NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions

NCERT Solutions For Class 7 Maths Chapter 12 PDF Free Download

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions are available here. Students can check for the solutions whenever they are facing difficulty, while solving the questions from NCERT Solutions for Class 7 Maths Chapter 12. These solutions for Chapter 12 are available in PDF format so that students can download and learn offline as well. This book is one of the top materials when it comes to providing a question bank to practice.

Chapter 12 – Algebraic Expressions contains 4 exercises, and the NCERT Solutions for Class 7 Maths available on this page provide solutions to the questions present in the exercises. Now, let us have a look at some of the concepts discussed in this chapter.

  • How are Expressions Formed
  • Terms of An Expression
  • Like and Unlike Terms
  • Monomials, Binomials, Trinomials and Polynomials
  • Addition and Subtraction of Algebraic Expressions
  • Finding The Value of An Expression
  • Using Algebraic Expressions – Formulas and Rules

Download the PDF of NCERT Solutions For Class 7 Maths Chapter 12 Algebraic Expressions

 

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Access answers to Maths NCERT Solutions For Class 7 Chapter 12 – Algebraic Expressions

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

= z2

(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:-

= x2 + y2

(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)

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

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Image 1

(b) 1 + x + x2

Solution:-

Expression: 1 + x + x2

Terms: 1, x, x2

Factors: 1; x; x,x

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Image 2

(c) y – y3

Solution:-

Expression: y – y3

Terms: y, -y3

Factors: y; -y, -y, -y

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Image 3

(d) 5xy2 + 7x2y

Solution:-

Expression: 5xy2 + 7x2y

Terms: 5xy2, 7x2y

Factors: 5, x, y, y; 7, x, x, y

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Image 4

(e) – ab + 2b2 – 3a2

Solution:-

Expression: -ab + 2b2 – 3a2

Terms: -ab, 2b2, -3a2

Factors: -a, b; 2, b, b; -3, a, a

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Image 5

(ii) Identify terms and factors in the expressions given below:

(a) – 4x + 5 (b) – 4x + 5y (c) 5y + 3y2 (d) xy + 2x2y2

(e) pq + q (f) 1.2 ab – 2.4 b + 3.6 a (g) ¾ x + ¼

(h) 0.1 p2 + 0.2 q2

Solution:-

Expressions is 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 is defined as, numbers we can multiply together to get another number.

Sl.No.

Expression

Terms

Factors

(a)

– 4x + 5

-4x

5

-4, x

5

(b)

– 4x + 5y

-4x

5y

-4, x

5, y

(c)

5y + 3y2

5y

3y2

5, y

3, y, y

(d)

xy + 2x2y2

xy

2x2y2

x, y

2, x, x, y, y

(e)

pq + q

pq

q

P, q

Q

(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 x

¼

¾, x

¼

(h)

0.1 p2 + 0.2 q2

0.1p2

0.2q2

0.1, p, p

0.2, q, q

3. Identify the numerical coefficients of terms (other than constants) in the following expressions:

(i) 5 – 3t2 (ii) 1 + t + t2 + t3 (iii) x + 2xy + 3y (iv) 100m + 1000n (v) – p2q2 + 7pq (vi) 1.2 a + 0.8 b (vii) 3.14 r2 (viii) 2 (l + b)

(ix) 0.1 y + 0.01 y2

Solution:-

Expressions is 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 – 3t2

– 3t2

-3

(ii)

1 + t + t2 + t3

t

t2

t3

1

1

1

(iii)

x + 2xy + 3y

x

2xy

3y

1

2

3

(iv)

100m + 1000n

100m

1000n

100

1000

(v)

– p2q2 + 7pq

-p2q2

7pq

-1

7

(vi)

1.2 a + 0.8 b

1.2a

0.8b

1.2

0.8

(vii)

3.14 r2

3.142

3.14

(viii)

2 (l + b)

2l

2b

2

2

(ix)

0.1 y + 0.01 y2

0.1y

0.01y2

0.1

0.01

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) 12xy2 + 25

(vii) 7x + xy2

Solution:-

Sl.No.

