Selina ICSE Mathematics Class 8 Textbook Solutions

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 19 Representing 3-D in 2-D

August 11, 2022January 27, 2019 by Veerendra

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 19 Representing 3-D in 2-D

Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 19 Representing 3-D in 2-D

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Mathematics Chapter 19 Representing 3-D in 2-D. You can download the Selina Concise Mathematics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Mathematics for Class 8 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

Selina Class 8 Maths Solutions Physics Chemistry Biology Geography History & Civics

Representing 3-D in 2-D Exercise 19 – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
If a polyhedron has 8 faces and 8 vertices, find the number of edges in it.
Solution:
Faces = 8
Vertices = 8
using Eulers formula,
F + V – E = 2
8 + 8 – E = 2
-E = 2 – 16
E= 14

Question 2.
If a polyhedron has 10 vertices and 7 faces, find the number of edges in it.
Solution:
Vertices = 10
Faces = 7
Using Eulers formula,
F + V – E = 2
7 + 10 – E = 2
-E = -15
E = 15

Question 3.
State, the number of faces, number of vertices and number of edges of:
(i) a pentagonal pyramid
(ii) a hexagonal prism
Solution:
(i) A pentagonal pyramid
Number of faces = 6
Number of vertices = 6
Number of edges = 10

(ii) A hexagonal prism
Number of faces = 8
Number of vertices = 12
Number of edges = 18

Question 4.
Verily Euler’s formula for the following three dimensional figures:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 19 Representing 3-D in 2-D - 1
Solution:
(i) Number of vertices = 6
Number of faces = 8
Number of edges = 12
Using Euler formula,
F + V – E = 2
8 + 6 – 12 = 2
2 = 2 Hence proved.

(ii) Number of vertices = 9
Number of faces = 8
Number of edges = 15
Using, Euler’s formula,
F + V – E = 2
9 + 8 – 15 = 2
2 = 2 Hence proved.

(iii) Number of vertices = 9
Number of faces = 5
Number of edges = 12
Using, Euler’s formula,
F + V – E = 2
9 + 5 – 12 = 2
2 = 2 Hence proved.

Question 5.
Can a polyhedron have 8 faces, 26 edges and 16 vertices?
Solution:
Number of faces = 8
Number of vertices = 16
Number of edges = 26
Using Euler’s formula
F + V – E
⇒ 8 + 16 – 26 ≠ -2
⇒ -2 ≠ 2
No, a polyhedron cannot have 8 faces, 26 edges and 16 vertices.

Question 6.
Can a polyhedron have:
(i) 3 triangles only ?
(ii) 4 triangles only ?
(iii) a square and four triangles ?
Solution:
(i) No.
(ii) Yes.
(iii) Yes.

Question 7.
Using Euler’s formula, find the values of x, y, z.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 19 Representing 3-D in 2-D - 2
Solution:

Question 8.
What is the least number of planes that can enclose a solid? What is the name of the solid.
Solution:
The least number of planes that can enclose a solid is 4.
The name of the solid is Tetrahedron.

Question 9.
Is a square prism same as a cube?
Solution:
Yes, a square prism is same as a cube.

Question 10.
A cubical box is 6 cm x 4 cm x 2 cm. Draw two different nets of it.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 19 Representing 3-D in 2-D - 4

Question 11.
Dice are cubes where the sum of the numbers on the opposite faces is 7. Find the missing numbers a, b and c.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 19 Representing 3-D in 2-D - 5
Solution:

Question 12.
Name the polyhedron that can be made by folding each of the following nets:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 19 Representing 3-D in 2-D - 7
Solution:
(i) Triangular prism. It has 3 rectangles and 2 triangles.
(ii) Triangular prism. It has 3 rectangles and 2 triangles.
(iii) Hexagonal pyramid as it has a hexagonal base and 6 triangles.

Question 13.
Draw nets for the following polyhedrons:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 19 Representing 3-D in 2-D - 8
Solution:
Net of hexagonal prism:

Net of pentagonal pyramid:

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions

August 11, 2022January 27, 2019 by Prasanna

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions (Including Operations on Algebraic Expressions)

Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions (Including Operations on Algebraic Expressions)

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Mathematics Chapter 11 Algebraic Expressions (Including Operations on Algebraic Expressions). You can download the Selina Concise Mathematics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Mathematics for Class 8 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

Selina Class 8 Maths Solutions Physics Chemistry Biology Geography History & Civics

Algebraic Expressions Exercise 11A – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Separate the constants and variables from the following :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 1
Solution:
Clearly constants are : -7, √5, 8 – 5

Question 2.
Write the number of terms in each of the following polynomials.
(i) 5x² + 3 x ax
(ii) ax ÷ 4 – 7
(iii) ax – by + y x z
(iv) 23 + a x b ÷ 2.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 3

