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NCERT solutions for class 7 maths chapter 12 Algebraic Expressions

Unlock the Power of Algebraic Expressions withour quest to facilitate your learning, we proudly offer the PDF solutions for NCERT Class 7 Maths Chapter 12 - "Algebraic Expressions." These solutions serve as a valuable resource, especially when students encounter challenges while working on questions from this chapter.Our solutions for Exercise 12.1 focus on providing a clear understanding of fundamental concepts in algebra. This exercise covers topics related to the formation of expressions, understanding terms within expressions, distinguishing between like and unlike terms, and exploring monomials, binomials, trinomials, and polynomials.

Exercise 12.1

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 numbers m and n.

(vii) Product of numbers y and z subtracted from 10.

(viii) Sum of numbers a and b subtracted from their product.

Answer :

(i) = Y – z

(ii) = ½ (x + y)

= (x + y)/2

(iii) = z × z

= z2

(iv) = ¼ (p × q)

= pq/4

(v) = x2 + y2

(vi) = 3mn + 5

(vii) = 10 – (y × z)

= 10 – yz

(viii) = (a × b) – (a + b)

= 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

(b) 1 + x + x2

(c) y – y3

(d) 5xy2 + 7x2y

(e) – ab + 2b2 – 3a2

(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

 

Answer :

(i)(a) Expression: x – 3

Terms: x, -3

Factors: x; -3

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

(b) 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) Expression: y – y3

Terms: y, -y3

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

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

(d) 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) 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) 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

 

 


Question 3 :

State whether a given pair of terms is of like or unlike terms.

(i) 1, 100

(ii) –7x, (5/2)x

(iii) – 29x, – 29y

(iv) 14xy, 42yx

(v) 4m2p, 4mp2

(vi) 12xz, 12x2z2

 

Answer :

(i) Like term.

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

(ii) Like term.

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

(iii) Unlike terms.

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

(iv) Like term.

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

(v) Unlike terms.

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

(vi) Unlike terms.

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

 


Question 4 :

Identify like terms in the following:

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

(b) 10pq, 7p, 8q, – p2q2, – 7qp, – 100q, – 23, 12q2p2, – 5p2, 41, 2405p, 78qp,13p2q, qp2, 701p2

 

Answer :

(a) When terms 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) When terms 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


Question 5 :

Classify into monomials, binomials and 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

Answer :

(i) Binomial.

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

(ii) Monomial.

An expression with only one term is called a monomial.

(iii) Trinomial.

An expression which contains three terms is called a trinomial.

(iv) Monomial.

An expression with only one term is called a monomial.

(v) Trinomial.

An expression which contains three terms is called a trinomial.

(vi) Binomial.

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

(vii) Binomial.

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

(viii) Monomial.

An expression with only one term is called a monomial.

(ix) Trinomial.

An expression which contains three terms is called a trinomial.

(x) Binomial.

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

(xi) Binomial.

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

(xii) Trinomial.

An expression which contains three terms is called a trinomial.

 


Question 6 :

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

 

Answer :

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 – 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

 


Question 7 :

(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

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

(i) 8 – xy2

(ii) 5y2 + 7x

(iii) 2x2y – 15xy2 + 7y2

 

Answer :

(a)

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)

Sl.No.

Expression

Terms

Coefficient of y2

(i)

8 – xy2

– xy2

– x

(ii)

5y2 + 7x

5y2

5

(iii)

2x2y – 15xy2 + 7y2

– 15xy2

7y2

– 15x

7

 


Exercise 12.2

Question 1 :

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

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

 

Answer :

(a) 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) 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

 


Question 2 :

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

 

Answer :

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

 


Question 3 :

Simplify combining like terms:

(i) 21b – 32 + 7b – 20b

(ii) – z2 + 13z2 – 5z + 7z3 – 15z

(iii) p – (p – q) – q – (q – p)

(iv) 3a – 2b – ab – (a – b + ab) + 3ab + b – a

(v) 5x2y – 5x2 + 3yx2 – 3y2 + x2 – y2 + 8xy2 – 3y2

(vi) (3y2 + 5y – 4) – (8y – y2 – 4)

Answer :

(i) 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) When terms 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 + 12z2 – 20z

(iii) When terms have the same algebraic factors, they are like terms.

Then,

= p – p + q – q – q + p

= p – q

(iv) 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) When terms 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) When terms 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

 


Question 4 :

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, – 10 p2q2, 15 + 9pq + 7p2q2

(ix) ab – 4a, 4b – ab, 4a – 4b

(x) x2 – y2 – 1, y2 – 1 – x2, 1 – x2 – y2

 

Answer :

(i) 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) 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) 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) 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) 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) 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) When terms 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) When terms 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) 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) When terms 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

 


Question 5 :

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

 

Answer :

(i) When terms have the same algebraic factors, they are like terms.

Then, we have to subtract the like terms

= y2 – (-5y2)

= y2 + 5y2

= 6y2

(ii) When terms have the same algebraic factors, they are like terms.

Then, we have to subtract the like terms

= -12xy – 6xy

= – 18xy

(iii) When terms 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) When terms 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) When terms 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

= 5m2 – 8mn + 8

(vi) When terms 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) When terms 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) When terms 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

 


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.

 

Answer :

(a) 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) 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

Question 1 :

If p = – 2, find the value of:

(i) 4p + 7

(ii) – 3p2 + 4p + 7

(iii) – 2p3 – 3p2 + 4p + 7

 

Answer :

(i) 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) 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) 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

 


Question 2 :

If a = 2, b = – 2, find the value of:

(i) a2 + b2

(ii) a2 + ab + b2

(iii) a2 – b2

 

Answer :

(i) 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) 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) 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

 


Question 3 :

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

 

Answer :

(i) 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) 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) 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) 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

 


Question 4 :

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

(i) 2x – 7

(ii) – x + 2

(iii) x2 + 2x + 1

(iv) 2x2 – x – 2

 

Answer :

(i) 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) From the question it is given that x = -1

Then, substitute the value of x in the question

= – (-1) + 2

= 1 + 2

= 3

(iii) 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) 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

 


Question 5 :

If m = 2, find the value of:

(i) m – 2

(ii) 3m – 5

(iii) 9 – 5m

(iv) 3m2 – 2m – 7

(v) (5m/2) – 4

Answer :

(i) From the question it is given that m = 2

Then, substitute the value of m in the question

= 2 -2

= 0

(ii) 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) 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) 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) 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

 


Question 6 :

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

2(a2 + ab) + 3 – ab

 

Answer :

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




Question 7 :

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

 

Answer :

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

 


Question 8 :

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

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

 

Answer :

(i) 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) 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

 


Question 9 :

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 – 5b

(v) 2a – 2b – 4 – 5 + a

 

Answer :

(i) 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) 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) 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) 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) 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

 


Question 10 :

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

 

Answer :

(i) 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) 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) 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) 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

 


Exercise 12.4

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.

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

 

Answer :

(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

 


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

10001

 

Answer :

(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




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