Expression

Terms

Coefficient of x

(i)

y2x + y

y2x

y2

(ii)

13y2 – 8yx

– 8yx

-8y

(iii)

x + y + 2

x

1

(iv)

5 + z + zx

x

zx

1

z

(v)

1 + x + xy

xy

y

(vi)

12xy2 + 25

12xy2

12y2

(vii)

7x + xy2

7x

xy2

7

y2

(b) Identify terms which contain y2 and give the coefficient of y2.

(i) 8 – xy2 (ii) 5y2 + 7x (iii) 2x2y – 15xy2 + 7y2

Solution:-

Sl.No.

Expression

Terms

Coefficient of x

(i)

8 – xy2

– xy2

– x

(ii)

5y2 + 7x

5y2

5

(iii)

2x2y – 15xy2 + 7y2

– 15xy2

7y2

– 15x

7

5. Classify into monomials, binomials and trinomials.

(i) 4y – 7z

Solution:-

Binomial.

An expression which contains two unlike terms is called a binomial.

(ii) y2

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) 4p2q – 4pq2

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) z2 – 3z + 8

Solution:-

Trinomial.

An expression which contains three terms is called a trinomial.

(x) a2 + b2

Solution:-

Binomial.

An expression which contains two unlike terms is called a binomial.

(xi) z2 + z

Solution:-

Binomial.

An expression which contains two unlike terms is called a binomial.

(xii) 1 + x + x2

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 term have the same algebraic factors, they are like terms.

(ii) –7x, (5/2)x

Solution:-

Like term.

When term 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 term have the same algebraic factors, they are like terms.

(v) 4m2p, 4mp2

Solution:-

Unlike terms.

The terms have different algebraic factors, they are unlike terms.

(vi) 12xz, 12x2z2

Solution:-

Unlike terms.

The terms have different algebraic factors, they are unlike terms.

7. Identify like terms in the following:

(a) – xy2, – 4yx2, 8x2, 2xy2, 7y, – 11x2, – 100x, – 11yx, 20x2y, – 6x2, y, 2xy, 3x

Solution:-

When term have the same algebraic factors, they are like terms.

They are,

– xy2, 2xy2

– 4yx2, 20x2y

8x2, – 11x2, – 6x2

7y, y

– 100x, 3x

– 11yx, 2xy

(b) 10pq, 7p, 8q, – p2q2, – 7qp, – 100q, – 23, 12q2p2, – 5p2, 41, 2405p, 78qp,

13p2q, qp2, 701p2

Solution:-

When term have the same algebraic factors, they are like terms.

They are,

10pq, – 7qp, 78qp

7p, 2405p

8q, – 100q

– p2q2, 12q2p2

– 23, 41

– 5p2, 701p2

13p2q, qp2


Exercise 12.2 Page: 239

1. Simplify combining like terms:

(i) 21b – 32 + 7b – 20b

Solution:-

When term 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) – z2 + 13z2 – 5z + 7z3 – 15z

Solution:-

When term have the same algebraic factors, they are like terms.

Then,

= 7z3 + (-z2 + 13z2) + (-5z – 15z)

= 7z3 + z2 (-1 + 13) + z (-5 – 15)

= 7z3 + z2 (12) + z (-20) + 7z3

= 7z3 + 12z2 – 20z + 7z3

(iii) p – (p – q) – q – (q – p)

Solution:-

When term 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 term 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) 5x2y – 5x2 + 3yx2 – 3y2 + x2 – y2 + 8xy2 – 3y2

Solution:-

When term have the same algebraic factors, they are like terms.

Then,

= 5x2y + 3yx2 – 5x2 + x2 – 3y2 – y2 – 3y2

= x2y (5 + 3) + x2 (- 5 + 1) + y2 (-3 – 1 -3) + 8xy2

= x2y (8) + x2 (-4) + y2 (-7) + 8xy2

= 8x2y – 4x2 – 7y2 + 8xy2

(vi) (3y2 + 5y – 4) – (8y – y2 – 4)

Solution:-

When term have the same algebraic factors, they are like terms.

