Prasanna

Variation Class 8 Maths Chapter 7 Practice Set 7.1 Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 7.1 8th Std Maths Answers Solutions Chapter 7 Variation.

Std 8 Maths Practice Set 7.1 Chapter 7 Solutions Answers

Question 1.
Write the following statements using the symbol of variation.

1. Circumference (c) of a circle is directly proportional to its radius (r).
2. Consumption of petrol (l) in a car and distance traveled by that car (d) are in direct variation.

Solution:

1. c ∝ r
2. l ∝ d

Question 2.
Complete the following table considering that the cost of apples and their number are in direct variation.

Solution:
The cost of apples (y) and their number (x) are in direct variation.
∴y ∝ x
∴y = kx …(i)
where k is the constant of variation

i. When, x = 1, y = 8
∴ Substituting, x = 1 and y = 8 in (i), we get y = kx
∴ 8 = k × 1
∴ k = 8
Substituting k = 8 in (i), we get
y = kx
∴ y = 8x …(ii)
This the equation of variation

ii. When,y = 56, x = ?
∴ Substituting y = 56 in (ii), we get
y = 8x
∴ 56 = 8x
∴ x = \(\frac { 56 }{ 8 }\)
∴ x = 7

iii. When, x = 12, y = ?
∴ Substituting x = 12 in (ii), we get
y = 8x
∴ y = 8 × 12
∴ y = 96

iv. When, y = 160, x = ?
∴ Substituting y = 160 in (ii), we get
y = 8x
∴ 160 = 8x
∴ x = \(\frac { 160 }{ 8 }\)
∴ x = 20

Question 3.
If m ∝ n and when m = 154, n = 7. Find the value of m, when n = 14.
Solution:
Given that,
m ∝ n
∴ m = kn …(i)
where k is constant of variation.
When m = 154, n = 7
∴ Substituting m = 154 and n = 7 in (i), we get
m = kn
∴ 154 = k × 7
∴ \(k=\frac { 154 }{ 7 }\)
∴ k = 22
Substituting k = 22 in (i), we get
m = kn
∴ m = 22n …(ii)
This is the equation of variation.
When n = 14, m = ?
∴ Substituting n = 14 in (ii), we get
m = 22n
∴ m = 22 × 14
∴ m = 308

Question 4.
If n varies directly as m, complete the following table.

Solution:
Given, n varies directly as m
∴ n ∝ m
∴ n = km …(i)
where, k is the constant of variation

i. When m = 3, n = 12
∴ Substituting m = 3 and n = 12 in (i), we get
n = km
∴ 12 = k × 3
∴ \(k=\frac { 12 }{ 3 }\)
∴ k = 4
Substituting, k = 4 in (i), we get
n = km
∴ n = 4m …(ii)
This is the equation of variation.

ii. When m = 6.5, n = ?
∴ Substituting, m = 6.5 in (ii), we get
n = 4m
∴ n = 4 × 6.5
∴ n = 26

iii. When n = 28, m = ?
∴ Substituting, n = 28 in (ii), we get
n = 4m
∴ 28 = 4m
∴ 28 = 4m
∴ \(m=\frac { 28 }{ 4 }\)
∴ m = 7

iv. When m = 1.25, n = ?
∴ Substituting m = 1.25 in (ii), we get
n = 4m
∴ n = 4 × 1.25
∴ n = 5

Question 5.
y varies directly as square root of x. When x = 16, y = 24. Find the constant of variation and equation of variation.
Solution:
Given, y varies directly as square root of x.
∴ y ∝ √4x
∴ y = k √x …(i)
where, k is the constant of variation.
When x = 16 ,y = 24.
∴ Substituting, x = 16 and y = 24 in (i), we get
y = k√x
∴24 = k√16
∴24 = 4k
∴ \(k=\frac { 24 }{ 4 }\)
∴ k = 6
Substituting k = 6 in (i), we get
y = k√x
∴ y = 6√x
This is the equation of variation
∴ The constant of variation is 6 and the equation of variation is y = 6√x .

Question 6.
The total remuneration paid to laborers, employed to harvest soybean is in direct variation with the number of laborers. If remuneration of 4 laborers is Rs 1000, find the remuneration of 17 laborers.
Solution:
Let, m represent total remuneration paid to laborers and n represent number of laborers employed to harvest soybean.
Since, the total remuneration paid to laborers, is in direct variation with the number of laborers.
∴ m ∝ n
∴ m = kn …(i)
where, k = constant of variation
Remuneration of 4 laborers is Rs 1000.
i. e., when n = 4, m = Rs 1000
∴ Substituting, n = 4 and m = 1000 in (i), we get m = kn
∴ 1000 = k × 4
∴ \(k=\frac { 1000 }{ 4 }\)
∴ k = 250
Substituting, k = 250 in (i), we get
m = kn
∴ m = 250 n …(ii)
This is the equation of variation
Now, we have to find remuneration of 17 laborers.
i. e., when n = 17, m = ?
∴ Substituting n = 17 in (ii), we get
m = 250 n
∴ m = 250 × 17
∴ m = 4250
∴ The remuneration of 17 laborers is Rs 4250.

