Class 9

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 4.2 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 4 Ratio and Proportion.

9th Standard Maths 1 Practice Set 4.2 Chapter 4 Ratio and Proportion Textbook Answers Maharashtra Board

Class 9 Maths Part 1 Practice Set 4.2 Chapter 4 Ratio and Proportion Questions With Answers Maharashtra Board

Question 1.
Using the property \(\frac { a }{ b }\) = \(\frac { ak }{ bk }\), fill in the blanks by substituting proper numbers in the following.

Solution:

Question 2.
Find the following ratios.
i. The ratio of radius to circumference of the circle.
ii. The ratio of circumference of circle with radius r to its area.
iii. The ratio of diagonal of a square to its side, if the length of side is 7 cm.
iv. The lengths of sides of a rectangle are 5 cm and 3.5 cm. Find the ratio of numbers denoting its perimeter to area.
Solution:
i. Let the radius of circle be r.
then, its circumference = 2πr
Ratio of radius to circumference of the circle

The ratio of radius to circumference of the circle is 1 : 2π.

ii. Let the radius of the circle is r.
∴ circumference = 2πr and area = πr²
Ratio of circumference to the area of circle

∴ The ratio of circumference of circle with radius r to its area is 2 : r.

iii. Length of side of square = 7 cm
∴ Diagonal of square = √2 x side
= √2 x 7
= 7 √2 cm
Ratio of diagonal of a square to its side

∴ The ratio of diagonal of a square to its side is √2 : 1.

iv. Length of rectangle = (l) = 5 cm,
Breadth of rectangle = (b) = 3.5 cm
Perimeter of the rectangle = 2(l + b)
= 2(5 + 3.5)
= 2 x 8.5
= 17 cm
Area of the rectangle = l x b
= 5 x 3.5
= 17.5 cm²
Ratio of numbers denoting perimeter to the area of rectangle

∴ Ratio of numbers denoting perimeter to the area of rectangle is 34 : 35.

Question 3.
Compare the following

Solution:

Question 4.
Solve.
ABCD is a parallelogram. The ratio of ∠A and ∠B of this parallelogram is 5 : 4. FInd the measure of ∠B. [2 Marksl
Solution:
Ratio of ∠A and ∠B for given parallelogram is 5 : 4
Let the common multiple be x.

m∠A = 5x°and m∠B=4x°
Now, m∠A + m∠B = 180° …[Adjacent angles of a parallelogram arc supplementary]
∴ 5x° + 4x°= 180°
∴ 9x° = 180°
∴ x° = 20°
∴ m∠B=4x°= 4 x 20° = 80°
∴ The measure of ∠B is 800.

ii. The ratio of present ages of Albert and Salim is 5 : 9. Five years hence ratio of their ages will be 3 : 5. Find their present ages.
Solution:
The ratio of present ages of Albert and Salim is 5 : 9
Let the common multiple be x.
∴ Present age of Albert = 5x years and
Present age of Salim = 9x years
After 5 years,
Albert’s age = (5x + 5) years and
Salim’s age = (9x + 5) years
According to the given condition,
Five years hence ratio of their ages will be 3 : 5
\(\frac{5 x+5}{9 x+5}=\frac{3}{5}\)
∴ 5(5x + 5) = 3(9x + 5)
∴ 25x + 25 = 27x + 15
∴ 25 – 15 = 27 x – 25 x
∴ 10 = 2x
∴ x = 5
∴ Present age of Albert = 5x = 5 x 5 = 25 years
Present age of Salim = 9x = 9 x 5 = 45 years
∴ The present ages of Albert and Salim are 25 years and 45 years respectively.

iii. The ratio of length and breadth of a rectangle is 3 : 1, and its perimeter is 36 cm. Find the length and breadth of the rectangle.
Solution:
The ratio of length and breadth of a rectangle is 3 : 1
Let the common multiple be x.
Length of the rectangle (l) = 3x cm
and Breadth of the rectangle (b) = x cm
Given, perimeter of the rectangle = 36 cm
Since, Perimeter of the rectangle = 2(l + b)
∴ 36 = 2(3x + x)
∴ 36 = 2(4x)
∴ 36 = 8x
∴ \(x=\frac{36}{8}=\frac{9}{2}=4.5\)
Length of the rectangle = 3x = 3 x 4.5 = 13.5 cm
∴ The length of the rectangle is 13.5 cm and its breadth is 4.5 cm.

iv. The ratio of two numbers is 31 : 23 and their sum is 216. Find these numbers.
Solution:
The ratio of two numbers is 31 : 23
Let the common multiple be x.
∴ First number = 31x and
Second number = 23x
According to the given condition,
Sum of the numbers is 216
∴ 31x + 23x = 216
∴ 54x = 216
∴ x = 4
∴ First number = 31x = 31 x 4 = 124
Second number = 23x = 23 x 4 = 92
∴ The two numbers are 124 and 92.

v. If the product of two numbers is 360 and their ratio is 10 : 9, then find the numbers.
Solution:
Ratio of two numbers is 10 : 9
Let the common multiple be x.
∴ First number = 10x and
Second number = 9x
According to the given condition,
Product of two numbers is 360
∴ (10x) (9x) = 360
∴ 90x² = 360
∴ x² = 4
∴ x = 2 …. [Taking positive square root on both sides]
∴ First number = 10x = 10x² = 20
Second number = 9x = 9x² = 18
∴ The two numbers are 20 and 18.