Then,

= 3y2 + 5y – 4 – 8y + y2 + 4

= 3y2 + y2 + 5y – 8y – 4 + 4

= y2 (3 + 1) + y (5 – 8) + (-4 + 4)

= y2 (4) + y (-3) + (0)

= 4y2 – 3y

2. Add:

(i) 3mn, – 5mn, 8mn, – 4mn

Solution:-

When term 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 term 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 term 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 term 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 term 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 term 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) 4x2y, – 3xy2, –5xy2, 5x2y

Solution:-

When term have the same algebraic factors, they are like terms.

Then, we have to add the like terms

= 4x2y + (-3xy2) + (-5xy2) + 5x2y

= 4x2y + 5x2y – 3xy2 – 5xy2

= x2y (4 + 5) + xy2 (-3 – 5)

= x2y (9) + xy2 (- 8)

= 9x2y – 8xy2

(viii) 3p2q2 – 4pq + 5, – 10 p2q2, 15 + 9pq + 7p2q2

Solution:-

When term have the same algebraic factors, they are like terms.

Then, we have to add the like terms

= 3p2q2 – 4pq + 5 + (- 10p2q2) + 15 + 9pq + 7p2q2

= 3p2q2 – 10p2q2 + 7p2q2 – 4pq + 9pq + 5 + 15

= p2q2 (3 -10 + 7) + pq (-4 + 9) + (5 + 15)

= p2q2 (0) + pq (5) + 20

= 5pq + 20

(ix) ab – 4a, 4b – ab, 4a – 4b

Solution:-

When term 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) x2 – y2 – 1, y2 – 1 – x2, 1 – x2 – y2

Solution:-

When term have the same algebraic factors, they are like terms.

Then, we have to add the like terms

= x2 – y2 – 1 + (y2 – 1 – x2) + (1 – x2 – y2)

= x2 – y2 – 1 + y2 – 1 – x2 + 1 – x2 – y2

= x2 – x2 – x2 – y2 + y2 – y2 – 1 – 1 + 1

= x2 (1 – 1- 1) + y2 (-1 + 1 – 1) + (-1 -1 + 1)

= x2 (1 – 2) + y2 (-2 +1) + (-2 + 1)

= x2 (-1) + y2 (-1) + (-1)

= -x2 – y2 -1

3. Subtract:

(i) –5y2 from y2

Solution:-

When term have the same algebraic factors, they are like terms.

Then, we have to subtract the like terms

= y2 – (-5y2)

= y2 + 5y2

= 6y2

(ii) 6xy from –12xy

Solution:-

When term have the same algebraic factors, they are like terms.

Then, we have to subtract the like terms

= -12xy – 6xy

= – 18xy

(iii) (a – b) from (a + b)

Solution:-

When term have the same algebraic factors, they are like terms.

Then, we have to subtract the like terms

= (a + b) – (a – b)

= a + b – a + b

= a – a + b + b

= a (1 – 1) + b (1 + 1)

= a (0) + b (2)

= 2b

(iv) a (b – 5) from b (5 – a)

Solution:-

When term have the same algebraic factors, they are like terms.

Then, we have to subtract the like terms

= b (5 -a) – a (b – 5)

= 5b – ab – ab + 5a

= 5b + ab (-1 -1) + 5a

= 5a + 5b – 2ab

(v) –m2 + 5mn from 4m2 – 3mn + 8

Solution:-

When term have the same algebraic factors, they are like terms.

Then, we have to subtract the like terms

= 4m2 – 3mn + 8 – (- m2 + 5mn)

= 4m2 – 3mn + 8 + m2 – 5mn

= 4m2 + m2 – 3mn – 5mn + 8

= 3m2 – 8mn + 8

(vi) – x2 + 10x – 5 from 5x – 10

Solution:-

When term have the same algebraic factors, they are like terms.

