Class 9

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 3.5 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 3 Polynomials.

Practice Set 3.5 Algebra 9th Std Maths Part 1 Answers Chapter 3 Polynomials

Question 1.
Find the value of the polynomial 2x – 2x³ + 7 using given values for x.
i. x = 3
ii. x = -1
iii. x = 0
Solution:
i. p(x) = 2x – 2x³ + 7
Put x = 3 in the given polynomial.
∴ p(3) = 2(3) – 2(3)³ + 7
= 6 – 2 x 27 + 7
= 6 – 54 + 7
∴ P(3) = – 41

ii. p(x) = 2x – 2x³ + 7
Put x = -1 in the given polynomial.
∴ p(- 1) = 2(- 1) – 2(-1)³ + 7
= – 2 – 2(-1) + 7
= -2 + 2 + 7
∴ p(-1) = 7

iii. p(x) = 2x – 2x³ + 7
Put x = 0 in the given polynomial.
∴ p(0) = 2(0) – 2(0)³ + 7
= 0 – 0 + 7
∴ P(0) = 7

Question 2.
For each of the following polynomial, find p(1), p(0) and p(- 2).
i. p(x) = x³
ii. p(y) = y² – 2y + 5
ii. p(y) = x⁴ – 2x² + x
Solution:
i. p(x) = x³
∴ p(1) = 1³ = 1
p(x) = x³
∴ p(0) = 0³ = 0
p(x) = x³
∴ p(-2) = (-2)³ = -8

ii. p(y) = y² – 2y + 5
∴ p(1) = 1² – 2(1) + 5
= 1 – 2 + 5
∴ P(1) = 4
p(y) = y² – 2y + 5
∴ p(0) = 0² – 2(0) + 5
= 0 – 0 + 5
∴ p(0) = 5
p(y) = y² – 2y + 5
∴ p(- 2) = (- 2)² – 2(- 2) + 5
= 4 + 4 + 5
∴ p(-2) = 13

iii. p(x) = x⁴ – 2x² – x
∴ p(1) = (1)⁴ – 2(1)² – 1
= 1 – 2 – 1
∴ p(1) = -2
∴ p(x) = x⁴ – 2x² – x
∴ p(0) = (0)⁴ – 2(0)² – 0
= 0 – 0 – 0
∴ p(0) = 0
p(x) = x⁴ – 2x² – x
∴ p(-2) = (-2)⁴ – 2(-2)² – (-2)
= 16 – 2(4) + 2
= 16 – 8 + 2
∴ p(-2) = 10

Question 3.
If the value of the polynomial m³ + 2m + a is 12 for m = 2, then find the value of a.
Solution:
p(m) = m³ + 2m + a
∴ p(2) = (2)³ + 2(2) + a
∴ 12 = 8 + 4 + a … [∵ p(2)= 12]
∴ 12 = 12 + a
∴ a = 12 – 12
∴ a = 0

Question 4.
For the polynomial mx² – 2x + 3 if p(-1) = 7, then find m.
Solution:
p(x) = mx² – 2x + 3
∴ p(- 1) = m (- 1)² – 2(- 1) + 3
∴ 7 = m(1) + 2 + 3 …[∵ p(-1) = 7]
∴ 7 = m + 5
∴ m = 7 – 5
∴ m = 2

Question 5.
Divide the first polynomial by the second polynomial and find the remainder using remainder theorem.
i. (x² – 1x + 9); (x + 1)
ii. (2x³ – 2x² + ax – a); (x – a)
iii. (54m³ + 18m² – 27m + 5); (m – 3)
Solution:
i. p(x) = x² – 7x + 9
Divisor = x + 1
∴ take x = – 1
∴ By remainder theorem,
∴ Remainder =p(-1)
p(x) = x² – 7x + 9
∴ p(-1) = (- 1)² – 7(- 1) + 9
= 1 + 7 + 9
∴ Remainder =17

ii. p(x) = 2x³ – 2x² + ax – a
Divisor = x – a
∴ take x = a
By remainder theorem,
Remainder = p(a)
p(x) = 2x³ – 2x² + ax – a
∴ p(a) = 2a³ – 2a² + a(a) – a
= 2a³– 2a² + a² – a
∴ Remainder = 2a³ – a² – a

iii. p(m) = 54m³ + 18m² – 27m + 5
Divisor = m – 3
∴ take m = 3
∴ By remainder theorem,
Remainder = p(3)
p(m) = 54m³ + 18m² – 27m + 5
∴ p(3) = 54(3)³ +18(3)² – 27(3) + 5
= 54(27) + 18(9) – 81 + 5
= 1458 + 162 – 81 + 5
∴ Remainder = 1544

Question 6.
If the polynomial y³ – 5y² + 7y + m is divided by y + 2 and the remainder is 50, then find the value of m.
Solution:
p(y) = y³ – 5y² + 7y + m
Divisor = y + 2
∴ take y = – 2
∴ By remainder theorem,
Remainder = p(- 2) = 50
P(y) = y³ – 5y² + 7y + m
∴ P(-2) = (- 2)³ – 5(- 2)² + 7(- 2) + m
∴ 50 = -8 – 5(4) – 14 + m
∴ 50 = -8 – 20 – 14 + m
∴ 50 = – 42 + m
∴ m = 50 + 42
∴ m = 92

Question 7.
Use factor theorem to determine whether x + 3 is a factor of x² + 2x – 3 or not.
Solution:
p(x) = x² + 2x – 3
Divisor = x + 3
∴ take x = – 3
∴ Remainder = p(-3)
p(x) = x² + 2x – 3
∴ p(-3) = (-3)² + 2(- 3) – 3
= 9 – 6 – 3
∴ p(-3) = 0
∴ By factor theorem, x + 3 is a factor of x² + 2x – 3.

Question 8.
If (x – 2) is a factor of x³ – mx² + 10x – 20, then find the value of m.
Solution:
p(x) = x³ – mx² + 10x – 20 x – 2 is a factor of x3 – mx2 + lOx – 20.
∴By factor theorem,
Remainder = p(2) = 0
p(x) = x³ – mx² + 10x – 20
∴ p(2) = (2)³ – m(2)² + 10(2) – 20
∴ 0 = 8 – 4m + 20 – 20
∴ 0 = 8 – 4m
∴ 4m = 8
∴ m = 2

Question 9.
By using factor theorem in the following examples, determine whether q(x) is a factor of p(x) or not.
i. p(x) = x³ – x² – x -1 ; q(x) = x – 1
ii. p(x) = 2x³ – x² – 45 ; q(x) = x – 3
Solution:
i. p(x) = x³ – x² – x – 1
Divisor = q(x) = x – 1
∴ take x = 1
Remainder = p(1)
p(x) = x³ – x² – x – 1
∴ P(1) = (1)³ – (1)² – 1 – 1
= 1 – 1 – 1 – 1
= -2 ≠ 0
∴ By factor theorem, x – 1 is not a factor of x³ – x² – x – 1.

ii. p(x) = 2x³ – x – 45
Divisor = q(x) = x – 3
take x = 3
Remainder = p(3)
p(x) = 2x³ – x² – 45
P(3) = 2(3)³ – (3)² – 45
= 2(27) – 9 – 45
= 54 – 9 – 45
= 0
∴ By factor theorem, x – 3 is a factor of 2x³ – x² – 45.

Question 10.
If (x³¹ + 31) is divided by (x + 1), then find the remainder.
Solution:
p(x) = x³¹ + 31
Divisor = x + 1
∴ take x = – 1
∴ By remainder theorem,
Remainder = p(-1)
p(x) =x³¹ + 31 …
∴ p(-1) = (-1)³¹ + 31
= -1 + 31 = 30
∴ Remainder = 30

Question 11.
Show that m – 1 is a factor of m²¹ – 1 and m²² – 1. [3 Marks]
Solution:
i. p(m) = m²¹ – 1
Divisor = m – 1
∴ take m = 1
Remainder = p(1)
p(m) = m²¹ – 1
∴ P(1) = 1²¹ – 1 = 1 – 1 = 0
∴ By factor theorem, m -1 is a factor of m²¹ -1.

ii. p(m) = m²² – 1
Divisor = m – 1
∴ take m = 1
Remainder = p(1)
p(m) = m²² – 1
∴ P(1) = 1²² – 1 = 1 – 1 = 0
∴ By factor theorem, m -1 is a factor of m²² – 1.

