Class 8

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 15.3 8th Std Maths Answers Solutions Chapter 15 Area.

Area Class 8 Maths Chapter 15 Practice Set 15.3 Solutions Maharashtra Board

Std 8 Maths Practice Set 15.3 Chapter 15 Solutions Answers

Question 1.
In the given figure, ☐ABCD is a trapezium, side AB || side DC, l(AB) = 13 cm, l(DC) = 9 cm, l(AD) = 8 cm, find the area ☐ABCD.

Solution:
☐ABCD is a trapezium, side AB || side DC,
l(AB) = 13 cm, l(DC) = 9 cm, l(AD) = 8 cm,
Area of a trapezium = \(\frac { 1 }{ 2 }\) x sum of lengths of parallel sides x height
∴ A (☐ABCD) = \(\frac { 1 }{ 2 }\) x [l(AB) + l(DC)] x l(AD)
= \(\frac { 1 }{ 2 }\) x (13 + 9) x 8
= \(\frac { 1 }{ 2 }\) x 22 x 8
= 11 x 8
= 88 sq.cm
∴ The area of ☐ABCD is 88 sq. cm.
[Note: The question is modified.]

Question 2.
Length of the two parallel sides of a trapezium are 8.5 cm and 11.5 cm respectively and its height is 4.2 cm, find its area.
Solution:
Length of the two parallel sides of a trapezium are 8.5 cm and 11.5 cm and its height is 4.2 cm.
Area of a trapezium
= \(\frac { 1 }{ 2 }\) x sum of lengths of parallel sides x height
= \(\frac { 1 }{ 2 }\) x (8.5 + 11.5) x 4.2
= \(\frac { 1 }{ 2 }\) x 20 x 4.2
= 10 x 4.2
= 42 sq. cm
∴ The area of the trapezium is 42 sq. cm.

Question 3.
☐PQRS is an isosceles trapezium. l(PQ) = 7 cm, seg PM ⊥ seg SR, l(SM) = 3 cm. Distance between two parallel sides is 4 cm, find the area of ☐PQRS.

Solution:
☐PQRS is an isosceles trapezium.
l(PQ) = 7 cm, seg PM ⊥ seg SR,
l(SM) = 3 cm, l(PM) = 4cm
Draw seg QN ⊥ seg SR.
In ☐PMNQ,
seg PQ || seg MN
∠PMN = ∠QNM = 90°
∴ ☐PMNQ is a rectangle.

Opposite sides of a rectangle are congruent.
∴ l(PM) = l(QN) = 4 cm and
l(PQ) = l(MN) = 7 cm
In ∆PMS, m∠PMS = 90°
∴ [l(PS)]² = [l(PM)]² + [l(SM)]² … [Pythagoras theorem]
∴ [l(PS)]² = (4)² + (3)²
∴ [l(PS)]² = 16 + 9 = 25
∴ l(PS) = √25 = 5 cm
…[Taking square root of both sides]
☐PQRS is an isosceles trapezium.
∴ l(PS) = l(QR) = 5 cm
In ∆QNR, m ∠QNR = 90°
∴ [l(QR)]² = [l(QN)]² + [l(NR)]²
… [Pythagoras theorem]
∴ (5)² = (4)² + [l(NR)]²
∴ 25 = 16 + [l(NR)]²
∴ [l(NR)]² = 25 – 16 = 9
∴ l(NR) = √9 = 3 cm
…[Taking square root of both sides]
l(SR) = l(SM) + l(MN) + l(NR)
= 3 + 7 + 3
= 13 cm
Area of a trapezium
= \(\frac { 1 }{ 2 }\) x sum of lengths of parallel sides x height
∴ A(☐PQRS) = \(\frac { 1 }{ 2 }\) x [l(PQ) + l(SR)] x l(PM)
= \(\frac { 1 }{ 2 }\) x (7+ 13) x 4
= \(\frac { 1 }{ 2 }\) x 20 x 4
= 40 sq.cm
∴ The area of ☐PQRS is 40 sq. cm.

Std 8 Maths Digest

• Practice Set 15.1 Class 8 Answers
• Practice Set 15.2 Class 8 Answers
• Practice Set 15.3 Class 8 Answers
• Practice Set 15.4 Class 8 Answers
• Practice Set 15.5 Class 8 Answers
• Practice Set 15.6 Class 8 Answers
• Practice Set 16.1 Class 8 Answers
• Practice Set 16.2 Class 8 Answers
• Practice Set 16.3 Class 8 Answers
• Practice Set 17.1 Class 8 Answers
• Practice Set 17.2 Class 8 Answers

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 15.2 8th Std Maths Answers Solutions Chapter 15 Area.

Area Class 8 Maths Chapter 15 Practice Set 15.2 Solutions Maharashtra Board

Std 8 Maths Practice Set 15.2 Chapter 15 Solutions Answers

Question 1.
Lengths of the diagonals of a rhombus are 15 cm and 24 cm, find its area.
Solution:
Lengths of the diagonals of a rhombus are 15 cm and 24 cm.
Area of a rhombus
= \(\frac { 1 }{ 2 }\) × product of lengths of diagonals
= \(\frac { 1 }{ 2 }\) × 15 × 24
= 15 × 12
= 180 sq.cm
∴ The area of the rhombus is 180 sq. cm.

Question 2.
Lengths of the diagonals of a rhombus are 16.5 cm and 14.2 cm, find its area.
Solution:
Lengths of the diagonals of a rhombus are 16.5 cm and 14.2 cm.
Area of a rhombus
= \(\frac { 1 }{ 2 }\) × product of lengths of diagonals
= \(\frac { 1 }{ 2 }\) × 16.5 × 14.2
= 16.5 × 7.1
= 117.15 sq cm
∴ The area of the rhombus is 117.15 sq. cm.