Question 5.
If a : b = 3 : 1 and b : c = 5 : 1, then find the value of [3 Marks each]

Solution:
Given, a : b = 3 : 1
∴ \(\frac { a }{ b }\) = \(\frac { 3 }{ 1 }\)
∴ a = 3b ….(i)
and b : c = 5 : 1
∴ \(\frac { b }{ c }\) = \(\frac { 5 }{ 1 }\)
b = 5c …..(ii)
Substituting (ii) in (i),
we get a = 3(5c)
∴ a = 15c …(iii)

Ratio and Proportion 9th Class Practice Set 4.1 Question 6. If \(\sqrt{0.04 \times 0.4 \times a}=0.4 \times 0.04 \times \sqrt{b}\) , then find the ratio \(\frac { a }{ b }\).
Solution:
\(\sqrt{0.04 \times 0.4 \times a}=0.4 \times 0.04 \times \sqrt{b}\) … [Given]
∴ 0.04 x 0.4 x a = (0.4)² x (0.04)² x b … [Squaring both sides]

9th Algebra Practice Set 4.2 Question 7. (x + 3) : (x + 11) = (x – 2) : (x + 1), then find the value of x.
Solution:
(x + 3) : (x + 11) = (x- 2) : (x+ 1)
\(\quad \frac{x+3}{x+11}=\frac{x-2}{x+1}\)
∴ (x + 3)(x +1) = (x – 2)(x + 11)
∴ x(x +1) + 3(x + 1) = x(x + 11) – 2(x + 11)
∴ x² + x + 3x + 3 = x² + 1 lx – 2x – 22
∴ x² + 4x + 3 = x² + 9x – 22
∴ 4x + 3 = 9x – 22
∴ 3 + 22 = 9x – 4x
∴ 25 = 5x
∴ x = 5

Class 9 Maths Digest

• Practice Set 4.1 Class 9 Answers
• Practice Set 4.2 Class 9 Answers
• Practice Set 4.3 Class 9 Answers
• Practice Set 4.4 Class 9 Answers
• Practice Set 4.5 Class 9 Answers
• Practice Set 4 Class 9 Answers
• Practice Set 5.1 Class 9 Answers
• Practice Set 5.2 Class 9 Answers
• Practice Set 5 Class 9 Answers

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 4.1 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 4 Ratio and Proportion.

9th Standard Maths 1 Practice Set 4.1 Chapter 4 Ratio and Proportion Textbook Answers Maharashtra Board

Class 9 Maths Part 1 Practice Set 4.1 Chapter 4 Ratio and Proportion Questions With Answers Maharashtra Board

Question 1.
From the following pairs of numbers, find the reduced form of ratio of first number to second number.
i. 72,60
ii. 38,57
iii. 52,78
Solution:
i. 72, 60
\(\text { Ratio }=\frac{72}{60}=\frac{12 \times 6}{12 \times 5}=\frac{6}{5}=6 : 5\)

ii. 38, 57
\(\text { Ratio }=\frac{38}{57}=\frac{19 \times 2}{19 \times 3}=\frac{2}{3}=2 : 3\)

iii. 52, 78
\(\text { Ratio }=\frac{52}{78}=\frac{26 \times 2}{26 \times 3}=\frac{2}{3}=2 : 3\)

Question 2.
Find the reduced form of the ratio of the first quantity to second quantity.
i. ₹ 700, ₹ 308
ii. ₹ 14, ₹ 12 and 40 paise
iii. 5 litres, 2500 ml
iv. 3 years 4 months, 5 years 8 months
v. 3.8 kg, 1900 gm
vi. 7 minutes 20 seconds, 5 minutes 6 seconds
Solution:
i. ₹ 700, ₹ 308
\( \text { Ratio }=\frac{700}{308}=\frac{28 \times 25}{28 \times 11}=\frac{25}{11}=25 : 11\)

ii. ₹ 14, ₹12 and 40 paise
₹ 14 = 14 x 100 paise = 1400 paise
₹ 12 and 40 paise = 12 x 100 paise + 40 paise
= (1200 + 40) paise
= 1240 paise

iii. 5 litres, 2500 ml
5 litres = 5 x 1000 ml = 5000ml

iv. 3 years 4 months, 5 years 8 months
3 years 4 months = 3×12 months + 4 months
= (36 + 4) months
= 40 months
5 years 8 months = 5 x 12 months + 8 months
= (60 + 8) months
= 68 months

v. 3.8 kg, 1900 gm
3.8 kg = 3.8 x 1000 gm = 3800 gm

vi. 7 minutes 20 seconds, 5 minutes 6 seconds
7 minutes 20 seconds = 7 x 60 seconds + 20 seconds
= (420 + 20) seconds
= 440 seconds
5 minutes 6 seconds = 5 x 60 seconds + 6 seconds
= (300 + 6) seconds
= 306 seconds

Question 3.
Express the following percentages as ratios
i. 75 : 100
ii. 44 : 100
iii. 6.25%
iv. 52: 100
v. 0.64%
Solution:

Question 4.
Three persons can build a small house in 8 days. To build the same house in 6 days, how many persons are required?
Solution:
Let the persons required to build a house in 6 days be x.
Days required to build a house and number of persons are in inverse proportion.
∴ 6 × x = 8 × 3
∴ 6 x = 24
∴ x = 4
∴ 4 persons are required to build the house in 6 days.

