Class 8

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 11.1 8th Std Maths Answers Solutions Chapter 11 Statistics.

Statistics Class 8 Maths Chapter 11 Practice Set 11.1 Solutions Maharashtra Board

Std 8 Maths Practice Set 11.1 Chapter 11 Solutions Answers

Question 1.
The following table shows the number of saplings planted by 30 students. Fill in the boxes and find the average number of saplings planted by each student.

No. of saplings (Scores) xi	No. of students (frequency) fi	fi × xi
1	4	4
2	6	__
3	12	__
4	8	__
	N = __	∑ fi × xi = __

Solution:

No. of saplings (Scores) xi	No. of students (frequency) fi	fi × xi
1	4	4
2	6	12
3	12	36
4	8	32
	N = __	∑ fi × xi = 84

Question 2.
The following table shows the electricity (in units) used by 25 families of Eklara village in a month of May. Complete the table and answer the following questions.

Electricity used (Units) xi	No. of families (frequency) fi	fi × xi
30	4
45	6
60	12
75	8
90	3
	N = __	∑ fi × xi =

i. How many families use 45 units electricity?
ii. State the score, the frequency of which is 5.
iii. Find N, and ∑ f_i × x_i .
iv. Find the mean of electricity used by each family in the month of May.
Solution:

Electricity used (Units) xi	No. of families (frequency) fi	fi × xi
30	7	210
45	2	90
60	8	480
75	5	375
90	3	270
	N = 25	∑ fi × xi = 1425

i. 2 families used 45 units of electricity.
ii. The score for which the frequency is 5 is 75
iii. N = 25 and ∑ f_i × x_i = 1425
iv.

The mean of electricity used by each.

Question 3.
The number of members in the 40 families in Bhilar are as follows:
1, 6, 5, 4, 3, 2, 7, 2, 3, 4, 5, 6, 4, 6, 2, 3, 2, 1, 4, 5, 6, 7, 3, 4, 5, 2, 4, 3, 2, 3, 5, 5, 4, 6, 2,3, 5, 6, 4, 2. Prepare a frequency table and And the mean of members of 40 families.
Solution:


∴ The mean of the members of 40 families is 3.9.

Question 4.
The number of Science and Mathematics projects submitted by Model high school, Nandpur in last 20 years at the state level science exhibition is:
2, 3 ,4, 1, 2, 3, 1, 5, 4, 2, 3, 1, 3, 5, 4, 3, 2, 2, 3, 2. Prepare a frequency table and find the mean of the data.
Solution:


∴ The mean of the given data is 2.75.

Maharashtra Board Class 8 Maths Chapter 11 Statistics Practice Set 11.1 Intext Questions and Activities

Question 1.
The number of pages of a book Ninad read for five consecutive days were 60, 50, 54, 46, 50. Find the average number of pages he read everyday. (Textbook pg. no. 67)
Solution:
\(\frac{60+[50]+[54]+[46]+50}{[5]}=\frac{260}{[5]}=[52]\)
∴ Average number of pages read daily is 52

Std 8 Maths Digest

• Practice Set 11.1 Class 8 Answers
• Practice Set 11.2 Class 8 Answers
• Practice Set 11.3 Class 8 Answers
• Practice Set 12.1 Class 8 Answers
• Practice Set 12.2 Class 8 Answers
• Practice Set 13.1 Class 8 Answers
• Practice Set 13.2 Class 8 Answers
• Practice Set 14.1 Class 8 Answers
• Practice Set 14.2 Class 8 Answers

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.

Quadrilateral: Constructions and Types Class 8 Maths Chapter 8 Practice Set 8.1 Solutions Maharashtra Board

Std 8 Maths Practice Set 8.1 Chapter 8 Solutions Answers

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.

Std 8 Maths Digest

• Practice Set 8.1 Class 8 Answers
• Practice Set 8.2 Class 8 Answers
• Practice Set 8.3 Class 8 Answers
• Practice Set 9.1 Class 8 Answers
• Practice Set 9.2 Class 8 Answers
• Practice Set 10.1 Class 8 Answers
• Practice Set 10.2 Class 8 Answers

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 8.2 8th Std Maths Answers Solutions Chapter 8 Quadrilateral: Constructions and Types.