Question 3.
If perimeter of a rhombus is 100 cm and length of one diagonal is 48 cm, what is the area of the quadrilateral?
Solution:
Let ₹ABCD be the rhombus. Diagonals AC and BD intersect at point E.

l(AC) = 48 cm …(i)
l(AE) = \(\frac { 1 }{ 2 }l(AC)\) …[Diagonals of a rhombus bisect each other]
= \(\frac { 1 }{ 2 }\) × 48 …[From (i)]
= 24 cm …(ii)
Perimeter of rhombus = 100 cm …[Given]
Perimeter of rhombus = 4 × side
∴ 100 = 4 × l(AD)
∴ l(AD) = \(\frac { 100 }{ 4 }\) = 25 cm …(iii)
In ∆ADE,
m∠AED = 90° …[Diagonals of a rhombus are perpendicular to each other]
∴ [l(AD)]² = [l(AE)]² + [l(DE)]² … [Pythagoras theorem]
∴ (25)² = (24)² + l(DE)² … [From (ii) and (iii)]
∴ 625 = 576 + l(DE)²
∴ l(DE)² = 625 – 576
∴ l(DE)² = 49
∴ l(DE) = √49
… [Taking square root of both sides]
l(DE) = 7 cm …(iv)
l(DE) = \(\frac { 1 }{ 2 } l(BD)\) ….[Diagonals of a rhombus bisect each other]
∴ 7 = \(\frac { 1 }{ 2 } l(BD)\) …[From (iv)]
∴ l(BD) = 7 × 2
= 14 cm …(v)
Area of a rhombus
= \(\frac { 1 }{ 2 }\) × product of lengths of diagonals
= \(\frac { 1 }{ 2 }\) × l(AC) × l(BD)
= \(\frac { 1 }{ 2 }\) × 48 × 14 … [From (i) and (v)]
= 48 × 7
= 336 sq.cm
∴ The area of the quadrilateral is 336 sq.cm.

Question 4.
If length of a diagonal of a rhombus is 30 cm and its area is 240 sq.cm, find its perimeter.
Solution:
Let ₹ABCD be the rhombus.
Diagonals AC and BD intersect at point E.
l(AC) = 30 cm …(i)
and A(₹ABCD) = 240 sq. cm .. .(ii)

Area of the rhombus = \(\frac { 1 }{ 2 }\) × product of lengths of diagonal
∴ 240 = \(\frac { 1 }{ 2 }\) × l(AC) x l(BD) …[From (ii)]
∴ 240 = \(\frac { 1 }{ 2 }\) × 30 × l(BD) …[From (i)]
∴ l(BD) = \(\frac { 240\times 2 }{ 30 }\)
∴ l(BD) = 8 × 2 = 16 cm …(iii)
Diagonals of a rhombus bisect each other.
∴ l(AE) = \(\frac { 1 }{ 2 }l(AC)\)
= \(\frac { 1 }{ 2 }\) × 30 … [From (i)]
= 15 cm …(iv)
and l(DE) = \(\frac { 1 }{ 2 }l(BD)\)
= \(\frac { 1 }{ 2 }\) × 16
= 8 cm
In ∆ADE,
m∠AED = 90°
…[Diagonals of a rhombus are perpendicular to each other]
∴[l(AD)]² = [l(AE)]² + [l(DE)]²
…[Pythagoras theorem]
∴l(AD)² = (15)² + (8)² … [From (iv) and (v)]
= 225 + 64
∴l(AD)² = 289
∴l(AD) = √289
…[Taking square root of both sides]
∴l(AD) = 17 cm
Perimeter of rhombus = 4 × side
= 4 × l(AD)
= 4 × 17
= 68 cm
∴The perimeter of the rhombus is 68 cm.

Std 8 Maths Digest

• Practice Set 15.1 Class 8 Answers
• Practice Set 15.2 Class 8 Answers
• Practice Set 15.3 Class 8 Answers
• Practice Set 15.4 Class 8 Answers
• Practice Set 15.5 Class 8 Answers
• Practice Set 15.6 Class 8 Answers
• Practice Set 16.1 Class 8 Answers
• Practice Set 16.2 Class 8 Answers
• Practice Set 16.3 Class 8 Answers
• Practice Set 17.1 Class 8 Answers
• Practice Set 17.2 Class 8 Answers

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 15.1 8th Std Maths Answers Solutions Chapter 15 Area.

Area Class 8 Maths Chapter 15 Practice Set 15.1 Solutions Maharashtra Board

Std 8 Maths Practice Set 15.1 Chapter 15 Solutions Answers

Question 1.
If base of a parallelogram is 18 cm and its height is 11 cm, find its area.
Solution:
Given, base = 18 cm, height = 11 cm
Area of a parallelogram = base × height
= 18 × 11
= 198 sq.cm
∴ Area of the parallelogram is 198 sq.cm.

Question 2.
If area of a parallelogram is 29.6 sq. cm and its base is 8 cm, find its height.
Solution:
Given, area of a parallelogram = 29.6 sq.cm,
base = 8 cm
Area of a parallelogram = base × height
∴ 29.6 = 8 × height
∴ height = \(\frac { 29.6 }{ 8 }\) = 3.7 cm
∴ Height of the parallelogram is 3.7 cm.

Question 3.
Area of a parallelogram is 83.2 sq.cm. If its height is 6.4 cm, find the length of its base.
Solution:
Given, area of a parallelogram = 83.2 sq.cm, height = 6.4 cm
Area of a parallelogram = base × height
∴ 83.2 = base × 6.4
∴ base = \(\frac { 83.2 }{ 6.4 }\) = 13 cm
∴ The length of the base of the parallelogram is 13 cm.

Maharashtra Board Class 8 Maths Chapter 15 Area Practice Set 15.1 Intext Questions and Activities

Question 1.
Draw a big enough parallelogram ABCD on a paper as shown in the figure.
Draw perpendicular AE on side BC.
Cut the right angled ∆AEB. Join it with the remaining part of ₹ABCD as shown in the figure.
The new figure formed is a rectangle.
The rectangle is formed from the parallelogram.
So, areas of both the figures are equal.
Base of parallelogram is one side (length) of the rectangle and its height is the other side (breadth) of the rectangle.
∴ Area of a parallelogram = base × height (Textbook pg. no.94)

Solution:
Draw a big enough parallelogram ABCD on a paper as shown in the figure.
Draw perpendicular AE on side BC.
Cut the right angled ∆AEB. Join it with the remaining part of ₹ABCD as shown in the figure.
The new figure formed is a rectangle.
The rectangle is formed from the parallelogram.
So, areas of both the figures are equal.
Base of parallelogram is one side (length) of the rectangle and its height is the other side (breadth) of the rectangle.
∴ Area of a parallelogram = Area of a rectangle = length × breadth = base × height

Std 8 Maths Digest

• Practice Set 15.1 Class 8 Answers
• Practice Set 15.2 Class 8 Answers
• Practice Set 15.3 Class 8 Answers
• Practice Set 15.4 Class 8 Answers
• Practice Set 15.5 Class 8 Answers
• Practice Set 15.6 Class 8 Answers
• Practice Set 16.1 Class 8 Answers
• Practice Set 16.2 Class 8 Answers
• Practice Set 16.3 Class 8 Answers
• Practice Set 17.1 Class 8 Answers
• Practice Set 17.2 Class 8 Answers

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 14.2 8th Std Maths Answers Solutions Chapter 14 Compound Interest.

