Mathematics NCERT Grade 7, Chapter 12: Algebraic Expressions- The chapters discuss concepts related to algebraic expressions.
After a short introduction, the first section discusses How are expressions formed?
• Algebraic expressions are formed from variables and constants.
​This is followed by the topics terms, factors, and coefficients.
• ​Expressions are made up of terms.
• A term is a product of factors.
• Coefficient is the numerical factor in the term
Before moving to terms like monomials, binomials, and polynomials, like and unlike terms are discussed.
• When terms have the same algebraic factors, they are like terms.
• When terms have different algebraic factors, they are unlike terms.
Any expression with one or more terms is called a polynomial.
• A one-term expression is called a monomial.
• A two-term expression is called binomial.
• An expression that contains three terms is called a trinomial.
Emphasis is given to important topics like the Addition and Subtraction of Algebraic Expressions.
2 sub parts are discussed in the following section:
• Adding and subtracting like terms: The sum of two or more like terms is a like term with a numerical coefficient equal to the sum of the numerical coefficients of all the like terms.
• The difference between two like terms is a like term with a numerical coefficient equal to the difference between the numerical coefficients of the two like terms.
• Adding and subtracting general algebraic expressions
• The like terms are added together while the unlike terms are left as they are.
The next part of the section explains about Finding the Value of an Expression where value of a variable is given depending on which value of the expression is calculated.

This is followed by the topic- Using Algebraic Expressions- Formulas and Rules where one understands how algebraic expressions can be used to write formulas and patterns. These patterns are related to numbers and geometrical patterns.
• Perimeter formulas
• Area formulas
• Rules for number patterns
• Some more number patterns
• Pattern in geometry
Unsolved, solved exercises, questions in different patterns will make the chapter more comprehensible to students.
All important points of the chapter are cited in the end under the title- What Have We Discussed?

#### Question 1:

Get the algebraicexpressions 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 z subtracted from 10.

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

(i) yz

(ii)

(iii) z2

(iv)

(v) x2 + y2

(vi) 5 + 3 (mn)

(vii) 10 − yz

(viii) 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) (e) − ab + 2b2 − 3a2

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

(a) − 4x + 5 (b) − 4x + 5y (c) 5y + 3y2

(d) (e) pq + q

(f) 1.2 ab − 2.4 b + 3.6 a (g)

(h) 0.1p2 + 0.2 q2

(i)

(a)

(b)

(c)

(d)

(e)

(ii)

 Row 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) 1.2ab − 2.4b + 3.6a 1.2ab − 2.4b 3.6a 1.2, a, b − 2.4, b 3.6, a (g) (h) 0.1p2 + 0.2q2 0.1p2 0.2q2 0.1, p, p 0.2, q, q

#### Question 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) 1.2a + 0.8b

(vii) 3.14 r2 (viii) 2 (l + b) (ix) 0.1y + 0.01 y2

 Row 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) 1.2a +0.8b 1.2a 0.8b 1.2 0.8 (vii) 3.14 r2 3.14 r2 3.14 (viii) 2(l + b) 2l 2b 2 2 (ix) 0.1y + 0.01y2 0.1y 0.01y2 0.1 0.01

#### 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) 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

(a)

 Row Expression Terms with x Coefficient of x (i) y2x + y y2x y2 (ii) 13y2 − 8yx − 8yx −8y (iii) x + y + 2 x 1 (iv) 5 + z + zx zx z (v) 1 + x + xy x xy 1 y (vi) 12xy2 + 25 12xy2 12y2 (vii) 7x+ xy2 7x xy2 7 y2

(b)

 Row Expression Terms with y2 Coefficient of y2 (i) 8 − xy2 −xy2 − x (ii) 5y2 + 7x 5y2 5 (iii) 2x2y + 7y2 −15xy2 7y2 −15xy2 7 −15x

#### Question 5:

Classify into monomials, binomials and trinomials.

(i) 4y7z (ii) y2 (iii) x + yxy

(iv) 100 (v) abab (vi) 5 − 3t

(vii) 4p2q − 4pq2 (viii) 7mn (ix) z2 − 3z + 8

(x) a2 + b2 (xi) z2 + z (xii) 1 + x + x2

The monomials, binomials, and trinomials have 1, 2, and 3 unlike terms in it respectively.

