NCERT Solutions for Class 7 Maths Chapter 12
NCERT Solutions for Class 7 Maths Chapter 12:
Here are the NCERT solutions for Class 7 Maths Chapter 12 Algebraic Expressions. When a student is having trouble answering a question from NCERT Solutions for Class 7 Maths Chapter 12, they may refer to the solutions.
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NCERT Solutions
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NCERT Solutions for Class 7 Maths Chapter 12 PDF
NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions download the PDF that is attached. If students want to get better at maths, they need to practise the ideas of arithmetic for each course. Using the NCERT Solutions for Class 7 Maths Chapter 12 might help you achieve higher results. Problems have been solved step-by-step by experts, who also offer concise explanations.
NCERT Solutions for Class 7 Maths Chapter 12 PDF
NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions
Below we have provided NCERT Solutions for Class 7 Maths Chapter 12 for students to help them understand the NCERT Solutions for Class 7 Maths Chapter 12 better and to score good marks in their examination.
Exercise 12.1 Page: 234
1. Get the algebraic expressions in the following cases using variables, constants and arithmetic operations.
(i) Subtraction of z from y.
Solution:-
= Y – z
(ii) One-half of the sum of numbers x and y.
Solution:-
= ½ (x + y)
= (x + y)/2
(iii) The number z multiplied by itself.
Solution:-
= z × z
= z
2
(iv) One-fourth of the product of numbers p and q.
Solution:-
= ¼ (p × q)
= pq/4
(v) Numbers x and y, both squared and added.
Solution:-
= x
2
+ y
2
(vi) Number 5 added to three times the product of numbers m and n.
Solution:-
= 3mn + 5
(vii) Product of numbers y and z subtracted from 10.
Solution:-
= 10 – (y × z)
= 10 – yz
(viii) Sum of numbers a and b subtracted from their product.
Solution:-
= (a × b) – (a + b)
= ab – (a + b)
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2. (i) Identify the terms and their factors in the following expressions.
Show the terms and factors by tree diagrams.
(a) x – 3
Solution:-
Expression: x – 3
Terms: x, -3
Factors: x; -3
(b) 1 + x + x
2
Solution:-
Expression: 1 + x + x
2
Terms: 1, x, x
2
Factors: 1; x; x,x
(c) y – y
3
Solution:-
Expression: y – y
3
Terms: y, -y
3
Factors: y; -y, -y, -y
(d) 5xy
2
+ 7x
2
y
Solution:-
Expression: 5xy
2
+ 7x
2
y
Terms: 5xy
2
, 7x
2
y
Factors: 5, x, y, y; 7, x, x, y
(e) – ab + 2b
2
– 3a
2
Solution:-
Expression: -ab + 2b
2
– 3a
2
Terms: -ab, 2b
2
, -3a
2
Factors: -a, b; 2, b, b; -3, a, a
(ii) Identify terms and factors in the expressions given below.
(a) – 4x + 5 (b) – 4x + 5y (c) 5y + 3y
2
(d) xy + 2x
2
y
2
(e) pq + q (f) 1.2 ab – 2.4 b + 3.6 a (g) ¾ x + ¼
(h) 0.1 p
2
+ 0.2 q
2
Solution:-
Expressions are defined as numbers, symbols and operators (such as +. – , × and ÷) grouped together that show the value of something.
In algebra, a term is either a single number or variable or numbers and variables multiplied together. Terms are separated by + or – signs or sometimes by division.