Maharashtra Board Class 8 Maths Chapter 7 Variation Practice Set 7.1 Intext Questions and Activities

Question 1.
If the rate of notebooks is Rs 240 per dozen, what is the cost of 3 notebooks?
Also find the cost of 9 notebooks, 24 notebooks and 50 notebooks and complete the following table. (Textbook pg. no. 35)

Solution:
As the number of notebooks increases their cost also increases.
∴ Number of notebooks and cost of notebooks are in direct proportion.

i.

∴ y = 3 × 20
∴ y = 60

ii.

∴ y = 9 × 20
∴ y = 180

iii.

∴ y = 24 × 20
∴ y = 480

iv.

∴ y = 50 × 20
∴ y = 1000

Maharashtra Board Class 8 Maths Solutions

• Variation Practice Set 7.1 Class 8 Maths Solutions
• Variation Practice Set 7.2 Class 8 Maths Solutions
• Variation Practice Set 7.3 Class 8 Maths Solutions
• Quadrilateral: Constructions and Types Practice Set 8.1 Class 8 Maths Solutions
• Quadrilateral: Constructions and Types Practice Set 8.2 Class 8 Maths Solutions
• Quadrilateral: Constructions and Types Practice Set 8.3 Class 8 Maths Solutions

Operations on Rational Numbers Class 7 Maths Chapter 5 Practice Set 23 Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 23 Answers Solutions Chapter 5 Operations on Rational Numbers.

Std 7 Maths Practice Set 23 Solutions Answers

Question 1.
Write three rational numbers that lie between the two given numbers.
i. \(\frac{2}{7}, \frac{6}{7}\)
ii. \(\frac{4}{5}, \frac{2}{3}\)
iii. \(-\frac{2}{3}, \frac{4}{5}\)
iv. \(\frac{7}{9},-\frac{5}{9}\)
v. \(\frac{-3}{4}, \frac{+5}{4}\)
vi. \(\frac{7}{8}, \frac{-5}{3}\)
vii. \(\frac{5}{7}, \frac{11}{7}\)
viii. \(0, \frac{-3}{4}\)
Solution:
i. \(\frac{2}{7}, \frac{6}{7}\)
The three numbers lying between \(\frac { 2 }{ 7 }\) and \(\frac { 6 }{ 7 }\) are \(\frac{3}{7}, \frac{4}{7}, \frac{5}{7}\)

ii. \(\frac{4}{5}, \frac{2}{3}\)
\(\frac{4}{5}=\frac{24}{30}, \frac{2}{3}=\frac{20}{30}\)
The three numbers between \(\frac { 4 }{ 5 }\) and \(\frac { 2 }{ 3 }\) are \(\frac{21}{30}, \frac{22}{30}, \frac{23}{30}\)

iii. \(-\frac{2}{3}, \frac{4}{5}\)
\(\frac{-2}{3}=\frac{-10}{15}, \frac{4}{5}=\frac{12}{15}\)
The three numbers between \(\frac { -2 }{ 3 }\) and \(\frac { 4 }{ 5 }\) are \(\frac{-9}{15}, \frac{-7}{15}, \frac{4}{15}\)

iv. \(\frac{7}{9},-\frac{5}{9}\)
The three numbers between \(\frac { 7 }{ 9 }\) and \(\frac { -5 }{ 9 }\) are \(\frac{6}{9}, 0, \frac{-4}{9}\)

v. \(\frac{-3}{4}, \frac{+5}{4}\)
The three numbers between \(\frac { -3 }{ 4 }\) and \(\frac { +5 }{ 4 }\) are \(\frac{-2}{4}, \frac{-1}{4}, \frac{3}{4}\)

vi. \(\frac{7}{8}, \frac{-5}{3}\)
\(\frac{7}{8}=\frac{21}{24}, \frac{-5}{3}=\frac{-40}{24}\)
The three numbers between \(\frac { 7 }{ 8 }\) and \(\frac { -5 }{ 3 }\) are \(\frac{17}{24}, \frac{11}{24}, \frac{-13}{24}\)

vii. \(\frac{5}{7}, \frac{11}{7}\)
The three numbers between \(\frac { 5 }{ 7 }\) and \(\frac { 11 }{ 7 }\) are \(\frac{6}{7}, \frac{8}{7}, \frac{9}{7}\)

viii. \(0, \frac{-3}{4}\)
The three numbers between 0 and \(\frac { -3 }{ 4 }\) are \(\frac{-1}{8}, \frac{-2}{8}, \frac{-5}{8}\)

Maharashtra Board Class 7 Maths Chapter 5 Operations on Rational Numbers Practice Set 23 Intext Questions and Activities

Question 1.
Answer the following questions: (Textbook pg. no. 36)

1. Write all the natural numbers between 2 and 9.
2. Write all the integers between -4, and 5.
3. Which rational numbers are there between \(\frac { 1 }{ 2 }\) and \(\frac { 3 }{ 4 }\) ?

Solution:

1. 3, 4, 5, 6, 7, 8
2. -3, -2, -1, 0, 1, 2, 3, 4
3. \(\frac{1}{2}=\frac{1 \times 2}{2 \times 2}=\frac{2}{4}=\frac{2 \times 10}{4 \times 10}=\frac{20}{40}\)
   \(\frac{3}{4}=\frac{3 \times 10}{4 \times 10}=\frac{30}{40}\)
   ∴ The rational numbers between \(\frac { 1 }{ 2 }\) and \(\frac { 3 }{ 4 }\) are \(\frac{21}{40}, \frac{22}{40}, \frac{25}{40}, \frac{27}{40}\) etc.

