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NCERT Solutions for Class 7 Maths
NCERT Solutions for Class 7 Maths are solved by experts in order to help students to obtain excellent marks in their annual examination. All the questions and answers that are present in the CBSE NCERT Books has been included in this page. We have provided all the Class 7 Maths NCERT Solutions with a detailed explanation i.e., we have solved all the question with step by step solutions in understandable language. So students having great knowledge over NCERT Solutions Class 7 Maths can easily make a grade in their board exams.
Chapter - 12 Algebraic Expressions
Question 1.
Get the algebraic expressions in the following cases using variables, constants and arithmetic operations.
- Subtraction of z from y.
- One-half of the sum of numbers x and y.
- The number z multiplied by itself
- One-fourth of the product of numbers p and q.
- Numbers x and y both squared and added.
- Number 5 added to three times the product of numbers m and n.
- Product of numbers y and z subtracted from 10.
- Sum of numbers a and b subtracted from their product.
Solution:
- y – z
- 12 ( x + y )
- z2
- 14 pq
- x2 + y2
- 3mn + 5
- 10 – yz
- ab – ( a + b )
Question 2.
(i) Identify the terms and their factors in the following expressions. Show the terms and factors by tree diagrams.
(a) x – 3
Solution:
(b) 1 + x + x2
Solution:
(c) y – y3
Solution:
(d) 5xy2 + 7x2y
Solution:
(e) – ab + 2b2 – 3a2
Solution:
(ii) Identify terms and factors in the expressions given below :
(a) -4x + 5
(b) -4x + 5y
(c) 5y + 3y2
(d) xy + 2x2 y2
(e) pq + q
(f) 1.2ab – 2.4b + 3.6a
(g) 34 x + 14
(h) 0.1p2 + 0.2q2
Solution:
Question 3.
Identify the numerical coefficients terms (other than constants) in the following expressions :
(i) 5 – 3t2
(ii) 1 + t + t2 + f3
(iii) x + 2xy + 3y
(iv) 100m + 1000n
(v) – p2q2 + 7 pq
(vi) 1.2a + 0.8b
(vii) 3.14r2
(viii) 2 (l + b)
(ix) 0.1y + 0.01y2
Solution:
Question 4.
(a) Identify terms that contain x and give the coefficient of x.
(i) y2x + y
(ii) 13y2 – 8yx
(iii) x + y + 2
(iv) 5 + z + zx
(v) 1 + x + xy
(vi) 12xy2 + 25
(vii) 7x + xy2
Solution:
(b) Identify terms that contain y2 and give the coefficient of y2.
(i) 8 – xy2
(ii) 5y2 + 7x
(iii) 2x2y – 15xy2 + 7y2
Solution:
Question 5.
Classify into monomials, binomials, and 4y trinomials
(i) 4y – 7z
(ii) y2
(iii) x + y – xy
(iv) 100
(v) ab – a – b
(vi) 5 – 3t
(vii) 4p2q – 4pq2
(viii) 7mn
(ix) z2 – 3z + 8
(x) a2 + b2
(xi) z2 + z
(xii) 1 + x + x2
Solution:
We know that an algebraic expression containing only one term is called a monomial. So, the monomials are : (ii), (iv), and (viii).
We know that an algebraic expression containing two terms is called a binomial. So, the binomials are : (i), (vi), (vii), (x) and (xi).
We know that an algebraic expression containing three terms is called a trinomial. So, the trinomial are : (iii), (v), (ix) and (xii).
Question 6.
State whether a given pair of terms j is of like or unlike terms:
- 1,100
- -7x, 52 x
- -29 x, -29 y
- 14 xy, 42 yx
- 4m2p, 4mp2
- 12xz, 12x2 z2.
Solution:
- Like
- Like
- Unlike
- Like
- Unlike
- Unlike
Question 7.
Identify like terms in the following:
(a) – xy2, -4yx2, 8x2, 2xy2, 7y, – 11x2, – 100x, – 11yx, 20x2y, – 6x2, y, 2xy, 3x
Solution:
In the given terms, like terms are grouped as under :
(b) 10pq, 7p, 8q, – p2q2, – 7qp, – 100q,- 23, 12q2p2, – 5p2, 41, 2405p, 78qp, 13p2q, qp2, 701p2
Solution:
In the given terms, like terms are grouped as under :
Question 1.
