## Selina Concise Mathematics class 7 ICSE Solutions – Fundamental Concepts (Including Fundamental Operations)

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POINTS TO REMEMBER

1. Constants and Variables : The numbers which has fixed value is called constant and same at English alphabet which can be assigned any value according to the requirement is called variables.
2. Term : A term is a number, (constant), a variable or a combination of numbers and variables.
3. Algebraic Expression : An algebraic expression is a collection of one or more terms, which are separated from each other by addition (+) or subtraction (-) signs.
4. Types of algebraic expressions :
(i) Monomial : It has only one term
(ii) Binomial : It has two terms
(iii) Trinomial : It has three terms
(iv) Multinomial : It has more than three terms
(v) Polynomial : It has two or more than two terms.
Note : An expression of the type $$\frac { 2 }{ 5 }$$ does not form a monomial unless JC is not equal to zero.
5. Product: When two or more quantities are multiplied together, the result is called their product.
6. Factors : Each of the quantities (numbers or variables) multiplied together to form a term is called a factor of the given term.
7. Co-efficient: In a monomial, any factor or group of factors of a term is called the co-efficient of the remaining part of the monomial.
8. Degree of a monomial: The degree of a monomial is the exponent of its variable or the sum of the exponents of its variables.
9. Degree of a polynomial: The degree of a polynomial is the degree of its highest degree term.
10. Like and unlike terms : Terms having the same literal co-efficients or alphabetic letters are called like terms ; whereas the terms with different literal co-efficients are called unlike terms.
11. Addition and subtraction : Addition and subtraction of only like terms is possible by adding or subtracting the numerical co-efficients.
12. Multiplication and division :
(A) Multiplication :
(i) Multiplications of monomials.
(a) Multiply the numerical co-efficient together
(ii) Multiply the literal co-efficients separately together.
(iii) Combine the like terms.
(B) Division :
(i) Dividing a polynomial by a monomial Divide each term of the polynomial by monomial and simplify each fractions.
(ii) While dividing one polynomial by another polynomial ; arrange the terms of both the dividend and the divisior both in descending or in ascending order of their powers and then divide.

SOME IMPORTANT POINTS

TYPES OF BRACKETS:
The name of different types of brackets and the order in which they are removed is shown below:
(a) ____ ; Bar (Vinculum) bracket
(b) ( ); Circular bracket .
(c) { } ; Curly bracket and then
(d) [ ]; square bracket

EXERCISE 11 (A)

Question 1.
Separate constant terms and variable terms from tile following :

Solution:

Constant is only 8 others are variables

Question 2.
Constant is only 8 others are variables
(i) 2x ÷ 15
(ii) ax+ 9
(iii) 3x2 × 5x
(iv) 5 + 2a-3b
(v) 2y – $$\frac { 7 }{ 3 }$$ z÷x
(vi) 3p x q ÷ z
(vii) 12z ÷ 5x + 4
(viii) 12 – 5z – 4
(ix) a3 – 3ab2 x c

Question 3.
Write the coefficient of:
(i) xy in – 3axy
(ii) z2 in p2yz2
(iii) mn in -mn
(iv) 15 in – 15p2

Solution:
(i) Co-efficient of xy in – 3 axy = – 3a
(ii) Co-efficient of z2 in p2yz2 = p2y
(iii) Co-efficient of mn in – mn = – 1
(iv) Co-efficient of 15 in – 15p2 is -p2

Question 4.
For each of the following monomials, write its degree :
(i) 7y
(ii) – x2y
(iii) xy2z
(iv) – 9y2z3
(v) 3 m3n4
(vi) – 2p2q3r4

Solution:
(i) Degree of 7y = 1
(ii) Degree of – x2y = 2+1=3
(iii) Degree of xy2z = 1 + 2 + 1 = 4
(iv) Degree of – 9y2z3 = 2 + 3 = 5
(v) Degree of 3m3n4 = 3 + 4 = 7
(vi) Degree of – 2p2q3r4 = 2 + 3 + 4 = 9