Factors are defined as numbers we can multiply together to get another number.
|
Sl.No.
|
Expression
|
Terms
|
Factors
|
|
(a)
|
– 4x + 5
|
-4x5
|
-4, x5
|
|
(b)
|
– 4x + 5y
|
-4x5y
|
-4, x5, y
|
|
(c)
|
5y + 3y
2
|
5y3y
2
|
5, y3, y, y
|
|
(d)
|
xy + 2x
2
y
2
|
xy2x
2
y
2
|
x, y2, x, x, y, y
|
|
(e)
|
pq + q
|
pqq
|
P, qQ
|
|
(f)
|
1.2 ab – 2.4 b + 3.6 a
|
1.2ab-2.4b
3.6a
|
1.2, a, b-2.4, b
3.6, a
|
|
(g)
|
¾ x + ¼
|
¾ x¼
|
¾, x¼
|
|
(h)
|
0.1 p
2
+ 0.2 q
2
|
0.1p
2
0.2q
2
|
0.1, p, p0.2, q, q
|
3. Identify the numerical coefficients of terms (other than constants) in the following expressions.
(i) 5 – 3t
2
(ii) 1 + t + t
2
+ t
3
(iii) x + 2xy + 3y (iv) 100m + 1000n (v) – p
2
q
2
+ 7pq (vi) 1.2 a + 0.8 b (vii) 3.14 r
2
(viii) 2 (l + b)
(ix) 0.1 y + 0.01 y
2
Solution:-
Expressions are defined as numbers, symbols and operators (such as +. – , × and ÷) grouped together that show the value of something.
In algebra, a term is either a single number or variable or numbers and variables multiplied together. Terms are separated by + or – signs or sometimes by division.
A coefficient is a number used to multiply a variable (2x means 2 times x, so 2 is a coefficient). Variables on their own (without a number next to them) actually have a coefficient of 1 (x is really 1x).
|
Sl.No.
|
Expression
|
Terms
|
Coefficients
|
|
(i)
|
5 – 3t
2
|
– 3t
2
|
-3
|
|
(ii)
|
1 + t + t
2
+ t
3
|
tt
2
t
3
|
11
1
|
|
(iii)
|
x + 2xy + 3y
|
x2xy
3y
|
12
3
|
|
(iv)
|
100m + 1000n
|
100m1000n
|
1001000
|
|
(v)
|
– p
2
q
2
+ 7pq
|
-p
2
q
2
7pq
|
-17
|
|
(vi)
|
1.2 a + 0.8 b
|
1.2a0.8b
|
1.20.8
|
|
(vii)
|
3.14 r
2
|
3.14
2
|
3.14
|
|
(viii)
|
2 (l + b)
|
2l2b
|
22
|
|
(ix)
|
0.1 y + 0.01 y
2
|
0.1y0.01y
2
|
0.10.01
|
4. (a) Identify terms which contain x and give the coefficient of x.
(i) y
2
x + y (ii) 13y
2
– 8yx (iii) x + y + 2
(iv) 5 + z + zx (v) 1 + x + xy (vi) 12xy
2
+ 25
(vii) 7x + xy
2
Solution:-
|
Sl.No.
|
Expression
|
Terms
|
Coefficient of x
|
|
(i)
|
y
2
x + y
|
y
2
x
|
y
2
|
|
(ii)
|
13y
2
– 8yx
|
– 8yx
|
-8y
|
|
(iii)
|
x + y + 2
|
x
|
1
|
|
(iv)
|
5 + z + zx
|
xzx
|
1z
|
|
(v)
|
1 + x + xy
|
xy
|
y
|
|
(vi)
|
12xy
2
+ 25
|
12xy
2
|
12y
2
|
|
(vii)
|
7x + xy
2
|
7xxy
2
|
7y
2
|
(b) Identify terms which contain y
2
and give the coefficient of y
2
.
(i) 8 – xy
2
(ii) 5y
2
+ 7x (iii) 2x
2
y – 15xy
2
+ 7y
2
Solution:-
|
Sl.No.
|
Expression
|
Terms
|
Coefficient of y
2
|
|
(i)
|
8 – xy
2
|
– xy
2
|
– x
|
|
(ii)
|
5y
2
+ 7x
|
5y
2
|
5
|
|
(iii)
|
2x
2
y – 15xy
2
+ 7y
2
|
– 15xy
2
7y
2
|
– 15x7
|
5. Classify into monomials, binomials and trinomials.
(i) 4y – 7z
Solution:-
Binomial.