Simplify combining like terms:
(i) 21b -32 + 7b- 206
(ii) -z2 + 13z2 -5z + 7z3 – 152
(iii) p – (p – q) – q – (q – p)
(iv) 3a – 2b — ab – (a – b + ab) + 3ab + 6 – a
(v) 5x2y – 5x2 + 3yx2 – 3y2 + x2 – y2 + 8xy2 -3y2
(vi) (3y2 + 5y – 4) – (8y – y2 – 4)
Solution:
(i) 21b – 32 + 7b – 206
Re-arranging the like terms, we get
216 + 7b – 206 – 32
= (21 + 7 – 20)b – 32
= 8b – 32 which is required.
(ii) -z2 + 13z2 – 5z – 15z
Re-arranging the like terms, we get
7z3 – z2 + 13z2 – 5z + 5z – 15z
= 7z3 + (-1 + 13)z2 + (-5 – 15)z
= 7z3 + 12z2 – 20z which is required.
(iii) p – (p – q) – q – (q – p)
=p – p + q – q – q + p
Re-arranging the like terms, we get
= p – q which is required.
(iv) 3a – 2b – ab – (a – b + ab) + 3ab + b – a
= 3a – 2b – ab – a + b – ab + 3ab + b – a
Re-arranging the like terms, we get
= 3a – a – a – 2b + b + b – ab – ab + 3ab
= a + ab which is required.
(v) 5x2y – 5x2 + 3yx2 – 3y2 + x2 – y2 + 8xy2 – 3y2
Re-arranging the like terms, we get
5x2y + 3x2y + 8xy2 – 5x2 + x2 – 3y2 – y2 – 3y2
= 8x2y + 8xy2 – 4x2 – 7y2 which is required.
(vi) (3y2 + 5y – 4) – (8y – y2 – 4)
= 3y2 + 5y – 4 – 8y + y2 + 4 (Solving the brackets)
Re-arranging the like terms, we get
= 3y2 + y2 + 5y – 8y – 4 + 4
= 4y2 – 3y which is required.
Ex 12.2 Class 7 Maths Question 2.
Add:
(i) 3mn, -5mn, 8mn, -4mn
(ii) t – 8tz, 3tz, -z, z – t
(iii) -7mn + 5, 12mn + 2, 9mn – 8, -2mn – 3
(iv) a + b – 3, b – a + 3, a – b + 3
(v) 14x + 10y – 12xy – 13, 18 – 7x – 10y + 8xy, 4xy
(vi) 5m – 7n, 3n – 4m + 2, 2m – 3mn – 5
(vii) 4x2y, -3xy2, -5xy2, 5x2y
(viii) 3p2q2 – 4pq + 5, -10p2q2, 15 + 9pq + 7p2q2
(ix) ab – 4a, 4b – ab, 4a – 4b
(x) x2 – y2 – 1, y2 – 1 – x2, 1 – x2 – y2
Solution:
(i) 3mn, -5mn, 8mn, -4mn
= (3 mn) + (-5 mn) + (8 mn) + (- 4 mn)
= (3 – 5 + 8 – 4)mn
= 2mn which is required.
(ii) t – 8tz, 3tz – z, z – t
t – 8tz + 3tz – z + z – t
Re-arranging the like terms, we get
t – t – 8tz + 3tz – z + z
⇒ 0 – 5 tz + 0
⇒ -5tz which is required.
(iii) -7mn + 5, 12mn + 2, 9mn – 8, -2mn – 3
= -7mn + 5 + 12mn + 2 + 9mn – 8 + (-2mn) – 3
Re-arranging the like terms, we get
-7mn + 12 mn + 9mn – 2 mn + 5 + 2 – 8 – 3
= 12 mn – 4 which is required.
(iv) a + b – 3, b – a + 3, a – b + 3
⇒ a + b – 3 + b – a + 3 + a – b + 3
Re-arranging the like terms, we get
a – a + a + b + b – b – 3 + 3 + 3
⇒ a + b + 3 which is required.