Question 3.
Separate monomials, binomials, trinomials and polynomials from the following algebraic expressions :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 4
Solution:

Question 4.
Write the degree of each polynomial given below :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 6
Solution:
(i) degree = 2 (Polynomial is xy + 7z)
(ii) degree = 3 (Polynomial is x² – 6x³ + 8 y)
(iii) degree = 8 (Polynomial is y – 6y² + 5y⁸)
(iv) degree = 3 (Polynomial is xyz – 3)
(v) degree = 4 (Polynomial is xy + yz² – xz³)
(vi) degree = 12 (Polynomial is x⁵y⁷ – 8x³y⁸ + 10x⁴, y⁴z⁴)

Question 5.
Write the coefficient of :
(i) ab in 7abx ,
(ii) 7a in 7abx ;
(iii) 5x² in 5x² – 5x ;
(iv) 8 in a² – 8ax + a ;
(v) 4xy in x² – 4xy + y².
Solution:
(i) The coefficient of ab in 7abx = 7x
(ii) The coefficient of 7a in 7abx = bx
(iii) The coefficient of 5x² in 5x² – 5x = 1
(iv) The coefficient of 8 in a² – 8ax + a = – ax
(v) The coefficient of 4xy in x² – 4xy + y² = -1

Question 6.
In \(\frac { 5 }{ 7 }\) xy²z³, write the coefficient of
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 7
Solution:

Question 7.
In each polynomial, given below, separate the like terms :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 9
Solution:

Algebraic Expressions Exercise 11B – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Evaluate :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 11
Solution:

Question 2.
Add :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 13
Solution:

Question 3.
Find the total savings of a boy who saves ₹ (4x – 6y) ; ₹ (6x + 2y) ; ₹ (4y – x) and ₹ (y – 2x) for four consecutive weeks.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 16

Question 4.
Subtract :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 17
Solution:

Question 5.
(i) Take away – 3x³ + 4x² – 5x+ 6 from 3x³ – 4x² + 5x – 6
(ii) Take m² + m + 4 from -m² + 3m + 6 and the result from m² + m + 1.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 19

Question 6.
Subtract the sum of 5y² + y – 3 and y² – 3y + 7 from 6y² + y – 2.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 20

Question 7.
What must be added to x⁴ – x³ + x² + x + 3 to obtain x⁴ + x² – 1 ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 21

Question 8.
(i) How much more than 2x² + 4xy + 2y² is 5x² + 10xy – y² ?
(ii) How much less 2a² + 1 is than 3a² – 6 ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 22

Question 9.
If x = 6a + 86 + 9c ; y = 2b – 3a – 6c and z = c – b + 3a ; find
(i) x + y + z
(ii) x – y + z
(iii) 2x – y – 3z
(iv) 3y – 2z – 5x
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 23

Question 10.
The sides of a triangle are x² – 3xy + 8, 4x² + 5xy – 3 and 6 – 3x² + 4xy. Find its perimeter.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 25

Question 11.
The perimeter of a triangle is 8y² – 9y + 4 and its two sides are 3y² – 5y and 4y² + 12. Find its third side.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 26

Question 12.
The two adjacent sides of a rectangle are 2x² – 5xy + 3z² and 4xy – x² – z². Find its perimeter.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 27

Question 13.
What must be subtracted from 19x⁴ + 2x³ + 30x – 37 to get 8x⁴ + 22x³ – 7x – 60 ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 28

Question 14.
How much smaller is 15x – 18y + 19z than 22x – 20y – 13z + 26 ?
Solution:
The required result is
(22x – 20y – 13z + 26) – (15x – 18y + 19z)
= 22x – 20y – 13z + 26 – 15x + 18y – 19z
= 7x – 2y – 32z + 26

Question 15.
How much bigger is 15x²y² – 18xy² – 10x²y than -5x² + 6x²y – 7xy ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 30

Algebraic Expressions Exercise 11C – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Multiply :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 31
Solution:

Question 2.
Multiply :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 37
Solution:

Question 3.
Simplify :
(i) (7x – 8) (3x + 2)
(ii) (px – q) (px + q)
(iii) (5a + 5b – c) (2b – 3c)
(iv) (4x – 5y) (5x – 4y)
(v) (3y + 4z) (3y – 4z) + (2y + 7z) (y + z)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 40

Question 4.
The adjacent sides of a rectangle are x² – 4xy + 7y² and x³ – 5xy². Find its area.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 41

Question 5.
The base and the altitude of a triangle are (3x – 4y) and (6x + 5y) respectively. Find its area.
Solution:
Reqd. Area = \(\frac { 1 }{ 2 }\) (base) x (altitude)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 42

Question 6.
Multiply -4xy³ and 6x²y and verify your result for x = 2 and y= 1.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 43