Question 12.
If x – 2 and x – \(\frac { 1 }{ 2 }\) both are the factors of the polynomial nx² – 5x + m, then show that m = n = 2.
Solution:
p(x) = nx² – 5x + m
(x – 2) is a factor of nx² – 5x + m.
∴ By factor theorem,
P(2) = 0
∴ p(x) = nx² – 5x + m
∴ p(2) = n(2)² – 5(2) + m
∴ 0 = n(4) – 10 + m
∴ 4n – 10 + m = 0 …(i)
Also, ( x = \(\frac { 1 }{ 2 }\) ) is a factor of nx² – 5x + m.
∴ By factor theorem,
p(\(\frac { 1 }{ 2 }\)) = 0
p(x) = nx² – 5x + m
∴ p(\(\frac { 1 }{ 2 }\)) = n(\(\frac { 1 }{ 2 }\))² – 5\(\frac { 1 }{ 2 }\) + m
0 = \(\frac { n }{ 4 }\) – \(\frac { 5 }{ 2 }\) + m
∴ 0 = n- 10 +4m … [Multiplying both sides by 4]
∴ n = 10 – 4m ……(ii)
Substituting n = 10 – 4m in equation (i),
4(10 – 4m) – 10 + m = 0
∴ 40 – 16m – 10 + m = 0
∴ -15m+ 30 = 0
∴ -15m = -30
∴ m = 2
Substituting m = 2 in equation (ii),
n = 10 – 4(2)
= 10 – 8
∴ n = 2
∴ m = n = 2

Question 13.
i. If p(x) = 2 + 5x, then find the value of p(2) + p(- 2) – p(1).
Solution:
p(x) = 2 + 5x
∴ P(2) = 2 + 5(2)
= 2 + 10
= 12
p(x) = 2 + 5x
P(- 2) = 2 + 5(- 2)
= 2 – 10 = – 8
p(x) = 2 + 5x
P(1) = 2 + 5(1)
= 2 + 5 = 7
∴ P(2) + P(- 2) – p(1) = 12 + (- 8) – 7
∴ P(2) + p(- 2) – p(1) = – 3

ii. If p(x) = 2x² – 5√3 x + 5, then find the value of p(5√3 ).
Solution:
p(x) = 2x² – 5√3 x + 5
∴ p(5√3) = 2(5√3)² – 5√3 (5√3 ) + 5
= 2 (25 x 3) – 25 x 3 + 5
= 150-75 + 5
∴ p( 5√3 ) = 80

Question 1.
1. Divide p(x) = 3x² + x + 7 by x + 2. Find the remainder.
2. Find the value of p(x) = 3x² + x + 7 when x = – 2.
3. See whether remainder obtained by division is same as the value of p(-2). Take one more example and verify. (Textbook pg. no. 50)
Solution:

∴ Remainder = 17

2. p(x) = 3x² + x + 7
Substituting x = – 2, we get
p(-2) = 3(2)² + (-2) + 7
= 12 – 2 + 7
∴ p(-2) = 17

3. Yes, remainder = p(-2)

Another Example:
If the polynomial t³ – 3t² + kt + 50 is divided by (t – 3), the remainder is 62. Find the value of k.
Solution:
When given polynomial is divided by (t – 3) the remainder is 62. It means the value of the polynomial when t = 3 is 62.
p(t) = t³ – 3t³ + kt + 50
By remainder theorem,
Remainder = p(3) = 33 – 3² + k x 3 + 50
= 27 – 3 x 9 + 3k + 50
= 27 – 27 + 3k + 50
= 3k + 50
But remainder is 62.
∴ 3k + 50 = 62
∴ 3k = 62 – 50
∴ 3k = 12
∴ k = 4

Question 2.
Verify that (x – 1) is a factor of the polynomial x³ + 4x – 5. (Textbook pg. no. 51)
Solution:
Here, p(x) = x³ + 4x – 5
Substituting x = 1 in p(x), we get
p(1) = (1)³ + 4(1) – 5
= 1 + 4 – 5
P(1) = 0
∴ By remainder theorem,
Remainder = 0
∴ (x -1) is the factor of x³ + 4x – 5.

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 3.4 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 3 Polynomials.

Practice Set 3.4 Algebra 9th Std Maths Part 1 Answers Chapter 3 Polynomials

Question 1.
For x = 0, find the value of the polynomial x² – 5x + 5.
Solution:
p(x) = x² – 5x + 5
Put x = 0 in the given polynomial.
∴ P(0) = (0)² – 5(0) + 5
= 0 – 0 + 5
∴ p(0) = 5

Question 2.
If p(y) = y² – 3√2 + 1, then find p( 3√2 ).
Solution:
p(y) = y² – 3√2 y + 1
Putp= 3√2 in the given polynomial.
∴ p( 3√2 ) = (3√2 )² – 3√2 (3√2 ) + 1
= 9 x 2 – 9 x 2 + 1
= 18 – 18 + 1
∴ p( 3√2 ) = 1

Question 3.
If p(m) = m³ + 2m² – m + 10, then P(a) + p(-a) = ?
Solution:
p(m) = m³ + 2m² – m + 10
Put m = a in the given polynomial.
∴ p(a) = a³ + 2a² – a + 10 …(i)
Put m = -a in the given polynomial.
p(-a) = (-a)³ + 2(-a)² – (-a) +10
∴ p (-a) = -a³ + 2a² + a + 10 …(ii)
Adding (i) and (ii),
p(a) + p(-a) = (a³ + 2a² – a + 10) + (-a³ + 2a² + a + 10)
= a³ – a³ + 2a² + 2a² – a + a + 10 + 10
∴ p(a) + p(-a) = 4a² + 20

Question 4.
If p(y) = 2y³ – 6y² – 5y + 7, then find p(2).
Solution:
p(y) = 2y³ – 6y² – 5y + 7
Put y = 2 in the given polynomial.
∴ p(2) = 2(2)³ – 6(2)² – 5(2) + 7
= 2 x 8 – 6 x 4 – 10 + 7
= 16 – 24 – 10 + 7
∴ P(2) = -11

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 3.3 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 3 Polynomials.

Practice Set 3.3 Algebra 9th Std Maths Part 1 Answers Chapter 3 Polynomials

Question 1.
Divide each of the following polynomials by synthetic division method and also by linear division method. Write the quotient and the remainder.
i. (2m² – 3m + 10) ÷ (m – 5)
ii. (x⁴ + 2x³ + 3x² + 4x + 5) ÷ (x + 2)
iii. (y³ – 216) ÷ (y – 6)
iv. (2x⁴ + 3x³ + 4x – 2x²) ÷ (x + 3)
v. (x⁴ – 3x² – 8) ÷ (x + 4)
vi. (y³ – 3y² + 5y – 1) ÷ (y – 1)
Solution:
i. Synthetic division:
(2m² – 3m + 10) ÷ (m – 5)
Dividend = 2m² – 3m + 10
∴ Coefficient form of dividend = (2, -3, 10)
Divisor = m – 5
∴ Opposite of -5 is 5.

Coefficient form of quotient = (2, 7)
∴ Quotient = 2m + 7,
Remainder = 45
Linear division method:
2m² – 3m + 10
To get the term 2m², multiply (m – 5) by 2m and add 10m,
= 2m(m – 5) + 10m- 3m + 10
= 2m(m – 5) + 7m + 10
To get the term 7m, multiply (m – 5) by 7 and add 35
= 2m(m – 5) + 7(m- 5) + 35+ 10
= (m – 5) (2m + 7) + 45
∴ Quotient = 2m + 7,
Remainder = 45

ii. Synthetic division:
(x⁴ + 2x³ + 3x² + 4x + 5) ÷ (x + 2)
Dividend = x⁴ + 2x³ + 3x² + 4x + 5
∴ Coefficient form of dividend = (1, 2, 3, 4, 5)
Divisor = x + 2
∴ Opposite of + 2 is -2.