Compound Interest Class 8 Maths Chapter 14 Practice Set 14.2 Solutions Maharashtra Board

Std 8 Maths Practice Set 14.2 Chapter 14 Solutions Answers

Compound Interest class 8 practice set 14.2 Question 1. On the construction work of a flyover bridge there were 320 workers initially. The number of workers were increased by 25% every year. Find the number of workers after 2 years.
Solution:
Here, P = Initial number of workers = 320
R = Increase in the number of workers per year = 25%
N = 2 years
A = Number of workers after 2 years

∴ The number of workers after 2 years would be 500.

Question 2.
A shepherd has 200 sheep with him. Find the number of sheeps with him after 3 years if the increase in number of sheeps is 8% every year.
Solution:
Here, P = Present number of sheeps = 200
R = Increase in number of sheeps per year = 8%
N = 3 years
A = Number of sheeps after 3 years

= \(\frac{0.32}{25} \times 27 \times 27 \times 27\)
= 0.0128 × 27 × 27 × 27
= 251.9424
= 252
∴ The number of sheeps with the shepherd after 2 years would be 252 (approx).

8th Class Math Practice Set 14.2 Question 3.
In a forest there are 40,000 trees. Find the expected number of trees after 3 years if the objective is to increase the number at the rate 5% per year.
Solution:
Here, P = Present number of trees in the forest = 40,000
R = Increase in the number of trees per year = 5%
N = 3 years
A = Number of trees after 3 years

= 5 × 21 × 21 × 21
= 5 × 9261
= 46,305
∴ The expected number of trees in the forest after 3 years is 46,305.

Std 8 Maths Practice Set 14.2 Question 4.
The cost price of a machine is Rs 2,50,000. If the rate of depreciation is 10% per year, find the depreciation in price of the machine after two years.
Solution:
Here, P = Cost price of machine = Rs 2,50,000
R = Rate of depreciation per year = 10%
N = 2 years
A = Depreciated price of the machine after 2 years

= 2,500 × 81
= Rs 2,02,500
Depreciation in price = Cost price (P) – Depreciated price (A)
= 2,50,000 – 2,02,500
= Rs 47,500
∴ The depreciation in price of the machine after 2 years would be Rs 47,500.

Question 5.
Find the compound interest if the amount of a certain principal after two years is Rs 4036.80 at the rate of 16 p.c.p.a.
Solution:
Here, A = Rs 4036.80, R = 16 p.c.p.a. and N = 2 years
i. \(\mathbf{A}=\mathbf{P}\left[1+\frac{\mathbf{R}}{100}\right]^{N}\)

ii. Interest = Amount (A) – Principal (P)
= 4036.80 – 3000
= Rs 1036.80
∴ The compound interest after 2 years would be Rs 1036.80.

Question 6.
A loan of Rs 15,000 was taken on compound interest. If the rate of compound interest is 12 p.c.p.a. find the amount to settle the loan after 3 years.
Solution:
Here, P = Rs 15,000, R = 12 p.c.p.a, and
N = 3 years

∴ The amount required to settle the loan after 3 years is Rs 21,073.92.

Practice Set 14.2 Class 8 Question 7.
A principal amounts to Rs 13,924 in 2 years by compound interest at 18 p.c.p.a. Find the principal.
Solution:
Here, A = Rs 13,924, R = 18 p.c.p.a., and N = 2 years

∴ P = 4 x 50 x 50
∴ P = Rs 10,000
∴ The principal is Rs 10,000.

Question 8.
The population of a suburb is 16,000. Find the rate of increase in the population if the population after two years is 17,640.
Solution:
Here, P = Population of a suburb = 16,000
N = 2 years
A = Increase in the population after 2 years = 17,640
R = Rate of increase in population


∴5 = R
i.e., R = 5%
∴The rate of increase in the population is 5 p.c.p.a.

Compound Interest Practice Set 14.2 Question 9.
In how many years Rs 700 will amount to Rs 847 at a compound interest rate of 10 p.c.p.a.
Solution:
Here, P = Rs 700, R = 10 p.c.p.a., A = Rs 847

∴Rs 700 will amount to Rs 847 in 2 years.

Practice Set 14.2 Question 10.
Find the difference between simple interest and compound interest on Rs 20,000 in 2 years at 8 p.c.p.a.
Solution:
Here, P = Rs 20,000, R = 8 p.c.p.a.,
N = 2 years
i. Simple interest (I)

Simple interest (I) = Rs 3200

ii. Compound Interest (I):


= 32 × 27 × 27
= Rs 23,328
Compound interest (I)
= Amount (A) – Principal (P)
= 23,328 – 20,000
= Rs 3328 ,..(ii)

iii. Difference
= Compound interest – Simple interest
= 3328 – 3200 … [Form (i) and (ii)]
= Rs 128
∴ The difference between compound interest and simple interest is Rs 128.
[Note: The question is modified as per the answer given in the textbook.]

Maharashtra Board Class 8 Maths Chapter 14 Compound Interest Practice Set 14.2 Intext Questions and Activities

8th Standard Maths Practice Set 14.2 Question 1.
Visit the bank nearer to your house and get the information regarding the different schemes and rates of interests. Make a chart and display in your class. (Textbook pg. no. 90)
Solution:
(Students should attempt this activity at their own.)

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 13.2 8th Std Maths Answers Solutions Chapter 13 Congruence of Triangles.

Congruence of Triangles Class 8 Maths Chapter 13 Practice Set 13.2 Solutions Maharashtra Board

Std 8 Maths Practice Set 13.2 Chapter 13 Solutions Answers

Congruence of Triangles Class 8th Practice Set 13.2 Question 1.
In each pair of triangles given below, parts shown by identical marks are congruent. State the test and the one-to-one correspondence of vertices by which triangles in each pair are congruent. Also state the remaining congruent parts.


