Category: Class 9

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 8.2 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 8 Trigonometry.

Practice Set 8.2 Geometry 9th Std Maths Part 2 Answers Chapter 8 Trigonometry

Question 1.
In the following table, a ratio is given in each column. Find the remaining two ratios in the column and complete the table.

Solution:
i. cos θ = \(\frac { 35 }{ 37 }\) …(i) )[Given]
In right angled ∆ABC,

∠C = θ.

Let the common multiple be k.
∴ BC = 35k and AC = 37k
Now, AC² = AB² + BC² …[Pythagoras theorem]
∴ (37k)² = AB²+ (35k)²
1369k² = AB² + 1225k²
AB2 = 1369k² – 1225k²
= 144k²
AB = 144k²
AB = \(\sqrt { 2ghK }\)² … [Taking square root of both sides]
= 12k

ii. sin θ = \(\frac { 11 }{ 61 }\) …..(i) [Given]
In right angled ∆ABC, ∠C = θ.


Let the common multiple be k.
AB = 11k and AC = 61k
Now, AC² = AB² + BC² …[Pythagoras theorem]
∴ (61k)² = (11k)² + BC²
∴ 3721k² = 121k² + BC²
∴ BC² = 3721k² – 121k² = 3600k²
BC = \(\sqrt { 3600{ k }^{ 2 } }\) .. .[Taking square root of both sides]
= 60k

iii. tan θ = 1 = \(\frac { 1 }{ 1 }\) ..(i) [Given]
In right angled ∆ABC,
∠C = θ.


Let the common multiple be k.
∴ AB = 1k and BC = 1k
Now, AC² = AB² + BC² …[Pythagoras theorem]
= K² + K²
= 2K²
∴ AC = \(\sqrt { 2{ k }\)

iv. sin θ = \(\frac { 1 }{ 2 }\) ..(i) [Given]
In right angled ∆ABC,
∠C = θ.


Let the common multiple be k.
∴ AB = 1k and BC = 2k
Now, AC² = AB² + BC² …[Pythagoras theorem]
∴ 2K² = K² + BC²
∴ 4K² = K² + BC²
∴ BC² = 4K² – K² = 3K²
∴ BC = \(\sqrt { 3{ k }^{ 2 } }\) .. .[Taking square root of both sides]
= \(\sqrt { 3{ k }\)

v. cos θ = \(\frac { 1 }{ \sqrt { 3 } } \) ..(i) [Given]
In right angled ∆ABC,
∠C = θ.


Let the common multiple be k.
∴ AB = 1k and BC = √3k
Now, AC² = AB² + BC² …[Pythagoras theorem]
∴ (√3K)² = AB² + K²
∴ 3K² = 3K² – K² = 2K²
∴ AB = \(\sqrt { 2{ k }^{ 2 } }\) .. .[Taking square root of both sides]
AB = √2K

vi. cos θ = \(\frac { 21 }{ \sqrt { 20 } } \) ..(i) [Given]
In right angled ∆ABC,
∠C = θ.


Let the common multiple be k.
∴ AB = 21k and BC = 20k
Now, AC² = AB² + BC² …[Pythagoras theorem]
= (21)K² + (20K)²
= 441K² – 400²
= 841K²
∴ AB = \(\sqrt { 841{ k }^{ 2 } }\) .. .[Taking square root of both sides]
= 29K

vii. tan θ = \(\frac { 8 }{ 15 } \) ..(i) [Given]
In right angled ∆ABC,
∠C = θ.


Let the common multiple be k.
∴ AB = 8k and BC = 15k
Now, AC² = AB² + BC² …[Pythagoras theorem]
= (8)K² + (15K)²
= 64K² – 225²
= 289K²
∴ AC = \(\sqrt { 289{ k }^{ 2 } }\) .. .[Taking square root of both sides]
= 17K

viii. sin θ = \(\frac { 3 }{ 5 } \) ..(i) [Given]
In right angled ∆ABC,
∠C = θ.



Let the common multiple be k.
∴ AB = 3k and AC = 5k
Now, AC² = AB² + BC² …[Pythagoras theorem]
∴ (5)K²= (3)K² + BC²
∴ 25K² = 9K² – 225²
∴ BC² = 25K² – 9K²
∴ BC = \(\sqrt { 16{ k }^{ 2 } }\) .. .[Taking square root of both sides]
= 4K

ix. tan θ = \(\frac { 1 }{ 2\sqrt { 2 } }\) ..(i) [Given]
In right angled ∆ABC,
∠C = θ.