Question 7.
Find the value of (3x³) x (-5xy²) x (2x²yz³) for x = 1, y = 2 and z = 3.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 44

Question 8.
Evaluate (3x⁴y²) (2x²y³) for x = 1 and y = 2.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 45

Question 9.
Evaluate (x⁵) x (3x²) x (-2x) for x = 1.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 46

Question 10.
If x = 2 and y = 1; find the value of (-4x²y³) x (-5x²y⁵).
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 47

Question 11.
Evaluate:
(i) (3x – 2)(x + 5) for x = 2.
(ii) (2x – 5y)(2x + 3y) for x = 2 and y = 3.
(iii) xz (x² + y²) for x = 2, y = 1 and z= 1.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 48

Question 12.
Evaluate:
(i) x(x – 5) + 2 for x = 1.
(ii) xy²(x – 5y) + 1 for x = 2 and y = 1.
(iii) 2x(3x – 5) – 5(x – 2) – 18 for x = 2.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 49

Question 13.
Multiply and then verify :
-3x²y² and (x – 2y) for x = 1 and y = 2.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 51

Question 14.
Multiply:
(i) 2x² – 4x + 5 by x² + 3x – 7
(ii) (ab – 1)(3 – 2ab)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 52

Question 15.
Simplify : (5 – x)(6 – 5x)(2 -x).
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 54

Algebraic Expressions Exercise 11D – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Divide :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 55
Solution:

Question 2.
Find the quotient and the remainder (if any) when :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 65
Solution:

Question 3.
The area of a rectangle is x³ – 8x² + 7 and one of its sides is x – 1. Find the length of the adjacent side.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 71

Question 4.
The product of two numbers-is 16x⁴ – 1. If one number is 2x – 1, find the other.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 72

Question 5.
Divide x⁶ – y⁶ by the product of x² + xy + y² and x – y.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 73

Simplification
(Using removal of brackets)
The signs for different types of brackets are :

____ ; Vinculum or bar brackets,
( ); Parenthesis or small brackets,
{ }; Curly brackets or middle brackets,
[ ]; Square brackets or big brackets.
In a combined operation, the brackets must be removed in the same order as written above:

Algebraic Expressions Exercise 11E – Selina Concise Mathematics Class 8 ICSE Solutions

Simplify :
Question 1.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 75

Question 2.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 77

Question 3.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 79

Question 4.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 81

Question 5.

Solution:

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 84

Question 6.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 86

Question 7.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 88

Question 8.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 90

Question 9.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 92

Question 10.

Solution:

Question 11.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 96

Question 12.

Solution:

Question 13.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 100

Question 14.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 102

Question 15.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 104

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage

August 11, 2022January 27, 2019 by Prasanna

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage

Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

Selina Class 8 Maths Solutions Physics Chemistry Biology Geography History & Civics

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Mathematics Chapter 7 Percent and Percentage. You can download the Selina Concise Mathematics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Mathematics for Class 8 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

Percent and Percentage Exercise 7A – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Evaluate :
(i) 55% of 160 + 24% of 50 – 36% of 150
(ii) 9.3% of 500 – 4.8% of 250 – 2.5% of 240
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 1

Question 2.
(i) A number is increased from 125 to 150 ; find the percentage increase.
(ii) A number is decreased from 125 to 100 ; find the percentage decrease.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 2

Question 3.
Find :
(i) 45 is what percent of 54 ?
(ii) 2.7 is what percent of 18 ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 3

Question 4.
(i) 252 is 35% of a certain number, find the number.
(ii) If 14% of a number is 315 ; find the number.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 4

Question 5.
Find the percentage change, when a number is changed from :
(i) 80 to 100
(ii) 100 to 80
(iii) 6.25 to 7.50
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 6

Question 6.
An auctioneer charges 8% for selling a house. If a house is sold for Rs.2, 30, 500; find the charges of the auctioneer.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 7

Question 7.
Out of 800 oranges, 50 are rotten. Find the percentage of good oranges.
Solution:
Total number of oranges = 800
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 8

Question 8.
A cistern contains 5 thousand litres of water. If 6% water is leaked. Find how many litres of water are left in the cistern.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 9

Question 9.
A man spends 87% of his salary. If he saves Rs.325 ; find his salary.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 10

Question 10.
(i) A number 3.625 is wrongly read as 3.265; find the percentage error.
(ii) A number 5.78 x 10³ is wrongly written as 5.87 x 10³; find the percentage error
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 11

Question 11.
In an election between two candidates, one candidate secured 58% of the votes polled and won the election by 18, 336 votes. Find the total number of votes polled and the votes secured by each candidate.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 12

Question 12.
In an election between two candidates, one candidate secured 47% of votes polled and lost the election by 12, 366 votes. Find the total votes polled and die votes secured by the winning candidate.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 14