Solution:
i. In ∆MST and ∆TBM,
∴ side MS ≅ side TB … [Given]
m∠MST = m∠TBM = 90° … [Given]
hypotenuse MT ≅ hypotenuse MT
…[Common side]
∴ ∆MST ≅ ∆TBM …[by hypotenuse-side test]
∴ side ST ≅ side BM …[Corresponding sides of congruent triangles]
∠SMT ≅ ∠BTM …[Corresponding sides of congruent triangles]
∠STM ≅ ∠BMT …[Corresponding sides of congruent triangles]

ii. In ∆PRQ and ∆TRS,
side PR ≅ side TR … [Given]
∠PRQ ≅ ∠TRS …[Vertically opposite angles]
side RQ ≅ side RS … [Given]
∴ ∆PRQ ≅ ∆TRS …[by SAS test]
∴ side PQ ≅ side TS …[Corresponding sides of congruent triangles]
∠RPQ ≅ ∠RTS …[Corresponding sides of congruent triangles]
∠PQR ≅ ∠TSR …[Corresponding sides of congruent triangles]

iii. In ∆DCH and ∆DCF,
∠DCH ≅ ∠DCF …[Given]
∠DHC ≅ ∠DFC …[Given]
side DC ≅ side DC …[Common side]
∴ ∆DCH ≅ ∆DCF …[by AAS test]
∴ side HC ≅ side FC …[Corresponding sides of congruent triangles]
side DH ≅ side DF…[Corresponding sides of congruent triangles]
∠HDC ≅ ∠FDC ….[Corresponding sides of congruent triangles]

Congruence of Triangles Practice Set 13.2 Question 2.
In the given figure, seg AD ≅ seg EC. Which additional information is needed to show that ∆ABD and ∆EBC will be congruent by AAS test?

Solution:
In ∆ABD and ∆CBE,
∴ seg AD ≅ seg CE …[Given]
∠ABD ≅ ∠CBE …[Vertically opposite angles]
∴ The necessary condition for the two triangles to be congruent by AAS test is
∠ADB ≅ ∠CEB, or
∠DAB ≅ ∠ECB

Maharashtra Board Class 8 Maths Chapter 13 Congruence of Triangles Practice Set 13.2 Intext Questions and Activities

Practice Set 13.2 Class 8 Question 1.
Draw ∆ABC and ∆LMN such that two pairs of their sides and the angles included by them are congruent.
Draw ∆ABC and ∆LMN, l(AB) = l(LM), l(BC) = l(MN), m∠ABC = m∠LMN.
Copy ∆ABC on a tracing paper. Place the paper on ∆LMN in such a way that point A coincides with point L, side AB overlaps side LM. What do you notice?(Textbook pg. no. 83)

Solution:
We notice that ∆ABC ≅ ∆LMN.

Congruence of Triangles Class 8 Solutions Question 2.
Draw ∆PQR and ∆XYZ such that l(PQ) = l(X Y), l(Q R) = l(YZ), l(RP) = l(ZX). Copy ∆PQR on a tracing paper. Place it on ∆XYZ observing the correspondence P ↔ X, Q ↔ Y, R ↔ Z. What do you notice? (Textbook pg. no. 84)

Solution:
We notice that ∆PQR ≅ ∆XYZ.

Congruence of Triangles Class 8 Question 3.
Draw ∆XYZ and ∆DEF such that, l(XZ) = l(DF), ∠X ≅ ∠D and ∠Z ≅ ∠F.
Copy ∆XYZ on a tracing paper and place it over ∆DEF. What do you notice?(Textbook pg. no. 84)

Solution:
We notice that ∆XYZ ≅ ∆DEF in the correspondence X ↔ D, Y ↔ E, Z ↔ F.

Question 4.
Draw two right angled triangles such that a side and the hypotenuse of one is congruent with the corresponding parts of the other. Copy one triangle on tracing paper and place it over the other. What do you notice? (Textbook pg. no. 84)

Solution:
We notice that the two triangles are congruent.
(Students should draw figures and verify the answers.)

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 14.1 8th Std Maths Answers Solutions Chapter 14 Compound Interest.

Compound Interest Class 8 Maths Chapter 14 Practice Set 14.1 Solutions Maharashtra Board

Std 8 Maths Practice Set 14.1 Chapter 14 Solutions Answers

Practice Set 14.1 Class 8 Question 1.
Find the amount and the compound interest.

No	Principal (Rs)	Rate (p.c.p.a.)	Duration (years)
i.	2000	5	2
ii.	5000	8	3
iii.	4000	7.5	2

Solution:
i. Here P = Rs 2000, R = 5 p.c.p.a. and N = 2 years

= 5 × 441
∴ A = Rs 2205
I = Amount (A) – Principal (P)
= 2205 – 2000
= Rs 205
∴ The amount is Rs 2205 and the compound interest is Rs 205.

ii. Here, P = Rs 5000, R = 8 p.c.p.a. and N = 3 years

∴ A = Rs 6298.56
I = Amount (A) – Principal (P)
= 6298.56 – 5000
= Rs 1298.56
∴ The amount is Rs 6298.56 and the compound interest is Rs 1298.56.

iii. Here, P = Rs 4000, R = 7.5 p.c.p.a. and N = 2 years

∴A = Rs 4622.50
I = Amount (A) – Principal (P)
= 4622.50 – 4000
= Rs 622.50
∴The amount is Rs 4622.50 and the compound interest is Rs 622.50.

Compound Interest Practice Set 14.1 Question 2.
Sameerrao has taken a loan of Rs 12500 at the rate of 12 p.c.p.a. for 3 years. If the interest is compounded annually then how many rupees should he pay to clear his loan?
Solution:
Here, P = Rs 12,500, R = 12 p.c.p.a. and
N = 3 years

= 0.8 × 28 × 28 × 28
= Rs 17,561.60
Sameerrao should pay Rs 17,561.60 to clear his loan.

8th Standard Maths Practice Set 14.1 Question 3.
To start a business Shalaka has taken a loan of Rs 8000 at a rate of \(10\frac { 1 }{ 2 }\) p.c.p.a. After two years how much compound interest will she have to pay?
Solution:
Here, P = Rs 8000, N = 2 years and


I = Amount (A) – Principal (P)
= 9768.20 – 8000
= Rs 1768.20
∴ After two years Shalaka will have to pay Rs 1768.20 as compound interest.