Class 7 Maths Solution Maharashtra Board

• Angles and Pairs of Angles Practice Set 15 Class 7 Maths Solution
• Angles and Pairs of Angles Practice Set 16 Class 7 Maths Solution
• Angles and Pairs of Angles Practice Set 17 Class 7 Maths Solution
• Angles and Pairs of Angles Practice Set 18 Class 7 Maths Solution
• Angles and Pairs of Angles Practice Set 19 Class 7 Maths Solution
• Angles and Pairs of Angles Practice Set 20 Class 7 Maths Solution
• Angles and Pairs of Angles Practice Set 21 Class 7 Maths Solution
• Operations on Rational Numbers Practice Set 22 Class 7 Maths Solution
• Operations on Rational Numbers Practice Set 23 Class 7 Maths Solution
• Operations on Rational Numbers Practice Set 24 Class 7 Maths Solution
• Operations on Rational Numbers Practice Set 25 Class 7 Maths Solution

Angles and Pairs of Angles Class 7 Maths Chapter 4 Practice Set 21 Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 21 Answers Solutions Chapter 4 Angles and Pairs of Angles.

Std 7 Maths Practice Set 21 Solutions Answers

Question 1.
∠ACD is an exterior angle of ∆ABC. The measures of ∠A and ∠B are equal. If m∠ACD = 140°, find the measures of the angles ∠A and ∠B.

Solution:
Let the measures of ∠A be x°.
m∠A = m∠B = x°
∠ACD is the exterior angle of ∆ABC
∴ m∠ACD = m∠A + m∠B
∴ 140 = x + x
∴ 140 = 2x
∴ 2x = 140
∴ x = \(\frac { 140 }{ 2 }\)
= 70
∴ The measures of the angles ∠A and ∠B is 70° each.

Question 2.
Using the measures of the angles given in the figure alongside, find the measures of the remaining three angles.

Solution:
m∠EOD = m∠AOB = 8y ….(vertically opposite angles)
∠FOL, ∠EOD and ∠COD form a straight angle.
∴ m∠FOE + m∠EOD + m∠COD = 180°
∴ 4y + 8y + 6y = 180
∴ 18y = 180
∴ y = \(\frac { 180 }{ 18 }\)
∴ y = 10
m∠EOD = 8y = 8 x 10 = 80°
m∠AOF = m∠COD ….(Vertically opposite angles)
= 6y = 6 x 10 = 60°
m∠BOC = m∠FOE ….(Vertically opposite angles)
= 4y = 4 x 10 = 40°
∴ The measures of ∠EOD, ∠AOF and ∠BOC are 80°, 60° and 40° respectively.

Question 3.
In the isosceles triangle ABC, ∠A and ∠B are equal. ∠ACD is an exterior angle of ∆ABC. The measures of ∠ACB and ∠ACD are (3x – 17)° and (8x + 10)° respectively. Find the measures of ∠ACB and ∠ACD. Also find the measures of ∠A and ∠B.

Solution:
Let the measure of ∠A be y°. A
∴ m∠A = m∠B = y°
∠ACB and ∠ACD form a pair of linear angles.
∴ m∠ACB + m∠ACD = 180°
∴ (3x – 17) + (8x + 10) = 180
∴ 3x + 8x – 17 + 10 = 180
∴ 11x – 7 = 180
∴ 11x – 7 + 7 = 180 + 7 …(Adding 7 on both sides.)
∴ 11x = 187
∴ x = \(\frac { 187 }{ 11 }\) = 17
m∠ACB = 3x – 17 = (3 x 17) – 17 = 51 – 17 = 34°
m∠ACD = 8x + 10 = 8 x 17 + 10 = 136 + 10 = 146°
Here ∠ACD is the exterior angle of ∆ABC and ∠A and ∠B are its remote interior angles.
∴ m∠ACD = m∠A + m∠B
∴ 146 = y + y
∴ 146 = 2y
∴ 2y = 146
∴ y = \(\frac { 146 }{ 2 }\) = 73
∴ The measures of ∠ACB, ∠ACD, ∠A and ∠B are 34°, 146°, 73° and 73° respectively.

Maharashtra Board Class 7 Maths Chapter 4 Angles and Pairs of Angles Practice Set 21 Intext Questions and Activities

Question 1.
Use straws or sticks to make all the kinds of angles that you have learnt about. (Textbook pg. no. 29)
Solution:
(Student should attempt the activity on their own)

Question 2.
Observe the table given below and draw your conclusions (Textbook pg. no. 31)

Solution:
i. 180°
ii. 360°
iii. 540°
iv. 720°
v. 180° x 5 = 900°
vi.  , 180° x 6 = 1080°

Class 7 Maths Solution Maharashtra Board

• Angles and Pairs of Angles Practice Set 15 Class 7 Maths Solution
• Angles and Pairs of Angles Practice Set 16 Class 7 Maths Solution
• Angles and Pairs of Angles Practice Set 17 Class 7 Maths Solution
• Angles and Pairs of Angles Practice Set 18 Class 7 Maths Solution
• Angles and Pairs of Angles Practice Set 19 Class 7 Maths Solution
• Angles and Pairs of Angles Practice Set 20 Class 7 Maths Solution
• Angles and Pairs of Angles Practice Set 21 Class 7 Maths Solution
• Operations on Rational Numbers Practice Set 22 Class 7 Maths Solution
• Operations on Rational Numbers Practice Set 23 Class 7 Maths Solution
• Operations on Rational Numbers Practice Set 24 Class 7 Maths Solution
• Operations on Rational Numbers Practice Set 25 Class 7 Maths Solution

Operations on Rational Numbers Class 7 Maths Chapter 5 Practice Set 22 Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 22 Answers Solutions Chapter 5 Operations on Rational Numbers.