Coefficient form of quotient = (1, 0, 3, -2)
∴ Quotient = x³ + 3x – 2,
Remainder = 9

Linear division method:
x⁴ + 2x³ + 3x² + 4x + 5
To get the term x⁴, multiply (x + 2) by x³ and subtract 2x³,
= x³(x + 2) – 2x³ + 2x³ + 3x² + 4x + 5
= x³(x + 2) + 3x² + 4x + 5
To get the term 3x², multiply (x + 2) by 3x and subtract 6x,
= x³(x + 2) + 3x(x + 2) – 6x + 4x + 5
= x³(x + 2) + 3x(x + 2) – 2x + 5
To get the term -2x, multiply (x + 2) by -2 and add 4,
= x³(x + 2) + 3x(x + 2) – 2(x + 2) + 4 + 5
= (x + 2) (x3 + 3x – 2) + 9
∴ Quotient = x³ + 3x – 2,
Remainder – 9

iii. Synthetic division:
(y³ – 216) ÷ (y – 6)
Dividend = y³ – 216
∴ Index form = y³ + 0y³ + 0y – 216
∴ Coefficient form of dividend = (1, 0, 0, -216)
Divisor = y – 6
∴ Opposite of – 6 is 6.

Coefficient form of quotient = (1, 6, 36)
∴ Quotient = y² + 6y + 36,
Remainder = 0

Linear division method:
y³ – 216
To get the term y³, multiply (y – 6) by y² and add 6y²,
= y²(y – 6) + 6y² – 216
= y²(y – 6) + 6ysup>2 – 216
To get the, term 6 y² multiply (y – 6) by 6y and add 36y,
= y²(y – 6) + 6y(y – 6) + 36y – 216
= y²(y – 6) + 6y(y – 6) + 36y – 216
To get the term 36y, multiply (y- 6) by 36 and add 216,
= y² (y – 6) + 6y(y – 6) + 36(y – 6) + 216 – 216
= (y – 6) (y² + 6y + 36) + 0
Quotient = y² + 6y + 36
Remainder = 0

iv. Synthetic division:
(2x⁴ + 3x³ + 4x – 2x²) ÷ (x + 3)
Dividend = 2x⁴ + 3x³ + 4x – 2x²
∴ Index form = 2x⁴ + 3x³ – 2x² + 4x + 0
∴ Coefficient form of the dividend = (2,3, -2,4,0)
Divisor = x + 3
∴ Opposite of + 3 is -3

Coefficient form of quotient = (2, -3, 7, -17)
∴ Quotient = 2x³ – 3x² + 7x – 17,
Remainder = 51

Linear division method:
2x⁴ + 3x³ + 4x – 2x² = 2x² + 3x³ – 2x² + 4x
To get the term 2x⁴, multiply (x + 3) by 2x³ and subtract 6x³,
= 2x³(x + 31 – 6x³ + 3x³ – 2x² + 4x
= 2x³(x + 3) – 3x³ – 2x² + 4x

To get the term – 3x³, multiply (x + 3) by -3x² and add 9x²,
= 2x³(x + 3) – 3x²(x + 3) + 9x² – 2x² + 4x
= 2x³(x + 3) – 3x²(x + 3) + 7x² + 4x

To get the term 7x², multiply (x + 3) by 7x and subtract 21x,
= 2x³(x + 3) – 3x²(x + 3) + 7x(x + 3) – 21x + 4x
= 2x³(x + 3) – 3x²(x + 3) + 7x(x + 3) – 17x

To get the term -17x, multiply (x + 3) by -17 and add 51,
= 2x³(x + 3) – 3x²(x + 3) + 7x(x+3) – 17(x + 3) + 51
= (x + 3) (2x³ – 3x² + 7x- 17) + 51
∴ Quotient = 2x³ – 3x² + 7x – 17,
Remainder = 51

v. Synthetic division:
(x⁴ – 3x² – 8) + (x + 4)
Dividend = x⁴ – 3x² – 8
∴ Index form = x⁴ + 0x³ – 3x² + 0x – 8
∴ Coefficient form of the dividend = (1,0, -3,0, -8)
Divisor = x + 4
∴ Opposite of + 4 is -4

∴ Coefficient form of quotient = (1, -4, 13, -52)
∴ Quotient = x³ – 4x² + 13x – 52,
Remainder = 200

Linear division method:
x⁴ – 3x² – 8
To get the term x⁴, multiply (x + 4) by x³ and subtract 4x³,
= x³(x + 4) – 4x³ – 3x² – 8
= x³(x + 4) – 4x³ – 3x² – 8
To get the term – 4x³, multiply (x + 4) by -4x² and add 16x²,
= x³(x + 4) – 4x² (x + 4) + 16x² – 3x² – 8
= x³(x + 4) – 4x² (x + 4) + 13x² – 8
To get the term 13x², multiply (x + 4) by 13x and subtract 52x,
= x³(x + 4) – 4x²(x + 4) + 13x(x + 4) – 52x – 8
= x³(x + 4) – 4x²(x + 4) + 13x(x + 4) – 52x – 8
To get the term -52x, multiply (x + 4) by – 52 and add 208,
= x³(x + 4) – 4x²(x + 4) + 13x(x + 4) – 52(x + 4) + 208 – 8
= (x + 4) (x³ – 4x² + 13x – 52) + 200
∴ Quotient = x³ – 4x² + 13x – 52,
Remainder 200

vi. Synthetic division:
(y³ – 3y² + 5y – 1) ÷ (y – 1)
Dividend = y³ – 3y² + 5y – 1
Coefficient form of the dividend = (1, -3, 5, -1)
Divisor = y – 1
∴Opposite of -1 is 1.

∴ Coefficient form of quotient = (1, -2, 3)
∴ Quotient = y² – 2y + 3,
Remainder = 2

Linear division method:
y³ -3y² + 5y – 1
To get the term y³ , multiply (y – 1) by y² and add y²
= y² (y – 1) + y² – 3y² + 5y – 1
= y² (y – 1) – 2y² + 5y – 1
To get the term -2y², multiply (y – 1) by -2y and subtract 2y,
= y² (y – 1) – 2y(y – 1) – 2y + 5y – 1
= y² (y – 1) – 2y(y – 1) + 3y – 1
To get the term 3y, multiply (y – 1) by 3 and add 3,
= y² (y – 1) – 2y(y – 1) + 3(y- 1) + 3 – 1
= (y – 1)(y² – 2y + 3) + 2
∴ Quotient = y² – 2y + 3,
Remainder = 2.

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 3.2 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 3 Polynomials.

Practice Set 3.2 Algebra 9th Std Maths Part 1 Answers Chapter 3 Polynomials

Question 1.
Use the given letters to write the answer.
i. There are ‘a’ trees in the village Lat. If the number of trees increases every year by ’b‘. then how many trees will there be after ‘x’ years?
ii. For the parade there are y students in each row and x such row are formed. Then, how many students are there for the parade in all ?
iii. The tens and units place of a two digit number is m and n respectively. Write the polynomial which represents the two digit number.
Solution:
i. Number of trees in the village Lat = a
Number of trees increasing each year = b
∴ Number of trees after x years = a + bx
∴ There will be a + bx trees in the village Lat after x years.

ii. Total rows = x
Number of students in each row = y
∴ Total students = Total rows × Number of students in each row
= x × y
= xy
∴ There are in all xy students for the parade.

iii. Digit in units place = n
Digit in tens place = m
∴ The two digit number = 10 x digit in tens place + digit in units place
= 10m + n
∴ The polynomial representing the two digit number is 10m + n.

