Class 8

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 10 Answers Solutions Chapter 3 HCF and LCM.

HCF and LCM Class 7 Practice Set 10 Answers Solutions Chapter 3

Question 1.
Which number is neither a prime number nor a composite number?
Solution:
1

Question 2.
Which of the following are pairs of co-primes?
i. 8,14
ii. 4,5
iii. 17,19
iv. 27,15
Solution:
i. Factors of 8: 1, 2, 4, 8
Factors of 14: 1, 2, 7, 14
∴ Common factors of 8 and 14: 1,2
∴ 8 and 14 are not a pair of co-prime numbers.

ii. Factors of 4: 1, 4
Factors of 5: 1, 5
∴ Common factors of 4 and 5: 1
∴ 4 and 5 are a pair of co-prime numbers.

iii. Factors of 17: 1, 17
Factors of 19: 1, 19
∴ Common factors of 17 and 19: 1
∴ 17 and 19 are a pair of co-prime numbers.

iv. Factors of 27: 1, 3, 9, 27
Factors of 15: 1, 3, 5, 15 .
∴ Common factors of 27 and 15 : 1,3
∴ 27 and 15 are not a pair of co-prime numbers.

Question 3.
List the prime numbers from 25 to 100 and say how many they are.
Solution:
29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.
There are 16 prime numbers from 25 to 100.

Question 4.
Write all the twin prime numbers from 51 to 100.
Solution:

1. 59 and 61
2. 71 and 73

Question 5.
Write 5 pairs of twin prime numbers from 1 to 50.
Solution:

1. 3,5
2. 5,7
3. 11,13
4. 17,19
5. 29,31
6. 41,43

Question 6.
Which are the even prime numbers?
Solution:
2

Maharashtra Board Class 7 Maths Chapter 3 HCF and LCM Practice Set 10 Intext Questions and Activities

Question 1.
Answer the following questions. (Textbook pg. no. 15)
i. Which is the smallest prime number?
ii. List the prime numbers from 1 to 50. How many are they?
iii. Identify the prime numbers in the list below.
17, 15 ,4, 3, 1, 2, 12, 23, 27, 35, 41, 43, 58, 51, 72, 79, 91, 97.
Solution:
i. 2 is the smallest prime number.
ii. There are 15 prime numbers from 1 to 50.
They are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47.
iii. [17], 15 ,4, [3], 1, [2], 12, [23], 27, 35, [41], [43], 58, 51, 72, [79], 91, [97].

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 10.1 8th Std Maths Answers Solutions Chapter 10 Division of Polynomials.

Practice Set 10.1 8th Std Maths Answers Chapter 10 Division of Polynomials

Question 1.
Divide and write the quotient and the remainder.
i. 21m² ÷ 7m
ii. 40a³ ÷ (-10a)
iii. (- 48p⁴) ÷ (- 9p²)
iv. 40m⁵ ÷ 30m³
v. (5x³ – 3x²) ÷ x²
vi. (8p³ – 4p²) ÷ 2p²
vii. (2y³ + 4y² + 3 ) ÷ 2y²
viii. (21x⁴ – 14x² + 7x) ÷ 7x³
ix. (6x⁵ – 4x⁴ + 8x³ + 2x²) ÷ 2x²
x. (25m⁴ – 15m³ + 10m + 8) ÷ 5m³
Solution:
i. 21m² ÷ 7m

∴ Quotient = 3m
Remainder = 0

ii. 40a³ ÷ (-10a)

∴ Quotient = -4a²
Remainder = 0

iii. (- 48p⁴) ÷ (- 9p²)