(i) 4y − 7z

Binomial

(ii) y2

Monomial

(iii) x + yxy

Trinomial

(iv) 100

Monomial

(v) abab

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

#### Question 6:

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

(i) 1, 100 (ii) (iii) − 29x, − 29y

(iv) 14xy, 42yx (v) 4m2p, 4mp2 (vi) 12xz, 12 x2z2

The terms which have the same algebraic factors are called like terms. However, when the terms have different algebraic factors, these are called unlike terms.

(i) 1, 100

Like

(ii) − 7x,

Like

(iii) −29x, −29y

Unlike

(iv) 14xy, 42yx

Like

(v) 4m2p, 4mp2

Unlike

(vi) 12xz, 12x2z2

Unlike

#### Question 7:

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

(a) −xy2, 2xy2

−4yx2, 20x2y

8x2, −11x2, −6x2

7y, y

−100x, 3x

−11xy, 2xy

(b) 10pq, −7qp, 78qp

7p, 2405p

8q, −100q

p2q2, 12p2q2

−23, 41

−5p2, 701p2

13p2q, qp2

#### Question 1:

Simplify combining like terms:

(i) 21b − 32 + 7b − 20b

(ii) − z2 + 13z2 − 5z + 7z3 − 15z

(iii) p − (pq) − q − (q p)

(iv) 3a − 2bab − (ab + ab) + 3ab + b − a

(v) 5x2y − 5x2 + 3y x2 − 3y2 + x2 y2 + 8xy2 −3y2

(vi) (3 y2 + 5y − 4) − (8yy2 − 4)

(i) 21b − 32 + 7b − 20b = 21b + 7b − 20b − 32

= b (21 + 7 − 20) −32

= 8b − 32

(ii) − z2 + 13z2 − 5z + 7z3 − 15z = 7z3z2 + 13z2 − 5z − 15z

= 7z3 + z2 (−1 + 13) + z (−5 − 15)

= 7z3 + 12z2 − 20z

(iii) p − (pq) − q − (qp) = pp + qqq + p

= p q

(iv) 3a − 2bab − (ab + ab) + 3ba + b a

= 3a − 2bab a + bab + 3ab + b a

= 3aaa − 2b + b + bab ab + 3ab

= a (3 − 1 − 1) + b (− 2 + 1 + 1) + ab (−1 −1 + 3)

= a + ab

(v) 5x2y − 5x2 + 3yx2 − 3y2 + x2y2 + 8xy2 − 3y2

= 5x2y + 3yx2 − 5x2 + x2 − 3y2y2 − 3y2 + 8xy2

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

= 8x2y − 4x2 − 7y2 + 8xy2

(vi) (3y2 + 5y − 4) − (8yy2 − 4)

= 3y2 + 5y − 4 − 8y + y2 + 4

= 3y2 + y2 + 5y − 8y − 4 + 4

= y2 (3 + 1) + y (5 − 8) + 4 (1 − 1)

= 4y2 − 3y

#### Question 2:

(i) 3mn, − 5mn, 8mn, −4mn

(ii) t − 8tz, 3tzz, zt

(iii) − 7mn + 5, 12mn + 2, 9mn − 8, − 2mn − 3

(iv) a + b − 3, ba + 3, ab + 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, 4bab, 4a − 4b

(x) x2 y2 − 1 , y2 − 1 − x2, 1− x2 y2

(i) 3mn + (−5mn) + 8mn + (−4mn) = mn (3 − 5 + 8 − 4)

= 2mn

(ii) (t − 8tz) + (3tzz) + (zt) = t − 8tz + 3tzz + zt

= t t − 8tz + 3tzz + z

= t (1 − 1) + tz (− 8 + 3) + z (− 1 + 1)

= −5tz

(iii) (− 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)

= 12mn − 4

(iv) (a + b − 3) + (ba + 3) + (ab + 3)

= a + b − 3 + ba + 3 + ab + 3

= aa + a + b + b b − 3 + 3 + 3

= a (1 − 1 + 1) + b (1 + 1 − 1) + 3 (− 1 + 1 + 1)