Let the common multiple be k.
∴ AB = 1k and AC = 2√2 k
Now, AC² = AB² + BC² …[Pythagoras theorem]
= K² + (2√2 k )²
= K² – 225²
= 25K² + 8K²
= 9K²
∴ AC = \(\sqrt { 9{ k }^{ 2 } }\) .. .[Taking square root of both sides]
= 3K

Question 2.
Find the values of:
i. 5 sin 30° + 3 tan 45°
ii. \(\frac { 4 }{ 5 }\)tan² 60° + 3 sin² 60°
iii. 2 sin 30° + cos 0° + 3 sin 90°
iv. \(\frac{\tan 60^{\circ}}{\sin 60^{\circ}+\cos 60^{\circ}}\)
v. cos² 45° + sin² 30°
vi. cos 60° x cos 30° + sin 60° x sin 30°
Solution:
i. sin 30° = \(\frac { 1 }{ 2 }\) and tan 45° = 1

ii. \(\frac { 4 }{ 5 }\)tan² 60° + 3 sin² 60°

iii. 2 sin 30° + cos 0° + 3 sin 90°
2 sin 30° + cos0° + 3 sin 90° = 2 (\(\frac { 1 }{ 2 }\)) + 1 + 3(1)
= 1 + 1 + 3
∴ 2 sin 30° + cos 0° + 3 sin 90° = 5

iv. \(\frac{\tan 60^{\circ}}{\sin 60^{\circ}+\cos 60^{\circ}}\)

v. cos² 45° + sin² 30°

vi. cos 60° x cos 30° + sin 60° x sin 30°

Question 3.
If sin θ = \(\frac { 4 }{ 5 }\) , then find cos θ.
Solution:
sin θ = \(\frac { 4 }{ 5 }\) .. .(i)[Given]
In right angled ∆ABC,
∠C = θ.


Let the common multiple be k.
∴ AB = 4k and AC = 5k
Now, AC² = AB² + BC² … [Pythagoras theorem]
∴ (5 k)² = (4k)² + BC²
∴ 25k² = 16k² + BC²
∴ BC² = 25k² – 16k² = 9k²
∴ BC = \(\sqrt { 9{ k }^{ 2 } }\) . .[Taking square root of both sides]
= 3k

Question 4.
If cos θ = \(\frac { 15 }{ 17 }\) , then find sin θ.
Solution:
cos θ = \(\frac { 15 }{ 17 }\) .. .(i)[Given]
In right angled ∆ABC,
∠C = θ.


Let the common multiple be k.
∴ BC = 15k and AC = 17k
Now, AC² = AB² + BC² … [Pythagoras theorem]
∴ (17 k)² = AB² + (15K)²
∴ 289k² = AB² + 225²
∴ AB² = 289k² – 225k²
= 64k²
∴ AB = \(\sqrt { 64{ k }^{ 2 } }\) . .[Taking square root of both sides]
= 8k

Maharashtra Board Class 9 Maths Chapter 8 Trigonometry Practice Set 8.2 Intext Questions and Activities

Question 1.
In right angled ∆PQR, ∠Q = 900. Therefore ∠P and ∠R are complementary angles of each other. Verify the following ratios.
i. sin θ = cos (90 – θ)
ii. cos θ = sin (90 – θ)
iii. sin 30° = cos (90° – 30°) = cos 60°
iv. cos 30° = sin (90° – 30°) = sin 60° (Textbook pg. no. 107)
Solution:
In ∆PQR, ∠Q = 90°, ∠P = θ
∴ ∠R = 90 – θ

i. sin θ = cos (90 – θ)

ii. cos θ = sin (90 – θ)

iii. Let ∠P = θ = 30°
∴ ∠R = 90° – 30°

sin 30° = cos (90° – 30°) … [From (i) and (ii)]
sin 30° = cos 60°

iv. cos 30° = sin (90° – 30°) = sin 60°

∴ cos 30° = sin (90° – 30°) .,.[From (i) and (ii)]
∴ cos 30° = sin 60°

Question 2.
In right angled ∆PQR, ∠Q = 90°, ∠R = θ and if sin θ = \(\frac { 5 }{ 13 }\), then find cos θ and tan θ. (Textbook pg. no. 110)
Solution:
i. Take the given trigonometric ratio as 13k equation (i).
sin θ = \(\frac { 5 }{ 13 }\) .. .(i)[Given]
By using the definition write the trigonometric ratio of sin O and take it as equation (ii).
In right angled ∆PQR, ∠R = θ


Let the common multiple be k.
∴ PQ = 5k and PR = 13k
Find QR by using Pythagoras theorem.
PR² = PQ² + QR² … [Pythagoras theorem]
∴ (13k)² = (5k)² + QR²
∴ 169k² = 25k² + QR²
∴ QR² = 169k² – 25k²
= 144k²
∴ QR = \(\sqrt { 144{ k }^{ 2 } }\) . . . [Taking square root of both sides]
= 12k

Question 3.
While solving the above Illustrative example, why the lengths of PQ and PR are taken 5k and 13k? (Textbook pg. no. 111)
Solution:
\(\frac { PQ }{ PR }\) = \(\frac { 5 }{ 13 }\) … [Given]
Here, the ratio of the lengths of sides PQ and PR is 5 : 13.
The actual lengths of the sides can be any multiple of the ratio. Hence, we consider the multiple k while solving.

Question 4.
While solving the above illustrative example, can we take the lengths of PQ and PR as 5 and 13? If so, then what changes are needed In the writing of the solution. (Tcxtbook pg. no. 111)
Solution:
Yes, we can take lengths of PQ and PR as 5 and 13.
In that case, we will have to take k = 1 and solve the problem accordingly.