Quadrilateral: Constructions and Types Class 8 Maths Chapter 8 Practice Set 8.2 Solutions Maharashtra Board

Std 8 Maths Practice Set 8.2 Chapter 8 Solutions Answers

Question 1.
Draw a rectangle ABCD such that l(AB) = 6.0 cm and l(BC) = 4.5 cm.
Solution:

Question 2.
Draw a square WXYZ with side 5.2 cm.
Solution:

Question 3.
Draw a rhombus KLMN such that its side is 4 cm and m∠K = 75°.
Solution:

Question 4.
If diagonal of a rectangle is 26 cm and one side is 24 cm, find the other side.
Solution:

Let ₹ABCD be the rectangle.
l(BC) = 24cm, l(AC) = 26cm
In ∆ABC,
m∠ABC = 90° …[Angle of a rectangle]
∴[l(AC)]² = [l(AB)]2 + [l(BC)]²
…[Pythagoras theorem]
∴ (26 )² = [l(AB)]² + (24)²
∴(26)² – (24)² = [l(AB)]²
∴(26 + 24) (26 – 24) = [l(AB)]²
…[∵ a² – b² = (a + b)(a – b)]
∴50 x 2 = [l(AB)]²
∴100 = [l(AB)]²
i.e. [l(AB)]² = 100
∴l(AB) = √100
…[Taking square root of both sides]
∴l(AB) =10 cm
∴The length of the other side is 10 cm.

Question 5.
Lengths of diagonals of a rhombus ABCD are 16 cm and 12 cm. Find the side and perimeter of the rhombus.
Solution:
In rhombus ABCD,

l(AC) = 16 cm and l(BD) = 12 cm.
Let the diagonals of rhombus ABCD intersect at point O.
l(AO) = \(\frac { 1 }{ 2 }\) l(AC)
…[Diagonals of a rhombus bisect each other]
∴l(AO) = \(\frac { 1 }{ 2 }\) × 16
∴l(AO) = 8 cm
Also, l(DO) = \(\frac { 1 }{ 2 }\) l(BD)
…[Diagonals of a rhombus bisect each other]
∴l(DO) = \(\frac { 1 }{ 2 }\) × 12
∴l(DO) = 6 cm
In ∆DOA,
m∠DOA = 90°
..[Diagonals of a rhombus are perpendicular to each other]
[l(AD)]² = [l(AO)]² + [l(DO)]²
…[Pythagoras theorem]

= (8)² + (6)²
= 64 + 36
∴[l(AD)]² = 100
∴l(AD) = √100
… [Taking square root of both sides]
∴l(AD) = 10 cm
∴l(AB) = l(BC) = l(CD) = l(AD) = 10 cm
…[Sides of a rhombus are congruent]
Perimeter of rhombus ABCD
= l(AB) + l(BC) + l(CD) + l(AD)
= 10+10+10+10
= 40 cm
∴The side and perimeter of the rhombus are 10 cm and 40 cm respectively.

Question 6.
Find the length of diagonal of a square with side 8 cm.
Solution:
Let ₹XYWZ be the square of side 8cm.

seg XW is a diagonal.
In ∆ XYW,
m∠XYW = 90°
… [Angle of a square]
∴ [l(XW)]² = [l(XY)]² + [l(YW)]²
…[Pythagoras theorem]
= (8)² + (8)²
= 64 + 64
∴ [l(XW)]² = 128
∴ l(XW) = √128
…[Taking square root of both sides]
= √64 × 2
= 8 √2 cm
∴ The length of the diagonal of the square is 8 √2 cm.

Question 7.
Measure of one angle of a rhombus is 50°, find the measures of remaining three angles.
Solution:
Let ₹ABCD be the rhombus.

m∠A = 50°
m∠C = m∠A
….[Opposite angles of a rhombus are congruent]
∴ m∠C = 50°
Also, m∠D = m∠B …(i)
….[Opposite angles of a rhombus are congruent]
In ₹ABCD,
m∠A + m∠B + m∠C + m∠D = 360°
….[Sum of the measures of the angles of a quadrilateral is 360°]
∴ 50° + m∠B + 50° + m∠D = 360°
∴ m∠B + m∠D + 100° = 360°
∴ m∠B + m∠D = 360° – 100°
∴ m∠B + m∠B = 260° …[From (i)]
∴ 2m∠B = 260°
∴ m∠B = \(\frac { 260 }{ 2 }\)
∴ m∠B = 130°
∴ m∠D = m∠B = 130° …[From (i)]
∴ The measures of the remaining angles of the rhombus are 130°, 50° and 130°.