Question 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 as x2
y2 is the term which has highest degree.
(ii) The degree of p3q2 – 6p2q5-p4q4 is 8 as p4 q4 is the term which has highest degree.
(iii) The degree of- 8mn6 + 5m3n is 7 as – 8mx6 is the term which has the highest degree.
(iv) The degree of 7 – 3x2 y + y2 is 3 as – 3x2y is the term which has the highest degree.
(v) The degree of 3x – 15 is 1 as 3x is the term which is highest degree.
(vi) The degree of 2y2 z + 9y z3 is 4 as 9yz3 has the highest degree.

Question 6.
Group the like term 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, – 3x2 and x2 are like terms
xy and – 2xy are like terms
(ii) ab, – 3ab, are like terms,
– a2b, 5a2b, – 8a2b are like terms
(iii) 7p, – 2p and 3p are like terms,
8pq, – 5pq are like terms.

Question 7.
Write numerical co-efficient of each of the followings :
(i) y
(ii) -y
(iii) 2x2y
(iv) – 8xy3
(v) 3py2
(vi) – 9a2b3

Solution:
(i) Co-efficient of y = 1
(ii) Co-efficient of-y = – 1
(iii) Co-efficient of 2x2y is = 2
(iv) Co-efficient of – 8xy3 is = – 8
(v) Co-efficient of Ipy2 is = 3
(vi) Co-efficient of – 9a2b3 is = – 9

Question 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:
-5x3y2z4
(i) Co-efficient of z2 is -5x3y2z2
(ii) Co-efficient of y2 is -5x3z4
(iii) Co-efficient of yz2 is -5x3yz2
(iv) Co-efficient of x3y is -5yz4
(v) Co-efficient of -xy2 is 5x2z4
(vi) Co-efficient of -5xy2z is x2z3
Degree of the given expression is 3 + 2 + 4 = 9

EXERCISE 11 (B)

Question 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 :

Question 2.
(i)- 9x, 3x and 4x
(ii) 23y2, 8y2 and – 12y2
(iii) 18pq – 15pq and 3pq

Solution:

Question 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:

Question 4.
(i) a + i and 2a + 3b
(ii) 2x + y and 3x – 4y
(iii)- 3a + 2b and 3a + b
(iv) 4 + x, 5 – 2x and 6x

Solution:

Question 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 and – 2x2 + 5xy + 9y2
and 3x2 -xy- 4y2
(vi) a2 + b2 + 2ab, 2b2 + c2 + 2bc
and 4c2-a2 + 2ac
(vii) 9ax – 6bx + 8, 4ax + 8bx – 7
and – 6ax – 46x – 3
(viii) abc + 2 ba + 3 ac, 4ca – 4ab + 2 bca
and 2ab – 3abc – 6ac
(ix) 4a2 + 5b2 – 6ab, 3ab, 6a2 – 2b2
and 4b2 – 5 ab
(x) x2 + x – 2, 2x – 3x2 + 5 and 2x2 – 5x + 7
(xi) 4x3 + 2x2 – x + 1, 2x3 – 5x2– 3x + 6, x2 + 8 and 5x3 – 7x

Solution:

Question 6.
Find the sum of:
(i) x and 3y
(ii) -2a and +5
(iii) – 4xand +7x
(iv) +4a and -7b
(v) x3+3x2y and 2y2
(vi) 11 and -by

Solution:

Question 7.
The sides of a triangle are 2x + 3y, x + 5y and 7x – 2y, find its perimeter.

Solution:

Question 8.
The two adjacent sides of a rectangle are 6a + 96 and 8a – 46. Find its, perimeter.

Solution

Question 9.
Subtract the second expression from the first:

Solution:

Question 10.
Subtract:

Solution:

Question 11.
Subtract – 5a2 – 3a + 1 from the sum of 4a2 + 3 – 8a and 9a – 7.