∴ Quotient = \(\frac { 16 }{ 3 }\) p²
Remainder = 0

iv. 40m⁵ ÷ 30m³

∴ Quotient = \(\frac { 4 }{ 3 }\) m²
Remainder = 0

v. (5x³ – 3x²) ÷ x²

∴ Quotient = 5x – 3
Remainder = 0

vi. (8p³ – 4p²) ÷ 2p²

∴ Quotient = 4p – 2
Remainder = 0

vii. (2y³ + 4y² + 3 ) ÷ 2y²

∴ Quotient = y + 2
Remainder = 3

viii. (21x⁴ – 14x² + 7x) ÷ 7x³

∴ Quotient = 3x
Remainder = -14x² + 7x

ix. (6x⁵ – 4x⁴ + 8x³ + 2x²) ÷ 2x²

∴ Quotient = 3x³ – 2x² + 4x + 1
Remainder = 0

x. (25m⁴ – 15m³ + 10m + 8) ÷ 5m³

∴ Quotient = 5m – 3
Remainder = 10m + 8

Maharashtra Board Class 8 Maths Chapter 10 Division of Polynomials Practice Set 10.1 Intext Questions and Activities

Question 1.
Fill in the blanks in the following examples. (Textbook pg. no. 61)

1. 2a + 3a = __
2. 7b – 4b = __
3. 3p × p² = __
4. 5m² × 3m² = __
5. (2x + 5y) × \(\frac { 3 }{ x }\) = __
6. (3x² + 4y) × (2x + 3y) = __

Solution:

1. 2a + 3a = 5a
2. 7b – 4b = 3b
3. 3p × p² = 3p³
4. 5m² × 3m² = 15m⁴
5. (2x + 5y) × \(\frac { 3 }{ x }\) = \(6+\frac { 15y }{ x }\)
6. (3x² + 4y) × (2x + 3y) = 6x³ + 9x²y + 8xy + 12y²

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 8.1 8th Std Maths Answers Solutions Chapter 8 Quadrilateral: Constructions and Types.

Practice Set 8.1 8th Std Maths Answers Chapter 8 Quadrilateral: Constructions and Types

Construct the following quadrilaterals of given measures.

Question 1.
In ∆MORE, l(MO) = 5.8 cm, l(OR) = 4.4 cm, m∠M = 58°, m∠O = 105°, m∠R = 90°.
Solution:

Question 2.
Construct ∆DEFG such that l(DE) = 4.5 cm, l(EF) = 6.5 cm, l(DG) = 5.5 cm, l(DF) = 7.2 cm, l(EG) = 7.8 cm.
Solution:

Question 3.
In ∆ABCD, l(AB) = 6.4 cm, l(BC) = 4.8 cm, m∠A = 70°, m∠B = 50°, m∠C = 140°.
Solution:

Question 4.
Construct ₹LMNO such that
l(LM) = l(LO) = 6 cm,
l(ON) = l(NM) = 4.5 cm, l(OM) = 7.5 cm.
Solution:

Maharashtra Board Class 8 Maths Chapter 8 Quadrilateral: Constructions and Types Practice Set 8.1 Intext Questions and Activities

Question 1.
Construction of a triangle:
Construct the triangles with given measures. (Textbook pg. no. 41)
i. ∆ABC: l(AB) = 5 cm, l(BC) = 5.5, l(AC) = 6 cm.
Solution:

Steps of construction:
Step 1 : As shown in the rough figure, draw seg BC of length 5.5 cm as the base.
Step 2 : By taking a distance of 5 cm on the compass and placing the metal tip of the compass on point B, draw an arc on one side of BC.
Step 3 : By taking a distance 6 cm on the compass and placing the metal tip of the t compass on point C and draw an arc ’ such that it intersects the previous arc. Name the point as A.
Step 4 : Draw segments AB and AC to get the triangle. ∆ABC is the required triangle.

ii. ∆DEF: m∠D = 35°, m∠F = 100°, l(DF) = 4.8 cm.
Solution:

Steps of construction:
Step 1 : As shown in the rough figure, draw seg DF of length 4.8 cm as the base.
Step 2 : Placing the centre of the protractor at point D, mark point P such that m∠PDF = 35°.
Step 3 : Placing the centre of the protractor at point F, mark point Q such that m∠QFD = 100°.
Step 4 : Draw ray DP and ray FQ. Name their point of intersection as E.
∆DEF is required triangle.

iii. ∆MNP: l(MP) = 6.2 cm, l(NP) = 4.5 cm, m∠P = 75°.
Solution:

Steps of construction:
Step 1 : As shown in the rough figure, draw seg PN of length 4.5 cm as the base.
Step 2 : Placing the centre of the protractor at point P, mark point Q such that m∠QPN = 75°.
Step 3 : By taking a distance of 6.2 cm on the compass and placing the metal tip at point P, draw an arc on ray PQ. Name the point as M.
Step 4 : Draw seg MN to get the triangle. ∆MNP is the required triangle.

iv. ∆XYZ: m∠Y = 90°, l(XY) = 4.2 cm, l(XZ) = 7 cm.
Solution:

Steps of construction:
Step 1 : As shown in the rough figure, draw seg XY of 4.2 cm as the base.
Step 2 : Placing the centre of the protractor at point Y, mark point Q such that m∠QYX = 90°.
Step 3 : By taking a distance of 7 cm on the compass and placing the metal tip on point X, draw an arc on ray YQ. Name the point as Z.
Step 4 : Draw seg XZ to get the triangle. ∆XYZ is the required triangle.

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

Practice Set 7.1 8th Std Maths Answers Chapter 7 Variation

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.

Number of apples (x)	1	4	__	12	__
Cost of apples (y)	8	32	56	__	160

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

Number of apples (x)	1	4	7	12	20
Cost of apples (y)	8	32	56	96	160

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.

m	3	5	6.5	__	1.25
n	12	20	__	28	__

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

m	3	5	6.5	7	1.25
n	12	20	26	28	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)

Number of notebooks (x)	12	3	9	24	50	1
Cost (In Rupees) (y)	240	__	__	__	__	20

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

Number of notebooks (x)	12	3	9	24	50	1
Cost (In Rupees) (y)	240	60	180	480	1000	20

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.

Practice Set 6.4 8th Std Maths Answers Chapter 6 Factorisation of Algebraic Expressions

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

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.

Practice Set 6.3 8th Std Maths Answers Chapter 6 Factorisation of Algebraic Expressions

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

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.

Practice Set 6.2 8th Std Maths Answers Chapter 6 Factorisation of Algebraic Expressions

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

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 6.1 8th Std Maths Answers Solutions Chapter 6 Factorisation of Algebraic Expressions.

Practice Set 6.1 8th Std Maths Answers Chapter 6 Factorisation of Algebraic Expressions

Question 1.
Factorize:
i. x² + 9x + 18
ii. x² – 10x + 9
iii. y² + 24y + 144
iv. 5y² + 5y – 10
v. p² – 2p – 35
vi. p² – 7p – 44
vii. m² – 23m + 120
viii. m² – 25m + 100
ix. 3x² + 14x + 15
x. 2x² + x – 45
xi. 20x² – 26x + 8
xii. 44x² – x – 3
Solution:
i. x² + 9x + 18
= x² + 6x + 3x + 18
= x (x + 6) + 3(x + 6)
= (x + 6) (x + 3)

ii. x² – 10x + 9
= x² – 9x – x + 9
= x (x – 9) – 1(x – 9)
= (x – 9)(x – 1)

iii. y² + 24y + 144
= y² + 12y + 12y + 144
= y(y + 12) + 12(y + 12)
= (y + 12)(y + 12)

iv. 5y² + 5y – 10
= 5(y² + y – 2)
… [Taking out the common factor 5]
= 5(y² + 2y – y – 2)
= 5[y(y + 2) – 1(y + 2)]
= 5 (p + 2)(y- 1)

v. p² – 2p – 35
= p² – 7p + 5p – 35
= p(p – 7) + 5(p – 7)
= (p – 7)(p + 5)

vi. p² – 7p – 44
= p² – 11p + 4p – 44
= p(p – 11) + 4(p – 11)
= (p – 11)(p + 4)

vii. m² – 23m + 120
= m² – 15m – 8m + 120
= m (m – 15) – 8 (m – 15)
= (m – 15) (m – 8)

viii. m² – 25m + 100
= m² – 20m – 5m + 100
= m(m – 20) – 5(m – 20)
= (m – 20) (m – 5)