Std 7 Maths Practice Set 22 Solutions Answers

Question 1.
Carry out the following additions of rational numbers:
i. \(\frac{5}{36}+\frac{6}{42}\)
ii. \(1 \frac{2}{3}+2 \frac{4}{5}\)
iii. \(\frac{11}{17}+\frac{13}{19}\)
iv. \(2 \frac{3}{11}+1 \frac{3}{77}\)
Solution:
i. \(\frac{5}{36}+\frac{6}{42}\)

ii. \(1 \frac{2}{3}+2 \frac{4}{5}\)

iii. \(\frac{11}{17}+\frac{13}{19}\)

iv. \(2 \frac{3}{11}+1 \frac{3}{77}\)

Question 2.
Carry out the following subtractions involving rational numbers.
i. \(\frac{7}{11}-\frac{3}{7}\)
ii. \(\frac{13}{36}-\frac{2}{40}\)
iii. \(1 \frac{2}{3}-3 \frac{5}{6}\)
iv. \(4 \frac{1}{2}-3 \frac{1}{3}\)
Solution:
i. \(\frac{7}{11}-\frac{3}{7}\)

ii. \(\frac{13}{36}-\frac{2}{40}\)

iii. \(1 \frac{2}{3}-3 \frac{5}{6}\)

iv. \(4 \frac{1}{2}-3 \frac{1}{3}\)

Question 3.
Multiply the following rational numbers.
i. \(\frac{3}{11} \times \frac{2}{5}\)
ii. \(\frac{12}{5} \times \frac{4}{15}\)
iii. \(\frac{(-8)}{9} \times \frac{3}{4}\)
iv. \(\frac{0}{6} \times \frac{3}{4}\)
Solution:
i. \(\frac{3}{11} \times \frac{2}{5}\)
\(=\frac{3 \times 2}{11 \times 5}=\frac{6}{55}\)

ii. \(\frac{12}{5} \times \frac{4}{15}\)
\(=\frac{4}{5} \times \frac{4}{5}=\frac{4 \times 4}{5 \times 5}=\frac{16}{25}\)

iii. \(\frac{(-8)}{9} \times \frac{3}{4}\)
\(=\frac{(-2)}{3} \times \frac{1}{1}=\frac{-2}{3}\)

iv. \(\frac{0}{6} \times \frac{3}{4}\)
\(=0 \times \frac{3}{4}=0\)

Question 4.
Write the multiplicative inverse of.
i. \(\frac{2}{5}\)
ii. \(\frac{-3}{8}\)
iii. \(\frac{-17}{39}\)
iv. 7
v. \(-7 \frac{1}{3}\)
Solution:
i. \(\frac{5}{2}\)
ii. \(\frac{-8}{3}\)
iii. \(\frac{-39}{17}\)
iv. \(\frac {1}{7}\)
v. \(\frac {-3}{22}\)

Question 5.
Carry out the divisions of rational numbers:
i. \(\frac{40}{12} \div \frac{10}{4}\)
ii. \(\frac{-10}{11} \div \frac{-11}{10}\)
iii. \(\frac{-7}{8} \div \frac{-3}{6}\)
iv. \(\frac{2}{3} \div(-4)\)
v. \(2 \frac{1}{5} \div 5 \frac{3}{6}\)
vi. \(\frac{-5}{13} \div \frac{7}{26}\)
vii. \(\frac{9}{11} \div(-8)\)
viii. \(5 \div \frac{2}{5}\)
Solution:
i. \(\frac{40}{12} \div \frac{10}{4}\)
\(=\frac{40}{12} \times \frac{4}{10}=\frac{4}{3}\)

ii. \(\frac{-10}{11} \div \frac{-11}{10}\)
\(=\frac{-10}{11} \times \frac{-10}{11}=\frac{100}{121}\)

iii. \(\frac{-7}{8} \div \frac{-3}{6}\)
\(=\frac{-7}{8} \times \frac{-6}{3}=\frac{-7}{4} \times \frac{-3}{3}=\frac{7}{4}\)

iv. \(\frac{2}{3} \div(-4)\)
\(=\frac{2}{3} \times \frac{-1}{4}=\frac{1}{3} \times \frac{-1}{2}=\frac{-1}{6}\)

v. \(2 \frac{1}{5} \div 5 \frac{3}{6}\)
\(=\frac{11}{5} \div \frac{33}{6}=\frac{11}{5} \times \frac{6}{33}=\frac{1}{5} \times \frac{6}{3}=\frac{2}{5}\)

vi. \(\frac{-5}{13} \div \frac{7}{26}\)
\(=\frac{-5}{13} \times \frac{26}{7}=\frac{-10}{7}\)

vii. \(\frac{9}{11} \div(-8)\)
\(=\frac{9}{11} \times \frac{-1}{8}=\frac{-9}{88}\)

viii. \(5 \div \frac{2}{5}\)
\(=\frac{5}{1} \times \frac{5}{2}=\frac{25}{2}\)