= a + b + 3

(v) (14x + 10y − 12xy − 13) + (18 − 7x − 10y + 8yx) + 4xy

= 14x + 10y − 12xy − 13 + 18 − 7x − 10y + 8yx + 4xy

= 14x − 7x + 10y − 10y − 12xy + 8yx + 4xy − 13 + 18

= x (14 − 7) + y (10 − 10) + xy (− 12 + 8 + 4) − 13 + 18

= 7x + 5

(vi) (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

= 3m − 4n − 3mn − 3

(vii) 4x2 y − 3xy2 − 5xy2 + 5x2y = 4x2 y + 5x2y − 3xy2 − 5xy2

= x2 y (4 + 5) + xy2 (− 3 − 5)

= 9x2y − 8xy2

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

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

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

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

= 5pq + 20

(ix) (ab − 4a) + (4b ab) + (4a − 4b)

= ab − 4a + 4b ab + 4a − 4b

= abab − 4a + 4a + 4b − 4b

= ab (1 − 1) + a (− 4 + 4) + b(4 − 4)

= 0

(x) (x2y2 − 1) + (y2 − 1 − x2) + (1 − x2y2)

= x2y2 − 1 + y2 − 1 − x2 + 1 − x2y2

= x2x2 x2 y2 + y2 y2 − 1 − 1 + 1

= x2(1 − 1 − 1) + y2 (−1 + 1 − 1) + (− 1 − 1 + 1)

= − x2y2 − 1

#### Question 3:

Subtract:

(i) − 5y2 from y2

(ii) 6xy from − 12xy

(iii) (ab) 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

(i) y2 − (−5y2) = y2 + 5y2 = 6y2

(ii) − 12xy − (6xy) = −18xy

(iii) (a + b) − (ab) = a + b a + b = 2b

(iv) b (5 − a) − a (b − 5) = 5babab + 5a

= 5a + 5b − 2ab

(v) (4m2 − 3mn + 8) − (− m2 + 5mn) = 4m2 − 3mn + 8 + m2 − 5 mn

= 4m2 + m2 − 3mn − 5 mn + 8

= 5m2 − 8mn + 8

(vi) (5x − 10) − (− x2 + 10x − 5) = 5x − 10 + x2 − 10x + 5

= x2 + 5x − 10x − 10 + 5

= x2 − 5x − 5

(vii) (3ab − 2a2 − 2b2) − (5a2− 7ab + 5b2)

= 3ab − 2a2 − 2b2 − 5a2 + 7ab − 5 b2

= 3ab + 7ab − 2a2 − 5a2 − 2b2 − 5 b2

= 10ab − 7a2 − 7b2

(viii) 4pq − 5q2 − 3p2 from 5p2 + 3q2pq

(5p2 + 3q2pq) − (4pq − 5q2− 3p2)

= 5p2 + 3q2 pq − 4pq + 5q2 + 3p2

= 5p2 + 3p2 + 3q2 + 5q2 pq − 4pq

= 8p2 + 8q2 − 5pq

#### Question 4:

(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?

(a) Let a be the required term.

a + (x2 + y2 + xy) = 2x2 + 3xy
a = 2x2 + 3xy − (x2 + y2 + xy)

a = 2x2 + 3xyx2y2xy

a = 2x2x2y2 + 3xyxy

= x2y2 + 2xy
​​​​​​​​​​​​​

(b) Let p be the required term.

(2a + 8b + 10) − p = − 3a + 7b + 16

p = 2a + 8b + 10 − (− 3a + 7b + 16)

= 2a + 8b + 10 + 3a − 7b − 16

= 2a + 3a + 8b − 7b + 10− 16

= 5a + b − 6

##### Video Solution for algebraic expressions (Page: 240 , Q.No.: 4)

NCERT Solution for Class 7 maths - algebraic expressions 240 , Question 4

#### Question 5:

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

x2y2 + 6xy + 20?

Let p be the required term.

(3x2 − 4y2 + 5xy + 20) − p = − x2y2 + 6xy + 20

p = (3x2 − 4y2 + 5xy + 20) − (− x2y2 + 6xy + 20)

= 3x2 − 4y2 + 5xy + 20 + x2 + y2 − 6xy − 20

= 3x2 + x2 − 4y2 + y2 + 5xy − 6xy + 20 − 20

= 4x2 − 3y2xy
​​​​​​​​​​​​​

##### Video Solution for algebraic expressions (Page: 240 , Q.No.: 5)

NCERT Solution for Class 7 maths - algebraic expressions 240 , Question 5

#### Question 6:

(a) From the sum of 3xy + 11 and − y − 11, subtract 3xy − 11.