Question 13.
The cost of a scooter depreciates every year by 15% of its value at the beginning of the year. If the present cost of the scooter is
₹ 8,000; find its cost:
(i) after one year
(ii) after 2 years
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 16

Question 14.
In an examination, the pass mark is 40%. If a candidate gets 65 marks and fails by 3 marks ; find the maximum marks.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 17

Question 15.
In an examination, a candidate secured 125 marks and failed by 15 marks. If the pass percentage was 35% ; find the maximum marks.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 18

Question 16.
In an objective type paper of 150 questions; John got 80% correct answers and Mohan got 64% correct answers.
(i) How many correct answers did each get?
(ii) What percent is Mohan’s correct answers to John’s correct answers ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 20

Question 17.
The number 8,000 is first increased by 20% and then decreased by 20%. Find the resulting number.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 21

Question 18.
The number 12,000 is first decreased by 25% and then increased by 25%. Find the resulting number.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 22

Question 19.
The cost of an article is first increased by 20% and then decreased by 30%, find the percentage change in the cost of the article.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 23

Question 20.
The cost of an article is first decreased by 25% and then further decreased by 40%. Find the percentage change in the cost of the article.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 25

Percent and Percentage Exercise 7B – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
A man bought a certain number of oranges ; out of which 13 percent were found rotten. He gave 75% of the remaining in charity and still has 522 oranges left. Find how many had he bought?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 26

Question 2.
5% pupil in a town died due to some diseases and 3% of the remaining left the town. If 2, 76, 450 pupil are still in the town; find the original number of pupil in the town.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 28

Question 3.
In a combined test in English and Physics ; 36% candidates failed in English ; 28% failed in Physics and 12% in both ; find:
(i) the percentage of passed candidates
(ii) the total number of candidates appeared, if 208 candidates have failed.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 30

Question 4.
In a combined test in Maths and Chemistry; 84% candidates passsed in Maths; 76% in Chemistry and 8% failed in both. Find :
(i) the percentage of failed candidates ;
(ii) if 340 candidates passed in the test ; then how many appeared ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 31

Question 5.
A’s income is 25% more than B’s. Find, B’s income is how much percent less than A’s.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 32

Question 6.
Mona is 20% younger than Neetu. How much percent is Neetu older than Mona ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 34

Question 7.
If the price of sugar is increased by 25% today; by what percent should it be decreased tomorrow to bring the price back to the original ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 35

Question 8.
A number increased by 15% becomes 391. Find the number.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 37

Question 9.
A number decreased by 23 % becomes 539. Find the number.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 38

Question 10.
Two numbers are respectively 20 percent and 50 percent more than a third number. What percent is the second of the first ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 40

Question 11.
Two numbers are respectively 20 percent and 50 percent of a third number. What percent is the second of the first ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 41

Question 12.
Two numbers are respectively 30 percent and 40 percent less than a third number. What percent is the second of the first ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 43

Percent and Percentage Exercise 7C – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
A bag contains 8 red balls, 11 blue balls and 6 green balls. Find the percentage of blue balls in the bag.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 44

Question 2.
Mohan gets Rs. 1, 350 from Geeta and Rs. 650 from Rohit. Out of the total money that Mohan gets from Geeta and Rohit. what percent does he get from Rohit ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 45

Question 3.
The monthly income of a man is Rs. 16, 000. 15 percent of it is paid as income-tax and 75% of the remainder is spent on rent, food, clothing, etc. How much money is still left with the man?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 46

Question 4.
A number is first increased by 20% and the resulting number is then decreased by 10%. Find the overall change in the number as percent.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 47

Question 5.
A number is increased by 10% and the resulting number is again increased by 20%. What is the overall percentage increase in the number ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 48

Question 6.
During 2003, the production of a factory decreased by 25%. But, during 2004, it (production) increased by 40% of what it was at the beginning of2004. Calculate the resulting change (increase or decrease) in production during these two years.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 49

Question 7.
Last year, oranges were available at Rs. 24 per dozen ; but this year, they are available at Rs. 50 per score. Find the percentage change in the price of oranges.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 50

Question 8.
In an examination, Kavita scored 120 out of 150 in Maths, 136 out of 200 in English and 108 out of 150 in Science. Find her percentage score in each subject and also on the whole (aggregate).
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 51

Question 9.
A is 25% older than B. By what percent is B younger than A ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 52

Question 10.
(i) Increase 180 by 25%.
(ii) Decrease 140 by 18%.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 54

Question 11.
In an election, three candidates contested and secured 29200, 58800 and 72000 votes. Find the percentage of votes scored by winning candidate.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 55

Question 12.
(i) A number when increased by 23% becomes 861 ; find the number.
(ii) A number when decreased by 16% becomes 798 ; find the number.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 56