Question 5.
Verify that the equation ‘sin² θ + cos² θ = 1’ is true when θ = 0° or θ = 90°.
(Textbook pg. no. 112)
Solution:
sin² θ + cos² θ = 1
i. lf θ = 0°,
LH.S. = sin² θ + cos² θ
= sin² 0° + cos² 0°
= 0 + 1 …[∵ sin 0° = 0, cos 0° = 1]
= R.H.S.
∴ sin² θ + cos² θ = 1

ii. If θ = 90°,
L.H.S.= sin² θ +cos² θ
= sin² 90° + cos² 90°
= 1 + 0 … [ ∵ sin 90° = 1, cos 90° = 0]
= 1
= R.H.S.
∴ sin² θ + cos² θ = 1

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 8.1 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 8 Trigonometry.

Practice Set 8.1 Geometry 9th Std Maths Part 2 Answers Chapter 8 Trigonometry

Question 1.
In the given figure, ∠R is the right angle of ∆PQR. Write the following ratios.

i. sin P
ii. cos Q
iii. tan P
iv. tan Q
Solution:

Question 2.
In the right angled ∆XYZ, ∠XYZ = 90° and a, b, c are the lengths of the sides as shown in the figure. Write the following ratios.

i. sin x
ii. tan z
iii. cos x
iv. tan x.
Solution:

Question 3.
In right angled ∆LMN, ∠LMN = 90°, ∠L = 50° and ∠N = 40°. Write the following ratios.

i. sin 50°
ii. cos 50°
iii. tan 40°
iv. cos 40°
Solution:

Question 4.
In the given figure, ∠PQR = 90°, ∠PQS = 90°, ∠PRQ = α and ∠QPS = θ. Write the following trigonometric ratios.
i. sin α, cos α , tan α
ii. sin θ, cos θ, tan θ

Solution:
i. In ∆PQR,

ii. In ∆PQS,

Maharashtra Board Class 9 Maths Chapter 8 Trigonometry Practice Set 8.1 Intext Questions and Activities

Question 1.
In the figure gIven below, ∆PQR is a right angled triangle. Write the names of sides opposite and adjacent to ∠P and ∠R. (Textbook pg no. 102)

Solution:
In right angled ∆PQR,
i. side opposite to ∠P = QR
ii. side opposite to ∠R = PQ
iii. side adjacent to ∠P = PQ
iv. side adjacent to ∠R = QR

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 6.1 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 6 Circle.

Practice Set 6.1 Geometry 9th Std Maths Part 2 Answers Chapter 6 Circle

Question 1.
Distance of chord AB from the centre of a circle is 8 cm. Length of the chord AB is 12 cm. Find the diameter of a circle.
Given: In a circle with centre O,
OA is radius and AB is its chord,
seg OP ⊥ chord AB, A-P-B
AB = 12 cm, OP =8 cm
To Find: Diameter of the circle
Solution:

i. AP = \(\frac { 1 }{ 2 }\) AB [Perpendicular drawn from the centre of a circle to the chord bisects the chord.]
∴ AP = \(\frac { 1 }{ 2 }\) x 12 = 6 cm ….(i)

ii. In ∆OPA, ∠OPA = 90°
∴ OA² = OP² + AP² [Pythagoras theorem]
= 8² + 6² [From (i)]
= 64 + 36
∴ OA² = 100
∴ OA = \(\sqrt { 100 }\) [Taking square root on both sides]
= 10 cm

iii. Radius (r) = 10 cm
∴ Diameter = 2r = 2 x 10 = 20 cm
∴ The diameter of the circle is 20 cm.

Question 2.
Diameter of a circle is 26 cm and length of a chord of the circle is 24 cm. Find the distance of the chord from the centre.
Solution:

Given: In a circle with centre O,
PO is radius and PQ is its chord,
seg OR ⊥ chord PQ, P-R-Q
PQ = 24 cm, diameter (d) = 26 cm
To Find: Distance of the chord from the centre (OR)
Solution:
Radius (OP) = \(\frac { d }{ 2 }\) = \(\frac { 26 }{ 2 }\) = 13 cm ……(i)
∴ PR = \(\frac { 1 }{ 2 }\) PQ [Perpendicular drawn from the centre of a circle to the chord bisects the chord.]
= \(\frac { 1 }{ 2 }\) x 24 = 12 cm …..(ii)

ii. In ∆ORP, ∠ORP = 90°
∴ OP²= OR² + PR² [Pythagoras theorem]
∴ 13² = OR² + 12² [From (i) and (ii)]
∴ 169 = OR² + 144
∴ OR² = 169 – 144
∴ OR² = 25
∴ OR = √25 = 5 cm [Taking square root on both sides]
∴ The distance of the chord from the centre of the circle is 5 cm.