An expression which contains two unlike terms is called a binomial.
(ii) y
2
Solution:-
Monomial.
An expression with only one term is called a monomial.
(iii) x + y – xy
Solution:-
Trinomial.
An expression which contains three terms is called a trinomial.
(iv) 100
Solution:-
Monomial.
An expression with only one term is called a monomial.
(v) ab – a – b
Solution:-
Trinomial.
An expression which contains three terms is called a trinomial.
(vi) 5 – 3t
Solution:-
Binomial.
An expression which contains two unlike terms is called a binomial.
(vii) 4p
2
q – 4pq
2
Solution:-
Binomial.
An expression which contains two unlike terms is called a binomial.
(viii) 7mn
Solution:-
Monomial.
An expression with only one term is called a monomial.
(ix) z
2
– 3z + 8
Solution:-
Trinomial.
An expression which contains three terms is called a trinomial.
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(x) a
2
+ b
2
Solution:-
Binomial.
An expression which contains two unlike terms is called a binomial.
(xi) z
2
+ z
Solution:-
Binomial.
An expression which contains two unlike terms is called a binomial.
(xii) 1 + x + x
2
Solution:-
Trinomial.
An expression which contains three terms is called a trinomial.
6. State whether a given pair of terms is of like or unlike terms.
(i) 1, 100
Solution:-
Like term.
When terms have the same algebraic factors, they are like terms.
(ii) –7x, (5/2)x
Solution:-
Like term.
When terms have the same algebraic factors, they are like terms.
(iii) – 29x, – 29y
Solution:-
Unlike terms.
The terms have different algebraic factors, they are unlike terms.
(iv) 14xy, 42yx
Solution:-
Like term.
When terms have the same algebraic factors, they are like terms.
(v) 4m
2
p, 4mp
2
Solution:-
Unlike terms.
The terms have different algebraic factors, they are unlike terms.
(vi) 12xz, 12x
2
z
2
Solution:-
Unlike terms.
The terms have different algebraic factors, they are unlike terms.
7. Identify like terms in the following.
(a) – xy
2
, – 4yx
2
, 8x
2
, 2xy
2
, 7y, – 11x
2
, – 100x, – 11yx, 20x
2
y, – 6x
2
, y, 2xy, 3x
Solution:-
When terms have the same algebraic factors, they are like terms.
They are,
– xy
2
, 2xy
2
– 4yx
2
, 20x
2
y
8x
2
, – 11x
2
, – 6x
2
7y, y
– 100x, 3x
– 11yx, 2xy
(b) 10pq, 7p, 8q, – p
2
q
2
, – 7qp, – 100q, – 23, 12q
2
p
2
, – 5p
2
, 41, 2405p, 78qp,
13p
2
q, qp
2
, 701p
2
Solution:-
When terms have the same algebraic factors, they are like terms.
They are,
10pq, – 7qp, 78qp
7p, 2405p
8q, – 100q
– p
2
q
2
, 12q
2
p
2
– 23, 41
– 5p
2
, 701p
2
13p
2
q, qp
2
Exercise 12.2 Page: 239
1. Simplify combining like terms.
(i) 21b – 32 + 7b – 20b
Solution:-
When terms have the same algebraic factors, they are like terms.
Then,
= (21b + 7b – 20b) – 32
= b (21 + 7 – 20) – 32
= b (28 – 20) – 32
= b (8) – 32
= 8b – 32
(ii) – z
2
+ 13z
2
– 5z + 7z
3
– 15z
Solution:-
When terms have the same algebraic factors, they are like terms.
Then,
= 7z
3
+ (-z
2
+ 13z
2
) + (-5z – 15z)
= 7z
3
+ z
2
(-1 + 13) + z (-5 – 15)
= 7z
3
+ z
2
(12) + z (-20)
= 7z
3
+ 12z
2
– 20z
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(iii) p – (p – q) – q – (q – p)
Solution:-
When terms have the same algebraic factors, they are like terms.