Question 13.
The price of sugar is increased by 20%. By what percent must the consumption of sugar be decreased so that the expenditure on sugar may remain the same ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 58

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 17 Special Types of Quadrilaterals

August 11, 2022January 27, 2019 by Prasanna

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 17 Special Types of Quadrilaterals

Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 17 Special Types of Quadrilaterals

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Mathematics Chapter 17 Special Types of Quadrilaterals. You can download the Selina Concise Mathematics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Mathematics for Class 8 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

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Special Types of Quadrilaterals Exercise 17 – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
In parallelogram ABCD, ∠A = 3 times ∠B. Find all the angles of the parallelogram. In the same parallelogram, if AB = 5x – 7 and CD = 3x +1 ; find the length of CD.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 17 Special Types of Quadrilaterals image - 1
Let ∠B = x
∠A = 3 ∠B = 3x
AD||BC
∠A + ∠B = 180°
3x + x = 180°
⇒ 4x = 180°
⇒ x = 45°
∠B = 45°
∠A = 3x = 3 x 45 = 135°
and ∠B = ∠D = 45°
opposite angles of || gm are equal.
∠A = ∠C = 135°
opposite sides of //gm are equal.
AB = CD
5x – 7 = 3x + 1
⇒ 5x – 3x = 1+7
⇒ 2x = 8
⇒ x = 4
CD = 3 x 4+1 = 13
Hence 135°, 45°, 135° and 45° ; 13

Question 2.
In parallelogram PQRS, ∠Q = (4x – 5)° and ∠S = (3x + 10)°. Calculate : ∠Q and ∠R.
Solution:
In parallelogram PQRS,
∠Q = (4x – 5)° and ∠S = (3x + 10)°
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 17 Special Types of Quadrilaterals image - 2
opposite ∠s of //gm are equal.
∠Q = ∠S
4x – 5 = 3x + 10
4x – 3x = 10+5
x = 15
∠Q = 4x – 5 =4 x 15 – 5 = 55°
Also ∠Q + ∠R = 180°
55° + ∠R = 180°
∠R = 180°-55° = 125°
∠Q = 55° ; ∠R = 125°

Question 3.
In rhombus ABCD ;
(i) if ∠A = 74° ; find ∠B and ∠C.
(ii) if AD = 7.5 cm ; find BC and CD.
Solution:
AD || BC
∠A + ∠B = 180°
74° + ∠B = 180°
∠B =180° – 74°= 106°
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 17 Special Types of Quadrilaterals image - 3
opposite angles of Rhombus are equal.
∠A = ∠C = 74°
Sides of Rhombus are equal.
BC = CD = AD = 7.5 cm
(i) ∠B = 106° ; ∠C = 74°
(ii) BC = 7.5 cm and CD = 7.5 cm Ans.

Question 4.
In square PQRS :
(i) if PQ = 3x – 7 and QR = x + 3 ; find PS
(ii) if PR = 5x and QR = 9x – 8. Find QS
Solution:
(i) sides of square are equal.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 17 Special Types of Quadrilaterals image - 4
PQ = QR
=> 3x – 7 = x + 3
=> 3x – x = 3 + 7
=> 2x = 10
x = 5
PS=PQ = 3x – 7 = 3 x 5 – 7 =8
(ii) PR = 5x and QS = 9x – 8

As diagonals of square are equal.
PR = QS
5x = 9x – 8
=> 5x – 9x = -8
=> -4x = -8
=> x = 2
QS = 9x – 8 = 9 x 2 – 8 =10

Question 5.
ABCD is a rectangle, if ∠BPC = 124°
Calculate : (i) ∠BAP (ii) ∠ADP
selina-concise-mathematics-class-8-icse-solutions-special-types-of-quadrilaterals-5
Solution:
Diagonals of rectangle are equal and bisect each other.
∠PBC = ∠PCB = x (say)
But ∠BPC + ∠PBC + ∠PCB = 180°
124° + x + x = 180°
2x = 180° – 124°
2x = 56°
=> x = 28°
∠PBC = 28°
But ∠PBC = ∠ADP [Alternate ∠s]
∠ADP = 28°
Again ∠APB = 180° – 124° = 56°
Also PA = PB
∠BAP = \(\frac { 1 }{ 2 }\) (180° – ∠APB)
= \(\frac { 1 }{ 2 }\) x (180°- 56°) = \(\frac { 1 }{ 2 }\) x 124° = 62°
Hence (i) ∠BAP = 62° (ii) ∠ADP =28°

Question 6.
ABCD is a rhombus. If ∠BAC = 38°, find :
(i) ∠ACB
(ii) ∠DAC
(iii) ∠ADC.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 17 Special Types of Quadrilaterals image - 7
Solution:
ABCD is Rhombus (Given)
AB = BC
∠BAC = ∠ACB (∠s opp. to equal sides)
But ∠BAC = 38° (Given)
∠ACB = 38°
In ∆ABC,
∠ABC + ∠BAC + ∠ACB = 180°
∠ABC + 38°+ 38° = 180°
∠ABC = 180° – 76° = 104°
But ∠ABC = ∠ADC (opp. ∠s of rhombus)
∠ADC = 104°
∠DAC = ∠DCA ( AD = CD)
∠DAC = \(\frac { 1 }{ 2 }\) [180° – 104°]
∠DAC = \(\frac { 1 }{ 2 }\) x 76° = 38°
Hence (i) ∠ACB = 38° (ii) ∠DAC = 38° (iii) ∠ADC = 104° Ans.