Question 5.
Convert the following ratios into percentages.
i. 15 : 25
ii. 47 : 50
iii. \(\frac { 7 }{ 10 }\)
iv. \(\frac { 546 }{ 600 }\)
v. \(\frac { 7 }{ 16 }\)
Solution:
Let 15 : 25 = x %

∴ 15 : 25 = 60 %

ii. Let 47 : 50 = x%

∴ 47 : 50 = 94 %

iii. Let \(\frac { 7 }{ 10 }\) = x %

iv. Let \(\frac { 546 }{ 600 }\) = x %

v. Let \(\frac { 7 }{ 16 }\) = x %

Question 6.
The ratio of ages of Abha and her mother is 2 : 5. At the time of Abha’s birth her mothers age was 27 years. Find the present ages of Abha and her mother.
Solution:
The ratio of ages of Abha and her mother is 2 : 5.
Let the common multiple be x.
∴ Present age of Abha = 2x years and
Present age of Abha’s mother = 5x years
According to the given condition, the age of Abha’s mother at the time of Abha’s birth = 27 years
∴ 5x – 2x = 27
∴ 3x = 27
∴ x = 9
∴ Present age of Abha = 2x = 2 x 9 = 18 years
∴ Present age of Abha’s mother = 5x =5 x 9 = 45 years
The present ages of Abha and her mother are 18 years and 45 years respectively.

Question 7.
Present ages of Vatsala and Sara are 14 years and 10 years respectively. After how many years the ratio of their ages will become 5 : 4?
Solution:
Present age of Vatsala = 14 years
Present age of Sara = 10 years
After x years,
Vatsala’s age = (14 + x) years
Sara’s age = (10 + x) years
According to the given condition,
After x years the ratio of their ages will become 5 : 4
∴ \(\frac { 14 + x }{ 10 + x }\) = \(\frac { 5 }{ 4 }\)
∴ 4(14 + x) = 5(10 + x)
∴ 56 + 4x = 50 + 5x
∴ 56 – 50 = 5x – 4x
∴ 6 = x
∴ x = 6
∴ After 6 years, the ratio of their ages will become 5 : 4.

Question 8.
The ratio of present ages of Rehana and her mother is 2 : 7. After 2 years, the ratio of their ages will be 1 : 3. What is Rehana’s present age ?
Solution:
The ratio of present ages of Rehana and her mother is 2 : 7
Let the common multiple be x.
∴ Present age of Rehana = 2x years and Present age of Rehana’s mother = 7x years
After 2 years,
Rehana’s age = (2x + 2) years
Age of Rehana’s mother = (7x + 2) years
According to the given condition,
After 2 years, the ratio of their ages will be 1 : 3
∴ \(\frac { 2x + 2 }{ 7x + 2 }\) = \(\frac { 1 }{ 3 }\)
∴ 3(2x + 2) = 1(7x + 2)
∴ 6x + 6 = 7x + 2
∴ 6 – 2 = 7x – 6x
∴ 4 = x
∴ x = 4
∴ Rehana’s present age = 2x = 2 x 4 = 8 years
∴ Rehana’s present age is 8 years.

Class 9 Maths Digest

• Practice Set 4.1 Class 9 Answers
• Practice Set 4.2 Class 9 Answers
• Practice Set 4.3 Class 9 Answers
• Practice Set 4.4 Class 9 Answers
• Practice Set 4.5 Class 9 Answers
• Practice Set 4 Class 9 Answers
• Practice Set 5.1 Class 9 Answers
• Practice Set 5.2 Class 9 Answers
• Practice Set 5 Class 9 Answers

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

9th Standard Maths 1 Problem Set 3 Chapter 3 Polynomials Textbook Answers Maharashtra Board

Class 9 Maths Part 1 Problem Set 3 Chapter 3 Polynomials Questions With Answers Maharashtra Board

Question 1.
Write the correct alternative answer for each of the following questions.

i. Which of the following is a polynomial?

Answer:
(D) √2x² + \(\frac { 1 }{ 2 }\)

ii. What is the degree of the polynomial √7 ?
(A) \(\frac { 1 }{ 2 }\)
(B) 5
(C) 2
(D) 0
Answer:
(D) 0

iii. What is the degree of the polynomial ?
(A) 0
(B) 1
(C) undefined
(D) any real number
Answer:
(C) undefined

iv. What is the degree of the polynomial 2x² + 5xsup>3 + 7?
(A) 3
(B) 2
(C) 5
(D) 7
Answer:
(A) 3

v. What is the coefficient form of x³ – 1 ?
(A) (1, -1)
(B) (3, -1)
(C) (1, 0, 0, -1)
(D) (1, 3, -1)
Answer:
(C) (1, 0, 0, -1)

vi. p(x) = x² – x + 3, then p (7√7) = ?
(A) 3
(B) 7√7
(C) 42√7+3
(D) 49√7
Answer:
(D) 49√7

vii. When x = – 1, what is the value of the polynomial 2x³ + 2x ?
(A) 4
(B) 2
(C) -2
(D) -4
Answer:
(A) 4

viii. If x – 1 is a factor of the polynomial 3x² + mx, then find the value of m.
(A) 2
(B) -2
(C) -3
(D) 3
Answer:
(C) -3

ix. Multiply (x² – 3) (2x – 7x³ + 4) and write the degree of the product.
(A) 5
(B) 3
(C) 2
(D) 0
Answer:
(A) 5

x. Which is the following is a linear polynomials?
(A)  x + 5
(B)  x² + 5
(C) x³ + 5
(D) x⁴ + 5
Answer:
(A)  x + 5