Maharashtra Board Class 8 Maths Chapter 8 Quadrilateral: Constructions and Types Practice Set 8.2 Intext Questions and Activities

Question 1.
Construct a rectangle PQRS by taking two convenient adjacent sides. Name the point of intersection of diagonals as T. Using divider and ruler, measure the following lengths.
i. lengths of opposite sides, seg QR and seg PS.
ii. lengths of seg PQ and seg SR.
iii. lengths of diagonals PR and QS.
iv. lengths of seg PT and seg TR, which are parts of the diagonal PR.
v. lengths of seg QT and seg TS, which are parts of the diagonal QS.
Observe the measures. Discuss about the measures obtained by your classmates. (Textbook pg. no. 44)
Solution:
Draw a rectangle PQRS such that, l(PQ) = 3 cm and l(QR) = 4 cm.

Steps of construction:
i. As shown in the rough figure, draw seg QR of length 4 cm.
ii. Placing the centre of the protractor at point Q, draw ray QW making an angle of 90° with seg QR.
iii. By taking a distance of 3 cm on the compass and placing it at point Q, draw an arc on ray QW. Name the point as P.
iv. Draw ray PV and ray RU making an angle of 90° with seg PQ and seg QR respectively.
v. Name the point of intersection of ray PV and ray RU as S.
₹PQRS is the required rectangle.

From the figure,
i. l(QR) = l(PS) = 4 cm
ii. l(PQ) = l(SR) = 3 cm
iii. l(PR) = l(QS) = 5 cm
iv. l(PT) = l(TR) = 2.5 cm
v. l(QT) = l(TS) = 2.5 cm

From the above measures, we can say that for any rectangle,
i. Opposite sides are congruent.
ii. Diagonals are congruent.
iii. Diagonals bisect each other.

Question 2.
Draw a square by taking convenient length of side. Name the point of intersection of its diagonals as E. Using the apparatus in a compass box, measure the following lengths.
i. lengths of diagonal AC and diagonal BD.
ii. lengths of two parts of each diagonal made by point E.
iii. all the angles made at the point E.
iv. parts of each angle of the square made by each diagonal, (e.g. ∠ADB and ∠CDB).
Observe the measures. Also observe the measures obtained by your classmates and discuss about them. (Textbook pg. no. 44)
Solution:
Draw a square ABCD such that its side is 5cm

Steps of construction:
i. As shown in the rough figure, draw seg BC of length 5 cm.
ii. Placing the centre of the protractor at point B, draw ray BP making an angle of 90° with seg BC.
iii. By taking a distance of 5 cm on the compass and placing it at point B, draw an arc on ray BP. Name the point as A.
iv. Placing the centre of the protractor at point C, draw ray CQ making an angle of 90° with seg BC.
v. By taking a distance of 5 cm on the compass and placing it at point C, draw an arc on ray CQ. Name the point as D.
vi. Draw seg AD.
₹ABCD is the required square.

From the figure,
i. l(AC) = l(BD) ≅ 7cm
ii. l(AE) = l(EC) ≅ 3.5cm,
l(BE) = l(ED) ≅ 3.5cm
iii. m∠AED = m∠BEC = m∠CED = m∠BEA = 90°
iv. Angles made by diagonal AC:
m∠BAC = m∠DAC = 45°
m∠BCA = m∠DCA = 45°
Angles made by diagonal BD:
m∠ABD = m∠CBD = 45°
m∠ADB = m∠CDB = 45°

From the above measures, we can say that for any square,
i. Diagonals are congruent.
ii. Diagonals bisect each other.
iii. Diagonals are perpendicular to each other.
iv. Diagonals bisect the opposite angles.

