NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Exercise 12.1
Ex 12.1 Class 7 Maths Question 1.
Get the algebraic expressions in the following cases using variables, constants and arithmetic operations:
(i) Subtraction of z from y.
(ii) One half of the sum of numbers x and y.
(iii) The number z multiplied by itself.
(iv) One-fourth of the product of numbers p and q.
(v) Numbers x and y both squared and added.
(vi) Number 5 added to three times the product of number m and n.
(vii) Product of numbers y and 2 subtracted from 10.
(viii) Sum of numbers a and b subtracted from their product.
Solution:
(i) Subtraction of z from y
Expression: y – z

(ii) One half of the sum of numbers x and y
Expression: 12(x+y) or x+y2

(iii) The number 2 multiplied by itself.
Expression: z × z = z²

(iv) One-fourth of the product of numbers p and q
Expression: 14pq or pq4

(v) Numbers x and y both squared and added
Expression: x² + y²

(vi) Number 5 added to three times the product of number m and n
Expression: 3mn + 5

(vii) Product of numbers y and z subtracted from 10
Expression: 10 – yz

(viii) Sum of numbers a and 6 subtracted from their product
Expression: Sum = a + b, Product = ab
∴ Required expression
= ab – (a + b)
= ab – a- b

Ex 12.1 Class 7 Maths Question 2.
(i) Identify the terms and their factors in the following expressions show the terms and factors by tree diagrams.
(a) x – 3
(b) 1 + x + x²
(c) y – y³
(d) 5xy² + 7x²y
(e) -ab + 2b² – 3a²
Solution:

(ii) Identify terms and factors in the expression
given below:
(a) -4x + 5
(b) -4x + 5y
(c) 5y + 3y²
(d) xy + 2x²y²
(e) pq + q
(f) 1.2ab – 2.4b + 3.6a
(g) 34x+14
(h) 0.1p² + 0.2q²
Solution:

Ex 12.1 Class 7 Maths Question 3.
Identify the numerical coefficients of terms (other than constants) in the following:
(i) 5 – 3t²
(ii) 1 + t + t² + t³
(iv) 100m + 1000n
(v) -p²q² + 7pq
(vi) 1.2 a + 0.86
(vii) 3.14r²
(viii) 2(l + b)
(ix) 0.1y + 0.01y²
Solution:

Ex 12.1 Class 7 Maths Question 4.
(a) Identify terms which 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) 12 xy² + 25
(vii) 7x + xy²
Solution:

(b) Identify terms which contain y² and give the coefficients of y².
(i) 8 – xy²
(ii) 5y² + 7x
(iii) 2x²y – 15xy² + 7y²
Solution:

Ex 12.1 Class 7 Maths Question 5.
Classify into monomials, binomials and trinomials:
(i) 4y – 7x
(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:
(i)4y – 7z – Binomial
(ii) y² – Monomial
(iii) x + y – xy – Trinomial
(iv) 100 Monomial
(v) ab – a – b – Trinomial
(vi) 5 – 3t – Binomial
(vii) 4p²q – 4pq² – Binomial
(viii) 7mn – Monomial
(ix) z² -3z + 8 – Trinomial
(x) a² + b² – Binomial
(xi) z² + z – Binomial
(xii) 1 + x + x² – Trinomial

Ex 12.1 Class 7 Maths Question 6.
State whether a given pair of terms is of like or unlike terms.
(i) 1, 100
(ii) -7x, 52x
(iii) -29x, -29y
(iv) 14xy, 42yx
(v) 4m²p, 4mp²
(vi) 12xz, 12 x²y²
Solution:
(i) 1, 100 – Like
(ii) -7x, 52x – Like
(iii) -29x, -29y – Unlike
(iv) 14xy, 42yx – Like
(v) 4m²p, 4mp² – Unlike
(vi) 12xz, 12x²z² – Unlike

Ex 12.1 Class 7 Maths Question 7.
Identify like terms in the following:
(a)-xy², -4yx², 8x², 2xy², 7y², -11x², -100x, -11yx, 20x²y, -6x², y, 2xy, 3x
(b) 10pq, 7p, 8q, -p²q², -7qp, -100q, -23, 12q²p², -5p², 41, 2405p, 78qp, 13p²q, qp², 701p²
Solution:
(a) Like terms are:
(i) -xy², 2xy²
(ii) -4yx², 20x²y
(iii)8x², -11x², -6x²
(iv) 7y, y
(v) -100x, 3x
(vi) -11yx, 2xy

(b) Like terms are:
(i) 10pq, – 7qp, 78qp
(ii) 7p, 2405p
(iii) 8q, -100q
(iv) -p²q², 12 q²p²
(v) -23, 41
(vi) -5p², 701p²
(vii) 13p²q, qp²

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Exercise 12.2
Ex 12.2 Class 7 Maths 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.

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Exercise 12.4
Ex 12.4 Class 7 Maths 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