)
We can write it as
= p – 2q – 3q + r
So we get
= p – 5q + r
11. 9a (2b – 3a + 7c)
Solution:
9a (2b – 3a + 7c)
We can write it as
= 18ab – 27a2 + 63ca
12. – 5m (- 2m + 3n – 7p)
Solution:
– 5m (- 2m + 3n – 7p)
We can write it as
= 10m2 – 15mn + 35mp
13. – 2x (x + y) + x2
Solution:
– 2x (x + y) + x2
We can write it as
= -2x2 – 2xy + x2
So we get
= – x2 – 2xy
14. b (2b – 1/b) – 2b (b – 1/b)
Solution:
b (2b – 1/b) – 2b (b – 1/b)
We can write it as
= 2b2 – 1 – 2b2 + 2
So we get
= 1
15. 8 (2a + 3b – c) – 10 (a + 2b + 3c)
Solution:
8 (2a + 3b – c) – 10 (a + 2b + 3c)
We can write it as
= 16a + 24b – 8c – 10a – 20b – 30c
So we get
= 6a + 4b – 38c
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