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 13.1 8th Std Maths Answers Solutions Chapter 13 Congruence of Triangles.

Congruence of Triangles Class 8 Maths Chapter 13 Practice Set 13.1 Solutions Maharashtra Board

Std 8 Maths Practice Set 13.1 Chapter 13 Solutions Answers

Congruence of Triangles Practice Set 13.1 Question 1.
In each pair of triangles in the following figures, parts bearing identical marks are congruent. State the test and correspondence of vertices by which triangles in each pair are congruent.
i.

ii.

iii.

iv.

v.

Solution:
i.

The two triangles are congruent by SAS test in the correspondence XWZ ↔ YWZ.

ii.

The two triangles are congruent by hypotenuse-side test in the correspondence KJI ↔ LJI.

iii.

The two triangles are congruent by SSS test in the correspondence HEG ↔ FGE.

iv.
The two triangles are congruent by ASA test is the correspondence SMA ↔ OPT.

v.

The two triangles are congruent by ASA test or SAS test or SAA test in the correspondence MTN ↔ STN.

Maharashtra Board Class 8 Maths Chapter 13 Congruence of Triangles Practice Set 13.1 Intext Questions and Activities

Practice Set 13.1 Question 1.
Write answers to the following questions referring to the given figure.

1. Which is the angle opposite to the side DE?
2. Which is the side opposite to ∠E?
3. Which angle is included by side DE and side DF?
4. Which side is included by ∠E and ∠F?
5. State the angles adjacent to side DE. (Textbook pg, no. 81)

Solution:

1. ∠DFE i.e. ∠F is the angle opposite to side DE.
2. Side DF is the side opposite to ∠E.
3. ∠EDF i.e. ∠D is included by side DE and side DF.
4. Side EF is included by ∠E and ∠F.
5. ∠DEF and ∠EDF i.e. ∠E and ∠D are adjacent to side DE.

Congruence of Triangles Class 8th Practice Set 13.1 Question 2.
In the given figure, parts of triangles indicated by identical marks are congruent.
a. Identify the one-to-one correspondence of vertices in which the two triangles are congruent and write the congruence.
b. State with reason, whether the statement, ∆XYZ ≅ ∆STU is right or wrong. (Textbook pg. no. 82)

Solution:
a. From the figure,
S ↔ X, T ↔ Z, U ↔ Y i.e.,
STU ↔ XZY, or SUT ↔ XYZ, or
TUS ↔ ZYX, or TSU ↔ ZXY, or
UTS ↔ YZX, or UST ↔ YXZ

∴ ∆STU ≅ ∆XZY, or ∆SUT ≅ ∆XYZ, or
∆TUS ≅ ∆ZYX, or ∆TSU ≅ ∆ZXY, or
∆UTS ≅ ∆YZX, or ∆UST ≅ ∆YXZ

b. If ∆XYZ ≅ ∆STU, then
∠Y ≅ ∠T, ∠Z ≅ ∠U,
seg XY ≅ seg ST, seg XZ ≅ seg SU
∴ But, all the above statements are wrong. The statement AXYZ ≅ ASTU is wrong.

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 12.2 8th Std Maths Answers Solutions Chapter 12 Equations in One Variable.

Equations in One Variable Class 8 Maths Chapter 12 Practice Set 12.2 Solutions Maharashtra Board

Std 8 Maths Practice Set 12.2 Chapter 12 Solutions Answers

Equation In One Variable Practice Set 12.2 Question 1.
Mother is 25 years older than her son. Find son’s age, if after 8 years ratio of son’s age to mother’s age will be \(\frac { 4 }{ 9 }\).
Solution:
Let the son’s present age be x years.
∴ Mother’s present age = (x + 25) years
After 8 years,
Son’s age = (x + 8) years
Mother’s age = (x + 25 + 8) = (x + 33) years
Since, the ratio of the son’s age to mother’s age after 8 years is \(\frac { 4 }{ 9 }\).
∴ \(\frac{x+8}{x+33}=\frac{4}{9}\)
∴ 9 (x + 8) = 4 (x + 33)
∴ 9x + 72 = 4x + 132
∴ 9x – 4x = 132 – 72
∴ 5x = 60
∴ x = \(\frac { 60 }{ 5 }\)
∴ x = 12
∴ Son’s present age is 12 years.

8th Std Maths Practice Set 12.2 Question 2.
The denominator of a fraction is greater than its numerator by 12. If the numerator is decreased by 2 and the denominator is increased by 7, the new fraction is equivalent to \(\frac { 1 }{ 2 }\) . Find the fraction.
Solution:
Let the numerator of the fraction be x.
The denominator of a fraction is greater than its numerator by 12.
∴ Denominator of the fraction = (x + 12)
∴ The required fraction = \(\frac { x }{ x+12 }\)
For the new fraction,
numerator is decreased by 2.
∴ The new numerator = (x – 2)
Also, denominator is increased by 7.
∴ The new denominator = (x + 12) + 7
= (x + 19)
Since, the new fraction is equivalent to \(\frac { 1 }{ 2 }\).
∴ \(\frac{x-2}{x+19}=\frac{1}{2}\)
∴ 2(x – 2) = 1(x + 19)
∴ 2x – 4 = x + 19
∴ 2x – x = 19 + 4
∴ x = 23
∴ The required fraction = \(\frac{x}{x+12}=\frac{23}{23+12}=\frac{23}{35}\)
∴ The required fraction is \(\frac { 23 }{ 35 }\)

Practice Set 12.2 Class 8 Question 3.
The ratio of the weights of copper and zinc in brass is 13:7. Find the weight of zinc in a brass utensil weighing 700 gm.
Solution:
Let the weight of zinc in the brass utensil be x gm.
Since, the ratio of the weights of copper to zinc in brass is 13:7.

∴ Weight of copper in the brass utensil = \(\left(\frac{13}{7} x\right)\) gm
The weight of the brass utensil = 700 gm
∴ \(\frac { 13 }{ 7 }x+x=700\)
∴ \(\frac { 13 }{ 7 }x\) x × 7 + x × 7 = 700 × 7
∴ 13x + 7x = 4900
∴ 20x = 4900
∴ \(x=\frac { 4900 }{ 20 }\)
∴ x = 245
∴ The weight of zinc in the brass utensil is 245 gm.