Question 3.
Draw a rhombus EFGH by taking convenient length of side and convenient measure of an angle.
Draw its diagonals and name their point of Intersection as M.
i. Measure the opposite angles of the quadrilateral and angles at the point M.
ii. Measure the two parts of every angle made by the diagonal.
iii. Measure the lengths of both diagonals. Measure the two parts of diagonals made by point M.
Observe the measures. Also observe the measures obtained by your classmates and discuss about them. (Textbook pg. no. 45)
Solution:
Draw a rhombus EFGH such that its side is 5 cm and m∠F = 60°.

Steps of construction:
i. As shown in the rough figure, draw seg FG of length 5 cm.
ii. Placing the centre of the protractor at point F, draw ray FX making an angle 60° with seg FG.
iii. By taking a distance of 5 cm on the compass and placing it at point F, draw an arc on ray FX. Name the point as E.
iv. By taking a distance of 5 cm on the compass and placing it at point E and point G, draw arcs. Name the point of intersection of arcs as H. ₹EFGH is the required rhombus.

From the figure,
i. Opposite angles:
m∠EFG = m∠GHE = 60°,
m∠FEH = m∠HGF = 120°
Angles at the point M:
m∠EMF = m∠FMG = m∠GMH = m∠HME = 90°

ii. Angles made by diagonal FH:
m∠EFH = m∠GFH = 30° m∠EHF = m∠GHF = 30°
Angles made by diagonal EG:
m∠FEG = m∠HEG = 60° m∠FGE = m∠HGE = 60°

iii. l(FH) ≈ 8.6 cm
l(EG) = 5 cm
l(FM) = l(HM) ≈ 4.3 cm
l(EM) = l(GM) ≈ 2.5 cm

From the above measures, we can say that for any rhombus,
i. Opposite angles are congruent.
ii. Diagonals bisect the opposite angles.
iii. Diagonals bisect each other and they are perpendicular to each other.

Std 8 Maths Digest

• Practice Set 8.1 Class 8 Answers
• Practice Set 8.2 Class 8 Answers
• Practice Set 8.3 Class 8 Answers
• Practice Set 9.1 Class 8 Answers
• Practice Set 9.2 Class 8 Answers
• Practice Set 10.1 Class 8 Answers
• Practice Set 10.2 Class 8 Answers

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

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

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.

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

Std 8 Maths Digest

• Practice Set 6.1 Class 8 Answers
• Practice Set 6.2 Class 8 Answers
• Practice Set 6.3 Class 8 Answers
• Practice Set 6.4 Class 8 Answers
• Practice Set 7.1 Class 8 Answers
• Practice Set 7.2 Class 8 Answers
• Practice Set 7.3 Class 8 Answers

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.

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

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

Std 8 Maths Digest

• Practice Set 6.1 Class 8 Answers
• Practice Set 6.2 Class 8 Answers
• Practice Set 6.3 Class 8 Answers
• Practice Set 6.4 Class 8 Answers
• Practice Set 7.1 Class 8 Answers
• Practice Set 7.2 Class 8 Answers
• Practice Set 7.3 Class 8 Answers

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.

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

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

Std 8 Maths Digest

• Practice Set 6.1 Class 8 Answers
• Practice Set 6.2 Class 8 Answers
• Practice Set 6.3 Class 8 Answers
• Practice Set 6.4 Class 8 Answers
• Practice Set 7.1 Class 8 Answers
• Practice Set 7.2 Class 8 Answers
• Practice Set 7.3 Class 8 Answers

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.

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

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

Std 8 Maths Digest

• Practice Set 6.1 Class 8 Answers
• Practice Set 6.2 Class 8 Answers
• Practice Set 6.3 Class 8 Answers
• Practice Set 6.4 Class 8 Answers
• Practice Set 7.1 Class 8 Answers
• Practice Set 7.2 Class 8 Answers
• Practice Set 7.3 Class 8 Answers

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.

Factorisation of Algebraic Expressions Class 8 Maths Chapter 6 Practice Set 6.1 Solutions Maharashtra Board

Std 8 Maths Practice Set 6.1 Chapter 6 Solutions Answers

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)

Std 8 Maths Digest

• Practice Set 6.1 Class 8 Answers
• Practice Set 6.2 Class 8 Answers
• Practice Set 6.3 Class 8 Answers
• Practice Set 6.4 Class 8 Answers
• Practice Set 7.1 Class 8 Answers
• Practice Set 7.2 Class 8 Answers
• Practice Set 7.3 Class 8 Answers

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 5.4 8th Std Maths Answers Solutions Chapter 5 Expansion Formulae.