(v) 14x + 10y – 12xy – 13, 18 – 7x – 10y + 8xy, 4xy
∴ 14x + 14y – 12xy – 13 + 18 – 7x – 10y + 8xy + 4xy
Re-arranging the like terms, we get
-12xy + 8xy + 4xy + 14x – 7x + 10y – 10y – 13 + 18
= 0 + 7x + 0 + 5
= 7x + 5 which is required
(vi) 5m – 7n, 3n – 4m + 2, 2m – 3mn – 5 5m -In + 3n — 4m + 2 + 2m – 3mn – 5
Re-arranging the like terms, we get
5m – 4m + 2m – 7n + 3n- 3mn + 2 – 5
= 3m – 4n – 3mn – 3 which is required.
(vii) 4x2y, -3xy2, -5xy2, 5x2y
Re-arranging the like terms and adding, we get
4x2y – 5xy2 – 3xy2 + 5x2y
=9x2y — 8xy2 which is required.
(viii) 3p2q2 – 4pq + 5, -10p2q2, 15 + 9pq + 7p2q2
= (3p2q2 – 4pq + 5) + (-10p2q2) + (15 + 9pq + 7p2q2)
= 3p2q2 – 4pq + 5 – 10p2q2 + (15 + 9pq + 7p2q2 )
= 3p2q2 + 7p2q2 – 10p2q2 – 4pq + 9pq + 5 + 15
= 10p2q2 – 10p2q2 + 5pq + 20
= 0 + 5pq + 20
= 5pq + 20 which is required.
(ix) ab – 4a, 4b – ab, 4a – 4b
= ab – 4a + 4b – ab + 4a – 4b
= 0 + 0 + 0 = 0 which is required.
(x) x2 – y2 – 1, y2 – 1 – x2, 1 – x2 – y2
= x2 – y2 – 1 + y2 – 1 – x2 + 1 – x2 – y2
= -x2 – y2 – 1
= -(x2 + y2 + 1) which is required.
Ex 12.2 Class 7 Maths Question 3.
Subtract:
(i) -5y2 from y2
(ii) 6xy from -12xy
(iii) (a – b) from (a + b)
(iv) a(b – 5) from b(5 – a)
(v) -m2 + 5mn from 4m2 – 3mn + 8
(vi) -x2 + 10x – 5 from 5x – 10
(vii) 5a2 – 7ab + 5b2 from 3ab – 2a2 – 2b2
(viii) 4pq – 5q2 – 3p2 from 5p2 + 3q2 – pq
Solution:
(i) -5y2 from y2 = y2 – (-5y2)
= y2 + 5y2 = 6y2
(ii) 6ry from -12ry = -12xy – 6xy = -18xy which is required.
(iii) (a – b) from (a + b)
= (a + b) – (a – b)
= a + b – a + b = 2b which is required
(iv) a(b – 5) from b(5 – a)
= b(5 – a) – a(b – 5)
= 5b – ab – ab + 5a
= 5a – 2ab + 5b
= 5a + 5b – 2ab which is required.
(v) -m2 + 5mn from 4m2 – 3mn + 8
= (4m2 – 3mn + 8) – (-m2 + 5mn)
= 4m2 – 3mn + 8 + m2 – 5mn
= 4m2 + m2 – 3mn – 5mn + 8
= 5m2 – 8mn + 8 which is required.
(vi) -x2 + 10x – 5 from 5x – 10
= (5x – 10) – (-x2 + 10x – 5)
= 5x – 10 + x2 – 10x + 5
= x2 + 5x – 10x – 10 + 5
= x2 – 5x – 5 which is required.
(vii) 5a2 – 7ab + 5b2 from 3ab – 2a2 – 2b2
= (3ab – 2a2 – 2b2) – (5a2 – 7ab + 5b2)
= 3ab – 2a2 – 2b2 – 5a2 + 7ab – 5b2
= 3ab + 7ab – 2a2 – 5a2 – 2b2 – 5b2
= 10ab – 7a2 – 7b2
which is required.
(viii) 4pq – 5q2 – 3p2 from 5p2 + 3q2 – pq
= (5p2 + 3q2 – pq) – (4pq – 5q2 – 3p2)
= 5p2 + 3q2 – pq – 4pq + 5q2 + 3p2
= 5p2 + 3p2 + 3q2 + 5q2 – pq – 4pq
= 8p2 + 8q2 – 5pq
which is required.
Ex 12.2 Class 7 Maths Question 4.