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

(a) (3xy + 11) + (− y − 11)

= 3xy + 11 − y − 11

= 3xy y + 11 − 11

= 3x − 2y

(3x − 2y) − (3xy − 11)

= 3x − 2y − 3x + y + 11

= 3x − 3x − 2y + y + 11

= − y + 11
​​​​​​​​​​​​​

(b) (4 + 3x) + (5 − 4x + 2x2) = 4 + 3x + 5 − 4x + 2x2

= 3x − 4x + 2x2 + 4 + 5

= − x + 2x2 + 9

(3x2 − 5x) + (− x2 + 2x + 5) = 3x2 − 5xx2 + 2x + 5

= 3x2x2 − 5x + 2x + 5

= 2x2 − 3x + 5

(− x + 2x2 + 9) − (2x2 − 3x + 5)

= − x + 2x2 + 9 − 2x2 + 3x − 5

= − x + 3x + 2x2 − 2x2 + 9 − 5

= 2x + 4

##### Video Solution for algebraic expressions (Page: 240 , Q.No.: 6)

NCERT Solution for Class 7 maths - algebraic expressions 240 , Question 6

#### Question 1:

If m = 2, find the value of:

(i) m − 2 (ii) 3m − 5 (iii) 9 − 5m

(iv) 3m2 − 2m − 7 (v)

(i) m − 2 = 2 − 2 = 0

(ii) 3m − 5 = (3 × 2) − 5 = 6 − 5 = 1

(iii) 9 − 5m = 9 − (5 × 2) = 9 −10 = −1

(iv) 3m2 − 2m − 7 = 3 × (2 × 2) − (2 × 2) − 7

= 12 − 4 − 7 = 1

(v)

#### Question 2:

If p = −2, find the value of:

(i) 4p + 7

(ii) −3p2 + 4p + 7

(iii) −2p3 − 3p2 + 4p + 7

(i) 4p + 7 = 4 × (−2) + 7 = − 8 + 7 = −1

(ii) − 3p2 + 4p + 7 = −3 (−2) × (−2) + 4 × (−2) + 7

= − 12 − 8 + 7 = −13

(iii) −2p3 − 3p2 + 4p + 7

= −2 (−2) × (−2) × (−2) − 3 (−2) × (−2) + 4 × (−2) + 7

= 16 − 12 − 8 + 7 = 3

#### Question 3:

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

(i) 2x − 7 (ii) − x + 2 (iii) x2 + 2x + 1

(iv) 2x2x − 2

(i) 2x − 7

= 2 × (−1) − 7 = −9

(ii) − x + 2 = − (−1) + 2 = 1 + 2 = 3

(iii) x2 + 2x + 1 = (−1) × (−1) + 2 × (−1) + 1

= 1 − 2 + 1 = 0

(iv) 2x2x − 2 = 2 (−1) × (−1) − (−1) − 2

= 2 + 1 − 2 = 1

#### Question 4:

If a = 2, b = − 2, find the value of:

(i) a2 + b2 (ii) a2 + ab + b2 (iii) a2b2

(i) a2 + b2

= (2)2 + (−2)2 = 4 + 4 = 8

(ii) a2 + ab + b2

= (2 × 2) + 2 × (−2) + (−2) × (−2)

= 4 − 4 + 4 = 4

(iii) a2b2

= (2)2 − (−2)2 = 4 − 4 = 0

#### Question 5:

When a = 0, b = − 1, find the value of the given expressions:

(i) 2a + 2b (ii) 2a2 + b2 + 1

(iii) 2a2 b + 2ab2 + ab (iv) a2 + ab + 2

(i) 2a + 2b = 2 × (0) + 2 × (−1) = 0 − 2 = −2

(ii) 2a2 + b2 + 1

= 2 × (0)2 + (−1) × (−1) + 1

= 0 + 1 + 1 = 2

(iii) 2a2b + 2ab2 + ab

= 2 × (0)2 × (−1) + 2 × (0) × (−1) × (−1) + 0 × (−1)