Hints:
v. x³ – 1 = x³ + 0x² + 0x – 1

vi. p(7√ 7) = (7√ 7)² (7√ 7) (7√ 7) + 3
= 3

vii. p(-1) = 2(-1)³ + 2(-1)
= -2 – 2 = -4

vii. p(1) = 0
∴ 3(1)² + m(1) = 0
∴ 3 + m =0
∴ m = -3

ix. Here, degree of first polynomial = 2 and
degree of second polynomial 3
∴ Degree of polynomial obtained by multiplication = 2 + 3 = 5

Question 2.
Write the degree of the polynomial for each of the following.
i. 5 + 3x⁴
ii. 7
iii. ax⁷ + bx⁹ (a, b are constants)
Answer:
i. 5 + 3x⁴
Here, the highest power of x is 4.
∴Degree of the polynomial = 4

ii. 7 = 7x°
∴ Degree of the polynomial = 0

iii. ax⁷ + bx⁹
Here, the highest power ofx is 9.
∴Degree of the polynomial = 9

Question 3.
Write the following polynomials in standard form. [1 Mark each]
i. 4x² + 7x⁴ – x³ – x + 9
ii. p + 2p³ + 10p² + 5p⁴ – 8
Answer:
i. 7x⁴ – x³ + 4x² – x + 9
ii. 5p⁴ + 2p³ + 10p² + p – 8

Question 4.
Write the following polynomial in coefficient form.
i. x⁴ + 16
ii. m⁵ + 2m² + 3m+15
Answer:
i. x⁴ + 16
Index form = x⁴ + 0x³ + 0x² + 0x + 16
∴ Coefficient form of the polynomial = (1,0,0,0,16)

ii. m⁵ + 2m² + 3m + 15
Index form = m⁵ + 0m⁴ + 0m³ + 2m² + 3m + 15
∴ Coefficient form of the polynomial = (1, 0, 0, 2, 3, 15)

Question 5.
Write the index form of the polynomial using variable x from its coefficient form.
i. (3, -2, 0, 7, 18)
ii. (6, 1, 0, 7)
iii. (4, 5, -3, 0)
Answer:
i. Number of coefficients = 5
∴ Degree = 5 – 1 = 4
∴Index form = 3x⁴ – 2x³ + 0x² + 7x + 18

ii. Number of coefficients = 4
∴Degree = 4 – 1 = 3
∴ Index form = 6x³ + x² + 0x + 7

iii. Number of coefficients = 4
∴ Degree = 4 – 1 = 3
∴ Index form = 4x³ + 5x² – 3x + 0

Question 6.
Add the following polynomials.
i. 7x⁴ – 2x³ + x + 10;
3x⁴ + 15x³ + 9x² – 8x + 2
ii. 3p³q + 2p²q + 7;
2p²q + 4pq – 2p³q
Solution:
i. (7x⁴ – 2x³ + x + 10) + (3x⁴ + 15x³ + 9x² – 8x + 2)
= 7x⁴ – 2x³ + x + 10 + 3x⁴ + 15x³ + 9x² – 8x + 2
= 7x⁴ + 3x⁴ – 2x³ + 1 5x³ + 9x² + x – 8x + 10 + 2
= 10x⁴ + 13x³ + 9x² – 7x + 12

ii. (3p³q + 2p²q + 7) + (2p²q + 4pq – 2p³q)
= 3p³q + 2p²q + 7 + 2p²q + 4pq – 2p³q
= 3p³q – 2p³q + 2p²q + 2p²q + 4pq + 7
= p³q + 4p²q + 4pq + 7

Question 7.
Subtract the second polynomial from the first.
i. 5x² – 2y + 9 ; 3x² + 5y – 7
ii. 2x² + 3x + 5 ; x² – 2x + 3
Solution:
i. (5x² – 2y + 9) – (3x² + 5y – 7)
= 5x² – 2y+ 9 – 3x² – 5y + 1
= 5x² – 3x² – 2y – 5y + 9 + 7
= 2x² – 1y + 16

ii. (2x²+ 3x + 5) – (x² – 2x + 3)
= 2x² + 3x + 5 – x² + 2x – 3
= 2x² – x² + 3x + 2x + 5 – 3
= x² + 5x + 2

Question 8.
Multiply the following polynomials.
i. (m³ – 2m + 3) (m⁴ – 2m² + 3m + 2)
ii. (5m³ – 2) (m² – m + 3)
Solution:
i. (m³ – 2m + 3) (m⁴ – 2m² + 3m + 2)
= m³(m⁴ – 2m² + 3m + 2) – 2m(m⁴ – 2m² + 3m + 2) + 3(m⁴ – 2m² + 3m + 2)
= m⁷ – 2m⁵ + 3m⁴ + 2m³ – 2m⁵ + 4m³ – 6m² – 4m + 3m⁴ – 6m² + 9m + 6
= m⁷ – 2m⁵ – 2m⁵ + 3m⁴ + 3m⁴ + 2m³ + 4m³ – 6m² – 6m² – 4m + 9m + 6
= m⁷ – 4m⁵ + 6m⁴ + 6m³ – 12m² + 5m + 6

ii. (5m³ – 2) (m² – m + 3)
= 5m³(m² – m + 3) – 2(m² – m + 3)
= 5m⁵ – 5m⁴ + 15m³ – 2m² + 2m – 6