(a) What should be added to x2 + xy + y2to obtain 2x2 + 3xy?
(b) What should be subtracted from 2a + 8b + 10 to get -3a + 7b + 16?
Solution:
(a) (2x2 + 3xy) – (x2 + xy + y2)
= 2x2 + 3xy – x2 – xy – y2
= 2x2 – x2 + 3xy – xy – y2
= x2 + 2xy – y2 is required expression.
(b) (2a + 8b + 10) – (-3a + 7b + 16)
= 2a + 8b + 10 + 3a – 7b – 16
= 2a + 3a + 8b – 7b + 10 – 16
= 5a + b – 6 is required expression.
Ex 12.2 Class 7 Maths Question 5.
What should be taken away from 3x2 – 4y2 + 5xy + 20 to obtain -x2 – y2 + 6xy + 20?
Solution:
Let A be taken away.
∴ (3x2 – 4y2 + 5 xy + 20)-A
= -x2 – y2 + 6xy + 20
⇒ A = (3x2 – 4y2 + 5xy + 20) – (-x22 – y2 + 6xy + 20)
= 3x2 – 4y2 + 5xy + 20 + x2 + y2 – 6xy – 20
= 3x2 + x2 – 4y2 + y2 + 5xy – 6xy + 20 – 20
= 4x2 – 3y2 – xy is required expression.
Ex 12.2 Class 7 Maths Question 6.
(a) From the sum of 3x – y + 11 and -y – 11, subtract 3x – y – 11.
(b) From the sum of 4 + 3x and 5 – 4x + 2x2, subtract the sum of 3x2 – 5x and -x2 + 2x + 5.
Solution:
(a) Sum of 3x – y + 11 and -y – 11
= (3x – y + 11) + (-y – 11)
= 3x – y + 11 – y – 11
∴ 3x – 2y – (3x – 2y) – (3x – y – 11)
= 3x – 2y – 3x + y + 11
= 3x – 3x – 2y + y + 11
= -y + 11 is required solution.
(b) Sum of (4 + 3x) and (5 – 4x + 2x2)
= 4 + 3x + 5 – 4x + 2x2
= 2x2 – 4x + 3x + 9 = 2x2 – x + 9
Sum of (3x2 – 5x) and (-x2 + 2x + 5)
= (3x2 – 5x) + (-x2 + 2x + 5)
= 3x2 – 5x – x2 + 2x + 5 = 2x2 – 3x + 5
Now (2x2 – x + 9) – (2x2 – 3x + 5)
= 2x2 – x + 9 – 2x2 + 3x – 5
= 2x2 – 2x2 + 3x – x + 4
= 2x + 4 is required expression.
Question 1.
If m = 2, find the value of:
(i) m – 2
(ii) 3m – 5
(iii) 9 – 5m
(iv) 3m2 – 2m – 7
(v) 5m2−4
Solution:
(i) m – 2
Putting m = 2, we get
2 – 2 = 0
(ii) 3m – 5
Putting m = 2, we get
3 × 2 – 5 = 6 – 5 = 1
(iii) 9 – 5m
Putting m = 2, we get
9 – 5 × 2 = 9 – 10 = -1
(iv) 3m2 – 2m – 7 Putting m = 2, we get
3(2)2 – 2(2) – 7 = 3 × 4 – 4 – 7
=12 – 4 – 7 = 12 – 11 = 1
(v) 5m2−4
Putting m = 2, we get
5×22−4=5−4=1
Ex 12.3 Class 7 Maths Question 2.
If p = -2, find the value of:
(i) 4p + 7
(ii) -3p2 + 4p + 7
(iii) -2p3 – 3p2 + 4p + 7
Solution:
(i) 4p + 7
Putting p = -2, we get 4(-2) + 7 = -8 + 7 = -1
(ii) -3p2 + 4p + l
Putting p = -2, we get
-3(-2)2 + 4(-2) + 7
= -3 × 4 – 8 + 7 = -12 – 8+ 7 = -13
(iii) -2p3 – 3p2 + 4p + 7
Putting p = -2, we get
– 2(-2)3 – 3(-2)2 + 4(-2) + 7
= -2 × (-8) – 3 × 4 – 8 + 7
= 16 – 12 – 8 + 7 = 3
Ex 12.3 Class 7 Maths Question 3.