Then, we have to subtract the like terms

= 5x – 10 – (-x2 + 10x – 5)

= 5x – 10 + x2 – 10x + 5

= x2 + 5x – 10x – 10 + 5

= x2 – 5x – 5

(vii) 5a2 – 7ab + 5b2 from 3ab – 2a2 – 2b2

Solution:-

When term have the same algebraic factors, they are like terms.

Then, we have to subtract the like terms

= 3ab – 2a2 – 2b2 – (5a2 – 7ab + 5b2)

= 3ab – 2a2 – 2b2 – 5a2 + 7ab – 5b2

= 3ab + 7ab – 2a2 – 5a2 – 2b2 – 5b2

= 10ab – 7a2 – 7b2

(viii) 4pq – 5q2 – 3p2 from 5p2 + 3q2 – pq

Solution:-

When term have the same algebraic factors, they are like terms.

Then, we have to subtract the like terms

= 5p2 + 3q2 – pq – (4pq – 5q2 – 3p2)

= 5p2 + 3q2 – pq – 4pq + 5q2 + 3p2

= 5p2 + 3p2 + 3q2 + 5q2 – pq – 4pq

= 8p2 + 8q2 – 5pq

4. (a) What should be added to x2 + xy + y2 to obtain 2x2 + 3xy?

Solution:-

Let us assume p be the required term

Then,

p + (x2 + xy + y2) = 2x2 + 3xy

p = (2x2 + 3xy) – (x2 + xy + y2)

p = 2x2 + 3xy – x2 – xy – y2

p = 2x2 – x2 + 3xy – xy – y2

p = x2 – 2xy – y2

(b) What should be subtracted from 2a + 8b + 10 to get – 3a + 7b + 16?

Solution:-

Let us assume x be the required term

Then,

2a + 8b + 10 – x = -3a + 7b + 16

x = (2a + 8b + 10) – (-3a + 7b + 16)

x = 2a + 8b + 10 + 3a – 7b – 16

x = 2a + 3a + 8b – 7b + 10 – 16

x = 5a + b – 6

5. What should be taken away from 3x2 – 4y2 + 5xy + 20 to obtain – x2 – y2 + 6xy + 20?

Solution:-

Let us assume a be the required term

Then,

3x2 – 4y2 + 5xy + 20 – a = -x2 – y2 + 6xy + 20

a = 3x2 – 4y2 + 5xy + 20 – (-x2 – y2 + 6xy + 20)

a = 3x2 – 4y2 + 5xy + 20 + x2 + y2 – 6xy – 20

a = 3x2 + x2 – 4y2 + y2 + 5xy – 6xy + 20 – 20

a = 4x2 – 3y2 – xy

6. (a) From the sum of 3x – y + 11 and – y – 11, subtract 3x – y – 11.

Solution:-

First we have to find out the sum of 3x – y + 11 and – y – 11

= 3x – y + 11 + (-y – 11)

= 3x – y + 11 – y – 11

= 3x – y – y + 11 – 11

= 3x – 2y

Now, subtract 3x – y – 11 from 3x – 2y

= 3x – 2y – (3x – y – 11)

= 3x – 2y – 3x + y + 11

= 3x – 3x – 2y + y + 11

= -y + 11

(b) From the sum of 4 + 3x and 5 – 4x + 2x2, subtract the sum of 3x2 – 5x and

–x2 + 2x + 5.

Solution:-

First we have to find out the sum of 4 + 3x and 5 – 4x + 2x2

= 4 + 3x + (5 – 4x + 2x2)

= 4 + 3x + 5 – 4x + 2x2

= 4 + 5 + 3x – 4x + 2x2

= 9 – x + 2x2

= 2x2 – x + 9 … [equation 1]

Then, we have to find out the sum of 3x2 – 5x and – x2 + 2x + 5

= 3x2 – 5x + (-x2 + 2x + 5)

= 3x2 – 5x – x2 + 2x + 5

= 3x2 – x2 – 5x + 2x + 5

= 2x2 – 3x + 5 … [equation 2]

Now, we have to subtract equation (2) from equation (1)

= 2x2 – x + 9 – (2x2 – 3x + 5)