Question 2.
Add the given polynomials.
i. x³ – 2x² – 9; 5x³ + 2x + 9
ii. -7m⁴+ 5m³ + √2 ; 5m⁴ – 3m³ + 2m² + 3m – 6
iii. 2y² + 7y + 5; 3y + 9; 3y² – 4y – 3
Solution:
i. (x³ – 2x² – 9) + (5x³ + 2x + 9)
= x³ – 2x² – 9 + 5x³ + 2x + 9
= x³ + 5x³ – 2x² + 2x – 9 + 9
= 6x³ – 2x² + 2x

ii. (-7m⁴ + 5m³ + √2 ) + (5m⁴ – 3m³ + 2m² + 3m – 6)
= -7m⁴ + 5m³ + √2 + 5m⁴ – 3m³ + 2m² + 3m – 6
= -7m⁴ + 5m⁴ + 5m³ – 3m³ + 2m² + 3m +√2 – 6
= -2m⁴ + 2m³ + 2m² + 3m + √2 – 6

iii. (2y² + 7y + 5) + (3y + 9) + (3y² – 4y – 3)
= 2y² + 7y + 5 + 3y + 9 + 3y² – 4y – 3
= 2y² + 3y² + 7y + 3y – 4y + 5 + 9 – 3
= 5y² + 6y + 11

Question 3.
Subtract the second polynomial from the first.
i. x² – 9x + √3 ; – 19x + √3 + 7x²
ii. 2ab² + 3a²b – 4ab; 3ab – 8ab² + 2a²b
Solution:
i. x² – 9x + √3 -(- 19x + √3 + 7x²)
= x² – 9x + √3 + 19x – √ 3 – 7x²
= x² – 7x² – 9x + 19x + √3 – √3
= – 6x² + 10x

ii. (2ab² + 3a²b – 4ab) – (3ab – 8ab² + 2a²b)
= 2ab² + 3a²b – 4ab – 3ab + 8ab² – 2a²b
= 2ab² + 8ab² + 3a²b – 2a²b – 4ab – 3ab
= 10ab² + a²b – 7ab

Question 4.
Multiply the given polynomials.
i. 2x; x² – 2x – 1
ii. x⁵ – 1; x³ + 2x² + 2
iii. 2y +1; y² – 2y + 3y
Solution:
i. (2x) x (x² – 2x – 1) = 2x³ – 4x² – 2x

ii. (x⁵ – 1) × (x³ + 2x² + 2)
= x⁵ (x³ + 2x² + 2) -1(x³ + 2x² + 2)
= x⁸ + 2x⁷ + 2x⁵ – x³ – 2x² – 2

iii. (2y + 1) × (y² – 2y³ + 3y)
= 2y(y² – 2y³ + 3y) + 1(y² – 2y³ + 3y)
= 2y³ – 4y⁴ + 6y² + y² – 2y³ + 3y
= -4y⁴ + 2y³ – 2y³ + 6y² + y² + 3y
= -4y⁴ + 7y² + 3y

Question 5.
Divide first polynomial by second polynomial and write the answer in the form ‘Dividend = Divisor x Quotient + Remainder’.
i. x³ – 64; x – 4
ii. 5x⁵ + 4x⁴ – 3x³ + 2x² + 2 ; x² – x
Solution:
i. x³ – 64 = x3 + 0x² + 0x – 64

∴ Quotient = x² + 4x + 16, Remainder = 0
Now, Dividend = Divisor x Quotient + Remainder
∴ x³ – 64 = (x – 4)(x² + 4x + 16) + 0

ii. 5x⁵ + 4x⁴ – 3x³ + 2x² + 2 = 5x⁵ + 4x⁴ – 3x³ + 2x + 0x + 2

∴ Quotient = 5x³ + 9x² + 6x + 8,
Remainder = 8x + 2
Now, Dividend = Divisor x Quotient + Remainder
∴ 5x⁵ + 4x⁴ – 3x³ + 2x² + 2 = (x² – x)(5x³ + 9x² + 6x + 8) + (8x + 2)

Question 6.
Write down the information in the form of algebraic expression and simplify.
There is a rectangular farm with length (2a² + 3b²) metre and breadth (a² + b²) metre. The farmer used a square shaped plot of the farm to build a house. The side of the plot was (a2 – b2) metre. What is the area of the remaining part of the farm? [4 Marks]
Solution:
Length of the rectangular farm = (2a² + 3b²) m
Breadth of the rectangular farm = (a² + b²) m
Area of the farm = length x breadth = (2a² + 3b²) x (a² + b²)
= 2a²(a² + b²) + 3b²(a² + b²)
= 2a² + 2a²b² + 3a²b² + 3b⁴
= (2a⁴ + 5a²b² + 3b⁴) sq. m … (i)
The farmer used a square shaped plot of the farm to build a house.
Side of the square shaped plot = (a² – b²) m
∴ Area of the plot = (side)²
= (a² – b²)²
= (a⁴ – 2a²b² + b⁴) sq m… .(ii)

∴ Area of the remaining farm = Area of the farm – Area of the plot
= (2a⁴ + 5a²b² + 3b⁴) – (a⁴ – 2a²b² + b⁴) … [From (i) and (ii)]
= 2a⁴ + 5a²b² + 3b⁴ – a⁴ + 2a²b² – b⁴
= 2a⁴ – a⁴ + 5a²b² + 2a²b² + 3b⁴ – b⁴
= a⁴ + 7a²b² + 2b⁴
∴ The area of the remaining farm is (a⁴ + 7a²b² + 2b⁴) sq. m.

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 3.1 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 3 Polynomials.

Practice Set 3.1 Algebra 9th Std Maths Part 1 Answers Chapter 3 Polynomials

Question 1.
State whether the given algebraic expressions are polynomials? Justify.
i. y + \(\frac { 1 }{ y }\)
ii. 2 – 5√x
iii. x² + 7x + 9
iv. 2m^-2 + 7m – 5
v. 10
Answer:
i. No, because power of v in the term 5√x is -1 (negative number).
ii. No, because the power of x in the term 5√x is
i. e. 0.5 (decimal number).
iii. Yes. All the coefficients are real numbers. Also, the power of each term is a whole number.
iv. No, because the power of m in the term 2m^-2 is -2 (negative number).
v. Yes, because 10 is a constant polynomial.

Question 2.
Write the coefficient of m³ in each of the given polynomial.
i. m³
ii. \(\sqrt [ -3 ]{ 2 }\) + m – √3m³
iii. \(\sqrt [ -2 ]{ 3 }\)m³ + 5m² – 7m -1
Answer:
i. 1
ii. -√3
iii. – \(\frac { 2 }{ 3 }\)

Question 3.
Write the polynomial in x using the given information. [1 Mark each]
i. Monomial with degree 7
ii. Binomial with degree 35
iii. Trinomial with degree 8
Answer:
i. 5x⁷
ii. x³⁵ – 1
iii. 3x⁸ + 2x⁶ + x⁵

Question 4.
Write the degree of the given polynomials.
i. √5
ii. x°
iii. x²
iv. √2m¹⁰ – 7
v. 2p – √7
vi. 7y – y³ + y⁵
vii. xyz +xy-z
viii. m³n⁷ – 3m⁵n + mn
Answer:
i. √5 = √5 x°
∴ Degree of the polynomial = 0

ii. x°
∴Degree of the polynomial = 0

iii. x²
∴Degree of the polynomial = 2

iv. √2m¹⁰ – 7
Here, the highest power of m is 10.
∴Degree of the polynomial = 10

v. 2p – √7
Here, the highest power of p is 1.
∴ Degree of the polynomial = 1

vi. 7y – y³ + y⁵
Here, the highest power of y is 5.
∴Degree of the polynomial = 5

vii. xyz + xy – z
Here, the sum of the powers of x, y and z in the term xyz is 1 + 1 + 1= 3,
which is the highest sum of powers in the given polynomial.
∴Degree of the polynomial = 3

viii. m³n⁷ – 3m⁵n + mn
Here, the sum of the powers of m and n in the term m³n⁷ is 3 + 7 = 10,
which is the highest sum of powers in the given polynomial.
∴ Degree of the polynomial = 10

Question 5.
Classify the following polynomials as linear, quadratic and cubic polynomial. [2 Marks]
i. 2x² + 3x +1
ii. 5p
iii. √2 – \(\frac { 1 }{ 2 }\)
iv. m³ + 7m² + \(\sqrt [ 5 ]{ 2 }\)m – √7
v. a²
vi. 3r³
Answer:
Linear polynomials: ii, iii
Quadratic polynomials: i, v
Cubic polynomials: iv, vi

Question 6.
Write the following polynomials in standard form.
i. m³ + 3 + 5m
ii. – 7y + y⁵ + 3y³ – \(\frac { 1 }{ 2 }\)+ 2y⁴ – y²
Answer:
i. m³ + 5m + 3
ii. y⁵ + 2y⁴ + 3y³ – y² – 7y – \(\frac { 1 }{ 2 }\)

Question 7.
Write the following polynomials in coefficient form.
i. x³ – 2
ii. 5y
iii. 2m⁴ – 3m² + 7
iv. – \(\frac { 2 }{ 3 }\)
Answer:
i. x³ – 2 = x³ + 0x² + 0x – 2
∴ Coefficient form of the given polynomial = (1, 0, 0, -2)

ii. 5y = 5y + 0
∴Coefficient form of the given polynomial = (5,0)

iii. 2m⁴ – 3m² + 7
= 2m⁴ + Om³ – 3m² + 0m + 7
∴ Coefficient form of the given polynomial = (2, 0, -3, 0, 7)

iv. – \(\frac { 2 }{ 3 }\)
∴Coefficient form of the given polynomial = (- \(\frac { 2 }{ 3 }\))

Question 8.
Write the polynomials in index form.
i. (1, 2, 3)
ii. (5, 0, 0, 0 ,-1)
iii. (-2, 2, -2, 2)
Answer:
i. Number of coefficients = 3
∴ Degree = 3 – 1 = 2
∴ Taking x as variable, the index form is x² + 2x + 3

ii. Number of coefficients = 5
∴ Degree = 5 – 1=4
∴ Taking x as variable, the index form is 5x⁴ + 0x³ + 0x² + 0x – 1

iii. Number of coefficients = 4
∴Degree = 4 – 1 = 3
∴Taking x as variable, the index form is -2x³ + 2x² – 2x + 2

Question 9.
Write the appropriate polynomials in the boxes.