If a = 2, b = -2, find the value of:
(i) a2 + b2
(ii) a2 + ab + b2
(iii) a2 – b2
Solution:
(i) a2 + b2
Putting a = 2 and b = -2, we get
(2)2 + (-2)2 = 4 + 4 = 8
(ii) a2 + ab + b2
Putting a = 2 and b = -2, we get
(2)2 + 2(-2) + (-2)2 = 4 – 4 + 4 = 4
(iii) a2 – b2
Putting a = 2 and b = -2, we get
(2)2 – (-2)2 = 4 – 4 = 0
Ex 12.3 Class 7 Maths Question 4.
When a = 0, b = -1, find the value of the given expressions:
(i) 2a + 2b
(ii) 2a2 + b2 + 1
(iii) 2a2b + 2ab2 + ab
(iv) a2 + ab + 2
Solution:
(i) 2a + 2b = 2(0) + 2(-1)
= 0 – 2 = -2 which is required.
(ii) 2a2 + b2 + 1
= 2(0)2 + (-1)2 + 1 =0 + 1 + 1 = 2 which is required.
(iii) 2a2b + 2ab2 + ab
= 2(0)2 (-1) + 2(0)(-1)2 + (0)(-1)
=0 + 0 + 0 = 0 which is required.
(iv) a2 + ab + 2
= (0)2 + (0)(-1) + 2
= 0 + 0 + 2 = 0 which is required.
Ex 12.3 Class 7 Maths Question 5.
Simplify the expressions and find the value if x is equal to 2.
(i) x + 7 +4(x – 5)
(ii) 3(x + 2) + 5x – 7
(iii) 6x + 5(x – 2)
(iv) 4(2x – 1) + 3x + 11
Solution:
(i) x + 7 + 4(x – 5) = x + 7 + 4x – 20 = 5x – 13
Putting x = 2, we get
= 5 × 2 – 13 = 10 – 13 = -3
which is required.
(ii) 3(x + 2) + 5x – 7 = 3x + 6 + 5x -7 = 8x – 1
Putting x = 2, we get
= 8 × 2 – 1 = 16 – 1 = 15
which is required.
(iii) 6x + 5(x – 2) = 6x + 5x – 10
= 11 × – 10
Putting x = 2, we get
= 11 × 2 – 10 = 22 – 10 = 12
which is required.
(iv) 4(2x – 1) + 3x + 11 = 8x – 4 + 3x + 11
= 11x + 7
Putting x = 2, we get
= 11 × 2 + 7 = 22+ 7 = 29
Ex 12.3 Class 7 Maths Question 6.
Simplify these expressions and find their values if x = 3, a = -1, b = -2.
(i) 3x – 5 – x + 9
(ii) 2 – 8x + 4x + 4
(iii) 3a + 5 – 8a + 1
(iv) 10 – 3b – 4 – 55
(v) 2a – 2b – 4 – 5 + a
Solution:
(i) 3x – 5 – x + 9 = 2x + 4
Putting x = 3, we get
2 × 3 + 4 = 6 + 4 = 10
which is required.
(ii) 2 – 8x + 4x + 4 = -8x + 4x + 2 + 4 = -4x + 6
Putting x = 2, we have
= -4 × 2 + 6 = -8 + 6 =-2
which is required.
(iii) 3a + 5 – 8a +1 = 3a – 8a + 5 + 1 = -5a + 6
Putting a = -1, we get
= -5(-1) + 6 = 5 + 6 = 11
which is required.
(iv) 10 – 3b – 4 – 5b = -3b – 5b + 10 – 4
= -8b + 6
Putting b = -2, we get
= -8(-2) + 6 = 16 + 6 = 22
which is required.
(v) 2a – 2b – 4 – 5 + a = 2a + a – 2b – 4 – 5
= 3a – 26 – 9
Putting a = -1 and b = -2, we get
= 3(-1) – 2(-2) – 9
= -3 + 4 – 9 = 1 – 9 = -8
which is required.
Ex 12.3 Class 7 Maths Question 7.
(i) If z = 10, find the value of z2 – 3(z – 10).
(ii) If p = -10, find the value of p2 -2p – 100.
Solution:
(i) z2 – 3(z – 10)
= z2 – 3z + 30
Putting z = 10, we get
= (10)2 – 3(10) + 30
= 1000 – 30 + 30 = 1000 which is required.