Maharashtra Board Class 7 Maths Chapter 5 Operations on Rational Numbers Practice Set 22 Intext Questions and Activities

Question 1.
Complete the table given below. (Textbook pg. no. 34)

Solution:

Question 2.
Discuss the characteristics of various groups of numbers in class and complete the table below. In front of each group, write the inference you make after carrying out the operations of addition, subtraction, multiplication and division, using a (✓) or a (x).
Remember that you cannot divide by zero. (Textbook pg. no. 35)

Solution:

Class 7 Maths Solution Maharashtra Board

• Angles and Pairs of Angles Practice Set 15 Class 7 Maths Solution
• Angles and Pairs of Angles Practice Set 16 Class 7 Maths Solution
• Angles and Pairs of Angles Practice Set 17 Class 7 Maths Solution
• Angles and Pairs of Angles Practice Set 18 Class 7 Maths Solution
• Angles and Pairs of Angles Practice Set 19 Class 7 Maths Solution
• Angles and Pairs of Angles Practice Set 20 Class 7 Maths Solution
• Angles and Pairs of Angles Practice Set 21 Class 7 Maths Solution
• Operations on Rational Numbers Practice Set 22 Class 7 Maths Solution
• Operations on Rational Numbers Practice Set 23 Class 7 Maths Solution
• Operations on Rational Numbers Practice Set 24 Class 7 Maths Solution
• Operations on Rational Numbers Practice Set 25 Class 7 Maths Solution

Factorisation of Algebraic Expressions Class 8 Maths Chapter 6 Practice Set 6.4 Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 6.4 8th Std Maths Answers Solutions Chapter 6 Factorisation of Algebraic Expressions.

Std 8 Maths Practice Set 6.4 Chapter 6 Solutions Answers

Question 1.
Simplify:
i. \(\frac{m^{2}-n^{2}}{(m+n)^{2}} \times \frac{m^{2}+m n+n^{2}}{m^{3}-n^{3}}\)
ii. \(\frac{a^{2}+10 a+21}{a^{2}+6 a-7} \times \frac{a^{2}-1}{a+3}\)
iii. \(\frac{8 x^{3}-27 y^{3}}{4 x^{2}-9 y^{2}}\)
iv. \(\frac{x^{2}-5 x-24}{(x+3)(x+8)} \times \frac{x^{2}-64}{(x-8)^{2}}\)
v. \(\frac{3 x^{2}-x-2}{x^{2}-7 x+12} \div \frac{3 x^{2}-7 x-6}{x^{2}-4}\)
vi. \(\frac{4 x^{2}-11 x+6}{16 x^{2}-9}\)
vii. \(\frac{a^{3}-27}{5 a^{2}-16 a+3} \div \frac{a^{2}+3 a+9}{25 a^{2}-1}\)
viii. \(\frac{1-2 x+x^{2}}{1-x^{3}} \times \frac{1+x+x^{2}}{1+x}\)
Solution:
i. \(\frac{m^{2}-n^{2}}{(m+n)^{2}} \times \frac{m^{2}+m n+n^{2}}{m^{3}-n^{3}}\)

ii. \(\frac{a^{2}+10 a+21}{a^{2}+6 a-7} \times \frac{a^{2}-1}{a+3}\)

iii. \(\frac{8 x^{3}-27 y^{3}}{4 x^{2}-9 y^{2}}\)

iv. \(\frac{x^{2}-5 x-24}{(x+3)(x+8)} \times \frac{x^{2}-64}{(x-8)^{2}}\)

v. \(\frac{3 x^{2}-x-2}{x^{2}-7 x+12} \div \frac{3 x^{2}-7 x-6}{x^{2}-4}\)

vi. \(\frac{4 x^{2}-11 x+6}{16 x^{2}-9}\)

vii. \(\frac{a^{3}-27}{5 a^{2}-16 a+3} \div \frac{a^{2}+3 a+9}{25 a^{2}-1}\)

viii. \(\frac{1-2 x+x^{2}}{1-x^{3}} \times \frac{1+x+x^{2}}{1+x}\)

Maharashtra Board Class 8 Maths Solutions

• Factorisation of Algebraic Expressions Practice Set 6.1 Class 8 Maths Solutions
• Factorisation of Algebraic Expressions Practice Set 6.2 Class 8 Maths Solutions
• Factorisation of Algebraic Expressions Practice Set 6.3 Class 8 Maths Solutions
• Factorisation of Algebraic Expressions Practice Set 6.4 Class 8 Maths Solutions

Angles and Pairs of Angles Class 7 Maths Chapter 4 Practice Set 20 Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 20 Answers Solutions Chapter 4 Angles and Pairs of Angles.

Std 7 Maths Practice Set 20 Solutions Answers

Question 1.
Lines AC and BD intersect at point P. m∠APD = 47° Find the measures of ∠APB, ∠BPC, ∠CPD.

Solution:
∠APD and ∠APB are angles in a linear pair.
∴m∠APD + m∠APB = 180°
∴47 + m∠APB = 180
∴47 + m∠APB – 47 = 180 – 47 ….(Subtracting 47 from both sides)
∴m∠APB = 133°
m∠CPD = m∠APB = 133° … .(Vertically opposite angles)
m∠BPC = m∠APD = 47° … .(Vertically opposite angles)
∴The measures of ∠APB, ∠BPC and ∠CPD are 133°, 47° and 133° respectively.