Question 9.
Divide polynomial 3x³ – 8x² + x + 7 by x – 3 using synthetic method and write the quotient and remainder.
Solution:
Dividend = 3x³ – 8x² + x + 7
∴ Coefficient form of dividend = (3, – 8, 1,7)
Divisor = x – 3
∴ Opposite of – 3 is 3

Coefficient form of quotient = (3, 1,4)
∴ Quotient = 3x² + x + 4 and
Remainder =19

Question 10.
For which value of m, x + 3 is the factor of the polynomial x³ – 2mx + 21?
Solution:
Here, p(x) = x³ – 2mx + 21
(x + 3) is a factor of x³ – 2mx + 21.
∴ By factor theorem,
Remainder = 0
∴ P(- 3) = 0
p(x) = x³ – 2mx + 21
∴ p(-3) = (-3)³ – 2(m)(-3) + 21
∴ 0 = – 27 + 6m + 21
∴ 6 + 6m = 0
∴ 6m = 6
∴ m = 1
∴ x + 3 is the factor of x³ – 2mx + 21 for m = 1.

Question 11.
At the end of the year 2016, the population of villages Kovad, Varud, Chikhali is 5x² – 3y², 7y² + 2xy and 9x² + 4xy respectively. At the beginning of the year 2017, x² + xy – y², 5xy and 3x² + xy persons from each of the three villages respectively went to another village for education, then what is the remaining total population of these three villages ?
Solution:
Total population of villages at the end of 2016 = (5x² – 3y²) + (7y² + 2xy) + (9x² + 4xy)
= 5x² + 9x² – 3y² + 7y² + 2xy + 4xy
= 14x² + 4y² + 6xy …….(i)
Total number of persons who went to other village at the beginning of 2017 = (x² + xy – y²) + (5xy) + (3x² + xy)
= x² + 3x² – y² + xy + 5xy + xy
= 4x² – y² + 7xy … (ii)
Remaining total population of villages = Total population at the end of 2016 – total number of persons who went to other village at the beginning of 2017
= 14x² + 4y² + 6xy – (4x² – y² + 7xy) … [From (i) and (ii)]
= 14x² + 4y² + 6xy – 4x² + y² – 7xy
= 14x² – 4x² + 4y² + y² + 6xy – 7xy = 1
= 10x² + 5y² – xy
∴ The remaining total population of the three villages is 10x² + 5y² – xy.

Question 12.
Polynomials bx² + x + 5 and bx³ – 2x + 5 are divided by polynomial x – 3 and the remainders are m and n respectively. If m – n = 0, then find the value of b.
Solution:
When polynomial bx² + x + 5 is divided by (x – 3), the remainder is m.
∴ By remainder theorem,
Remainder = p(3) = m
p(x) = bx² + x + 5
∴ p(3) = b(3)² + 3 + 5
∴m = b(9) + 8
m = 9b + 8 …(i)
When polynomial bx³ – 2x + 5 is divided by x – 3 the remainder is n
∴ remainder = p(3) = n
p(x) = bx³ – 2x + 5
∴ P(3)= b(3)³ – 2(3) + 5
∴ n = b(27) – 6 + 5
∴ n = 27b – 1 …(ii)
Now, m – n = 0 …[Given]
∴ m = n
∴ 9b + 8 = 27b – 1 …[From (i) and (ii)]
∴ 8 + 1 = 27b – 9b
∴ 9 = 18b
∴ b = \(\frac { 1 }{ 2 }\)

Question 13.
Simplify.
(8m² + 3m – 6) – (9m – 7) + (3m² – 2m + 4)
Solution:
(8m² + 3m – 6) – (9m – 7) + (3m² – 2m + 4)
= 8m² + 3m – 6 – 9m + 7 + 3m² – 2m + 4
= 8m² + 3m² + 3m – 9m – 2m – 6 + 7 + 4
= 11m² – 8m + 5

Question 14.
Which polynomial is to be subtracted from x² + 13x + 7 to get the polynomial 3x² + 5x – 4?
Solution:
Let the required polynomial be A.
∴ (x² + 13x + 7) – A = 3x² + 5x – 4
∴ A = (x² + 13x + 7) – (3x² + 5x – 4)
= x² + 13x + 7 – 3x² – 5x + 4
= x² – 3x² + 13x – 5x + 7+4
= -2x² + 8x + 11
∴ – 2x² + 8x + 11 must be subtracted from x² + 13x + 7 to get 3x² + 5x – 4.

Question 15.
Which polynomial is to be added to 4m + 2n + 3 to get the polynomial 6m + 3n + 10?
Solution:
Let the required polynomial be A.
∴ (4m + 2n + 3) + A = 6m + 3n + 10
∴ A = 6m + 3n + 10 – (4m + 2n + 3)
= 6m + 3n + 10 – 4m – 2n – 3
= 6m – 4m + 3n – 2n + 10 – 3
= 2m + n + 7
∴ 2m + n + 7 must be added to 4m + 2n + 3 to get 6m + 3n + 10.

