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Page No 234:

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.

Answer:

(i) yz

(ii)

(iii) z2

(iv)

(v) x2 + y2

(vi) 5 + 3 (mn)

(vii) 10 − yz

(viii) ab − (a + b)

Page No 234:

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

Answer:

(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



Page No 235:

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

Answer:

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

Page No 235:

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

Answer:

(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

Page No 235:

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

Answer:

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

Page No 235:

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

Answer:

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

Page No 235:

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

Answer:

(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



Page No 239:

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)

Answer:

(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

Page No 239:

Question 2:

Add:

(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

Answer:

(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



Page No 240:

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

Answer:

(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

Page No 240:

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?

Answer:

(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 math - Algebraic Expressions 240 , Question 4

Page No 240:

Question 5:

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

x2y2 + 6xy + 20?

Answer:

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 math - Algebraic Expressions 240 , Question 5

Page No 240:

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.

Answer:

(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 math - Algebraic Expressions 240 , Question 6



Page No 242:

Question 1:

If m = 2, find the value of:

(i) m − 2 (ii) 3m − 5 (iii) 9 − 5m

(iv) 3m2 − 2m − 7 (v)

Answer:

(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)

Page No 242:

Question 2:

If p = −2, find the value of:

(i) 4p + 7

(ii) −3p2 + 4p + 7

(iii) −2p3 − 3p2 + 4p + 7

Answer:

(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

Page No 242:

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

Answer:

(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

Page No 242:

Question 4:

If a = 2, b = − 2, find the value of:

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

Answer:

(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

Page No 242:

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

Answer:

(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 math - Algebraic Expressions 242 , Question 5

Page No 242:

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

Answer:

(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

Page No 242:

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

Answer:

(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

Page No 242:

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) 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 math - Algebraic Expressions 242 , Question 8

Page No 242:

Question 9:

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

Answer:

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 math - Algebraic Expressions 242 , Question 9

Page No 242:

Question 10:

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

2 (a2 + ab) + 3 − ab

Answer:

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



Page No 246:

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 −

, , .

Answer:

(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



Page No 247:

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

-

Answer:

The given table can be completed as follows.

S.No.

Expression

Terms

1st

2nd

3rd

4th

5th

10th

100th

(i)

2n − 1

1

3

5

7

9

-

19

-

199

-

(ii)

3n + 2

2

5

8

11

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

-

10,001-

-



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