Answer:
i. Quadratic polynomial: x²; 2x² + 5x + 10; 3x² + 5x
ii. Cubic polynomial: x³ + x² + x + 5; x³ + 9
iii. Linear polynomial: x + 7
iv. Binomial: x + 7; x³ + 9; 3x² + 5x
v. Trinomial: 2x² + 5x + 10
vi. Monomial: x²

Question 1.
Write an example of a monomial, a binomial and a trinomial having variable x and degree 5. ( Textbook pg. no. 3)
Answer:
Monomial: x⁵
Binomial: x⁵ + x
Trinomial: 2x⁵ – x² + 5

Question 2.
Give example of a binomial in two variables having degree 5. (Textbook pg. no. 38)
Answer:
x³y² + xy

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Problem Set 2 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 2 Real Numbers.

Problem Set 2 Algebra 9th Std Maths Part 1 Answers Chapter 2 Real Numbers

Question 1.
Choose the correct alternative answer for the questions given below. [1 Mark each]

i. Which one of the following is an irrational number?

Answer:
√5

ii. Which of the following is an irrational number?
(A) 0.17
(B) \(1.\overline { 513 }\)
(C) \(0.27\overline { 46 }\)
(D) 0.101001000……..
Answer:
(D) 0.101001000……..

iii. Decimal expansion of which of the following is non-terminating recurring?

Answer:
(C) \(\frac { 3 }{ 11 }\)

iv. Every point on the number line represents which of the following numbers?
(A) Natural numbers
(B) Irrational numbers
(C) Rational numbers
(D) Real numbers
Answer:
(D) Real numbers

v. The number [/latex]0.\dot { 4 }[/latex] in \(\frac { p }{ q }\) form is ……

Answer:
(A) \(\frac { 4 }{ 9 }\)

vi. What is √n , if n is not a perfect square number ?
(A) Natural number
(B) Rational number
(C) Irrational number
(D) Options A, B, C all are correct.
Answer:
(C) Irrational number

vii. Which of the following is not a surd ?

Answer:
(C) \(\sqrt [ 3 ]{ \sqrt { 64 } }\)

viii. What is the order of the surd \(\sqrt [ 3 ]{ \sqrt { 5 } }\) ?
(A) 3
(B) 2
(C) 6
(D) 5
Answer:
(C) 6

ix. Which one is the conjugate pair of 2√5 + √3 ?
(A) -2√5 + √3
(B) -2√5 – √3
(C) 2√3 – √5
(D) √3 + 2√5
Answer:
(A) -2√5 + √3

x. The value of |12 – (13 + 7) x 4| is ____ .
(A) – 68
(B) 68
(C) – 32
(D) 32
Answer:
(B) 68

Hints:
ii. Since the decimal expansion is neither terminating nor recurring, 0.101001000…. is an irrational number.

iii. \(\frac { 3 }{ 11 }\)
Denominator =11 = 1 x 11
Since, the denominator is other than prime factors 2 or 5.
∴ the decimal expansion of \(\frac { 3 }{ 11 }\) will be non terminating recurring.

v. Let x = [/latex]0.\dot { 4 }[/latex]
∴10 x = [/latex]0.\dot { 4 }[/latex]
∴10 – x = [/latex]4.\dot { 4 }[/latex] – [/latex]0.\dot { 4 }[/latex]
∴9x = 4
∴ x = \(\frac { 4 }{ 9 }\)

vii. \(\sqrt[3]{61}\) = 4, which is not an irrational number.

viii. \(\sqrt[3]{\sqrt{5}}=\sqrt[3 \times 2]{5}=\sqrt[6]{5}\)
∴ Order = 6

ix. The conjugate of 2√5 + √3 is 2√5 – √3 or -2√5 + √3

x. |12 – (13+7) x 4| = |12 – 20 x 4|
= |12 – 80|
= |-68|
= 68

Question 2.
Write the following numbers in \(\frac { p }{ q }\) form.
i. 0.555
ii. \(29.\overline { 568 }\)
iii. 9.315315…..
iv. 357.417417…..
v . \(30.\overline { 219 }\)
Solution:

ii. Let x = \(29.\overline { 568 }\) …(i)
x = 29.568568…
Since, three numbers i.e. 5, 6 and 8 are repeating after the decimal point.
Thus, multiplying both sides by 1000,
1000x = 29568.568568…
1000 x= \(29568.\overline { 568 }\) …(ii)
Subtracting (i) from (ii),
1000x – x = \(29568.\overline { 568 }\) – \(29.\overline { 568 }\)
∴ 999x = 29539

iii. Let x = 9.315315 … = \(9.\overline { 315 }\) …(i)
Since, three numbers i.e. 3, 1 and 5 are repeating after the decimal point.
Thus, multiplying both sides by 1000,
1000x = 9315.315315…
∴1000x = \(9315.\overline { 315 }\) …(ii)
Subtracting (i) from (ii),
1000x – x = \(9315.\overline { 315 }\) – \(9.\overline { 315 }\)
∴ 999x = 9306

iv. Let x = 357.417417… = \(357.\overline { 417 }\) …(i)
Since, three numbers i.e. 4, 1 and 7 are repeating after the decimal point.
Thus, multiplying both sides by 1000,
1000x = 357417.417417…
∴ 1000x = 357417.417 …(ii)
Subtracting (i) from (ii),
1000x – x = \(357417.\overline { 417 }\) – \(357.\overline { 417 }\)
∴ 999x = 357060

v. Let x = \(30.\overline { 219 }\) …(i)
∴ x = 30.219219
Since, three numbers i.e. 2, 1 and 9 are repeating after the decimal point.
Thus, multiplying both sides by 1000,
1000x= 30219.219219…
∴ 1000x = \(30219.\overline { 219 }\) …(ii)
Subtracting (i) from (ii),
1000x – x = \(30219.\overline { 219 }\) – \(30.\overline { 219 }\)
∴ 999x = 30189

Question 3.
Write the following numbers in its decimal form.

Solution:
i. \(\frac { -5 }{ 7 }\)

ii. \(\frac { 9 }{ 11 }\)

iii. √5

iv. \(\frac { 121 }{ 13 }\)

v. \(\frac { 29 }{ 8 }\)

Question 4.
Show that 5 + √7 is an irrational number. [3 Marks]
Solution:
Let us assume that 5 + √7 is a rational number. So, we can find co-prime integers ‘a’ and ‘b’ (b ≠ 0) such that

Since, ‘a’ and ‘b’ are integers, \(\sqrt [ a ]{ b }\) – 5 is a rational number and so √7 is a rational number.
∴ But this contradicts the fact that √7 is an irrational number.
Our assumption that 5 + √7 is a rational number is wrong.
∴ 5 + √7 is an irrational number.

Question 5.
Write the following surds in simplest form.

Solution:

Question 6.
Write the simplest form of rationalising factor for the given surds.

Solution:

Now, 4√2 x √2 = 4 x 2 = 8, which is a rational number.
∴ √2 is the simplest form of the rationalising factor of √32 .

Now, 5√2 x √2 = 5 x 2 = 10, which is a rational number.
∴ √2 is the simplest form of the rationalising factor of √50 .