Question 2.
Lines PQ and RS intersect at point M. m∠PMR = x°.What are the measures of ∠PMS, ∠SMQ and ∠QMR?
Solution:
∠PMR and ∠PMS are angles in a linear pair.
∴ m∠PMR + m∠PMS = 180°
∴ x + m∠PMS = 180
∴ m∠PMS = (180-x)°
m∠QMR = m∠PMS = (180 – x)° … .(Vertically opposite angles)
m∠SMQ = m∠PMR = x° …. (Vertically opposite angles)
∴The measures of ∠PMS, ∠SMQ and ∠QMR are (180 – x)°, x° and (180 – x)° respectively.

Class 7 Maths Solution Maharashtra Board

• Angles and Pairs of Angles Practice Set 15 Class 7 Maths Solution
• Angles and Pairs of Angles Practice Set 16 Class 7 Maths Solution
• Angles and Pairs of Angles Practice Set 17 Class 7 Maths Solution
• Angles and Pairs of Angles Practice Set 18 Class 7 Maths Solution
• Angles and Pairs of Angles Practice Set 19 Class 7 Maths Solution
• Angles and Pairs of Angles Practice Set 20 Class 7 Maths Solution
• Angles and Pairs of Angles Practice Set 21 Class 7 Maths Solution
• Operations on Rational Numbers Practice Set 22 Class 7 Maths Solution
• Operations on Rational Numbers Practice Set 23 Class 7 Maths Solution
• Operations on Rational Numbers Practice Set 24 Class 7 Maths Solution
• Operations on Rational Numbers Practice Set 25 Class 7 Maths Solution

Angles and Pairs of Angles Class 7 Maths Chapter 4 Practice Set 19 Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 19 Answers Solutions Chapter 4 Angles and Pairs of Angles.

Std 7 Maths Practice Set 19 Solutions Answers

Question 1.
Draw the pairs of angles as described below. If that is not possible, say why.
i. Complementary angles that are not adjacent.
ii. Angles in a linear pair which are not supplementary.
iii. Complementary angles that do not form a linear pair.
iv. Adjacent angles which are not in linear pair.
v. Angles which are neither complementary nor adjacent.
vi. Angles in a linear pair which are complementary.
Solution:
i.

ii. Sum of angles in a linear pair is 180°.
i.e. they are supplementary .
∴ Angles in a linear pair which are not supplementary cannot be drawn.

iii.

iv.

v.

vi. Angles in linear pair have their sum as 180° But, complementary angles have their sum as 90°.
∴ Angles in a linear pair which are complementary cannot be drawn.

Note: Problem No. i, iii, iv, and v have more than one answers students may draw angles other than the once given.

Maharashtra Board Class 7 Maths Chapter 4 Angles and Pairs of Angles Practice Set 19 Intext Questions and Activities

Question 1.
Observe the adjacent figure and answer the following questions: (Textbook pg. no. 29)

1. Write the names of the angles in the figure alongside.
2. What type of a pair of angles is it?
3. Which arms of the angles are not the common arms?
4. m∠PQR = __.
5. m∠RQS = __.

Solution:

1. ∠PQR and ∠RQS
2. Angles in a linear pair
3. Ray QP and ray QS
4. 125
5. 55
   Here, m∠PQR + m∠RQS = 125° + 55°
   = 180°
   ∴The adjacent angles ∠PQR and ∠RQS are supplementary.

Class 7 Maths Solution Maharashtra Board

• Angles and Pairs of Angles Practice Set 15 Class 7 Maths Solution
• Angles and Pairs of Angles Practice Set 16 Class 7 Maths Solution
• Angles and Pairs of Angles Practice Set 17 Class 7 Maths Solution
• Angles and Pairs of Angles Practice Set 18 Class 7 Maths Solution
• Angles and Pairs of Angles Practice Set 19 Class 7 Maths Solution
• Angles and Pairs of Angles Practice Set 20 Class 7 Maths Solution
• Angles and Pairs of Angles Practice Set 21 Class 7 Maths Solution
• Operations on Rational Numbers Practice Set 22 Class 7 Maths Solution
• Operations on Rational Numbers Practice Set 23 Class 7 Maths Solution
• Operations on Rational Numbers Practice Set 24 Class 7 Maths Solution
• Operations on Rational Numbers Practice Set 25 Class 7 Maths Solution

Angles and Pairs of Angles Class 7 Maths Chapter 4 Practice Set 18 Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 18 Answers Solutions Chapter 4 Angles and Pairs of Angles.

Std 7 Maths Practice Set 18 Solutions Answers

Question 1.
Name the pairs of opposite rays in the figure alongside.

Solution:

1. Ray PL and ray PM
2. Ray PN and ray PT

Question 2.
Are the ray PM and PT opposite rays? Give reasons for your answer.

Solution:
No.
Ray PM and Ray PT do not form a straight line and hence are not opposite rays.