Question 1.
Read the following passage, write the appropriate amount in the boxes and discuss.
Govind, who is a dry land farmer from Shiralas has a 5 acre field. His family includes his wife, two children and his old mother. He borrowed one lakh twenty five thousand rupees from the bank for one year as agricultural loan at 10 p.c.p.a. He cultivated soyabean in x acres and cotton and tur in y acres. The expenditure he incurred was as follows :
He spent ₹10,000 on seeds. The expenses for fertilizers and pesticides for the soyabean crop was ₹ 2000x and ₹ 4000x² were spent on wages and cultivation of land. He spent ₹ 8000y on fertilizers and pesticides and ₹ 9000y2 for wages and cultivation of land for the cotton and tur crops.

Let us write the total expenditure on all the crops by using variables x and y.
₹ 10000 + 2000x + 4000×2 + 8000y + 9000y²
He harvested 5x² quintals soyabean and sold it at ₹ 2800 per quintal. The cotton crop yield was \(\frac { 5 }{ 3 }\) y² quintals which fetched ₹ 5000 per quintal.
The tur crop yield was 4y quintals and was sold at ₹ 4000 per quintal. Write the total income in rupees that was obtained by selling the entire farm produce, with the help of an expression using variables x and y. (Textbook pg. no. 44)
Answer:
Total income = income on soyabean crop + income on cotton crop + income on tur crop
= ₹ (5x² x 2800) + ₹(\(\frac { 5 }{ 3 }\) y² x 5000) + ₹ (4y x 4000)
= ₹ ( 14000x² + \(\frac { 25000 }{ 3 }\)y² + 16000y)

Question 2.
We have seen the example of expenditure and income (in terms of polynomials) of Govind who is a dry land farmer. He has borrowed rupees one lakh twenty-five thousand from the bank as an agriculture loan and repaid the said loan at 10 p.c.p.a. He had spent ₹ 10,000 on seeds. The expenses on soyabean crop was ₹ 2000x for fertilizers and pesticides and ₹ 4000x² was spent on wages and cultivation. He spent ₹ 8000y on fertilizers and pesticides and ₹9000y² on cultivation and wages for cotton and tur crop.
His total income was
₹ (14000x² + \(\frac { 25000 }{ 3 }\)y² + 16000y)
By taking x = 2, y = 3 write the income expenditure account of Govind’s farming. (Textbook pg. no. 52)
Solution:
–           Credit (Income)
₹ 1,25,000   Bank loan
₹ 56000      Income from soyabean
₹ 75000      Income from cotton
₹ 48000      Income from tur
₹ 304000     Total income

–                     Debit (Expenses)
₹ 1,37,000       loan paid with interest for seeds
₹ 10000          For seeds
₹ 4000            Fertilizers and pesticides for soyabean
₹ 16000         Wages and cultivation charges for soyabean
₹ 24000          Fertilizers and pesticides for cotton & tur
₹ 81000         Wages and cultivation charges for cotton & tur
₹ 272000       Total expenditure

Class 9 Maths Digest

• Practice Set 3.1 Class 9 Answers
• Practice Set 3.2 Class 9 Answers
• Practice Set 3.3 Class 9 Answers
• Practice Set 3.4 Class 9 Answers
• Practice Set 3.5 Class 9 Answers
• Practice Set 3.6 Class 9 Answers
• Practice Set 3 Class 9 Answers

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 2.5 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 2 Real Numbers.

9th Standard Maths 1 Practice Set 2.5 Chapter 2 Real Numbers Textbook Answers Maharashtra Board

Class 9 Maths Part 1 Practice Set 2.5 Chapter 2 Real Numbers Questions With Answers Maharashtra Board

Question 1.
Find the value.
i. | 15 – 2|
ii. | 4 – 9|
iii. | 7| x | -4|
Solution:
i. |15 – 2| = |13| = 13
ii. |4 – 9| = |-5| = 5
iii. |7| x |- 4| = 7 x 4 = 28

Question 2.
Solve.

Solution:
i. |3x – 5| = 1
∴ 3x – 5 = 1 or 3x – 5 = -1
∴ 3x = 1 + 5 or 3x = -1 + 5
∴ 3x = 6 or 3x = 4

ii. |7 – 2x| = 5
∴ 7 – 2x = 5 or 7 – 2x = -5
∴ 7 – 5 = 2x or 7 + 5 = 2x
∴ 2x = 2 or 2x = 12
∴ x = \(\frac { 2 }{ 2 }\) or x = \(\frac { 12 }{ 2 }\)
∴ x = 1 or x = 6

∴ 8 – x = 10 or 8 – x = -10 .. [Multiplying both the sides by 2]
∴ 8 – 10 = x or 8 + 10 = x
∴ x = -2 or x = 18

Class 9 Maths Digest

• Practice Set 1.1 Class 9 Answers
• Practice Set 1.2 Class 9 Answers
• Practice Set 1.3 Class 9 Answers
• Practice Set 1.4 Class 9 Answers
• Problem Set 1 Class 9 Answers
• Practice Set 2.1 Class 9 Answers
• Practice Set 2.2 Class 9 Answers
• Practice Set 2.3 Class 9 Answers
• Practice Set 2.4 Class 9 Answers
• Practice Set 2.5 Class 9 Answers
• Problem Set 2 Class 9 Answers

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 2.4 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 2 Real Numbers.

9th Standard Maths 1 Practice Set 2.4 Chapter 2 Real Numbers Textbook Answers Maharashtra Board

Class 9 Maths Part 1 Practice Set 2.4 Chapter 2 Real Numbers Questions With Answers Maharashtra Board

Question 1.
Multiply.