Now, 3√3 x √3 = 3 x 3 = 9, which is a rational number.
∴ √ 3 is the simplest form of the rationalising factor of √27 .

= 6, which is a rational number.
∴ √10 is the simplest form of the rationalising factor of \(\sqrt [ 3 ]{ 5 }\) √10 .

Now, 18√2 x √2 = 18 x 2 = 36, which is a rational number.
∴ √2 is the simplest form of the rationalising factor of 3√72.

vi. 4√11
4√11 x √11 = 4 x 11 = 44, which is a rational number.
∴ √11 is the simplest form of the rationalising factor of 4√11.

Question 7.
Simplify.

Solution:

Question 8.
Rationalize the denominator.

Solution:

Question 1.
Draw three or four circles of different radii on a card board. Cut these circles. Take a thread and measure the length of circumference and diameter of each of the circles. Note down the readings in the given table. (Textbook pg.no.23 )

Solution:
i. 14,44,3.1
ii. 16,50.3,3.1
iii. 11,34.6,3.1
From table, we observe that the ratio \(\sqrt [ c ]{ d }\) is nearly 3.1 which is constant. This ratio is denoted by π (pi).

Question 2.
To find the approximate value of π, take the wire of length 11 cm, 22 cm and 33 cm each. Make a circle from the wire. Measure the diameter and complete the following table.

Verify that the ratio of circumference to the diameter of a circle is approximately \(\sqrt [ 22 ]{ 7 }\). (Textbook pg. no. 24)
Solution:
i. 3.5, \(\sqrt [ 22 ]{ 7 }\)
ii. 7, \(\sqrt [ 22 ]{ 7 }\)
iii. 10.5, \(\sqrt [ 22 ]{ 7 }\)
∴ The ratio of circumference to the diameter of each circle is \(\sqrt [ 22 ]{ 7 }\).

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 2.2 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 2 Real Numbers.

Practice Set 2.2 Algebra 9th Std Maths Part 1 Answers Chapter 2 Real Numbers

Question 1.
Show that 4√2 is an irrational number.
Solution:
Let us assume that 4√2 is a rational number .
So, we can find co-prime integers ‘a’ and ‘b’ (b ≠ 0) such that
4√2 = \(\frac { a }{ b }\)
∴ √2 = \(\frac { a }{ 4b }\)
Since, a and b are integers, \(\frac { a }{ 4b }\) is a rational number and so √2 is a rational number.

Alternate Proof:
Let us assume that 4√2 is a rational number.
So, we can find co-prime integers ‘a’ and ‘b’ (b ≠ 0) such that

Since, 32 divides a², so 32 divides ‘a’ as well.
So, we write a = 32c, where c is an integer.
∴ a² = (32c)² … [Squaring both the sides]
∴ 32b² = 32 x 32c² …[From(i)]
∴ b² = 32c²
∴ c² = \(\frac { { b }^{ 2 } }{ 32 }\)
Since, 32 divides b², so 32 divides ‘b’.
∴ 32 divides both a and b.
a and b have at least 32 as a common factor.
But this contradicts the fact that a and b have no common factor other than 1.
∴ Our assumption that 4√2 is a rational number is wrong.
∴ 4√2 is an irrational number.

Question 2.
Prove that 3 + √5 is an irrational number.
Solution:
Let us assume that 3 + √5 is a rational number.
So, we can find co-prime integers ‘a’ and ‘b’ (b ≠ 0) such that

Since, a and b are integers, \(\frac { a }{ b }\) – 3 is a rational
number and so √5 is a rational number.
But this contradicts the fact that √5 is an irrational number.
∴ Our assumption that 3 – √5 is a rational number is wrong.
3 + √5 is an irrational number.

Question 3.
Represent the numbers √5 and √10 on a number line.
Solution:
i. Draw a number line and take point A at 2.
Draw AB perpendicular to the number line such that AB = 1 unit.
In ∆OAB, m∠OAB = 90°
∴ (OB)² = (OA)² + (AB)² … [Pythagoras theorem]
= (2)² + (1)²
∴ (OB)² = 5
∴ OB = √5 units. … [Taking square root of both sides]
With O as centre and radius equal to OB, draw an arc to intersect the number line at C.

The coordinate of the point C is √5 .

ii. Draw a number line and take point Pat 3.
Draw PR perpendicular to the number line such that PR = 1 unit.
In ∆OPR, m∠OPR = 90°
∴ (OR)² = (OP)² + (PR)² … [Pythagoras theorem]
= (3)² + (1)²
∴ (OR)² = 10
∴ OR= √10units. … [Taking square root of both sides]
With O as centre and radius equal to OR, draw an arc to intersect the number line at Q.

The coordinate of the point Q is √10 .

Question 4.
Write any three rational numbers between the two numbers given below.
i. 0.3 and – 0.5
ii. – 2.3 and – 2.33
iii. 5.2 and 5.3
iv. – 4.5 and – 4.6
Solution:
i. 0.3 = 0.30 and -0.5 = -0.50
We know that,
0. 30 >0.29 >….. >0.10>.. > – 0.10>…. > -0.30>…> -0.50
∴ the three rational numbers between 0.3 and -0.5 are -0.3, -0.1 and 0.1.

Alternate Method:
A rational number between two rational numbers a and b


∴ the three rational numbers between 0.3 and -0.5 are -0.3, -0.1 and 0.1.

ii. -2.3 = -2.300 and -2.33 = -2.330
We know that,
-2.300 > -2.301>… > -2.310>…> -2.320>…> -2.330
∴ the three rational numbers between -2.3 and -2.33 are -2.310, -2.320 and -2.325.

iii. 5.2 = 5.20 and 5.3 = 5.30
We know that,
5.20 < 5.21 < 5.22 < 5.23 < … < 5.30
∴ the three rational numbers between 5.2 and 5.3 are 5.21, 5.22 and 5.23.

iv. -4.5 = -4.50 and -4.6 = -4.60 We know that,
-4.50 > -4.51 > -4.52 >… > – 4.55 >…>- 4.60
∴ the three rational numbers between -4.5 and -4.6 are -4.51, -4.52 and -4.55.

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 2.3 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 2 Real Numbers.

Practice Set 2.3 Algebra 9th Std Maths Part 1 Answers Chapter 2 Real Numbers

Question 1.
State the order of the surds given below.

Answer:
i. 3, ii. 2, iii. 4, iv. 2, v. 3

Question 2.
State which of the following are surds Justify. [2 Marks each]

Answer:
i. \(\sqrt [ 3 ]{ 51 }\) is a surd because 51 is a positive rational number, 3 is a positive integer greater than 1 and \(\sqrt [ 3 ]{ 51 }\) is irrational.

ii. \(\sqrt [ 4 ]{ 16 }\) is not a surd because

= 2, which is not an irrational number.

iii. \(\sqrt [ 5 ]{ 81 }\) is a surd because 81 is a positive rational number, 5 is a positive integer greater than 1 and \(\sqrt [ 5 ]{ 81 }\) is irrational.

iv. \(\sqrt { 256 }\) is not a surd because

= 16, which is not an irrational number.

v. \(\sqrt [ 3 ]{ 64 }\) is not a surd because

= 4, which is not an irrational number.

vi. \(\sqrt { \frac { 22 }{ 7 } }\) is a surd because \(\frac { 22 }{ 7 }\) is a positive rational number, 2 is a positive integer greater than 1 and \(\sqrt { \frac { 22 }{ 7 } }\) is irrational.

Question 3.
Classify the given pair of surds into like surds and unlike surds. [2 Marks each]

Solution:
If the order of the surds and the radicands are same, then the surds are like surds.

Here, the order of 2\(\sqrt { 13 }\) and 5\(\sqrt { 13 }\) is same and their radicands are also same.
∴ \(\sqrt { 52 }\) and 5\(\sqrt { 13 }\) are like surds.

Here, the order of 2\(\sqrt { 17 }\) and 5\(\sqrt { 3 }\) is same but their radicands are not.
∴ \(\sqrt { 68 }\) and 5\(\sqrt { 3 }\) are unlike surds.

Here, the order of 12\(\sqrt { 2 }\) and 7\(\sqrt { 2 }\) is same and their radicands are also same.
∴ 4\(\sqrt { 18 }\) and 7\(\sqrt { 2 }\) are like surds.