Question 3.
Radius of a circle is 34 cm and the distance of the chord from the centre is 30 cm, find the length of the chord.
Given: in a circle with centre A,
PA is radius and PQ is chord,
seg AM ⊥ chord PQ, P-M-Q
AP = 34 cm, AM = 30 cm
To Find: Length of the chord (PQ)
Solution:

I. In ∆AMP, ∠AMP = 90°
∴ AP² = AM² + PM² [Pythagoras theorem]
34² = 30² + PM²
∴ PM² = 34² – 30²
∴ PM² (34 – 30)(34 + 30) [a² – b² = (a – b)(a + b)]
= 4 x 64
∴ PM = \(\sqrt { 4\times64 }\) ………(i) [Taking square root on both sides]
= 2 x 8 = 16cm

ii. Now, PM = \(\frac { 1 }{ 2 }\)(PQ) [Perpendicular drawn from the centre of a circle to the chord bisects the chord.]
16 = \(\frac { 1 }{ 2 }\)(PQ) [From (i)]
∴ PQ = 16 x 2
= 32cm
∴ The length of the chord of the circle is 32cm.

Question 4.
Radius of a circle with centre O is 41 units. Length of a chord PQ is 80 units, find the distance of the chord from the centre of the circle.
Given: In a circle with centre O,
OP is radius and PQ is its chord,
seg OM ⊥ chord PQ, P-M-Q
OP = 41 units, PQ = 80 units,
To Find: Distance of the chord from the centre of the circle(OM)
Solution:

i. \(\frac { 1 }{ 2 }\)PM = (PQ) [Perpendicular drawn from the centre of a circle to the chord bisects the chord.]
= \(\frac { 1 }{ 2 }\)(80) = 40 Units ….(i)

ii. In ∆OMP, ∠OMP = 90°
∴ OP² = OM² + PM² [Pythagoras theorem]
∴ 41² = OM² + 40² [From (i)]
∴ OM² = 41² – 40²
= (41 -40) (41 +40) [a² – b² = (a – b) (a + b)]
= (1)(81)
∴ OM² = 81 OM = √81 = 9 units [Taking square root on both sides] [From (i)]
∴ The distance of the chord from the centre of the circle is 9 units.

Question 5.
In the adjoining figure, centre of two circles is O. Chord AB of bigger circle intersects the smaller circle in points P and Q. Show that AP = BQ.

Given: Two concentric circles having centre O.
To prove: AP = BQ
Construction: Draw seg OM ⊥ chord AB, A-M-B
Solution:
Proof:
For smaller circle,
seg OM ⊥ chord PQ [Construction, A-P-M, M-Q-B]
∴ PM = MQ …..(i) [Perpendicular drawn from the centre of the circle to the chord bisects the chord.]
For bigger circle,
seg OM ⊥ chord AB [Construction]
∴ AM = MB [Perpendicular drawn from the centre of the circle to the chord bisects the chord.]
∴ AP + PM = MQ + QB [A-P-M, M-Q-B]
∴ AP + MQ = MQ + QB [From (i)]
∴ AP = BQ

Question 6.
Prove that, if a diameter of a circle bisects two chords of the circle then those two chords are parallel to each other.
Solution:

Given: O is the centre of the circle.
seg PQ is the diameter.
Diameter PQ bisects the chords AB and CD in points M and N respectively.
To prove: chord AB || chord CD.
Proof:
Diameter PQ bisects the chord AB in point M [Given]
∴ seg AM ≅ seg BM
∴ seg OM ⊥ chord AB [Segment joining the centre of a circle and the midpoint of its chord is perpendicular to the chord, P-M-O, O-N-Q]
∴ ∠OMA = 90° …..(i)
Also, diameter PQ bisects the chord CD in point N [Given]
∴ seg CN ≅ seg DN
seg ON ⊥ chord CD [Segment joining the centre of a circle and the midpoint of its chord is perpendicular to the chord, P-M-O, O-N-Q]
∴ ∠ONC = 90° …..(ii)
Now, ∠OMA + ∠ONC = 90° + 90° [From (i) and (ii)]
= 180°
But, ∠OMA and ∠ONC form a pair of interior angles on lines AB and CD when seg MN is their transversal.
∴ chord AB || chord CD [Interior angles test]

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 1.3 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 1 Sets.

Practice Set 1.3 Algebra 9th Std Maths Part 1 Answers Chapter 1 Sets

Question 1.
If A = {a, b, c, d, e}, B = {c, d, e, f}, C = {b, d}, D = {a, e}, then which of the following statements are true and which are false?
i. C ⊆ 3
ii. A ⊆ D
iii. D ⊆ B
iv. D ⊆ A
V. B ⊆ A
vi. C ⊆ A
Ans:
i. C = {b, d}, B = {c, d, e ,f}
C ⊆ B
False
Since, all the elements of C are not present in B.

ii. A = {a, b, c, d, e}, D = {a, e}
A ⊆ D
False
Since, all the elements of A are not present in D.

iii. D = {a, e}, B = {c, d, e, f}
D ⊆ B
False
Since, all the elements of D are not present in B.

iv. D = {a, e}, A = {a, b, c, d, e}
D ⊆ A
True
Since, all the elements of D are present in A.

v. B = {c, d, e, f}, A = {a, b, c, d, e}
B ⊆ A
False
Since, all the elements of B are not present in A.

vi. C = {b, d}, A= {a, b, c, d, e}
C ⊆A
True
Since, all the elements of C are present in A.