Question 7.
ABCD is a rhombus. If ∠BCA = 35°. find ∠ADC.
Solution:
Given : Rhombus ABCD in which ∠BCA = 35°
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 17 Special Types of Quadrilaterals image - 8
To find : ∠ADC
Proof : AD || BC
∠DAC = ∠BCA (Alternate ∠s)
But ∠BCA = 35° (Given)
∠DAC = 35°
But ∠DAC = ∠ACD ( AD = CD) & ∠DAC +∠ACD + ∠ADC = 180°
35°+ 35° + ∠ADC = 180°
∠ADC = 180° – 70° = 110°
Hence ∠ADC = 110°

Question 8.
PQRS is a parallelogram whose diagonals intersect at M.
If ∠PMS = 54°, ∠QSR = 25° and ∠SQR = 30° ; find :
(i) ∠RPS
(ii) ∠PRS
(iii) ∠PSR.
Solution:
Given : ||gm PQRS in which diagonals PR & QS intersect at M.
∠PMS = 54° ; ∠QSR = 25° and ∠SQR=30°
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 17 Special Types of Quadrilaterals image - 9
To find : (i) ∠RPS (ii) ∠PRS (iii) ∠PSR
Proof : QR || PS
=> ∠PSQ = ∠SQR (Alternate ∠s)
But ∠SQR = 30° (Given)
∠PSQ = 30°
In ∆SMP,
∠PMS + ∠ PSM +∠MPS = 180° or 54° + 30° + ∠RPS = 180°
∠RPS = 180°- 84° = 96°
Now ∠PRS + ∠RSQ = ∠PMS
∠PRS + 25° =54°
∠PRS = 54° – 25° = 29°
∠PSR = ∠PSQ + ∠RSQ = 30°+25° = 55°
Hence (i) ∠RPS = 96° (ii) ∠PRS = 29° (iii) ∠PSR = 55°

Question 9.
Given : Parallelogram ABCD in which diagonals AC and BD intersect at M.
Prove : M is mid-point of LN.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 17 Special Types of Quadrilaterals image - 10
Proof : Diagonals of //gm bisect each other.
MD = MB
Also ∠ADB = ∠DBN (Alternate ∠s)
& ∠DML = ∠BMN (Vert. opp. ∠s)
∆DML = ∆BMN
LM = MN
M is mid-point of LN.
Hence proved.

Question 10.
In an Isosceles-trapezium, show that the opposite angles are supplementary.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 17 Special Types of Quadrilaterals image - 11
Given : ABCD is isosceles trapezium in which AD = BC
To Prove : (i) ∠A + ∠C = 180°
(ii) ∠B + ∠D = 180°
Proof : AB || CD.
=> ∠A + ∠D = 180°
But ∠A = ∠B [Trapezium is isosceles)]
∠B + ∠D = 180°
Similarly ∠A + ∠C = 180°
Hence the result.

Question 11.
ABCD is a parallelogram. What kind of quadrilateral is it if :
(i) AC = BD and AC is perpendicular to BD?
(ii) AC is perpendicular to BD but is not equal to it ?
(iii) AC = BD but AC is not perpendicular to BD ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 17 Special Types of Quadrilaterals image - 12

Question 12.
Prove that the diagonals of a parallelogram bisect each other.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 17 Special Types of Quadrilaterals image - 14
Given : ||gm ABCD in which diagonals AC and BD bisect each other.
To Prove : OA = OC and OB = OD
Proof : AB || CD (Given)
∠1 = ∠2 (alternate ∠s)
∠3 = ∠4 = (alternate ∠s)
and AB = CD (opposite sides of //gm)
∆COD = ∆AOB (A.S.A. rule)
OA = OC and OB = OD
Hence the result.

Question 13.
If the diagonals of a parallelogram are of equal lengths, the parallelogram is a rectangle. Prove it.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 17 Special Types of Quadrilaterals image - 15
Given : //gm ABCD in which AC = BD
To Prove : ABCD is rectangle.
Proof : In ∆ABC and ∆ABD
AB = AB (Common)
AC = BD (Given)
BC = AD (opposite sides of ||gm)
∆ABC = ∆ABD (S.S.S. Rule)
∠A = ∠B
But AD // BC (opp. sides of ||gm are ||)
∠A + ∠B = 180°
∠A = ∠B = 90°
Similarly ∠D = ∠C = 90°
Hence ABCD is a rectangle.