ix. 3x² + 14x + 15 3 × 15 = 45
= 3x² + 9x + 5x + 15
= 3x(x + 3) + 5(x + 3)
= (x + 3) (3x + 5)

x. 2x² + x – 45 2 × (- 45) = -90
= 2x² + 10x – 9x – 45
= 2x(x + 5) – 9 (x + 5)
= (x + 5) (2x – 9)

xi. 20x² – 26x + 8
= 2(10x² – 13x + 4) 10 × 4 = 40
… [Taking out the common factor 2]
= 2(10x² – 8x – 5x + 4)
= 2[2x(5x – 4) – 1(5x – 4)]
= 2 (5x – 4) (2x – 1)

xii. 44x² – x – 3 44 × (-3) = -132
= 44x² – 12x + 11x – 3
= 4x(11x – 3) + 1(11x – 3)
= (11x – 3) (4x + 1)

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 4.1 8th Std Maths Answers Solutions Chapter 4 Altitudes and Medians of a Triangle.

Practice Set 4.1 8th Std Maths Answers Chapter 4 Altitudes and Medians of a Triangle

Question 1.
In ∆LMN, ___ is an altitude and __ is a median, (write the names of appropriate segments.)

Solution:
In ∆LMN, seg LX is an altitude and seg LY is a median.

Question 2.
Draw an acute angled ∆PQR. Draw all of its altitudes. Name the point of concurrence as ‘O’.
Solution:

Question 3.
Draw an obtuse angled ∆STV. Draw its medians and show the centroid.
Solution:

Question 4.
Draw an obtuse angled ∆LMN. Draw its altitudes and denote the ortho centre by ‘O’.
Solution:

Question 5.
Draw a right angled ∆XYZ. Draw its medians and show their point of concurrence by G.
Solution:

Question 6.
Draw an isosceles triangle. Draw all of its medians and altitudes. Write your observation about their points of concurrence.
Solution:

The point of concurrence of medians i.e. G and that of altitudes i.e. O lie on the same line PS which is the perpendicular bisector of seg QR.

Question 7.
Fill in the blanks.
Point G is the centroid of ∆ABC.
i. If l(RG) = 2.5, then l(GC) = ___
ii. If l(BG) = 6, then l(BQ) = ____
iii. If l(AP) = 6, then l(AG) = ___ and l(GP) = ___.

Solution:
The centroid of a triangle divides each median in the ratio 2:1.
i. Point G is the centroid and seg CR is the median.
∴ \(\frac{l(\mathrm{GC})}{l(\mathrm{RG})}=\frac{2}{1}\)
∴ \(\frac{l(\mathrm{GC})}{2.5}=\frac{2}{1}\) ……[∵ l(RG) = 2.5]
∴ l(GC) × 1 = 2 × 2.5
∴ l(GC) = 5

ii. Point G is the centroid and seg BQ is the median.
∴ \(\frac{l(\mathrm{BG})}{l(\mathrm{GQ})}=\frac{2}{1}\)
∴ \(\frac{6}{l(\mathrm{GQ})}=\frac{2}{1}\) …..[∵ l(BG) = 6]
∴ 6 × 1 = 2 × l(GQ)
∴ \(\frac { 6 }{ 2 }\) = l(GQ)
∴ 3 = l(GQ)
i.e. l(GQ) = 3
Now, l (BQ) = l(BG) + l(GQ)
∴ l(BQ) = 6 + 3
∴ l(BQ) = 9

iii. Point G is the centroid and seg AP is the median.
∴ \(\frac{l(\mathrm{AG})}{l(\mathrm{GP})}=\frac{2}{1}\)
∴ l(AG) = 2 l(GP) …..(i)
Now, l(AP) = l(AG) + l(GP) … (ii)
∴ l(AP) = 2l(GP) + l(GP) … [From (i)]
∴ l(AP) = 3l(GP)
∴ 6 = 3l(GP) ..[∵ l(AP) = 6]
∴ \(\frac { 6 }{ 3 }\) = l(GP)
∴ 2 = l(GP)
i.e. l(GP) = 2
l(AP) = l(AG) + l(GP) …[from (ii)]
∴ 6 = l(AG) + 2
∴ l(AG) = 6 – 2
∴ l(AG) = 4

Maharashtra Board Class 8 Maths Chapter 4 Altitudes and Medians of a Triangle Practice Set 4.1 Intext Questions and Activities

Question 1.
Draw a line. Take a point outside the line. Draw a perpendicular from the point to the line with the help of a set-square (Textbook pg. no, 19)
Solution:
Step 1: Draw a line l and a point P lying outside it.