Question 2.
Take the set of natural numbers from 1 to 20 as universal set and show set X and Y using Venn diagram. [2 Marks each]
i. X= {x |x ∈ N, and 7 < x < 15}
ii. Y = { y | y ∈ N, y is a prime number from 1 to 20}
Answer:
i. U = {1, 2, 3, 4, …….., 18, 19, 20}
x = {x | x ∈ N, and 7 < x < 15}
∴ x = {8, 9, 10, 11, 12, 13, 14}

ii. U = {1, 2, 3, 4, …… ,18, 19, 20}
Y = { y | y ∈ N, y is a prime number from 1 to 20}
∴ Y = {2, 3, 5, 7, 11, 13, 17, 19}

Question 3.
U = {1, 2, 3, 7, 8, 9, 10, 11, 12} P = {1, 3, 7,10}, then
i. show the sets U, P and P’ by Venn diagram.
ii. Verify (P’)’ = P
Solution:
i. Here, U = {1,2, 3, 7, 8,9, 10, 11, 12} P = {1, 3, 7, 10}
∴ P’ = {2, 8, 9, 11, 12}

II. Here, U = {1, 2, 3, 7, 8, 9, 10, 11, 12}
P = {1, 3, 7, 10} ….(i)
∴ P’= {2, 8, 9, 11, 12}
Also, (P’)’ = {1,3,7, 10} …(ii)
∴ (P’)’ = P … [From (i) and (ii)]

Question 4.
A = {1, 3, 2, 7}, then write any three subsets of A.
Solution:
Three subsets of A:
i. B = {3}
ii. C = {2, 1}
iii. D= {1, 2, 7}
[Note: The above problem has many solutions. Students may write solutions other than the ones given]

Question 5.
i. Write the subset relation between the sets.
P is the set of all residents in Pune.
M is the set of all residents in Madhya Pradesh.
I is the set of all residents in Indore.
B is the set of all residents in India.
H is the set of all residents in Maharashtra.

ii. Which set can be the universal set for above sets ?
Solution:
i.
a. The residents of Pune are residents of India.
∴ P ⊆ B
b. The residents of Pune are residents of Maharashtra.
∴ P ⊆ H
c. The residents of Madhya Pradesh are residents of India.
∴ M ⊆ B
d. The residents of Indore are residents of India.
∴ I ⊆ B
e. The residents of Indore are residents of Madhya Pradesh.
∴ I ⊆ M
f. The residents of Maharashtra are residents of India.
∴ H ⊆B

ii. The residents of Pune, Madhya Pradesh, Indore and Maharashtra are all residents of India.
∴ B can be the Universal set for the above sets.

Question 6.
Which set of numbers could be the universal set for the sets given below?
i. A = set of multiples of 5,
B = set of multiples of 7,
C = set of multiples of 12

ii. P = set of integers which are multiples of 4.
T = set of all even square numbers.
Answer:
i. A = set of multiples of 5
∴ A = {5, 10, 15, …}
B = set of multiples of 7
∴ B = {7, 14, 21,…}
C = set of multiples of 12
∴ C = {12, 24, 36, …}
Now, set of natural numbers, whole numbers, integers, rational numbers are as follows:
N = {1, 2, 3, …}, W = {0, 1, 2, 3, …}
I = {…,-3, -2, -1, 0, 1, 2, 3, …}
Q = { \(\frac { p }{ q }\) | p,q ∈ I,q ≠ 0}
Since, set A, B and C are the subsets of sets N, W , I and Q.
∴ For set A, B and C we can take any one of the set from N, W, I or Q as universal set.

ii. P = set of integers which are multiples of 4.
P = {4, 8, 12,…}
T = set of all even square numbers T = {2², 4², 6², …]
Since, set P and T are the subsets of sets N, W, I and Q.
∴ For set P and T we can take any one of the set from N, W, I or Q as universal set.

Question 7.
Let all the students of a class form a Universal set. Let set A be the students who secure 50% or more marks in Maths. Then write the complement of set A.
Answer:
Here, U = all the students of a class.
A = Students who secured 50% or more marks in Maths.
∴ A’= Students who secured less than 50% marks in Maths.

Question 1.
If A = {1, 3, 4, 7, 8}, then write all possible subsets of A.
i. e. P = {1, 3}, T = {4, 7, 8}, V = {1, 4, 8}, S = {1, 4, 7, 8}
In this way many subsets can be written. Write five more subsets of set A. (Textbook pg. no, 8)
Answer:
B = { },
E = {4},
C = {1, 4},
D = {3, 4, 7},
F = {3, 4, 7,8}

Question 2.
Some sets are given below.
A ={…,-4, -2, 0, 2, 4, 6,…}
B = {1, 2, 3,…}
C = {…,-12, -6, 0, 6, 12, 18, }
D = {…, -8, -4, 0, 4, 8,…}
I = {…,-3, -2, -1, 0, 1, 2, 3, 4, }
Discuss and decide which of the following statements are true.
a. A is a subset of sets B, C and D.
b. B is a subset of all the sets which are given above. (Textbook pg. no. 9)
Solution:
a. All elements of set A are not present in set B, C and D.
∴ A ⊆ B,
∴ A ⊆ C,
∴ A ⊆ D
∴ Statement (a) is false.

b. All elements of set B are not present in set A, C and D.
∴ B ⊆ A,
∴ B ⊆ C,
∴ B ⊆ D
∴ Statement (b) is false.