(ii) p2 – 2p – 100
Putting p = -10, we get
(-10)2 – 2(-10) – 100
= 100 + 20 – 100 = 20 which is required.
Ex 12.3 Class 7 Maths Question 8.
What should be the value of a if the value of 2x2 + x – a equals to 5, when x = 0?
Solution:
2x2 + x – a = 5
Putting x = 0, we get
2(0)2 + (0) – a = 5
0 + 0 – a = 5
-a = 5
⇒ a = -5 which is required value.
Ex 12.3 Class 7 Maths Question 9.
Simplify the expression and find its value when a = 5 and b = -3.
2(a2 + ab) + 3 – ab
Solution:
2(a2 + ab) + 3 – ab = 2a2 + 2ab + 3 – ab
= 2a2 + 2ab – ab + 3
= 2ab + ab + 3
Putting, a = 5 and b = -3, we get
= 2(5)2 + (5)(-3) + 3
= 2 × 25 – 15 + 3
= 50 – 15 + 3
= 53 – 15 = 38
Hence, the required value = 38.
Question 1.
Observe the patterns of digits made from line segments of equal length. You will find such segmented digits on the display of electronic watches or calculators.
If the number of digits formed is taken to be n, the number of segments required to form n digits is given by the algebraic expression appearing on the right of each pattern. How many segments are required to form 5, 10, 100 digits of the kind 6, 4, 3.
Solution:
(i) The number of line segments required to form
n digits is given by the expressions.
For 5 figures, the number of line segments = 5 × 5 + 1 = 25 + 1 = 26
For 10 figures, the number of line segments = 5 × 10 + 1
= 50 + 1 = 51
For 100 figures, the number of line segments = 5 × 100 + 1
= 500 + 1 = 501
(ii) 
For 5 figures, the number of line segments =3 ×5 + 1
= 15 + 1 = 16
For 10 figures, the number of line segments = 3 × 10 + 1
= 30 + 1 = 31
For 100 figures, the number of line segments = 3 × 100 + 1
= 300 + 1 = 301
(iii) 
For 5 figures, the number of line numbers = 5 × 5 + 2
= 25 + 2 = 27
For 10 figures, the number of line segments = 5 × 10 + 2
= 50 + 2 = 52
For 100 figures, the number of line segments = 5 × 100 + 2
= 500 + 2 = 502
Ex 12.4 Class 7 Maths Question 2.
Use the given algebraic expression to complete the table of number patterns:
| S.No. | Expression | Terms | |||||||||
| 1st | 2nd | 3rd | 4th | 5th | … | 10th | … | 100th | … | ||
| (i) | 2n – 1 | 1 | 3 | 5 | 7 | 9 | – | 19 | – | – | – |
| (ii) | 3n + 2 | 5 | 8 | 11 | 14 | – | – | – | – | – | – |
| (iii) | 4n + 1 | 5 | 9 | 13 | 17 | – | – | – | – | – | – |
| (iv) | 7n + 20 | 27 | 34 | 41 | 48 | – | – | – | – | – | – |
| (v) | n2 + 1 | 2 | 5 | 10 | 17 | – | – | – | – | 10,001 | – |
Solution:
(i) Given expression is 2n – 1
For n = 100, 2 × 100 – 1
= 200 – 1 = 199
(ii) Given expression is 3n + 2
For n = 5, 3 × 5 + 2 = 15 + 2 = 17
For n = 10, 3 × 10 + 2 = 30 + 2 = 32
For n = 100, 3 × 100 + 2 = 300 + 2 = 302
(iii) Given expression is 4n + 1
For n = 5, 4 × 5 + 1 = 20 + 1 = 21
For n = 10, 4 × 10 + 1 = 40 + 1 = 41
For n = 100, 4 × 100 + 1 = 400 + 1 = 401
(iv) Given expression is 7n + 20
For n = 5, 7 × 5 + 20 = 35 + 20 = 55
For n = 10, 7 × 10 + 20 = 70 + 20 = 90
For n = 100, 7 × 100 + 20 = 700 + 20 = 720
(v) Given expression is n2 + 1
For n = 5, 52 + 1 = 25 + 1 = 26
For n = 10, 102 + 1 = 100 + 1 = 101
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