Solution:

Question 12.
By how much does 8x3 – 6x2 + 9x – 10 exceed 4x3 + 2x2 + 7x -3 ?

Solution:

Question 13.
What must be added to 2a3 + 5a – a2 – 6 to get a2 – a – a3 + 1 ?

Solution:

Question 14.
What must be subtracted from a2 + b2 + lab to get – 4ab + 2b2 ?

Solution:

Question 15.
Find the excess of 4m2 + 4n2 + 4pover m2+ 3n2 – 5p2

Solution:

Question 16.
By how much is 3x3 – 2x2y + xy2 -y3 less than 4x3 – 3x2y – 7xy2 +2y3

Solution:

Question 17.
Subtract the sum of 3a2 – 2a + 5 and a2 – 5a – 7 from the sum of 5a2 -9a + 3 and 2a – a2 – 1

Solution:

Question 18.
The perimeter of a rectangle is 28x3+ 16x2 + 8x + 4. One of its sides is 8x2 + 4x. Find the other side

Solution:

Question 19.
The perimeter of a triangle is 14a2 + 20a + 13. Two of its sides are 3a2 + 5a + 1 and a2 + 10a – 6. Find its third side.

Solution:

Question 20.

Solution:

Question 21.

Solution:

Question 22.
Simplify:

Solution:

EXERCISE 11 (C)

Question 1.
Multiply:

Solution:

Question 2.
Copy and complete the following multi-plications :

Solution:

Question 3.
Evaluate :

Solution:

Question 4.
Evaluate:

Solution:

Question 5.
Evaluate :

Solution:

Question 6.
Multiply:

Solution:

Question 7.
Multiply:

Solution:

EXERCISE 11 (D)

Question 1.
Divide:

Solution:

Question 2.
Divide :

Solution:

Question 3.
The area of a rectangle is 6x2– 4xy – 10y2 square unit and its length is 2x + 2y unit. Find its breadth

Solution:

Question 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:

Question 5.
Divide:

Solution:

EXERCISE 11 (E)

Simplify
Question 1.

Solution:

Question 2.

Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.

Solution:

Question 6.

Solution:

Question 7.

Solution:

Question 8.

Solution:

Question 9.

Solution:

Question 10.

Solution:

Question 11.

Solution:

Question 12.

Solution:

Question 13.

Solution:

Question 14.

Solution:

Question 15.

Solution:

Question 16.

Solution:

Question 17.

Solution:

Question 18.

Solution:

Question 19.

Solution:

Question 20.

Solution:

Question 21.

Solution:

Question 22.

Solution:

Question 23.

Solution:

Question 24.

Solution:

Question 25.

Solution:

Question 26.

Solution:

EXERCISE 11 (F)

Enclose the given terms in brackets as required :

Question 1.
x – y – z = x-{…….)

Solution:
x – y – z = x – (y + z)

Question 2.
x2 – xy2 – 2xy – y2 = x2 – (…….. )

Solution:
x– xy– 2xy – y2
= x2 – (xy2 + 2xy + y2)

Question 3.
4a – 9 + 2b – 6 = 4a – (…….. )

Solution:
4a – 9 + 2b – 6
= 4a – (9 – 2b + 6)

Question 4.
x2 -y2 + z2 + 3x – 2y = x2 – (…….. )

Solution:
x2 – y2 + z2 + 3x – 2y
= x2 – (y2 – z2 – 3x + 2y)

Question 5.
– 2a2 + 4ab – 6a2b2 + 8ab2 = – 2a (……… )

Solution:
– 2a2 + 4ab – 6a2b2 + 8ab2
= – 2a (a – 2b + 3ab2 – 4b2)

Simplify :

Question 6.
2x – (x + 2y- z)

Solution:
2x-(x + 2y-z) = 2x – x – 2y + z
= x – 2y + z

Question 7.
p + q – (p – q) + (2p – 3q)