Question 3.
Suppose U = {1, 3, 9, 11, 13, 18, 19}, and B = {3, 9, 11, 13}. Find (B’)’ and draw the inference. (Textbook pg. no. 10)

Solution:
U = {1, 3, 9, 11, 13, 18, 19},
B= {3, 9, 11, 13} ….(i)
∴ B’= {1, 18, 19}
(B’)’= {3, 9, 11, 13} ….(ii)
∴ (B’)’ = B … [From (i) and (ii)]
∴ Complement of a complement is the given set itself.

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 1.2 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 1 Sets.

Practice Set 1.2 Algebra 9th Std Maths Part 1 Answers Chapter 1 Sets

Question 1.
Decide which of the following are equal sets and which are not ? Justify your answer.
A= {x | 3x – 1 = 2}
B = {x | x is a natural number but x is neither prime nor composite}
C = {x | x e N, x < 2}
Solution:
A= {x | 3x – 1 = 2}
Here, 3x – 1 = 2
∴ 3x = 3
∴ x = 1
∴ A = {1} …(i)

B = {x | x is a natural number but x is neither prime nor composite}
1 is the only number which is neither prime nor composite,
∴ x = 1
∴ B = {1} …(ii)

C = {x | x G N, x < 2}
1 is the only natural number less than 2.
∴ x = 1
∴ C = {1} …(iii)
∴ The element in sets A, B and C is identical. … [From (i), (ii) and (iii)]
∴ A, B and C are equal sets.

Question 2.
Decide whether set A and B are equal sets. Give reason for your answer.
A = Even prime numbers
B = {x | 7x – 1 = 13}
Solution:
A = Even prime numbers
Since 2 is the only even prime number,
∴ A = {2} …(i)
B= {x | 7x – 1 = 13}
Here, 7x – 1 = 13
∴ 7x = 14
∴ x = 2
∴ B = {2} …(ii)
∴ The element in set A and B is identical. … [From (i) and (ii)]
∴ A and B are equal sets.

Question 3.
Which of the following are empty sets? Why?
i. A = {a | a is a natural number smaller than zero}
ii. B = {x | x² = 0}
iii. C = {x | 5x – 2 = 0, x ∈N}
Solution:
i. A = {a| a is a natural number smaller than zero}
Natural numbers begin from 1.
∴ A = { }
∴ A is an empty set.

ii. B = {x | x² = 0}
Here, x² = 0
∴ x = 0 … [Taking square root on both sides]
∴ B = {0}
∴B is not an empty set.

iii. C = {x | 5x – 2 = 0, x ∈ N}
Here, 5x – 2 = 0
∴ 5x = 2
∴ x = \(\frac { 2 }{ 5 }\)
Given, x ∈ N
But, x = \(\frac { 2 }{ 5 }\) is not a natural number.
∴ C = { }
∴ C is an empty set.

Question 4.
Write with reasons, which of the following sets are finite or infinite.
i. A = {x | x<10, xisa natural number}
ii. B = {y | y < -1, y is an integer}
iii. C = Set of students of class 9 from your school.
iv. Set of people from your village.
v. Set of apparatus in laboratory
vi. Set of whole numbers
vii. Set of rational number
Solution:
i. A={x| x < 10, x is a natural number}
∴ A = {1,2, 3,4, 5,6, 7, 8, 9}
The number of elements in A are limited and can be counted.
∴A is a finite set.

ii. B = (y | y < -1, y is an integer}
∴ B = { …,-4, -3, -2}
The number of elements in B are unlimited and uncountable.
∴ B is an infinite set.

iii. C = Set of students of class 9 from your school.
The number of students in a class is limited and can be counted.
∴ C is a finite set.

iv. Set of people from your village.
The number of people in a village is limited and can be counted.
∴ Given set is a finite set.

v. Set of apparatus in laboratory
The number of apparatus in the laboratory are limited and can be counted.
∴ Given set is a finite set.

vi. Set of whole numbers
The number of elements in the set of whole numbers are unlimited and uncountable.
∴ Given set is an infinite set.

vii. Set of rational number
The number of elements in the set of rational numbers are unlimited and uncountable.
∴ Given set is an infinite set.

Question 1.
If A = {1, 2, 3} and B = {1, 2, 3, 4}, then A ≠ B verify it. (Textbook pg. no. 6)
Answer:
Here, 4 ∈ B but 4 ∉ A
∴ A and B are not equal sets,
i.e. A ≠ B

Question 2.
A = {x | x is prime number and 10 < x < 20} and B = {11,13,17,19}. Here A = B. Verify. (Textbook pg. no. 6)
Answer:
A = {x | x is prime number and 10 < x < 20}
∴ A = {11, 13, 17, 19}
B = {11, 13, 17, 19}
∴ All the elements in set A and B are identical.
∴ A and B are equal sets, i.e. A = B

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 1.4 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 1 Sets.