= 2x2 – x + 9 – 2x2 + 3x – 5

= 2x2 – 2x2 – x + 3x + 9 – 5

= 2x + 4


Exercise 12.3 Page: 242

1. If m = 2, find the value of:

(i) m – 2

Solution:-

From the question it is given that m = 2

Then, substitute the value of m in the question

= 2 -2

= 0

(ii) 3m – 5

Solution:-

From the question it is given that m = 2

Then, substitute the value of m in the question

= (3 × 2) – 5

= 6 – 5

= 1

(iii) 9 – 5m

Solution:-

From the question it is given that m = 2

Then, substitute the value of m in the question

= 9 – (5 × 2)

= 9 – 10

= – 1

(iv) 3m2 – 2m – 7

Solution:-

From the question it is given that m = 2

Then, substitute the value of m in the question

= (3 × 22) – (2 × 2) – 7

= (3 × 4) – (4) – 7

= 12 – 4 -7

= 12 – 11

= 1

(v) (5m/2) – 4

Solution:-

From the question it is given that m = 2

Then, substitute the value of m in the question

= ((5 × 2)/2) – 4

= (10/2) – 4

= 5 – 4

= 1

2. If p = – 2, find the value of:

(i) 4p + 7

Solution:-

From the question it is given that p = -2

Then, substitute the value of p in the question

= (4 × (-2)) + 7

= -8 + 7

= -1

(ii) – 3p2 + 4p + 7

Solution:-

From the question it is given that p = -2

Then, substitute the value of p in the question

= (-3 × (-2)2) + (4 × (-2)) + 7

= (-3 × 4) + (-8) + 7

= -12 – 8 + 7

= -20 + 7

= -13

(iii) – 2p3 – 3p2 + 4p + 7

Solution:-

From the question it is given that p = -2

Then, substitute the value of p in the question

= (-2 × (-2)3) – (3 × (-2)2) + (4 × (-2)) + 7

= (-2 × -8) – (3 × 4) + (-8) + 7

= 16 – 12 – 8 + 7

= 23 – 20

= 3

3. Find the value of the following expressions, when x = –1:

(i) 2x – 7

Solution:-

From the question it is given that x = -1

Then, substitute the value of x in the question

= (2 × -1) – 7

= – 2 – 7

= – 9

(ii) – x + 2

Solution:-

From the question it is given that x = -1

Then, substitute the value of x in the question

= – (-1) + 2

= 1 + 2

= 3

(iii) x2 + 2x + 1

Solution:-

From the question it is given that x = -1

Then, substitute the value of x in the question

= (-1)2 + (2 × -1) + 1

= 1 – 2 + 1

= 2 – 2

= 0

(iv) 2x2 – x – 2

Solution:-

From the question it is given that x = -1

Then, substitute the value of x in the question

= (2 × (-1)2) – (-1) – 2

= (2 × 1) + 1 – 2

= 2 + 1 – 2

= 3 – 2

= 1

4. If a = 2, b = – 2, find the value of:

(i) a2 + b2

Solution:-

From the question it is given that a = 2, b = -2

Then, substitute the value of a and b in the question

= (2)2 + (-2)2

= 4 + 4

= 8

(ii) a2 + ab + b2

Solution:-

From the question it is given that a = 2, b = -2

Then, substitute the value of a and b in the question

= 22 + (2 × -2) + (-2)2

= 4 + (-4) + (4)

= 4 – 4 + 4

= 4

(iii) a2 – b2

Solution:-

From the question it is given that a = 2, b = -2

Then, substitute the value of a and b in the question

= 22 – (-2)2

= 4 – (4)

= 4 – 4

= 0

5. When a = 0, b = – 1, find the value of the given expressions:

(i) 2a + 2b

Solution:-

From the question it is given that a = 0, b = -1

Then, substitute the value of a and b in the question

= (2 × 0) + (2 × -1)