Practice Set 12.2 8th Class Question 4.
Find three consecutive whole numbers whose sum is more than 45 but less than 54.
Solution:
Let the three consecutive whole numbers be (x – 1), x and (x + 1).
∴ Sum of the three numbers
= (x – 1) + x + (x + 1)
= 3x
Given that, the sum of the three numbers is greater than 45 and less than 54.
When the sum of the three numbers is 45,
3x = 45
∴ x = \(\frac { 45 }{ 3 }\)
∴ x = 15
When the sum of the three numbers is 54,
∴ 3x = 54
∴ x = \(\frac { 54 }{ 3 }\)
∴ x = 18
∴ the value of x is greater than 15 and less than 18.
∴ the value of x is either 16 or 17

Case I:
If the value of x is 16, then the three consecutive whole numbers are
(16 – 1), 16,(16 + 1)i.e., 15, 16, 17

Case II:
If the value of x is 17, then the three consecutive whole numbers are (17 – 1), 17, (17 + 1) i.e., 16, 17, 18.
∴ The three consecutive whole numbers are 15, 16, 17 or 16, 17, 18.

Practice Set 12.2 8th Standard Question 5.
In a two-digit number, digit at the ten’s place is twice the digit at unit’s place. If the number obtained by interchanging the digits is added to the original number, the sum is 66. Find the number.
Solution:
Let the digit at unit’s place be x.
The digit at the ten’s place is twice the digit at unit’s place.
∴ The digit at ten’s place = 2x

	Digit in units place	Digit in tens place	Number
Original Number	x	2x	(2x × 10) + x = 20x + x = 21x
New Number	2x	x	(x × 10) + 2x = 10x + 2x = 12x

Since, the sum of the original number and the new number is 66.
∴ 21x + 12x = 66
∴ 33x = 66
∴ x = \(\frac { 66 }{ 33 }\)
∴ x = 2
∴ Original number = 21x = 21 × 2 = 42
∴ the original number is 42.

8th Standard Maths Practice Set 12.2 Question 6.
Some tickets of Rs 200 and some of Rs 100, of a drama in a theatre were sold. The number of tickets of Rs 200 sold was 20 more than the number of tickets of Rs 100. The total amount received by the theatre by sale of tickets was Rs 37000. Find the number of Rs 100 tickets sold.
Solution:
Let the number of tickets sold of Rs 100 be x.
The number of tickets of Rs 200 sold was 20 more than the number of tickets of Rs 100.
∴ Number of tickets sold of Rs 200 = (x + 20)
∴ Total amount received by the theatre through the sale of tickets = 100 × x + 200 × (x + 20)
= 100x + 200x + 4000
= 300x + 4000
Since, the total amount received by the theatre through the sale of tickets = Rs 37000
∴ 300x + 4000 = 37000
∴ 300x = 37000 – 4000
∴ 300x = 33000
∴ \(x=\frac { 33000 }{ 300 }\)
∴ x = 110
∴ 110 tickets of Rs 100 were sold.

8th Maths Practice Set 12.2 Question 7.
Of the three consecutive natural numbers, five times the smallest number is 9 more than four times the greatest number, find the numbers.
Solution:
Let the three consecutive natural numbers be (x – 1), x and (x + 1).
Here, the smallest number is (x – 1) and the greatest number is (x + 1).
Since, five times the smallest number is 9 more than four times the greatest number.
∴ 5 × (x – 1) = [4 × (x + 1)] + 9
∴ 5x – 5 = 4x + 4 + 9
∴ 5x – 5 = 4x + 13
∴ 5x – 4x = 13 + 5
∴ x = 18 .
∴ the three numbers are (18 – 1), 18, (18 + 1)
i. e., 17, 18, 19
∴ The three consecutive natural numbers are 17,18 and 19.

Raju Sold A Bicycle to Amit at 8 Question 8.
Raju sold a bicycle to Amit at 8% profit. Amit repaired it spending Rs 54. Then he sold the bicycle to Nikhil for Rs 1134 with no loss and no profit. Find the cost price of the bicycle for which Raju purchased it.
Solution:
Let the cost price at which Raju purchased the bicycle be Rs x.
Since, Raju sold the bicycle at 8% profit to Amit.
∴ Selling price of bicycle for Raju = x + 8% of x

Since, Amit spent Rs 54 on repairing the bicycle and then sold it to Nikhil for Rs 1134, at no loss and no profit.
∴ Selling price of bicycle + repairing cost = Rs 1134

∴ The cost price of the bicycle at which Raju purchased it is Rs 1000.

Class 8 Maths Practice Set 12.2 Question 9.
A cricket player scored 180 runs in the first match and 257 runs in the second match. Find the number of runs he should score in the third match so that the average of runs in the three matches be 230.
Solution:
Let the number of runs required by the cricket player to score in the third match be x.
Number of runs scored by the player in first match = 180
Number of runs scored in second match = 257
∴ Total runs scored by the player = 180 + 257 + x = 437 + x
Average of runs in the three matches = \(\frac { 437+x }{ 3 }\)
Since, the average of runs should be 230.
\(\frac { 437+x }{ 3 }=230\)
∴ 437 + x = 230 × 3
∴ 437 + x = 690
∴ x = 690 – 437
∴ x = 253
∴ The cricket player should score 253 runs in the third match.

8th Class Math Practice Set 12.2 Question 10.
Sudhir’s present age is 5 more than three times the age of Viru. Anil’s age is half the age of Sudhir. If the ratio of the sum of Sudhir’s and Viru’s age to three times Anil’s age is 5:6, then find Viru’s age.
Solution:
Let Viru’s present age be x years.
Sudhir’s present age is 5 more than three times the age of Viru.
∴ Sudhir’s present age = (3x + 5) years
Anil’s age is half the age of Sudhir.
∴ Anil’s present age = \(\left(\frac{3 x+5}{2}\right)\) years
Since, the ratio of the sum of Sudhir’s and Viru’s age to three times Anil’s age is 5:6.

∴ 2 × (24x + 30) = 45x + 75
∴ 48x + 60 = 45x + 75
∴ 48x – 45x = 75 – 60
∴ 3x = 15
∴ x = \(\frac { 15 }{ 3 }\)
∴ x = 5
∴ Viru’s present age is 5 years.