Solution:
p + q – (p – q) + (2p- 3q)
= p + q – p + q + 2p – 3q = 2p – q

Question 8.
9x – (-4x + 5)

Solution:
9x – (-4x + 5) = 9x + 4x – 5
= 13x- 5

Question 9.
6a – (- 5a – 8b) + (3a + b)

Solution:
6a – (- 5a – 8b) + (3a + b)
= 6a + 5a + 8b + 3a + b
= 6a + 5a + 3a + 8b + b
= 14a + 9b

Question 10.
(p – 2q) – (3q – r)

Solution:
(p-2q) – (3q – r) =p – 2q – 3q + r =p – 5q + r

Question 11.
9a (2b – 3a + 7c)

Solution:
9a (2b – 3a + 7c)
= 18ab – 27a2 + 63ca

Question 12.
-5m (-2m + 3n – 7p)

Solution:
-5m (-2m + 3n- 7p)
= – 5m x (-2m) + (-5m) (3n) – (-5m) (7p)
= 10m2 – 15mn + 35 mp.

Question 13.
-2x (x + y) + x2

Solution:
– 2x (x + y) + x2
= -2x x x + (-2x)y + x2
= – 2x2 – 2xy + x2
= – 2x2 + x2 – 2xy = – x2 – 2xy

Question 14.

Solution:

Question 15.
8 (2a + 3b – c) – 10 (a + 2b + 3c)

Solution:
8 (2a + 3b -c)- 10 (a + 2b + 3c)
= 16a + 24b – 8c – 10a – 20b- 30c
= 16a – 10a + 24b – 20b – 8c – 30c
= 6a + 4b – 38c

Question 16.

Solution:

Question 17.
5 x (2x + 3y) – 2x (x – 9y)

Solution:
5x (2x + 3y) – 2x (x – 9y)
= 10x2 + 15xy – 2x2 + 18xy
= 10x– 2x2+ 15xy+ 18xy
= 8x2 + 33 xy

Question 18.
a + (b + c – d)

Solution:
a + (b + c – d) = a + (b + c – d)
= a + b + c – d

Question 19.
5 – 8x – 6 – x

Solution:
5 – 8x – 6 – x
= 5 – 6 –  8x – x
= -1 -7x

Question 20.
2a + (6- $$\overline { a-b }$$ )

Solution:
2a + (6 – $$\overline { a-b }$$ )
= 2a + (b – a + b)
= 2a + b – a + b
= a + 2b

Question 21.
3x + [4x – (6x – 3)]

Solution:
3x + [4x – (6x – 3)]
= 3x + [4x – 6x + 3]
= 3x + 4x – 6x + 3
= 3x + 4x – 6x + 3
= 7x – 6x + 3= x + 3

Question 22.
5b – {6a + (8 – b – a)}

Solution:
5b- {6a + 8- 6-a}
= 5b – 6a – 8 + b + a
= -6a + a + 5b +b – 8
= -5a + 6b-8

Question 23.
2x-[5y- (3x -y) + x]

Solution:
2x – [5y- (3x – y) + x]
= 2x – {5y – 3x +y + x}
= 2x – 5y + 3x -y – x
= 2x + 3x – x – 5y – y
= 4x – 6y

Question 24.
6a – 3 (a + b – 2)

Solution:
6a – 3 (a + b – 2)
=
6a – 3a – 3b + 6
= 3a -3b + 6

Question 25.
8 [m + 2n-p – 7 (2m -n + 3p)]

Solution:
8 [m + 2n-p -1 (2m – n + 3p)]
8 [m + 2n-p- 14m + 7n-21p]
= 8m+ 16n -8p- 112m + 56n – 168p
= 8m – 112m + 16n + 56n -8p – 168p
= -104m + 72n – 176p

Question 26.
{9 – (4p – 6q)} – {3q – (5p – 10)}

Solution:
{9 – {4p – 6q)} – {3q – (5p – 10)}
{9 – 4p + 6q} – {3q -5p+ 10}
= 9 – 4p + 6q – 3q + 5p – 10
= 9 – 4p +
5p + 6q – 3q – 10
= p + 3q – 1