= 0 – 2

= -2

(ii) 2a2 + b2 + 1

Solution:-

From the question it is given that a = 0, b = -1

Then, substitute the value of a and b in the question

= (2 × 02) + (-1)2 + 1

= 0 + 1 + 1

= 2

(iii) 2a2b + 2ab2 + ab

Solution:-

From the question it is given that a = 0, b = -1

Then, substitute the value of a and b in the question

= (2 × 02 × -1) + (2 × 0 × (-1)2) + (0 × -1)

= 0 + 0 +0

= 0

(iv) a2 + ab + 2

Solution:-

From the question it is given that a = 0, b = -1

Then, substitute the value of a and b in the question

= (02) + (0 × (-1)) + 2

= 0 + 0 + 2

= 2

6. Simplify the expressions and find the value if x is equal to 2

(i) x + 7 + 4 (x – 5)

Solution:-

From the question it is given that x = 2

We have,

= x + 7 + 4x – 20

= 5x + 7 – 20

Then, substitute the value of x in the equation

= (5 × 2) + 7 – 20

= 10 + 7 – 20

= 17 – 20

= – 3

(ii) 3 (x + 2) + 5x – 7

Solution:-

From the question it is given that x = 2

We have,

= 3x + 6 + 5x – 7

= 8x – 1

Then, substitute the value of x in the equation

= (8 × 2) – 1

= 16 – 1

= 15

(iii) 6x + 5 (x – 2)

Solution:-

From the question it is given that x = 2

We have,

= 6x + 5x – 10

= 11x – 10

Then, substitute the value of x in the equation

= (11 × 2) – 10

= 22 – 10

= 12

(iv) 4(2x – 1) + 3x + 11

Solution:-

From the question it is given that x = 2

We have,

= 8x – 4 + 3x + 11

= 11x + 7

Then, substitute the value of x in the equation

= (11 × 2) + 7

= 22 + 7

= 29

7. Simplify these expressions and find their values if x = 3, a = – 1, b = – 2.

(i) 3x – 5 – x + 9

Solution:-

From the question it is given that x = 3

We have,

= 3x – x – 5 + 9

= 2x + 4

Then, substitute the value of x in the equation

= (2 × 3) + 4

= 6 + 4

= 10

(ii) 2 – 8x + 4x + 4

Solution:-

From the question it is given that x = 3

We have,

= 2 + 4 – 8x + 4x

= 6 – 4x

Then, substitute the value of x in the equation

= 6 – (4 × 3)

= 6 – 12

= – 6

(iii) 3a + 5 – 8a + 1

Solution:-

From the question it is given that a = -1

We have,

= 3a – 8a + 5 + 1

= – 5a + 6

Then, substitute the value of a in the equation

= – (5 × (-1)) + 6

= – (-5) + 6

= 5 + 6

= 11

(iv) 10 – 3b – 4 – 5b

Solution:-

From the question it is given that b = -2

We have,

= 10 – 4 – 3b – 5b

= 6 – 8b

Then, substitute the value of b in the equation

= 6 – (8 × (-2))

= 6 – (-16)

= 6 + 16

= 22

(v) 2a – 2b – 4 – 5 + a

Solution:-

From the question it is given that a = -1, b = -2

We have,

= 2a + a – 2b – 4 – 5

= 3a – 2b – 9

Then, substitute the value of a and b in the equation

= (3 × (-1)) – (2 × (-2)) – 9

= -3 – (-4) – 9

= – 3 + 4 – 9

= -12 + 4

= -8

8. (i) If z = 10, find the value of z3 – 3(z – 10).