Expansion Formulae Class 8 Maths Chapter 5 Practice Set 5.4 Solutions Maharashtra Board

Std 8 Maths Practice Set 5.4 Chapter 5 Solutions Answers

Question 1.
Expand:
i. (2p + q + 5)²
ii. (m + 2n + 3r)²
iii. (3x + 4y – 5p)²
iv. (7m – 3n – 4k)²
Solution:
i. (2p + q + 5)² = (2p)² + (q)² + (5)² + 2(2p) (q) + 2(q) (5) + 2(2p) (5)
… [(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac]
= 4p² + q² + 25 + 4pq + 10q + 20p

ii. (m + 2n + 3r)² = (m)² + (2n)² + (3r)² + 2(m) (2n) + 2(2n) (3r) + 2(m) (3r)
… [(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac]
= m² + 4n² + 9r² + 4mn + 12nr + 6mr

iii. (3x + 4y – 5p)² = (3x)² + (4y)² + (- 5p)² + 2(3x) (4y) + 2(4y) (- 5p) + 2(3x) (- 5p)
… [(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac]
= 9x + 16y² + 25p² + 24xy – 40py – 30px

iv. (7m – 3n – 4k)² = (7m)² + (- 3n)² + (- 4k)² + 2(7m) (- 3n) + 2 (- 3n) (- 4k) + 2 (7m) (- 4k)
… [(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac]
= 49m² + 9n² + 16k² – 42mn + 24nk – 56km

Question 2.
Simplify:
i. (x – 2y + 3)² + (x + 2y – 3)²
ii. (3k – 4r – 2m)² – (3k + 4r – 2m)²
iii. (7a – 6b + 5c)² + (7a + 6b – 5c)²
Solution:
i. (x – 2y + 3)² + (x + 2y – 3)²
= [(x)² + (- 2y)² + (3)² + 2 (x) (- 2y) + 2 (- 2y) (3) + 2 (x) (3)] + [(x)² + (2y)² + (- 3)² + 2 (x) (2y) + 2 (2y) (- 3) + 2 (x) (- 3)]
… [(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac]
= x² + 4y² + 9 – 4xy – 12y + 6x + x² + 4y² + 9 + 4xy – 12y – 6x
= x + x² + 4y² + 4y² + 9 + 9 – 4xy + 4xy – 12y – 12y + 6x – 6x
= 2x² + 8y² + 18 – 24y

ii. (3k – 4r – 2m)² – (3k + 4r – 2m)²
= [(3k)² + (- 4r)² + (- 2m)² + 2 (3k) (- 4r) + 2 (- 4r) (- 2m) + 2 (3k) (- 2m)] – [(3k)² + (4r)² + (- 2m)² + 2 (3k) (4r) + 2 (4r) (- 2m) + 2 (3k) (- 2m)]
… [(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac]
= (9k² + 16r² + 4m² – 24kr + 16rm – 12km) – (9k² + 16r² + 4m² + 24kr – 16rm – 12km)
= 9k² + 16r² + 4m² – 24kr + 16rm – 12km – 9k² – 16r² – 4m² – 24kr + 16rm + 12km
= 9k² – 9k² + 16r² – 16r² + 4m² – 4m² – 24kr – 24kr + 16rm + 16rm – 12km + 12km
= 32rm – 48kr

iii. (7a – 6b + 5c)² + (7a + 6b – 5c)²
= [(7a)² + (- 6b)² + (5c)² + 2(7a) (-6b) + 2(-6b) (5c) + 2(7a) (5c)] + [(7a)² + (6b)² + (- 5c)² + 2 (7a) (6b) + 2 (6b) (- 5c) + 2 (7a) (- 5c)]
… [(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac]
= 49a² + 36b² + 25c² – 84ab – 60bc + 70ac + 49a² + 36b² + 25c² + 84ab – 60bc – 70ac
= 49a² + 49a² + 36b² + 36b² + 25c² + 25c² – 84ab + 84ab – 60bc – 60bc + 70ac – 70ac
= 98a² + 72b² + 50c² – 120bc