Question 27.
2 [a – 3 {a + 5 {a – 2) + 7}]

Solution:
2 [a – 3 {a + 5 {a – 2) + 7}]
= 2 [a- 3 {a + 5a- 10 + 7}]
= 2 [a -3a- 15a + 30 -21]
= 2a-6a- 30a + 60-42
= 2a- 36a + 60-42
= -34a + 18

Question 28.
5a – [6a – {9a – (10a – $$\overline { 4a-3a }$$  )}]

Solution:
5a – [6a – {9a – (10a – 4a + 3a)}]
= 5a – [6a – {9a – (10a – 4a + 3a)}]
= 5a – [6a – {9a – 10a + 4a – 3a}]
= 5a- [6a – 9a + 10a – 4a + 3a]
= 5a – 6a + 9a – 10a + 4a – 3a
= 5a + 9a + 4a – 6a – 10a – 3a
= 18a – 19a = – a

Question 29.
9x + 5 – [4x – {3x – 2 (4x – 3)}]

Solution:
9x + 5 – [4x – {3x – 2 (4x – 3)}]
= 9x + 5 – [4x – {3x – 8x + 6}]
= 9x + 5 – [4x – 3x + 8x – 6]
= 9x + 5-4x + 3x-8x + 6
= 9x + 3x-4x-8x + 5 + 6
= 12x- 12x+ 11 = 11

Question 30.
(x + y – z)x + (z + x – y)y – (x + y – z)z

Solution:
(x + y – z)x + (z + x -y )y – (x + y -z)z
= x+ xy – zx + yz + xy -y– zx – yz + z2
= x2 -y2 + z2 + 2xy – 2zx

Question 31.
-1 [a-3 {b -4 (a-b-8) + 4a} + 10]

Solution:
– 1 [a – 3 {b – 4(a – b – 8) + 4a} + 10]
= -1 [a-3 {b-4{a-b-8) + 4a} + 10]
= -1[a-3 {b-4a + Ab +32 + 4a} + 10]
= -1 [a-3b+ 12a- 126-96- 12a + 10]
= -a + 3b – 12a + 12b + 96 + 12a – 10
= -a-12a + 12a+ 3b+ 12b-96-10
= – a + 15b – 106

Question 32.

Solution:

Question 33.
10 – {4a – (7 – $$\overline { a-5 }$$) – (5a – $$\overline { 1+a }$$)}

Solution:

10 – {4a – (7 – $$\overline { a-5 }$$) – (5a – $$\overline { 1+a }$$)}
= 10 – {4a – (7 – a + 5) – (5a – 1 – a)}
= 10- {4a -(12 -a) -(4a- 1)}
= 10 – {4a – 12 + a- 4a + 1}
= 10 – 4a + 12 – a + 4a- 1
= 10 + 12 – 1 – 4a – a + 4a
= 21 -a

Question 34.
7a- [8a- (11a-(12a- $$\overline { 6a-5a }$$)}]

Solution:
7a – [8a – {1 la – (12a $$\overline { 6a-5a }$$)}]
= 7a-[8a-{11a-(12a-6a + 5a)}]
= 7a -[8a -{11a -(17a -6a)}]
= 7a- [8a- {11a-(11a)}]
= 7a- [8a- {11a- 11a}]
= 7a – 8a = -a

Question 35.

Solution:

Question 36.
x-(3y- $$\overline { 4z-3x }$$ +2z- $$\overline { 5y-7x }$$)

Solution:
x-(3y- $$\overline { 4z-3x }$$ +2z- $$\overline { 5y-7x }$$)
= x – (3y – 4z + 3x  + 2z -5y + 7x)
= x-(-2y-2z+10x)
= x + 2y + 2z- 10x
= -9x + 2y + 2z