Maharashtra Board Class 7 Maths Chapter 4 Angles and Pairs of Angles Practice Set 18 Intext Questions and Activities

Question 1.
Observe the adjacent figure and answer the following questions. (Textbook pg. no. 28)

1. Name the rays in the figure alongside.
2. Name the origin of the rays
3. Name the angle in the given figure

Solution:

1. Ray BA and ray BC
2. Point B
3. ∠ABC or ∠CBA

Question 2.
Observe the adjacent figure and answer the following questions. (Textbook pg. no. 28)

1. Name the angle in the figure alongside.
2. Name the rays whose origin is point B

Solution:

1. ∠ABC or ∠CBA
2. Ray BA and ray BC

Class 7 Maths Solution Maharashtra Board

• Angles and Pairs of Angles Practice Set 15 Class 7 Maths Solution
• Angles and Pairs of Angles Practice Set 16 Class 7 Maths Solution
• Angles and Pairs of Angles Practice Set 17 Class 7 Maths Solution
• Angles and Pairs of Angles Practice Set 18 Class 7 Maths Solution
• Angles and Pairs of Angles Practice Set 19 Class 7 Maths Solution
• Angles and Pairs of Angles Practice Set 20 Class 7 Maths Solution
• Angles and Pairs of Angles Practice Set 21 Class 7 Maths Solution
• Operations on Rational Numbers Practice Set 22 Class 7 Maths Solution
• Operations on Rational Numbers Practice Set 23 Class 7 Maths Solution
• Operations on Rational Numbers Practice Set 24 Class 7 Maths Solution
• Operations on Rational Numbers Practice Set 25 Class 7 Maths Solution

Factorisation of Algebraic Expressions Class 8 Maths Chapter 6 Practice Set 6.3 Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 6.3 8th Std Maths Answers Solutions Chapter 6 Factorisation of Algebraic Expressions.

Std 8 Maths Practice Set 6.3 Chapter 6 Solutions Answers

Question 1.
Factorize
i. y³ – 27
ii. x³ – 64y³
iii. 27m³ – 216n³
iv. 125y³ – 1
v. \(8 p^{3}-\frac{27}{p^{3}}\)
vi. 343a³ – 512b³
vii. 64x³ – 729y³
viii. \(16 a^{3}-\frac{128}{b^{3}}\)
Solution:
i. y³ – 27
= y³ – (3)³
Here, a = y and b = 3
∴ y³ – 27 = (y – 3)[y² + y(3) + (3)2]
…[∵ a³ – b³ = (a – b) (a² + ab + b²)]
= (y – 3)(y² + 3y + 9)

ii. x³ – 64y³
= x³ – (4y)³
Here, a = x and b = 4y
∴ x³ – 64y³ = (x – 4y)[x² + x(4y) + (4y)²]
…[∵ a³ – b³ = (a – b)(a² + ab + b²)]
= (x – 4y)(x² + 4xy + 16y²)

iii. 27m³ – 216n³
= 27 (m³ – 8n³)
… [Taking out the common factor 27]
= 27 [m³ – (2n)³]
Here, a = m and b = 2n
∴ 27m³ – 216n³
= 27 {(m – 2n) [m² + m(2n) + (2n)²]}
….[∵ a³ – b³ = (a – b) (a² + ab + b²)]
= 27 (m – 2n)(m² + 2mn + 4n²)

iv. 125y³ – 1
= (5y)³ – 1³
Here, a = 5y and b = 1
∴ 125y³ – 1 = (5y – 1) [(5y)² + (5y)(1) + (1)²]
…[∵ a³ – b³ = (a – b)(a² + ab + b²)]
= (5y – 1) (25y² + 5y + 1)

v. \(8 p^{3}-\frac{27}{p^{3}}\)

vi. 343a³ – 512b³
= (7a)³ – (8b)³
Here, A = 7a and B = 8b
∴ 343a³ – 512b³
= (7a – 8b) [(7a)² + (7a)(8b) + (8b)²]
…[∵ A³ – B³ = (A – B)(A² + AB + B²)]
= (7a – 8b) (49a² + 56ab + 64b²)

vii. 64x³ – 729y³
= (4x)³ – (9y)³
Here, a = 4x and b = 9y
∴ 64x³ – 729y³
= (4x – 9y) [(4x)² + (4x) (9y) + (9y)²]
…[∵ a³ – b³ = (a – b)(a² + ab + b²)]
= (4x – 9y) (16x² + 36xy + 81y²)

viii. \(16 a^{3}-\frac{128}{b^{3}}\)

Question 2.
Simplify:
i. (x + y)³ – (x – y)³
ii. (3a + 5b)³ – (3a – 5b)³
iii. (a + b)³ – a³ – b³
iv. p³ – (p + 1)³
v. (3xy – 2ab)³ – (3xy + 2ab)³
Solution:
i. (x + y)³ – (x – y)³
Here, a = x + y and b = x – y
(x + y)³ – (x – y)³
= [(x + y) – (x – y)] [(x + y)² + (x + y) (x – y) + (x – y)]
…[a³ – b³ = (a – b)(a² + ab + b²)]
= (x + y – x + y) [(x² + 2xy + y²) + (x² – y²) + (x² – 2xy + y²)]
= 2y(x² + x² + x² + 2xy – 2xy + y² – y² + y²)
= 2y (3x² + y²)
= 6x²y + 2y³

ii. (3a + 5b)³ – (3a – 5b)³
Here, A = 3a + 5b and B = 3a – 5b
= [(3a + 5b) – (3a – 5b)] [(3a + 5b)² + (3a + 5b) (3a – 5b) + (3a – 5b)²]
…[∵ A³ – B³ = (A – B)(A² + AB + B²)]
= (3a + 5b – 3a + 5b) [(9a² + 30ab + 25b²) + (9a² – 25b²) + (9a² – 30ab + 25b²)]
= 10b (9a² + 9a² + 9a² + 30ab – 30ab + 25b² – 25b² + 25b²)
= 10b (27a² + 25b²)
= 270a²b + 250b³