Maharashtra Board Class 8 Maths Chapter 5 Expansion Formulae Practice Set 5.4 Intext Questions and Activities

Question 1.
Fill in the boxes with appropriate terms in the steps of expansion. (Textbook pg. no. 27)
(2p + 3m + 4n)²
= (2p)² + (3m)² + __ + 2 × 2p × 3m + 2 × __ × 4n + 2 × 2p × __
= __ + 9m² + __ + 12pm + __ + __
Solution:
(2p + 3m + 4n)²
= (2p)² + (3m)² + (4n)² + 2 x 2p x 3m + 2 x 3m x 4n + 2 x 2p x 4n
= 4p² + 9m² + 16n² + 12pm + 24mn + 16pn

Std 8 Maths Digest

• Practice Set 3.1 Class 8 Answers
• Practice Set 3.2 Class 8 Answers
• Practice Set 3.3 Class 8 Answers
• Practice Set 4.1 Class 8 Answers
• Practice Set 5.1 Class 8 Answers
• Practice Set 5.2 Class 8 Answers
• Practice Set 5.3 Class 8 Answers
• Practice Set 5.4 Class 8 Answers

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 5.3 8th Std Maths Answers Solutions Chapter 5 Expansion Formulae.

Expansion Formulae Class 8 Maths Chapter 5 Practice Set 5.3 Solutions Maharashtra Board

Std 8 Maths Practice Set 5.3 Chapter 5 Solutions Answers

Question 1.
Expand:
i. (2m – 5)³
ii. (4 – p)³
iii. (7x – 9y)³
iv. (58)³
v. (198)³
vi. \(\left(2 p-\frac{1}{2 p}\right)^{3}\)
vii. \(\left(1-\frac{1}{a}\right)^{3}\)
viii. \(\left(\frac{x}{3}-\frac{3}{x}\right)^{3}\)
Solution:
i. Here, a = 2m and b = 5
(2m – 5)³
= (2m)³ – 3(2m)² (5) + 3(2m) (5)² – (5)³
… [(a – b)³ = a³ – 3a²b + 3ab² – b³]
= 8m³ – 3(4m²)(5) + 3(2m)(25) – 125
= 8m³ – 60m² + 150m – 125

ii. Here, a = 4 and b = p
(4 – p)³ = (4)³ – 3(4)²(p) + 3(4)(p)² – (p)³
… [(a – b)³ = a³ – 3a²b + 3ab² – b³]
= 64 – 3(16)(p) + 3(4)(p²) – p³
= 64 – 48p + 12p² – p³

iii. Here, a = 7x and b = 9y
(7x – 9y)³
= (7x)³ – 3(7x)² (9y) + 3 (7x)(9y)² – (9y)³
…[(a – b)³ = a³ – 3a²b + 3ab² – b³]
= 343x³ – 3(49x²)(9y) + 3(7x)(81y²) – 729y³
= 343x³ – 1323x²y + 1701xy² – 729y³

iv. (58)³ = (60 – 2)³
Here, a = 60 and b = 2
(58)³ = (60)³ – 3(60)²(2) + 3(60)(2)² – (2)³
… [(a – b)³ = a³ – 3a²b + 3ab² – b³]
= 216000 – 3(3600)(2) + 3(60)(4) – 8
= 216000 – 21600 + 720 – 8
=195112

v. (198)³ = (200 – 2)³
Here, a = 200 and b = 2
(198)³ = (200)³ – 3(200)²(2) + 3(200)(2)² – (2)³
… [(a – b)³ = a³ – 3a²b + 3ab² – b³]
= 8000000 – 3(40000)(2) + 3(200)(4) – 8
= 8000000 – 240000 + 2400 – 8
= 7762392

vi. Here, a = 2p and b = \(\frac { 1 }{ 2p }\)

vii. Here, A = 1 and B = \(\frac { 1 }{ a }\)

viii. Here, a = \(\frac { x }{ 3 }\) and b = \(\frac { 3 }{ x }\)