Then,
= p – p + q – q – q + p
= p – q
(iv) 3a – 2b – ab – (a – b + ab) + 3ab + b – a
Solution:-
When terms have the same algebraic factors, they are like terms.
Then,
= 3a – 2b – ab – a + b – ab + 3ab + b – a
= 3a – a – a – 2b + b + b – ab – ab + 3ab
= a (1 – 1- 1) + b (-2 + 1 + 1) + ab (-1 -1 + 3)
= a (1 – 2) + b (-2 + 2) + ab (-2 + 3)
= a (1) + b (0) + ab (1)
= a + ab
(v) 5x
2
y – 5x
2
+ 3yx
2
– 3y
2
+ x
2
– y
2
+ 8xy
2
– 3y
2
Solution:-
When terms have the same algebraic factors, they are like terms.
Then,
= 5x
2
y + 3yx
2
– 5x
2
+ x
2
– 3y
2
– y
2
– 3y
2
= x
2
y (5 + 3) + x
2
(- 5 + 1) + y
2
(-3 – 1 -3) + 8xy
2
= x
2
y (8) + x
2
(-4) + y
2
(-7) + 8xy
2
= 8x
2
y – 4x
2
– 7y
2
+ 8xy
2
(vi) (3y
2
+ 5y – 4) – (8y – y
2
– 4)
Solution:-
When terms have the same algebraic factors, they are like terms.
Then,
= 3y
2
+ 5y – 4 – 8y + y
2
+ 4
= 3y
2
+ y
2
+ 5y – 8y – 4 + 4
= y
2
(3 + 1) + y (5 – 8) + (-4 + 4)
= y
2
(4) + y (-3) + (0)
= 4y
2
– 3y
2. Add:
(i) 3mn, – 5mn, 8mn, – 4mn
Solution:-
When terms have the same algebraic factors, they are like terms.
Then, we have to add the like terms.
= 3mn + (-5mn) + 8mn + (- 4mn)
= 3mn – 5mn + 8mn – 4mn
= mn (3 – 5 + 8 – 4)
= mn (11 – 9)
= mn (2)
= 2mn
(ii) t – 8tz, 3tz – z, z – t
Solution:-
When terms have the same algebraic factors, they are like terms.
Then, we have to add the like terms.
= t – 8tz + (3tz – z) + (z – t)
= t – 8tz + 3tz – z + z – t
= t – t – 8tz + 3tz – z + z
= t (1 – 1) + tz (- 8 + 3) + z (-1 + 1)
= t (0) + tz (- 5) + z (0)
= – 5tz
(iii) – 7mn + 5, 12mn + 2, 9mn – 8, – 2mn – 3
Solution:-
When terms have the same algebraic factors, they are like terms.
Then, we have to add the like terms.
= – 7mn + 5 + 12mn + 2 + (9mn – 8) + (- 2mn – 3)
= – 7mn + 5 + 12mn + 2 + 9mn – 8 – 2mn – 3
= – 7mn + 12mn + 9mn – 2mn + 5 + 2 – 8 – 3
= mn (-7 + 12 + 9 – 2) + (5 + 2 – 8 – 3)
= mn (- 9 + 21) + (7 – 11)
= mn (12) – 4
= 12mn – 4
(iv) a + b – 3, b – a + 3, a – b + 3
Solution:-
When terms have the same algebraic factors, they are like terms.
Then, we have to add the like terms.
= a + b – 3 + (b – a + 3) + (a – b + 3)
= a + b – 3 + b – a + 3 + a – b + 3
= a – a + a + b + b – b – 3 + 3 + 3
= a (1 – 1 + 1) + b (1 + 1 – 1) + (-3 + 3 + 3)
= a (2 -1) + b (2 -1) + (-3 + 6)
= a (1) + b (1) + (3)
= a + b + 3
(v) 14x + 10y – 12xy – 13, 18 – 7x – 10y + 8xy, 4xy
Solution:-
When terms have the same algebraic factors, they are like terms.