Practice Set 1.4 Algebra 9th Std Maths Part 1 Answers Chapter 1 Sets

Question 1.
If n(A) = 15, n(A ∪ B) = 29, n(A ∩ B) = 7, then n(B) = ?
Solution:
Here, n(A) = 15, n(A ∪ B) = 29, n(A ∩ B) = 7
n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
∴ 29 = 15 + n(B) – 7
∴ 29 – 15 + 7 = n(B)
∴ n(B) = 21

Question 2.
In a hostel there are 125 students, out of which 80 drink tea, 60 drink coffee and 20 drink tea and coffee both. Find the number of students who do not drink tea or coffee.
Solution:
i. Let U be the set of students in the hostel, T be the set of students who drink tea and C be the set of students who drink coffee.
n(U) = 125, n(T) = 80, n(C) = 60,
number of students who drink Tea and Coffee = n(T ∩ C) = 20

ii. n(T ∪ C) = n(T) + n(C) – n(T ∩ C)
= 80 + 60 – 20
∴ n(T ∪ C) = 120
∴ 120 students drink tea or coffee
Also, there are 125 students in the hostel.

iii. Number of students who do not drink tea or coffee = n(U) – n(T ∪ C)
= 125 – 120
= 5
∴ 5 students do not drink tea or coffee.

Alternate Method:
Let U be the set of students in the hostel, T be the set of students who drink tea and C be the set of students who drink coffee.

From Venn diagram,
Student who drinks tea or coffee = n(T ∪ C) = 60 + 20 + 40 = 120
∴ The number of students who do not drink tea or coffee = n(U) – n(T ∪ C)
= 125 – 120 = 5
∴ 5 students do not drink tea or coffee.

Question 3.
In a competitive exam 50 students passed in English, 60 students passed in Mathematics and 40 students passed in both the subjects. None of them failed in both the subjects. Find the number of students who passed at least in one of the subjects ?
Solution:
Let U be the set of students who appeared for the exam,
E be the set of students who passed in English and
M be the set of students who passed in Maths.
∴ n(E) = 50, n(M) = 60,
40 students passed in both the subjects
∴ n(M ∩ E) = 40
Since, none of the students failed in both subjects
∴ Total students = n(E ∪M)
= n(E) + n(M) – n(E ∩ M)
= 50 + 60 – 40
= 70
∴ The number of students who passed at least in one of the subjects is 70.

Alternate Method:
Let U be the set of students who appeared for the exam,
E be the set of students who passed in English and M be the set of students who passed in Maths.

Since, none of the students failed in both subjects
∴ Total student = n(E ∪M)
= 10 + 40 + 20
= 70
∴ The number of students who passed at least in one of the subjects is 70.

Question 4.
A survey was conducted to know the hobby of 220 students of class IX. Out of which 130 students informed about their hobby as ’rock climbing and 180 students informed about their hobby as sky watching. There are 110 students who follow both the hobbies. Then how many students do not have any of the two hobbies? How many of them follow the hobby of rock climbing only? How many students follow the hobby of sky watching only?
Solution:
i. Let U be the set of students of class IX,
R be the set of students who follow the hobby of rock climbing and
S be the set of students who follow the hobby of sky watching.
∴ n (U) = 220, n (R) = 130, n (S) = 180,
110 students follow both the hobbies
∴ n (R ∩ S) = 110

ii. n(R ∪ S)=n (R) + n (S) – n (R ∩ S)
= 130 + 180 – 110
∴n (R ∪ S) = 200
∴ 200 students follow the hobby of rock climbing or sky watching.

iii. Total number of students = 220.
Number of students who do not follow the hobby of rock climbing or sky watching
= n (U) – n (R ∪ S)
= 220 – 200
= 20

iv. Number of students who follow the hobby of rock climbing only
= n (R) – n(R ∩ S)
= 130 – 110
= 20

v. Number of students who follow the hobby of sky watching only
= n (S) – n (R ∩ S)
= 180 – 110
= 70

Alternate Method:
Let U be the set of students of class IX,
R be the set of students who follow the hobby of rock climbing and
S be the set of students who follow the hobby of sky watching.

From the Venn diagram
i. Students who follow the hobby of rock climbing or sky watching
= n(R ∪ S)
= 20 + 110 + 70
= 200

ii. Number of students who do not follow the hobby of rock climbing or sky watching
= n (U) – n(R ∪S)
= 220 – 200
= 20

iii. Number of students who follow the hobby of rock climbing only
= n (R) – n(R ∩S)
= 130 – 110
= 20

iv. Number of students who follow the hobby of sky watching only
= n (S) – n (R ∩ S)
= 180 – 110
= 70

Question 5.
Observe the given Venn diagram and write the following sets.