Maharashtra Board Class 8 Maths Chapter 12 Equations in One Variable Practice Set 12.2 Intext Questions and Activities

8th Math Practice Set 12.2 Question 1.
Write correct numbers in the boxes given. (Textbook pg. no. 78)
length is 3 times the breadth

Perimeter of the rectangle = 40
2(__x + __x) = 40
2 × __ x = 40
__ x = 40
x = __
∴ Breadth of rectangle = __ cm and Length of rectangle = __ cm
Solution:
length is 3 times the breadth

Perimeter of the rectangle = 40
∴ 2(3x + 1x) = 40
∴ 2 × 4x = 40
∴ 8x = 40
∴ x = 5
∴ Breadth of rectangle = 5 cm and Length of rectangle = 15 cm

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 12.1 8th Std Maths Answers Solutions Chapter 12 Equations in One Variable.

Equations in One Variable Class 8 Maths Chapter 12 Practice Set 12.1 Solutions Maharashtra Board

Std 8 Maths Practice Set 12.1 Chapter 12 Solutions Answers

Equation in One Variable Practice Set 12.1 Question 1. Each equation is followed by the values of the variable. Decide whether these values are the solutions of that equation.
i. x – 4 = 3, x = – 1, 7, – 7
ii. 9m = 81, m = 3, 9, -3
iii. 2a + 4 = 0, a = 2, – 2, 1
iv. 3 – y = 4, y = – 1, 1, 2
Solution:
i. x – 4 = 3 ….(i)
Substituting x = – 1 in L.H.S. of equation (i),
L.H.S. = (-1) – 4
= – 5
R.H.S. = 3
∴ L.H.S. ≠ R.H.S.
∴ x = – 1 is not the solution of the given equation.

Substituting x = 7 in L.H.S. of equation (i),
L.H.S. = (7) – 4
= 3
R.H.S. = 3
∴ L.H.S. = R.H.S.
∴ x = 7 is the solution of the given equation.

Substituting x = – 7 in L.H.S. of equation (i),
L.H.S. = (- 7) – 4
= -11
R.H.S. = 3
∴ L.H.S. ≠ R.H.S.
∴ x = – 7 is not the solution of the given equation.

ii. 9m = 81 …(i)
Substituting m = 3 in L.H.S. of equation (i),
L.H.S. = 9 × (3)
= 27
R.H.S. = 81
∴L.H.S. ≠ R.H.S.
∴m = 3 is not the solution of the given equation.

Substituting m = 9 in L.H.S. of equation (i),
L.H.S. = 9 × (9)
= 81
R.H.S. = 81
∴L.H.S. = R.H.S.
∴m = 9 is the solution of the given equation.

Substituting m = – 3 in L.H.S. of equation (i),
L.H.S. = 9 × (- 3)
= -27
R.H.S. = 81
∴L.H.S. ≠ R.H.S.
∴m = – 3 is not the solution of the given equation.

iii. 2a + 4 = 0 …..(i)
Substituting a = 2 in L.H.S. of equation (i),
L.H.S. = 2 (2) + 4
= 4 + 4
= 8
R.H.S. = 0
∴L.H.S. ≠ R.H.S.
∴a = 2 is not the solution of the given equation.

Substituting a = – 2 in L.H.S. of equation (i),
L.H.S. = 2 (-2)+ 4
= -4 + 4
= 0
R.H.S. = 0
∴L.H.S. = R.H.S.
∴a = – 2 is the solution of the given equation.

Substituting a = 1 in L.H.S. of equation (i),
L.H.S. = 2(1)+ 4
= 2 + 4
= 6
R.H.S. = 0
∴ L.H.S. ≠ R.H.S.
∴a = 1 is not the solution of the given equation.

iv. 3 – y = 4 …(i)
Substituting y = -1 in L.H.S. of equation (i),
L.H.S. = 3 – (- 1)
= 3 + 1
= 4
R.H.S. = 4
∴L.H.S. = R.H.S.
∴y = – 1 is the solution of the given equation.

Substituting y = 1 in L.H.S. of equation (i),
L.H.S. = 3-(1)
= 2
R.H.S. = 4
∴L.H.S. ≠ R.H.S.
∴y = 1 is not the solution of the given equation.

Substituting y = 2 in L.H.S. of equation (i),
L.H.S. = 3-(2)
= 1
R.H.S. = 4
∴L.H.S. ≠ R.H.S.
∴y = 2 is not the solution of the given equation.

Practice Set 12.1 Question 2.
Solve the following equations:
i. 17p – 2 = 49
ii. 2m + 7 = 9
iii. 3x + 12 = 2x – 4
iv. 5 (x – 3) = 3 (x + 2)
v. \(\frac { 9x }{ 8 }+1=10\)
vi. \(\frac{y}{7}+\frac{y-4}{3}=2\)
vii. 13x – 5 = \(\frac { 3 }{ 2 }\)
viii. 3 (y + 8) = 10 (y – 4) + 8
ix. \(\frac{x-9}{x-5}=\frac{5}{7}\)
x. \(\frac{y-4}{3}+3 y=4\)
xi. \(\frac{b+(b+1)+(b+2)}{4}=21\)
Solution:
i. 17p – 2 = 49
∴ 17p – 2 + 2 = 49 + 2
…[Adding 2 on both the sides]
∴ 17p = 51
∴ \(\frac{17 p}{17}=\frac{51}{17}\) …[Dividing both the sides by 17]
p = 3

ii. 2m + 7 = 9
∴ 2m + 7 – 7 = 9 – 7
…[Subtracting 7 from both the sides]
∴ 2m = 2
∴ \(\frac{2 m}{2}=\frac{2}{2}\) [Dividing both the sides by 2]
∴ m = 1

iii. 3x + 12 = 2x – 4
∴ 3x + 12 – 12 = 2x – 4 – 12
…[Subtracting 12 from both the sides]
∴ 3x = 2x – 16
∴ 3x – 2x = 2x – 16 – 2x
…[Subtracting 2x from both the sides]
∴ x = – 16

iv. 5 (x – 3) = 3 (x + 2)
∴ 5x – 15 = 3x + 6
∴ 5x – 15 + 15 = 3x + 6 + 15
…[Adding 15 on both the sides]
∴ 5x = 3x + 21
∴ 5x – 3x = 3x + 21 – 3x
…[Subtracting 3x from both the sides]
∴ 2x = 21
∴ \(\frac{2 x}{2}=\frac{21}{2}\) …[Dividing both the sides by 2]
∴ \(x=\frac{21}{2}\)