Then, we have to add the like terms.
= 14x + 10y – 12xy – 13 + (18 – 7x – 10y + 8xy) + 4xy
= 14x + 10y – 12xy – 13 + 18 – 7x – 10y + 8xy + 4xy
= 14x – 7x + 10y– 10y – 12xy + 8xy + 4xy – 13 + 18
= x (14 – 7) + y (10 – 10) + xy(-12 + 8 + 4) + (-13 + 18)
= x (7) + y (0) + xy(0) + (5)
= 7x + 5
(vi) 5m – 7n, 3n – 4m + 2, 2m – 3mn – 5
Solution:-
When terms have the same algebraic factors, they are like terms.
Then, we have to add the like terms.
= 5m – 7n + (3n – 4m + 2) + (2m – 3mn – 5)
= 5m – 7n + 3n – 4m + 2 + 2m – 3mn – 5
= 5m – 4m + 2m – 7n + 3n – 3mn + 2 – 5
= m (5 – 4 + 2) + n (-7 + 3) – 3mn + (2 – 5)
= m (3) + n (-4) – 3mn + (-3)
= 3m – 4n – 3mn – 3
(vii) 4x
2
y, – 3xy
2
, –5xy
2
, 5x
2
y
Solution:-
When terms have the same algebraic factors, they are like terms.
Then, we have to add the like terms.
= 4x
2
y + (-3xy
2
) + (-5xy
2
) + 5x
2
y
= 4x
2
y + 5x
2
y – 3xy
2
– 5xy
2
= x
2
y (4 + 5) + xy
2
(-3 – 5)
= x
2
y (9) + xy
2
(- 8)
= 9x
2
y – 8xy
2
(viii) 3p
2
q
2
– 4pq + 5, – 10 p
2
q
2
, 15 + 9pq + 7p
2
q
2
Solution:-
When terms have the same algebraic factors, they are like terms.
Then, we have to add the like terms.
= 3p
2
q
2
– 4pq + 5 + (- 10p
2
q
2
) + 15 + 9pq + 7p
2
q
2
= 3p
2
q
2
– 10p
2
q
2
+ 7p
2
q
2
– 4pq + 9pq + 5 + 15
= p
2
q
2
(3 -10 + 7) + pq (-4 + 9) + (5 + 15)
= p
2
q
2
(0) + pq (5) + 20
= 5pq + 20
(ix) ab – 4a, 4b – ab, 4a – 4b
Solution:-
When terms have the same algebraic factors, they are like terms.
Then, we have to add the like terms.
= ab – 4a + (4b – ab) + (4a – 4b)
= ab – 4a + 4b – ab + 4a – 4b
= ab – ab – 4a + 4a + 4b – 4b
= ab (1 -1) + a (4 – 4) + b (4 – 4)
= ab (0) + a (0) + b (0)
= 0
(x) x
2
– y
2
– 1, y
2
– 1 – x
2
, 1 – x
2
– y
2
Solution:-
When terms have the same algebraic factors, they are like terms.
Then, we have to add the like terms.
= x
2
– y
2
– 1 + (y
2
– 1 – x
2
) + (1 – x
2
– y
2
)
= x
2
– y
2
– 1 + y
2
– 1 – x
2
+ 1 – x
2
– y
2
= x
2
– x
2
– x
2
– y
2
+ y
2
– y
2
– 1 – 1 + 1
= x
2
(1 – 1- 1) + y
2
(-1 + 1 – 1) + (-1 -1 + 1)
= x
2
(1 – 2) + y
2
(-2 +1) + (-2 + 1)
= x
2
(-1) + y
2
(-1) + (-1)
= -x
2
– y
2
-1