iii. (a + b)³ – a³ – b³
= a³ + 3a²b + 3ab² + b³ – a³ – b³
= 3a²b + 3ab²

iv. p³ – (p + 1)³
= p³ – (p³ + 3p² + 3p + 1) …[∵ (a + b)³ = a³ + 3a²b + 3ab² + b³]
= p³ – p³ – 3p² – 3p – 1
= – 3p² – 3p – 1

v. (3xy – 2ab)³ – (3xy + 2ab)³
Here, A = 3xy – 2ab and B = 3xy + 2ab
∴ (3xy – 2ab)³ – (3xy + 2ab)³
= [(3xy – 2ab) – (3xy + 2ab)] [(3xy – 2ab)² + (3xy – 2ab) (3xy + 2ab) + (3xy + 2ab)²]
…[∵ A³ – B³ = (A – B) (A² + AB + B²)]
= (3xy – 2ab – 3xy – 2ab) [(9x²y² – 12xyab + 4a²b²) + (9x²y² – 4a²b²) + (9x²y² + 12xyab + 4a²b²)]
= (- 4ab) (9x²y² + 9x²y² + 9x²y² – 12xyab + 12xyab + 4a²b² – 4a²b² + 4a²b²)
= (- 4ab) (27 xy² + 4a²b²)
= -108x²y²ab – 16a³b³

Maharashtra Board Class 8 Maths Solutions

• Factorisation of Algebraic Expressions Practice Set 6.1 Class 8 Maths Solutions
• Factorisation of Algebraic Expressions Practice Set 6.2 Class 8 Maths Solutions
• Factorisation of Algebraic Expressions Practice Set 6.3 Class 8 Maths Solutions
• Factorisation of Algebraic Expressions Practice Set 6.4 Class 8 Maths Solutions

Factorisation of Algebraic Expressions Class 8 Maths Chapter 6 Practice Set 6.2 Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 6.2 8th Std Maths Answers Solutions Chapter 6 Factorisation of Algebraic Expressions.

Std 8 Maths Practice Set 6.2 Chapter 6 Solutions Answers

Question 1.
Factorise:
i. x³ + 64y³
ii. 125p³ + q³
iii. 125k³ + 27m³
iv. 2l³ + 432m³
v. 24a³ + 81b³
vi. \(y^{3}+\frac{1}{8 y^{3}}\)
vii. \(\mathrm{a}^{3}+\frac{8}{\mathrm{a}^{3}}\)
viii. \(1+\frac{\mathrm{q}^{3}}{125}\)
Solution:
i. x³ + 64y³
= x³ + (4y)³
Here, a = x and b = 4y
∴ x³ + 64y³ = (x + 4y) [x² – x(4y) + (4y)²]
….[∵ a³ + b³ = (a + b)(a² – ab + b²)]
= (x + 4y)(x² – 4xy + 16y²)

ii. 125p³ + q³
= (5p)³ + q³
Here, a = 5p and b = q
∴ 125p³ + q³ = (5p + q)[(5p)² – (5p)(q) + q²]
…[∵ a³ + b³ = (a + b)(a² – ab + b²)]
= (5p + q)(25p² – 5pq + q²)

iii. 125k³ + 27m³
= (5k)³ + (3m)³
Here, a = 5k and b = 3m
∴ 125k³ + 27m³
= (5k + 3m) [(5k)² – (5k)(3m) + (3m)²]
…[∵ a³ + b³ = (a + b)(a² – ab + b²)]
= (5k + 3m)(25k² – 15km + 9m²)

iv. 2l³ + 432m³
= 2 (l³ + 216m³)
… [Taking out the common factor 2]
= 2[l³ + (6m)³]
Here, a = l and b = 6m
2l³ + 432m³ = 2 {(l + 6m)[l² – l(6m) + (6m)²]}
…[∵ a³ + b³ = (a + b)(a² – ab + b²)]
= 2(l + 6m)(l² – 6lm + 36m²)

v. 24a³ + 81b³
…[Taking out the common factor 3]
= 3 [(2a)³ + (3b)³]
Here, A = 2a and B = 3b
∴ 24a³ + 81b³
= 3 {(2a + 3b) [(2a)² – (2a)(3b) + (3b)²]}
…[∵ A³ + B³ = (A + B) (A² – AB + B²)]
= 3(2a + 3b)(4a² – 6ab + 9b²)

vi. \(y^{3}+\frac{1}{8 y^{3}}\)

vii. \(\mathrm{a}^{3}+\frac{8}{\mathrm{a}^{3}}\)

viii. \(1+\frac{\mathrm{q}^{3}}{125}\)

Maharashtra Board Class 8 Maths Solutions

• Factorisation of Algebraic Expressions Practice Set 6.1 Class 8 Maths Solutions
• Factorisation of Algebraic Expressions Practice Set 6.2 Class 8 Maths Solutions
• Factorisation of Algebraic Expressions Practice Set 6.3 Class 8 Maths Solutions
• Factorisation of Algebraic Expressions Practice Set 6.4 Class 8 Maths Solutions