Step 2: By placing a set-square on line l, draw a perpendicular to the line from point P.

Question 2.
Draw an acute angled ∆ABC and all its altitudes. Observe the location of the orthocentre. (Textbook pg. no. 20)
Solution:

Point O is the orthocentre.
Orthocentre lies in the interior of ∆ABC.

Question 3.
Draw a right angled triangle and draw all its altitudes. Write the point of concurrence. (Textbook: pg, no. 20)
Solution:

Point Q is the orthocentre.
The point of concurrence of altitudes PQ, QR and QS is Q.

Question 4.
i. Draw an obtuse angled triangle and all its altitudes.
ii. Do they intersect each other?
Draw the lines containing the altitudes. Observe that these lines are concurrent. (Textbook pg. no. 20)
Solution:
i.

Point O is the orthocentre.

ii. Yes, all the altitudes intersect at point O in the exterior of ∆PQR.

Question 5.
Draw three different triangles; a right angled triangle, an obtuse angled triangle and an acute angled triangle. Draw the medians of the triangles. Note that the centroid of each of them is in the interior of the triangle. (Textbook pg. no. 21)
Solution:
i. Right angled triangle:

ii. Obtuse angled triangle:

iii. Acute angled triangle:

Question 6.
Draw a sufficiently large ∆ABC.
Draw medians; seg AR, seg BQ and seg CP of ∆ABC.
Name the point of concurrence as G.
Measure the lengths of segments from the figure and fill in the boxes in the following table.

l(AG) =	l(GR) =	l(AG): l(GR) =
l(BG) =	l(GQ) =	l(BG): l(GQ) =
l(CG) =	l(GP) =	l(CG): l(GP) =

Observe that all of these ratios are nearly 2 : 1 (Textbook pg. no. 21)
Solution:

l(AG) = 2.9	l(GR) = 1.4	l(AG): l(GR) = \(\frac{2.8}{1.4}=\frac{2}{1}\)
l(BG) = 2.4	l(GQ) = 1.2	l(BG): l(GQ) = \(\frac{2.4}{1.2}=\frac{2}{1}\)
l(CG) = 2.8	l(GP) = 1.4	l(CG): l(GP) = \(\frac{2.8}{1.4}=\frac{2}{1}\)

Question 7.
As shown in the given figure, a student drew ∆ABC using five parallel lines of a notebook. Then he found the centroid G of the triangle. How will you decide whether the location of G he found, is correct. (Textbook pg. no. 21)

Solution:
Draw seg AP ⊥ seg PE and seg EQ ⊥ seg QC.

Side AP || side EQ and AC is their transversal.
∴ ∠PAE ≅ ∠QEC …(i) [Corresponding angles]
In ∆ APE and ∆ EQC,
∠PAE ≅ ∠QEC …[From (i)]
∠APE ≅ ∠EQC
… [Each angle is of measure 90°]
side PE ≅ side QC
…. [Perpendicular distance between parallel lines]
∴ ∆ APE ≅ ∆ EQC … [By AAS test]
∴ AE = EC
… [Corresponding sides of congruent triangles]
∴ E is the midpoint of AC.
∴ seg BE is the median.
Similarly, seg CF is the median.
Since, the medians of a triangle are concurrent.
∴ G is the centroid of ∆ABC.

Question 8.
Draw an equilateral triangle. Find its circumcentre (C), incentre (I), centroid (G) and orthocentre (O). Write your observation. (Textbook pg. no. 22)
Solution:

From the figure, circumcentre (C), incentre (I), centroid (G) and orthocentre (O) of an equilateral triangle are the same.