Solution:

Question 2.
Rationalize the denominator.

Solution:

Class 9 Maths Digest

• Practice Set 1.1 Class 9 Answers
• Practice Set 1.2 Class 9 Answers
• Practice Set 1.3 Class 9 Answers
• Practice Set 1.4 Class 9 Answers
• Problem Set 1 Class 9 Answers
• Practice Set 2.1 Class 9 Answers
• Practice Set 2.2 Class 9 Answers
• Practice Set 2.3 Class 9 Answers
• Practice Set 2.4 Class 9 Answers
• Practice Set 2.5 Class 9 Answers
• Problem Set 2 Class 9 Answers

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 3.6 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 3 Polynomials.

9th Standard Maths 1 Practice Set 3.6 Chapter 3 Polynomials Textbook Answers Maharashtra Board

Class 9 Maths Part 1 Practice Set 3.6 Chapter 3 Polynomials Questions With Answers Maharashtra Board

Question 1.
Find the factors of the polynomials given below:
i. 2x² + x – 1
ii. 2m² + 5m – 3
iii. 12x² + 61x + 77
iv. 3y² – 2y – 1
v. √3x² + 4x + √3
vi. \(\frac { 1 }{ 2 }\)x² – 3x + 4
Solution:
i. 2x² + x – 1
= 2x² + 2x – x – 1
= 2x(x + 1)- 1(x + 1)
= (x + 1)(2x – 1)

ii. 2m² + 5m – 3
= 2m² + 6m – m – 3
= 2m(m + 3) – 1(m + 3)
= (m + 3)(2m – 1)

iii. 12x² + 61x + 77
= 12x² + 28x + 33x + 77
= 4x(3x + 7) 4 + 11(3x + 7)
= (3x + 7)(4x + 11)

iv. 3y² – 2y – 1
= 3y² – 3y + y – 1
= 3y(y – 1) + 1 (y – 1)
= (y – 1)(3y + 1)

v. √3×2 + 4x + √3
= √3×2 + 3x + x + √3
= √3×2 + √3 x √3x + x + √3
= √3x(x + √3) + 1 ( x + √3 )
= (x + √3)(√3x + 1)

vi. \(\frac { 1 }{ 2 }\) x² – 3x + 4
= \(\frac { 1 }{ 2 }\) x² – 2x – x + 4
= \(\frac{1}{2} x^{2}-\frac{2 \times 2}{2} x-x+4\)
= \(\frac { 1 }{ 2 }\) x(x – 4) – 1 (x – 4)
= (x – 4) (\(\frac { 1 }{ 2 }\) x – 1)

Alternate method
\(\frac { 1 }{ 2 }\) x² – 3x + 4 = \(\frac { 1 }{ 2 }\) (x² – 6x + 8)
= \(\frac { 1 }{ 2 }\) (x² – 4x – 2x + 8)
= \(\frac { 1 }{ 2 }\) [x(x – 4) – 2(x – 4)]
= \(\frac { 1 }{ 2 }\) (x – 2)(x – 4)

Question 2.
Factorize the following polynomials.
i. (x² – x)² – 8(x² – x) + 12
iii. (x² – 6x)² – 8(x² – 6x + 8) – 64
v. (y + 2) (y – 3) (y + 8) (y + 3) + 56
vii. (x – 3) (x – 4)² (x – 5) – 6
Solution:
i. (x² – x)² – 8(x² – x) + 12
= m² – 8m + 12 …[Putting x² – x = m]
= m² – 6m – 2m + 12
= m(m – 6) – 2(m – 6)
= (m – 6)(m – 2)
= (x² – x- 6) (x² – x- 2) …[Replacing m = x² -x]
= (x² – 3x + 2x – 6) (x² – 2x + x – 2)
= [x(x – 3) + 2(x – 3)] [x(x – 2) + 1 (x-2)]
= (x – 3) (x + 2) (x – 2) (x + 1)

ii. (x – 5)² – (5x – 25) – 24
= (x – 5)² – (5x – 25) – 24
= (x – 5)² – 5(x – 5) – 24
= m² – 5m – 24 … [Putting x – 5 = m]
= m² – 8m + 3m – 24
= m(m – 8) + 3(m – 8)
= (m – 8) (m + 3)
= (x – 5 – 8) (x – 5 + 3) … [Replacing m = x – 5]
= (x – 13) (x – 2)

iii. (x² – 6x)² – 8(x² – 6x + 8) – 64
= m² – 8(m + 8)-64 …[Putting x² – 6x = m]
= m² – 8m – 64 – 64
= m² – 8m – 128
= m² – 16m + 8m- 128
= m(m – 16) + 8(m – 16)
= (m – 16)(m + 8)
= (x² – 6x – 16) (x² – 6x + 8) … [Replacing m = x² – 6x]
= (x² – 8x + 2x – 16) (x² – 4x – 2x + 8)
= [x(x – 8) + 2(x – 8)] [x(x – 4) – 2(x – 4)]
= (x – 8) (x + 2) (x – 4) (x – 2)

iv. (x²– 2x + 3) (x² – 2x + 5) – 35
= (m + 3) (m + 5) – 35 … [Putting x² – 2x = m]
= m (m + 5) + 3(m + 5) – 35
= m² + 5m + 3m + 15 – 35
= m² + 8m – 20
= m² + 10m – 2m – 20
= m(m + 10) – 2(m + 10)
= (m + 10) (m – 2)
= (x² – 2x + 10) (x² – 2x – 2) … [Replacing m = x² – 2x]