Here, the order of 38\(\sqrt { 3 }\) and 6\(\sqrt { 3 }\) is same and their radicands are also same.
∴ 19\(\sqrt { 12 }\) and 6\(\sqrt { 3 }\) are like surds.

v. 5\(\sqrt { 22 }\), 7\(\sqrt { 33 }\)
Here, the order of 5\(\sqrt { 22 }\) and 7\(\sqrt { 33 }\) is same but their radicands are not.
∴ 5\(\sqrt { 22 }\) and 7\(\sqrt { 33 }\) are unlike surds.

Here, the order of 5√5 and 5√3 is same but their radicands are not.
∴ 5√5 and √75 are unlike surds.

Question 4.
Simplify the following surds.

Solution:

Question 5.
Compare the following pair of surds.


Solution:

Question 6.
Simplify.

Solution:

Question 7.
Multiply and write the answer in the simplest form.

Solution:

Question 8.
Divide and write form.

Solution:

Question 9.
Rationalize the denominator.

Solution:

Question 1.
\(\sqrt { 9+16 }\) ? + \(\sqrt { 9 }\) + \(\sqrt { 16 }\) (Texbookpg. no. 28)
Solution:

Question 2.
\(\sqrt { 100+36 }\) ? \(\sqrt { 100 }\) + \(\sqrt { 36 }\) (Textbook pg. no. 28)
Solution:

Question 3.
Follow the arrows and complete the chart by doing the operations given. (Textbook pg. no. 34)

Solution:

Question 4.
There are some real numbers written on a card sheet. Use these numbers and construct two examples each of addition, subtraction, multiplication and division. Solve these examples. (Textbook pg. no. 34)

Solution:

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Problem Set 1 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 1 Sets.

Problem Set 1 Algebra 9th Std Maths Part 1 Answers Chapter 1 Sets

Question 1.
Choose the correct alternative answer for each of the following questions.
i. M= {1, 3, 5}, N= {2, 4, 6}, then M ∩ N = ?
(A) {1, 2, 3, 4, 5, 6}
(B) {1, 3, 5}
(C) φ
(D) {2, 4, 6}
Answer:
(C) φ

ii. P = {x | x is an odd natural number, 1< x ≤ 5}. How to write this set in roster form?
(A) {1, 3, 5}
(B) {1, 2, 3, 4, 5}
(C) {1, 3}
(D) {3, 5}
Answer:
(D) {3, 5}

iii. P= {1, 2, ………. , 10}. What type of set Pis?
(A) Null set
(B) Infinite set
(C) Finite set
(D) None of these
Answer:
(C) Finite set

iv. M ∪ N = {1, 2, 3, 4, 5, 6} and M = {1, 2, 4}, then which of the following represent set N ?
(A) {1, 2, 3}
(B) {3, 4, 5, 6}
(C) {2, 5, 6}
(D) {4, 5, 6}
Answer:
(B) {3, 4, 5, 6}

v. If P ⊆ M, then which of the following set represent P ∩ (P ∪ M)?
(A) P
(B) M
(C) P ∪ M
(D) P’ ∩ M
Answer:
(A) P

vi. Which of the following sets are empty sets?
(A) Set of intersecting points of parallel lines.
(B) Set of even prime numbers.
(C) Month of an english calendar having less than 30 days.
(D) P = {x | x ∈ I , – 1 < x < 1}
Answer:
(A) Set of intersecting points of parallel lines.

Hints:
v. Here, P ⊆ M
∴ P ∪ M = M
∴ P ∩ (P ∪ M) = P ∩ M
= P … [∵ P ⊆M]

Question 2.
Find the correct option for the given question.
i. Which of the following collections is a set ?
(A) Colors of the rainbow
(B) Tall trees in the school campus.
(C) Rich people in the village
(D) Easy examples in the book
Answer:
(A) Colors of the rainbow

ii. Which of the following set represent N ∩W?
(A) {1, 2, 3,….}
(B) {0, 1, 2, 3,….}
(C) {0}
(D) { }
Answer:
(A) {1, 2, 3,….}

iii. P = {x | x is an odd natural number, 1< x < 5}. How to write this set in roster form?
(A) {1, 3, 5}
(B) {1, 2, 3, 4, 5}
(C) {1, 3}
(D) {3, 5}
Answer:
(B) {1, 2, 3, 4, 5}

iv. If T = {1, 2, 3, 4, 5} and M = {3,4, 7, 8}, then T ∪ M = ?
(A) {1, 2, 3, 4, 5,7}
(B) {1, 2, 3, 7, 8}
(C) {1, 2, 3, 4, 5, 7, 8}
(D) {3, 4}
Answer:
(C) {1, 2, 3, 4, 5, 7, 8}

Hints:
i. The elements of options B, C and D cannot be definitely and clearly decided.
ii. The common elements of N and W are 1 2, 3,….

Question 3.
Out of 100 persons in a group, 72 persons speak English and 43 persons speak French. Each one out of 100 persons speak at least one language. Then how many speak only English? How many speak only French ? How many of them speak English and French both?
Solution:
i. Let U be the set of all the persons,
E be the set of persons who speak English and
F be the set of persons who speak French.
∴ n(E) = 72, n(F) = 43
Since, each one out of 100 persons speak at least one language
∴ n(U) = n(E ∪ F)= 100,

ii. n (E ∪ F) = n (E) + n (F) – n(E ∩ F)
100 = 72 + 43 – n (E ∩ F)
n (E ∩ F) = 72 + 43 – 100
∴ n(E ∩ F) = 15
Number of people who speak English and French = 15

iii. Number of people who speak only English = n(E) – n(E ∩ F)
= 72 – 15 = 57

iv. Number of
people who speak only French = n(F) – n(E ∩ F)
= 43 – 15 = 28

Alternate Method:
Let U be the set of all the persons,
E be the set of persons who speak English,
F be the set of persons who speak French and x people speak both the languages.

Since, each one out of 100 persons speak at least one language.
∴ n(U) = n(E ∪ F) = 100
∴ 72 – x + x + 43 – x = 100
∴ 115 – x= 100
∴ x = 115 – 100= 15.
Number of people who speak English and French = 15
Number of people who speak only English = 72 – x = 72 – 15 = 57
Number of people who speak only French = 43 – x = 43 – 15 = 28

Question 4.
70 trees were planted by Parth and 90 trees were planted by Pradnya on the occasion of Tree Plantation Week. Out of these 25 trees were planted by both of them together. How many trees were planted by Parth or Pradnya?
Solution:
i. Let P be the trees planted by Parth and Q be the trees planted by Pradnya
∴ n(P) = 70 and n(Q) = 90
Total number of trees planted by Parth and Pradnya = n(P ∩ Q) = 25

ii. Number of trees planted by Parth or Pradnya = n(P ∪ Q)
= n(P) + n(Q) – n(P ∩ Q)
= 70 + 90 – 25 = 135
∴ A total of 135 trees were planted by Parth or Pradnya.

Alternate Method:
Let P be the trees planted by Parth and Q be the trees planted by Pradnya

From Venn diagram
∴ Total trees planted by parth or pradnya = n(P ∪ Q)
= 45 + 25 + 65
= 135
A total of 135 trees were planted by Parth or Pradnya.