Question 2.
Simplify:
i. (2a + b)³ – (2a – b)³
ii. (3r – 2k)³ + (3r + 2k)³
iii. (4a – 3)³ – (4a + 3)³
iv. (5x – 7y)³ + (5x + 7y)³
Solution:
i. (2a + b)³ – (2a – b)³
= [(2a)³ + 3(2a)²(b) + 3 (2a)(b)² + (b)³] – [(2a)³ – 3(2a)²(b) + 3 (2a)(b)² – (b)³]
… [(a + b)³ = a³ + 3a²b + 3ab² + b³, (a – b)³ = a³ – 3a²b + 3ab² – b³]
= (8a³ + 12a²b + 6ab² + b³) – (8a³ – 12a²b + 6ab² – b³)
= 8a³ + 12a²b + 6ab² + b³ – 8a³ + 12a²b – 6ab² + b³
= 8a³ – 8a³ + 12a²b + 12a²b + 6ab² – 6ab² + b³ + b³
= 24a²b + 2b³

ii. (3r – 2k)³ + (3r + 2k)³
= [(3r)³ – 3(3r)²(2k) + 3(3r)(2k)² – (2k)³] + [(3r)³ + 3(3r)²(2k) + 3(3r)(2k)² + (2k)³]
… [(a – b)³ = a³ – 3a²b + 3ab² – b³, (a + b)³ = a³ + 3a²b + 3ab² + b³]
= (27r³ – 54r²k + 36rk² – 8k³) + (27r³ + 54r²k + 36rk² + 8k³)
= 27r³ – 54r²k + 36rk² – 8k³ + 27r³ + 54r²k + 36rk² + 8k³
= 27r³ + 27r³ – 54r²k + 54r²k + 36rk² + 36rk² – 8k³ + 8k³
= 54r³ + 72rk²

iii. (4a – 3)³ – (4a + 3)³
= [(4a)³ – 3(4a)² (3) + 3(4a)(3)² – (3)³] – [(4a)³ + 3(4a)²(3) + 3(4a)(3)² + (3)³]
… [(a – b)³ = a³ – 3a²b + 3ab² – b³, (a + b)³ = a³ + 3a²b + 3ab² + b³]
= (64a³ – 144a² + 108a – 27) – (64a³ + 144a² + 108a + 27)
= 64a³ – 144a² + 108a – 27 – 64a³ -144a² – 108a – 27
= 64a³ – 64a³ – 144a² – 144a² + 108a – 108a – 27 – 27
= -288a² – 54

iv. (5x – 7y)³ + (5x + 7y)³
= [(5x)³ – 3(5x)²(7y) + 3(5x)(7y)² – (7y)³] + [(5x)³ + 3(5x)² (7y) + 3(5x) (7y)² +(7y)³]
… [(a – b)³ = a³ – 3a²b + 3ab² – b³, (a + b)³ = a³ + 3a²b + 3ab² + b³]
= (125x³ – 525x²y + 735xy² – 343y³) + (125x³ + 525x²y + 735xy² + 343y³)
= 125x³ – 525x²y + 735xy² – 343y³ + 125x³ + 525x²y + 735xy² + 343y³
= 125x³ + 125x³ – 525x²y + 525x²y + 735xy² + 735xy² – 343y³ + 343y³
= 250x³ + 1470xy²

Maharashtra Board Class 8 Maths Chapter 5 Expansion Formulae Practice Set 5.3 Intext Questions and Activities

Question 1.
Make two cubes of side a and of side b each. Make six parallelopipeds; three of them measuring a × a × b and the remaining three measuring b × b × a. Arrange all these solid figures properly and make a cube of side (a + b). (Textbook pg. no. 25)
Solution:
(a + b)³ = a³ + 3a²b + 3ab² + b³
= a × a × a + 3 × a × a × b + 3 × a × b × b + b × b × b

Std 8 Maths Digest

• Practice Set 3.1 Class 8 Answers
• Practice Set 3.2 Class 8 Answers
• Practice Set 3.3 Class 8 Answers
• Practice Set 4.1 Class 8 Answers
• Practice Set 5.1 Class 8 Answers
• Practice Set 5.2 Class 8 Answers
• Practice Set 5.3 Class 8 Answers
• Practice Set 5.4 Class 8 Answers