v. (y + 2) (y – 3) (y + 8) (y + 3) + 56
= (y + 2)(y + 3)(y – 3)(y + 8) + 56
= (y² + 3y + 2y + 6) (y² + 8y – 3y – 24) + 56
= (y² + 5y + 6) (y² + 5y – 24) + 56
= (m + 6) (m – 24) + 56 … [Putting y2 + 5y = m]
= m (m – 24) + 6 (m – 24) + 56
= m² – 24m + 6m – 144 + 56
= m² – 18m – 88
= m² – 22m + 4m – 88
= m(m – 22) + 4(m – 22)
= (m – 22) (m + 4)
= (y² + 5y – 22)(y² + 5y + 4) … [Replacing m = y² + 5y]
= (y² + 5y – 22) (y² + 4y + y + 4)
= (y² + 5y – 22) [y(y + 4) + 1(y + 4)]
= (y² + 5y – 22) (y + 4) (y + 1)

vi. (y² + 5y) (y² + 5y – 2) – 24
= (m)(m – 2) – 24 … [Putting y² + 5y = m]
= m² – 2m – 24
= m² – 6m + 4m – 24
= m(m – 6) + 4(m – 6)
= (m – 6) (m + 4)
= (y² + 5y – 6) (y² + 5y + 4) … [Replacing m = y² + 5y]
= (y² + 6y – y – 6) (y² + 4y + y + 4)
= [y(y + 6) – 1(y + 6)] [y(y + 4) + 1(y + 4)]
= (y + 6) (y – 1) (y + 4) (y + 1)

vii. (x – 3) (x – 4)2 (x – 5) – 6
= (x – 3) (x – 5) (x – 4)² – 6
= (x² – 5x – 3x + 15) (x² – 8x + 16) – 6
= (x² – 8x + 15) (x² – 8x + 16) – 6
= (m + 15) (m+ 16) – 6 … [Putting x² – 8x = m]
= m (m + 16) + 15 (m + 16) – 6
= m² + 16m + 15m + 240 – 6
= m² + 31m + 234
= m² + 18m + 13m + 234
= m(m + 18) + 13(m + 18)
= (m + 18) (m + 13)
= (x² – 8x + 18) (x² – 8x + 13) … [Replacing m = x² – 8x]

Class 9 Maths Digest

• Practice Set 3.1 Class 9 Answers
• Practice Set 3.2 Class 9 Answers
• Practice Set 3.3 Class 9 Answers
• Practice Set 3.4 Class 9 Answers
• Practice Set 3.5 Class 9 Answers
• Practice Set 3.6 Class 9 Answers
• Practice Set 3 Class 9 Answers

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.

9th Standard Maths 1 Practice Set 3.5 Chapter 3 Polynomials Textbook Answers Maharashtra Board

Class 9 Maths Part 1 Practice Set 3.5 Chapter 3 Polynomials Questions With Answers Maharashtra Board

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.

Class 9 Maths Digest

• Practice Set 3.1 Class 9 Answers
• Practice Set 3.2 Class 9 Answers
• Practice Set 3.3 Class 9 Answers
• Practice Set 3.4 Class 9 Answers
• Practice Set 3.5 Class 9 Answers
• Practice Set 3.6 Class 9 Answers
• Practice Set 3 Class 9 Answers

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.

9th Standard Maths 1 Practice Set 3.4 Chapter 3 Polynomials Textbook Answers Maharashtra Board

Class 9 Maths Part 1 Practice Set 3.4 Chapter 3 Polynomials Questions With Answers Maharashtra Board

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

Class 9 Maths Digest

• Practice Set 3.1 Class 9 Answers
• Practice Set 3.2 Class 9 Answers
• Practice Set 3.3 Class 9 Answers
• Practice Set 3.4 Class 9 Answers
• Practice Set 3.5 Class 9 Answers
• Practice Set 3.6 Class 9 Answers
• Practice Set 3 Class 9 Answers

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.

9th Standard Maths 1 Practice Set 3.3 Chapter 3 Polynomials Textbook Answers Maharashtra Board

Class 9 Maths Part 1 Practice Set 3.3 Chapter 3 Polynomials Questions With Answers Maharashtra Board

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.

Class 9 Maths Digest

• Practice Set 3.1 Class 9 Answers
• Practice Set 3.2 Class 9 Answers
• Practice Set 3.3 Class 9 Answers
• Practice Set 3.4 Class 9 Answers
• Practice Set 3.5 Class 9 Answers
• Practice Set 3.6 Class 9 Answers
• Practice Set 3 Class 9 Answers

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.

9th Standard Maths 1 Practice Set 3.2 Chapter 3 Polynomials Textbook Answers Maharashtra Board

Class 9 Maths Part 1 Practice Set 3.2 Chapter 3 Polynomials Questions With Answers Maharashtra Board

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.

Class 9 Maths Digest

• Practice Set 3.1 Class 9 Answers
• Practice Set 3.2 Class 9 Answers
• Practice Set 3.3 Class 9 Answers
• Practice Set 3.4 Class 9 Answers
• Practice Set 3.5 Class 9 Answers
• Practice Set 3.6 Class 9 Answers
• Practice Set 3 Class 9 Answers