Question 9.
Draw an isosceles triangle. Locate its centroid, orthocentre, circumcentre and incentre. Verify that they are collinear. (Textbook pg. no. 22)
Solution:

From the figure, centroid (G), orthocentre (O), circumcentre (C) and incentre (I) of an isosceles triangle lie on the same line AD.
∴ they are collinear.

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 3.1 8th Std Maths Answers Solutions Chapter 3 Indices and Cube Root.

Practice Set 3.1 8th Std Maths Answers Chapter 3 Indices and Cube Root

Question 1.
Express the following numbers in index form.
i. Fifth root of 13
ii. Sixth root of 9
iii. Square root of 256
iv. Cube root of 17
v. Eighth root of 100
vi. Seventh root of 30
Solution:
i. \((13)^{\frac{1}{5}}\)
ii. \((9)^{\frac{1}{6}}\)
iii. \((256)^{\frac{1}{2}}\)
iv. \((17)^{\frac{1}{3}}\)
v. \((100)^{\frac{1}{8}}\)
vi. \((30)^{\frac{1}{7}}\)

Question 2.
Write in the form ‘nth root of a’ in each of the following numbers.
i. \((81)^{\frac{1}{4}}\)
ii. \((49)^{\frac{1}{2}}\)
iii. \((15)^{\frac{1}{5}}\)
iv. \((512)^{\frac{1}{9}}\)
v. \((100)^{\frac{1}{19}}\)
vi. \((6)^{\frac{1}{7}}\)
Solution:
i. Fourth root of 81.
ii. Square root of 49.
iii. Fifth root of 15.
iv. Ninth root of 512.
v. Nineteenth root of 100.
vi. Seventh root of 6.

Maharashtra Board Class 8 Maths Chapter 3 Indices and Cube Root Practice Set 3.1 Intext Questions and Activities

Question 1.
Using laws of indices, write proper numbers in the following boxes. (Textbook pg, no. 14)
i. \({ 3 }^{ 5 }\times { 3 }^{ 2 }={ 3 }^{ \left( \right) }\)
ii. \({ 3 }^{ 7 }\div { 3 }^{ 9 }={ 3 }^{ \left( \right) }\)
iii. \(({ 3 }^{ 4 })^{ 5 }={ 3 }^{ \left( \right) }\)
iv. \(5^{ -3 }=\frac { 1 }{ { 5 }^{ \left( \right) } }\)
v. \(5^{ 0 }=\left( \right) \)
vi. \(5^{ 1 }=\left( \right) \)
vii. \((5\times 7)^{ 2 }={ 5 }^{ \left( \right) }\times { 7 }^{ \left( \right) }\)
viii. \({ \left( \frac { 5 }{ 7 } \right) }^{ 3 }=\frac { { \left( \right) }^{ 3 } }{ { \left( \right) }^{ 3 } } \)
ix. \({ \left( \frac { 5 }{ 7 } \right) }^{ -3 }={ \left( \frac { \left( \right) }{ \left( \right) } \right) }^{ 3 }\)
Solution:
i. \({ 3 }^{ 5 }\times { 3 }^{ 2 }={ 3 }^{ 7 }\)
ii. \({ 3 }^{ 7 }\div { 3 }^{ 9 }={ 3 }^{ -2 }\)
iii. \(({ 3 }^{ 4 })^{ 5 }={ 3 }^{ 20 }\)
iv. \(5^{ -3 }=\frac { 1 }{ { 5 }^{ 3 } } \)
v. \(5^{ 0 }=1\)
vi. \(5^{ 1 }=5\)
vii. \((5\times 7)^{ 2 }={ 5 }^{ 2 }\times { 7 }^{ 2 }\)
viii. \({ \left( \frac { 5 }{ 7 } \right) }^{ 3 }=\frac { { 5 }^{ 3 } }{ { 7 }^{ 3 } } \)
ix. \({ \left( \frac { 5 }{ 7 } \right) }^{ -3 }={ \left( \frac { 7 }{ 5 } \right) }^{ 3 }\)