Question 14.
In parallelogram ABCD, E is the mid-point of AD and F is the mid-point of BC. Prove that BFDE is a parallelogram.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 17 Special Types of Quadrilaterals image - 16
Given : //gm ABCD in which E and F are mid-points of AD and BC respectively.
To Prove : BFDE is a ||gm.
Proof : E is mid-point of AD. (Given)
DE = \(\frac { 1 }{ 2 }\) AD
Also F is mid-point of BC (Given)
BF = \(\frac { 1 }{ 2 }\) BC
But AD = BC (opp. sides of ||gm)
BF = DE
Again AD || BC
=> DE || BF
Now DE || BF and DE = BF
Hence BFDE is a ||gm.

Question 15.
In parallelogram ABCD, E is the mid-point of side AB and CE bisects angle BCD. Prove that :
(i) AE = AD,
(ii) DE bisects and ∠ADC and
(iii) Angle DEC is a right angle.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 17 Special Types of Quadrilaterals image - 17
Given : ||gm ABCD in which E is mid-point of AB and CE bisects ZBCD.
To Prove : (i) AE = AD
(ii) DE bisects ∠ADC
(iii) ∠DEC = 90°
Const. Join DE
Proof : (i) AB || CD (Given)
and CE bisects it.
∠1 = ∠3 (alternate ∠s) ……… (i)
But ∠1 = ∠2 (Given) …………. (ii)
From (i) & (ii)
∠2 = ∠3
BC = BE (sides opp. to equal angles)
But BC = AD (opp. sides of ||gm)
and BE = AE (Given)
AD = AE
∠4 = ∠5 (∠s opp. to equal sides)
But ∠5 = ∠6 (alternate ∠s)
=> ∠4 = ∠6
DE bisects ∠ADC.
Now AD // BC
=> ∠D + ∠C = 180°
2∠6+2∠1 = 180°
DE and CE are bisectors.
∠6 + ∠1 = \(\frac { { 180 }^{ 0 } }{ 2 }\)
∠6 + ∠1 = 90°
But ∠DEC + ∠6 + ∠1 = 180°
∠DEC + 90° = 180°
∠DEC = 180° – 90°
∠DEC = 90°
Hence the result.

Question 16.
In the following diagram, the bisectors of interior angles of the parallelogram PQRS enclose a quadrilateral ABCD.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 17 Special Types of Quadrilaterals image - 18
Show that:
(i) ∠PSB + ∠SPB = 90°
(ii) ∠PBS = 90°
(iii) ∠ABC = 90°
(iv) ∠ADC = 90°
(v) ∠A = 90°
(vi) ABCD is a rectangle
Thus, the bisectors of the angles of a parallelogram enclose a rectangle.
Solution:
Given : In parallelogram ABCD bisector of angles P and Q, meet at A, bisectors of ∠R and ∠S meet at C. Forming a quadrilateral ABCD as shown in the figure.
To prove :
(i) ∠PSB + ∠SPB = 90°
(ii) ∠PBS = 90°
(iii) ∠ABC = 90°
(iv) ∠ADC = 90°
(v) ∠A = 9°
(vi) ABCD is a rectangle
Proof : In parallelogram PQRS,
PS || QR (opposite sides)
∠P +∠Q = 180°
and AP and AQ are the bisectors of consecutive angles ∠P and ∠Q of the parallelogram
∠APQ + ∠AQP = \(\frac { 1 }{ 2 }\) x 180° = 90°
But in ∆APQ,
∠A + ∠APQ + ∠AQP = 180° (Angles of a triangle)
∠A + 90° = 180°
∠A = 180° – 90°
(v) ∠A = 90°
Similarly PQ || SR
∠PSB + SPB = 90°
(ii) and ∠PBS = 90°
But, ∠ABC = ∠PBS (Vertically opposite angles)
(iii) ∠ABC = 90°
Similarly we can prove that
(iv) ∠ADC = 90° and ∠C = 90°
(vi) ABCD is a rectangle (Each angle of a quadrilateral is 90°)
Hence proved.

Question 17.
In parallelogram ABCD, X and Y are midpoints of opposite sides AB and DC respectively. Prove that:
(i) AX = YC
(ii) AX is parallel to YC
(iii) AXCY is a parallelogram.
Solution:
Given : In parallelogram ABCD, X and Y are the mid-points of sides AB and DC respectively AY and CX are joined
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 17 Special Types of Quadrilaterals image - 19
To prove :
(i) AX = YC
(ii) AX is parallel to YC
(iii) AXCY is a parallelogram
Proof : AB || DC and X and Y are the mid-points of the sides AB and DC respectively
(i) AX = YC ( \(\frac { 1 }{ 2 }\) of opposite sides of a parallelogram)
(ii) and AX || YC
(iii) AXCY is a parallelogram (A pair of opposite sides are equal and parallel)
Hence proved.