i. A
ii. B
iii. A ∪ B
iv. U
v. A’
vi. B’
vii. (A ∪B )’
Ans:
i. A = {x, y, z, m, n}
ii. B = {p, q, r, m, n}
iii. A ∪ B = {x, y, z, m, n, p, q, r }
iv. U = {x, y, z, m, n, p, q, r, s, t}
v. A’ = {p, q, r, s, t}
vi. B’ = {x, y, z, s, t}
vii. (A ∪ B )’ = {s, t}

Question 1.
Take different examples of sets and verify the above mentioned properties. (Textbook pg.no. 12)
Solution:
i. Let A = {3, 5}, B= {3, 5, 8, 9, 10}
A ∩ B = B ∩ A = {3, 5}

ii. Let A = {3, 5}, B = {3, 5, 8, 9, 10}
Since, all elements of set A are present in set B.
∴ A ⊆ B
Also, A ∩ B = {3, 5} = A
∴ If A ⊆ B, then A ∩B = A.

iii. Let A = {2, 3, 8, 10}, B = {3,8}
A ∩ B = {3, 8} = B
Also, all the elements of set B are present in set A
∴ B ⊆ A
∴ If A ∩ B = B, then B ⊆ A.

iv. Let A = {2, 3, 8, 10}, B = {3, 8}, A ∩B = {3, 8}
Since, all the elements of set A n B are present in set A and B
A ∩ B ⊆ A and A ∩B ⊆B

v. Let U= {3, 4, 6, 8}, A = {6, 4}
∴ A’ = {3, 8}
∴ A ∩ A’= { } = φ

vi. A ∩ φ = { } = φ

vii. Let A = {6, 4}
∴ A ∩ A = {6, 4}
∴ A ∩ A = A

Question 2.
Observe the set A, B, C given by Venn diagrams and write which of these are disjoint sets. (Textbook pg. no. 12)

Solution:
Here, A = {1, 2, 3, 4, 5, 6, 7}
B = {3, 6, 8, 9, 10, 11, 12}
C = {10, 11, 12}
Now, A ∩ C = φ
∴ A and C are disjoint sets.

Question 3.
Let the set of English alphabets be the Universal set. The letters of the word ‘LAUGH’ is one set and the letter of the word ‘CRY’ is another set. Can we say that these are two disjoint sets? Observe that intersection of these two sets is empty. (Textbook pg. no. 13)

Solution:
Let A = {L, A, U, G, H}
B = {C, R, Y}
Now, A ∩ B = φ
∴ A and B are disjoint sets.

Question 4.
Fill in the blanks with elements of that set.
U = {1, 3, 5, 8, 9, 10, 11, 12, 13, 15}
A = {1,11, 13}
B = {8,5, 10, 11, 15}
A’ = { }
B’ = { }
A ∩ B = { }
A’ ∩ B’ = { }
A ∪ B = { }
A’ ∪ B’= { }
(A ∩ B)’ = { }
(A ∪ B)’ = { }
Verify: (A ∩ B)’ = A’ u B’, (A u B)’ = A’ ∩ B’ (Textbook pg. no, 18)
Solution:
U = {1, 3, 5, 8, 9, 10, 11, 12, 13, 15}
A = {1, 11, 13}
B = {8, 5, 10, 11, 15}
A’ = {3, 5, 8, 9, 10, 12, 15}
B’ = {1, 3, 9, 12, 13}
A∩ B= {11}
A’ ∩ B’= {3, 9, 12} …(i)
A ∪ B = {1, 5, 8, 10, 11, 13, 15}
A’ ∪ B’ = { 1, 3, 5, 8, 9, 10, 12, 13, 15} …(ii)
(A ∩ B)’= { 1, 3, 5, 8, 9, 10, 12, 13,15} …(iii)
(A ∪ B)’ = {3, 9, 12} ,..(iv)
(A ∩ B)’ = A’ ∪ B’ … [From (ii) and (iii)]
(A ∪ B)’ = A’ ∩ B’ … [From (i) and (iv)]

Question 5.
A = {1,2,3, 5, 7,9,11,13}
B = {1,2,4, 6, 8,12,13}
Verify the above rule for the given set A and set B. (Textbook pg. no. 14)
Solution:
A = {1, 2, 3, 5, 7, 9, 11, 13}
B = {1, 2, 4, 6, 8, 12, 13}
A ∪ B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13}
A ∩ B= {1, 2, 13}
n(A) = 8, n(B) = 7,
n(A ∪ B) = 12, n(A ∩ B) = 3
n(A ∩ B) = 12 …(i)
n(A) + n(B) – n(A ∩ B) = 8 + 7 – 3 = 12 …(ii)
∴ n(A ∪ B) = n(A) + n(B) – n(A ∩ B) … [From (i) and (ii)]

Question 6.
Verify the above rule for the given Venn diagram. (Textbook pg. no. 14)

Solution:
n(A) = 5 , n(B) = 6
n(A ∪ B) = 9 , n(A ∩ B) = 2
Now, n(A ∪ B) = 9 …(i)
n(A) + n(B) – n(A ∩ B) = 5 + 6 – 2 = 9 …(ii)
∴ n(A ∪ B) = n(A) + n(B) – n(A ∩ B). …[From (i) and (ii)]