Solution:-

From the question it is given that z = 10

We have,

= z3 – 3z + 30

Then, substitute the value of z in the equation

= (10)3 – (3 × 10) + 30

= 1000 – 30 + 30

= 1000

(ii) If p = – 10, find the value of p2 – 2p – 100

Solution:-

From the question it is given that p = -10

We have,

= p2 – 2p – 100

Then, substitute the value of p in the equation

= (-10)2 – (2 × (-10)) – 100

= 100 + 20 – 100

= 20

9. What should be the value of a if the value of 2x2 + x – a equals to 5, when x = 0?

Solution:-

From the question it is given that x = 0

We have,

2x2 + x – a = 5

a = 2x2 + x – 5

Then, substitute the value of x in the equation

a = (2 × 02) + 0 – 5

a = 0 + 0 – 5

a = -5

10. Simplify the expression and find its value when a = 5 and b = – 3.

2(a2 + ab) + 3 – ab

Solution:-

From the question it is given that a = 5 and b = -3

We have,

= 2a2 + 2ab + 3 – ab

= 2a2 + ab + 3

Then, substitute the value of a and b in the equation

= (2 × 52) + (5 × (-3)) + 3

= (2 × 25) + (-15) + 3

= 50 – 15 + 3

= 53 – 15

= 38


Exercise 12.4 Page: 246

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.

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Image 6

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 NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Image 7

Solution:-

(a) From the question it is given that the numbers of segments required to form n digits of the kind
NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Image 8is (5n + 1)

Then,

The number of segments required to form 5 digits = ((5 × 5) + 1)

= (25 + 1)

= 26

The number of segments required to form 10 digits = ((5 × 10) + 1)

= (50 + 1)

= 51

The number of segments required to form 100 digits = ((5 × 100) + 1)

= (500 + 1)

= 501

(b) From the question it is given that the numbers of segments required to form n digits of the kind
NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Image 9is (3n + 1)

Then,

The number of segments required to form 5 digits = ((3 × 5) + 1)

= (15 + 1)

= 16

The number of segments required to form 10 digits = ((3 × 10) + 1)

= (30 + 1)

= 31

The number of segments required to form 100 digits = ((3 × 100) + 1)

= (300 + 1)

= 301

(c) From the question it is given that the numbers of segments required to form n digits of the kind
NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Image 10is (5n + 2)

Then,

The number of segments required to form 5 digits = ((5 × 5) + 2)

= (25 + 2)

= 27

The number of segments required to form 10 digits = ((5 × 10) + 2)

= (50 + 2)

= 52

The number of segments required to form 100 digits = ((5 × 100) + 1)

= (500 + 2)

= 502

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

10001

Solution:-

(i) From the table (2n – 1)

Then, 100th term =?

Where n = 100

= (2 × 100) – 1

= 200 – 1

= 199

(ii) From the table (3n + 2)

5th term =?

Where n = 5

= (3 × 5) + 2

= 15 + 2

= 17

Then, 10th term =?

Where n = 10

= (3 × 10) + 2

= 30 + 2

= 32

Then, 100th term =?

Where n = 100

= (3 × 100) + 2

= 300 + 2

= 302

(iii) From the table (4n + 1)

5th term =?

Where n = 5

= (4 × 5) + 1

= 20 + 1

= 21

Then, 10th term =?

Where n = 10

= (4 × 10) + 1

= 40 + 1

= 41

Then, 100th term =?

Where n = 100

= (4 × 100) + 1

= 400 + 1

= 401

(iv) From the table (7n + 20)

5th term =?

Where n = 5

= (7 × 5) + 20

= 35 + 20

= 55

Then, 10th term =?

Where n = 10

= (7 × 10) + 20

= 70 + 20

= 90

Then, 100th term =?

Where n = 100

= (7 × 100) + 20

= 700 + 20

= 720

(v) From the table (n2 + 1)

5th term =?

Where n = 5

= (52) + 1

= 25+ 1

= 26

Then, 10th term =?

Where n = 10

= (102) + 1

= 100 + 1

= 101

So the table is completed below.

S. No.

Expression

Terms

1st

2nd

3rd

4th

5th

10th

100th

(i)

2n – 1

1

3

5

7

9

19

199

(ii)

3n + 2

5

8

11

14

17

32

302

(iii)

4n + 1

5

9

13

17

21

41

401

(iv)

7n + 20

27

34

41

48

55

90

720

(v)

n2 + 1

2

5

10

17

26

101

10001


Also, access exercises of NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions

Exercise 12.1 Solutions

Exercise 12.2 Solutions

Exercise 12.3 Solutions

Exercise 12.4 Solutions

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