v. \(\frac { 9x }{ 8 }+1=10\)

vi. \(\frac{y}{7}+\frac{y-4}{3}=2\)

vii. 13x – 5 = \(\frac { 3 }{ 2 }\)

viii. 3 (y + 8) = 10 (y – 4) + 8
∴ 3y + 24 = 10y – 40 + 8
∴ 3y + 24 = 10y – 32
∴ 3y + 24 – 24 = 10y – 32 – 24
…[Subtracting 24 from both the sides]
∴ 3y = 10y – 56
∴ 3y – 10y = 10y – 56
…[Subtracting 10y from both the sides]
∴ – 7y = – 56
∴ \(\frac{-7 y}{-7}=\frac{-56}{-7}\)…[Dividing both the sides by – 7]
∴ y = 8

ix. \(\frac{x-9}{x-5}=\frac{5}{7}\)
∴\(\frac{x-9}{x-5} \times 7(x-5)=\frac{5}{7} \times 7(x-5)\)
…[Multiplying both the sides by 7 (x – 5)]
∴7 (x – 9) = 5 (x – 5)
∴7x – 63 = 5x – 25
∴7x – 63 + 63 = 5x – 25 + 63
…[Adding 63 on both the sides]
∴7x = 5x + 38
∴7x – 5x = 5x + 38 – 5x
…[Subtracting 5x from both the sides]
∴ 2x = 38
∴\(\frac{2 x}{2}=\frac{38}{2}\) …[Dividing both the sides by 2]
∴x = 19

x. \(\frac{y-4}{3}+3 y=4\)
∴\(\frac{y-4}{3} \times 3+3 y \times 3=4 \times 3\)
…[Multiplying both the sides by 3]
∴y – 4 + 9y = 12
∴10y – 4 = 12
∴10y – 4 + 4=12 + 4
…[Adding 4 on both the sides]
∴10y = 16
∴\(\frac{10 y}{10}=\frac{16}{10}\)…[Dividing both the sides by 10]
∴y = \(\frac { 8 }{ 5 }\)

xi. \(\frac{b+(b+1)+(b+2)}{4}=21\)
∴\(\frac{b+(b+1)+(b+2)}{4} \times 4=21 \times 4\)
…[Multiplying both the sides by 4]
∴b + b + 1 + b + 2 = 84
∴3b + 3 = 84
∴3b + 3 – 3 = 84 – 3
…[ Subtracting 3 from both the sides]
∴3b = 81
∴\(\frac{3 b}{3}=\frac{81}{3}[/latex …[Dividing both the sides by 3]
∴b = 27

Maharashtra Board Class 8 Maths Chapter 12 Equations in One Variable Practice Set 12.1 Intext Questions and Activities

Std 8 Maths Practice Set 12.1 Question 1.
Fill in the boxes to solve the following equations. (Textbook pg. no. 75)
i. x + 4 = 9
∴x + 4 – __ = 9 – __
… [Subtracting 4 from both the sides]
∴ x = __

ii. x – 2 = 7
∴x – 2 + __ = 7 + __
… [Adding 2 on both the sides]
∴x = __

iii. [latex]\frac { x }{ 3 }=4\)
∴\(\frac { x }{ 3 }\) × __ = 4 ×__
∴x = __

iv. 4x = 24
∴ __ = __
∴x = __
Solution:
i. x + 4 = 9
∴x + 4 – 4 = 9 – 4
… [Subtracting 4 from both the sides]
∴ x = 5

ii. x – 2 = 7
∴x – 2 + 2 = 7 + 2
… [Adding 2 on both the sides]
∴x = 9

iii. \(\frac { x }{ 3 }=4\)
∴\(\frac { x }{ 3 }\) × 3 = 4 × 3
… [Multiplying both the sides by 3]
∴x = 12

iv. 4x = 24
∴ \(\frac{4 x}{[4]}=\frac{24}{[4]}\)
… [Dividing both the sides by 4]
∴x = 6

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 10.2 8th Std Maths Answers Solutions Chapter 10 Division of Polynomials.

Division of Polynomials Class 8 Maths Chapter 10 Practice Set 10.2 Solutions Maharashtra Board

Std 8 Maths Practice Set 10.2 Chapter 10 Solutions Answers

Division of Polynomials Class 8 Practice Set 10.2 Question 1. Divide and write the quotient and the remainder.
i. (y² + 10y + 24) ÷ (y + 4)
ii. (p² + 7p – 5) ÷ (p + 3)
iii. (3x + 2x² + 4x³) ÷ (x – 4)
iv. (2m³ + m² + m + 9) ÷ (2m – 1)
v. (3x – 3x² – 12 + x⁴ + x³) ÷ (2 + x²)
vi. (a⁴ – a³ + a² – a + 1) ÷ (a³ – 2)
vii. (4x⁴ – 5x³ – 7x + 1) ÷ (4x – 1)
Solution:
i. (y² + 10y + 24) ÷ (y + 4)

∴ Quotient = y + 6
Remainder = 0

ii. (p² + 7p – 5) ÷ (p + 3)

∴ Quotient = p + 4
Remainder = -17

iii. (3x + 2x² + 4x³) ÷ (x – 4)
Write the dividend in descending order of their indices.
3x + 2x² + 4x³ = 4x³ + 2x² + 3x

∴ Quotient = 4x² + 18x + 75
Remainder = 300

iv. (2m³ + m² + m + 9) ÷ (2m – 1)

∴ Quotient = m² + m + 1
Remainder = 10

v. (3x – 3x² – 12 + x⁴ + x³) ÷ (2 + x²)
Write the dividend in descending order of their indices.
(x⁴ + x³ – 3x² + 3x – 12) ÷ (x² + 2)

∴ Quotient = x² + x – 5
Remainder = x – 2

vi. (a⁴ – a³ + a² – a + 1) ÷ (a³ – 2)

∴ Quotient = a – 1
Remainder = a² + a – 1

vii. (4x⁴ – 5x³ – 7x + 1) ÷ (4x – 1)
Write the dividend in descending order of their indices.
(4x⁴ – 5x³ – 7x + 1) = (4x⁴ – 5x³ + 0x² – 7x + 1)

∴ Quotient = \(x^{3}-x^{2}-\frac{x}{4}-\frac{29}{16}\)
Remainder = \(\frac { -13 }{ 16 }\)

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