Question 5.
If n(A) = 20, n(B) = 28 and n(A ∪ B) = 36, then n(A ∩ B) = ?
Solution:
n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
∴ 36 = 20 + 28 – n(A ∩ B)
∴ n(A ∩ B) = 20 + 28 – 36
∴ n(A ∩ B) = 12

Question 6.
In a class, 8 students out of 28 have a dog as their pet animal at home, 6 students have a cat as their pet animal, 10 students have dog and cat both, then how many students do not have dog or cat as their pet animal at home?
Solution:
i. Let U be the set of all the students, then n(U) = 28
Let D be the set of students who have dog as pet and C be the set of students who have cat as pet.
10 students have dog and cat as their pet animal
n(D ∩ C) = 10

From venn diagram,

ii. Number of students who have cat or dog as pet
= n(D ∪ C)
= 8 + 10 + 6
= 24

iii. Number of students who do not have dog or cat as pet = n (U) – n(D ∪ C)
= 28 – 24
= 4

Question 7.
Represent the union of two sets by Venn diagram for each of the following.
i. A = {3, 4, 5, 7},B = {1, 4, 8} l Marks
ii. P {a, b, c, e, f, Q = {l, m, n, e, b) I Markj
iii. X = {x x is a prime number between 80 and 100}
Y = { y | y is an odd number between 90 and 100}
Solution:
i. A = {3, 4, 5, 7}, B = {1, 4, 8}

ii. P = {a, b, c, e, f}, Q = {l, m, n, e, b}

iii. X = {x | x is a prime number between 80 and 100}
∴ X = {83, 89, 97}
Y = {y | y is an odd number between 90 and 100}
∴ Y = {91, 93, 95, 97, 99}

Question 8.
Write the subset relations between the following sets.
X = set of all quadrilaterals.
Y = set of all rhombuses.
S = set of all squares.
T = set of all parallelograms.
V = set of all rectangles. [3 Marks]
Solution:
i. Rhombus, square, parallelogram and rectangle all are quadrilaterals.
∴ Y ⊆ X,S ⊆ X,T ⊆ X,V ⊆ X

ii. Every square is a rhombus, parallelogram and rectangle.
∴ S ⊆ Y, S ⊆ T, S ⊆ V

iii. Every rhombus and rectangle is a parallelogram.
∴ Y ⊆ T, V ⊆ T

Question 9.
If M is any set, then write M ∪Φ and M ∩ Φ.
Solution:
Let M = {2, 3, 4, 8} and Φ = { }
∴ M ∪ Φ = {2, 3, 4, 8}
∴ M ∪ Φ = M Also, M ∩ Φ = { }
∴ M ∩ Φ = i(i

Question 10.
Observe the Venn diagram and write the given sets U, A, B, A ∪ B and A ∩ B.

U = {1,2, 3,4, 5, 7, 8, 9, 10, 11, 13}
A = {1, 2, 3, 5,7}
B = {1, 5, 8, 9, 10}
A ∪ B = {1,2, 3, 5, 7, 8, 9, 10}
A ∩ B = {1, 5}

Question 11.
If n(A) = 7, n(B) = 13, n(A ∩ B) = 4, then n(A ∪ B) = ?
Solution:
n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
= 7 + 13 – 4
n(A ∪ B) = 16

Question 1.
Set of students in a class and set of students in the same class who can swim, are shown by the Venn diagram.

Observe the diagram and draw Venn diagrams for the following subsets.
i. a. Set of students in a class
b. Set of students who can ride bicycles in the same class

ii. A set of fruits is given as follows.
U = {guava, orange, mango, jackfruit, chickoo, jamun, custard apple, papaya, plum}
Show these subsets.
A = fruit with one seed
B = fruit with more than one seed. (Textbook pg. no. 8)
Solution:

ii. A = {mango, jamun, plum}
B = {guava, orange, jackfruit, chickoo, custard apple, papaya}

Question 2.
Every student should take 9 triangular sheets of paper and one plate. Numbers from 1 to 9 should, be written on each triangle. Everyone should keep some numbered triangles in the plate. Now the triangles in each plate form a subset of the set of numbers from 1 to 9.

Look at the plates of Sujata, Hameed, Mukta, Nandini, Joseph with the numbered triangles. Guess the thinking behind selecting these numbers. Hence write the subsets in set builder form. (Textbook pg, no. 9)
Solution:
Sujata:
S = {x | x = 2n- 1, n ∈ N, x < 9}
Hameed:
f H = {x | x = 2n, n ∈ N, x < 9}
Mukta:
M = {x | x = n², n ∈ N, x ≤ 9}
Nandini:
N = {x | x ∈ N, x ≤ 9}
Joseph:
J = {x | x is a prime number between 1 and 9}

Question 3.
Collect the following information from 20 families nearby your house.
i. Number of families subscribing for Marathi Newspaper.
ii. Number of families subscribing for English Newspaper.
iii. Number of families subscribing for both English as well as Marathi Newspaper.
Show the collected information using Venn diagram. (Textbook pg.no. 18)
[Students should attempt the above activity on their own.]

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 2.1 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 2 Real Numbers.

Practice Set 2.1 Algebra 9th Std Maths Part 1 Answers Chapter 2 Real Numbers

Question 1.
Classify the decimal form of the given rational numbers into terminating and non-terminating recurring type.

Solution:
i. Denominator = 5 = 1 x 5
Since, 5 is the only prime factor denominator.
the decimal form of the rational number \(\frac { 13 }{ 5 }\) will be terminating type.

ii. Denominator = 11 = 1 x 11
Since, the denominator is other than prime factors 2 or 5.
∴ the decimal form of the rational number \(\frac { 2 }{ 11 }\) will be non-terminating recurring type.

iii. Denominator = 16
= 2 x 2 x 2 x 2
Since, 2 is the only prime factor in the denominator.
∴ the decimal form of the rational number \(\frac { 29 }{ 16 }\) will be terminating type.

iv. Denominator = 125
= 5 x 5 x 5
Since, 5 is the only prime factor in the denominator.
the decimal form of the rational number \(\frac { 17 }{ 125 }\) will be terminating type.

v. Denominator = 6
= 2 x 3
Since, the denominator is other than prime factors 2 or 5.
∴ the decimal form of the rational number \(\frac { 11 }{ 6 }\) will be non-terminating recurring type.

Question 2.
Write the following rational numbers in decimal form.





Solution:
i. \(\frac { -5 }{ 7 }\)

ii. \(\frac { 9 }{ 11 }\)

iii. √5

iv. \(\frac { 121 }{ 13 }\)

v. \(\frac { 29 }{ 8 }\)

Question 3.
Write the following rational numbers in \(\frac { p }{ q }\) form.

Solution:
i. Let x = \(0.\dot { 6 }\) …(i)
∴ x = 0.666…
Since, one number i.e. 6 is repeating after the decimal point.
Thus, multiplying both sides by 10,
10x = 6.666…
∴ 10 x 6.6 …(ii)
Subtracting (i) from (ii),
10x – x = 6.6 – 0.6
∴ 9x = 6

ii. Let x = \(0.\overline { 37 }\)
∴ x = 0.3737…
Since, two numbers i.e. 3 and 7 are repeating after the decimal point.
Thus, multiplying both sides by 100,
100x = 37.3737……
∴ 100x = \(37.\overline { 37 }\) ……(ii)
Subtracting (i) from (ii),
100x – x = \(37.\overline { 37 }\) – \(0.\overline { 37 }\)
∴ 99x = 37

iii. Letx = \(3.\overline { 17 }\) …(i)
∴ x = 3.1717…
Since, two numbers i.e. 1 and 7 are repeating after the decimal point.
Thus, multiplying both sides by 100,
100x = 317.1717…
∴ 100x= 317.17 …(ii)
Subtracting (i) from (ii),
100x – x = \(317.\overline { 17 }\) – \(3.\overline { 17 }\)
∴ 99x = 314

iv. Let x = \(15.\overline { 89 }\) …….. (i)
∴ x = 15.8989…
Since, two numbers i.e. 8 and 9 are repeating after the decimal point.
Thus, multiplying both sides by 100,
100x= 1589.8989…
∴ 100x = \(1589.\overline { 89 }\) …(ii)
Subtracting (i) from (ii),
100x – x = \(1589.\overline { 89 }\) – \(15.\overline { 89 }\)
∴ 99x = 1574

v. Let x = \(2.\overline { 514 }\)
∴ x = 2.514514…
Since, three numbers i.e. 5, 1 and 4 are repeating after the decimal point.
Thus, multiplying both sides by 1000,
1000x = 2514.514514…
1000x = \(2514.\overline { 514 }\) ….(ii)
Subtracting (i) from (ii),
1000x – x = \(2514.\overline { 514 }\) – \(2.\overline { 514 }\)
∴ 999x = 2512

Question 1.
How to convert 2.43 in \(\frac { p }{ q }\) form ? (Textbook pg. no. 20)
Solution:
Let x = 2.43
In 2.43, the number 4 on the right side of the decimal point is not recurring.
So, in order to get only recurring digits on the right side of the decimal point, we will multiply 2.43 by 10.
∴ 10x = 24.3 …(i)
∴ 10x = 24.333…
Here, digit 3 is the only recurring digit. Thus, by multiplying both sides by 10, 100x = 243.333…
∴ 100x= 243.3 …(ii)
Subtracting (i) from (ii),
100x – 10x = 243.3 – 24.3
∴ 90x = 219