= 0 + 0 + 0 = 0

(iv) a2 + ab + 2

= (0)2 + 0 × (−1) + 2

= 0 + 0 + 2 = 2
​​​​​​​​​​​​​

##### Video Solution for algebraic expressions (Page: 242 , Q.No.: 5)

NCERT Solution for Class 7 maths - algebraic expressions 242 , Question 5

#### Question 6:

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

(i) x + 7 + 4 (x − 5) = x + 7 + 4x − 20

= x + 4x + 7 − 20

= 5x − 13

= (5 × 2) − 13

= 10 − 13 = −3

(ii) 3 (x + 2) + 5x − 7 = 3x + 6 + 5x − 7

= 3x + 5x + 6 − 7 = 8x − 1

= (8 × 2) − 1 = 16 − 1 =15

(iii) 6x + 5 (x − 2) = 6x + 5x − 10

= 11x − 10

= (11 × 2) − 10 = 22 − 10 = 12

(iv) 4 (2x − 1) + 3x + 11 = 8x − 4 + 3x + 11

= 11x + 7

= (11 × 2) + 7

= 22 + 7 = 29

#### Question 7:

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

(i) 3x − 5 − x + 9 = 3xx − 5 + 9

= 2x + 4 = (2 × 3) + 4 = 10

(ii) 2 − 8x + 4x + 4 = 2 + 4 − 8x + 4x

= 6 − 4x = 6 − (4 × 3) = 6 − 12 = −6

(iii) 3a + 5 − 8a + 1 = 3a − 8a + 5 + 1

= − 5a + 6 = −5 × (−1) + 6

= 5 + 6 = 11

(iv) 10 − 3b − 4 − 5b = 10 − 4− 3b − 5b

= 6 − 8b = 6 − 8 × (−2)

= 6 + 16 = 22

(v) 2a − 2b − 4 − 5 + a = 2a + a − 2b − 4 − 5

= 3a − 2b − 9s

= 3 × (−1) − 2 (−2) − 9

= − 3 + 4 − 9 = −8

#### 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

(i) z3 − 3 (z − 10) = z3 − 3z + 30

= (10 × 10 × 10) − (3 × 10) + 30

= 1000 − 30 + 30 = 1000

(ii) p2 − 2p − 100

= (−10) × (−10) − 2 (−10) − 100

= 100 + 20 − 100 = 20
​​​​​​​​​​​​​​

##### Video Solution for algebraic expressions (Page: 242 , Q.No.: 8)

NCERT Solution for Class 7 maths - algebraic expressions 242 , Question 8

#### Question 9:

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

2x2 + xa = 5, when x = 0

(2 × 0) + 0 − a = 5

0 − a = 5

a = −5
​​​​​​​​​​​​​​​

##### Video Solution for algebraic expressions (Page: 242 , Q.No.: 9)

NCERT Solution for Class 7 maths - algebraic expressions 242 , Question 9

#### Question 10:

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

2 (a2 + ab) + 3 − ab

2 (a2 + ab) + 3 − ab = 2a2 + 2ab + 3 − ab

= 2a2 + 2abab + 3

= 2a2 + ab + 3

= 2 × (5 × 5) + 5 × (−3) + 3

= 50 − 15 + 3 = 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.

(a)

(b)

(c)

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 −

, , .

(a) It is given that the number of segments required to form n digits of the kind

is (5n + 1).

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

= 25 + 1 = 26

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

= 50 + 1 = 51

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

= 500 + 1 = 501

(b) It is given that the number of segments required to form n digits of the kind is (3n + 1).

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

= 15 + 1 = 16

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

= 30 + 1 = 31

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

= 300 + 1 = 301

(c)It is given that the number of segments required to form n digits of the kind is (5n + 2).

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

= 25 + 2 = 27

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

= 50 + 2 = 52

Number of segments required to form 100 digits = (5 × 100 + 2)

= 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 2 5 8 11 - - - - - - (iii) 4n + 1 5 9 13 17 - - - - - - (iv) 7n + 20 27 34 41 48 - - - - - - (v) n2 + 1 2 5 10 17 - - - - 10, 001 -