Question 18.
The given figure shows parallelogram ABCD. Points M and N lie in diagonal BD such that DM = BN.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 17 Special Types of Quadrilaterals image - 20
Prove that:
(i) ∆DMC = ∆BNA and so CM = AN
(ii) ∆AMD = ∆CNB and so AM CN
(iii) ANCM is a parallelogram.
Solution:
Given : In parallelogram ABCD, points M and N lie on the diagonal BD such that DM = BN
AN, NC, CM and MA are joined
To prove :
(i) ∆DMC = ∆BNA and so CM = AN
(ii) ∆AMD = ∆CNB and so AM = CN
(iii) ANCM is a parallelogram
Proof :
(i) In ∆DMC and ∆BNA.
CD = AB (opposite sides of ||gm ABCD)
DM = BN (given)
∠CDM = ∠ABN (alternate angles)
∆DMC = ∆BNA (SAS axiom)
CM =AN (c.p.c.t.)
Similarly, in ∆AMD and ∆CNB
AD = BC (opposite sides of ||gm)
DM = BN (given)
∠ADM = ∠CBN – (alternate angles)
∆AMD = ∆CNB (SAS axiom)
AM = CN (c.p.c.t.)
(iii) CM = AN and AM = CN (proved)
ANCM is a parallelogram (opposite sides are equal)
Hence proved.

Question 19.
The given figure shows a rhombus ABCD in which angle BCD = 80°. Find angles x and y.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 17 Special Types of Quadrilaterals image - 21
Solution:
In rhombus ABCD, diagonals AC and BD bisect each other at 90°
∠BCD = 80°
Diagonals bisect the opposite angles also ∠BCD = ∠BAD (Opposite angles of rhombus)
∠BAD = 80° and ∠ABC = ∠ADC = 180° – 80° = 100°
Diagonals bisect opposite angles
∠OCB or ∠PCB = \(\frac { { 80 }^{ 0 } }{ 2 }\) = 40°
In ∆PCM,
Ext. CPD = ∠OCB + ∠PMC
110° = 40° + x
=> x = 110° – 40° = 70°
and ∠ADO = \(\frac { 1 }{ 2 }\) ∠ADC = \(\frac { 1 }{ 2 }\) x 100° = 50°
Hence x = 70° and y = 50°

Question 20.
Use the information given in the alongside diagram to find the value of x, y and z.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 17 Special Types of Quadrilaterals image - 22
Solution:
ABCD is a parallelogram and AC is its diagonal which bisects the opposite angle
Opposite sides of a parallelogram are equal
3x + 14 = 2x + 25
=> 3x – 2x = 25 – 14
=> x = 11
∴ x = 11 cm
∠DCA = ∠CAB (Alternate angles)
y + 9° = 24
y = 24° – 9° = 15°
∠DAB = 3y° + 5° + 24° = 3 x 15 + 5 + 24° = 50° + 24° = 74°
∠ABC =180°- ∠DAB = 180° – 74° = 106°
z = 106°
Hence x = 11 cm, y = 15°, z = 106°

Question 21.
The following figure is a rectangle in which x : y = 3 : 7; find the values of x and y.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 17 Special Types of Quadrilaterals image - 23
Solution:
ABCD is a rectangle,
x : y = 3 : 1
In ∆BCE, ∠B = 90°
x + y = 90°
But x : y = 3 : 7
Sum of ratios = 3 + 7=10

Hence x = 27°, y = 63°

Question 22.
In the given figure, AB // EC, AB = AC and AE bisects ∠DAC. Prove that:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 17 Special Types of Quadrilaterals image - 25
(i) ∠EAC = ∠ACB
(ii) ABCE is a parallelogram.
Solution:
ABCE is a quadrilateral in which AC is its diagonal and AB || EC, AB = AC
BA is produced to D
AE bisects ∠DAC
To prove:
(i) ∠EAC = ∠ACB
(ii) ABCE is a parallelogram
Proof:
(i) In ∆ABC and ∆ZAEC
AC=AC (common)
AB = CE (given)
∠BAC = ∠ACE (Alternate angle)
∆ABC = ∆AEC (SAS Axiom)
(ii) ∠BCA = ∠CAE (c.p.c.t.)
But these are alternate angles
AE || BC
But AB || EC (given)
∴ ABCD is a parallelogram

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 2 Exponents (Powers)

August 11, 2022January 27, 2019 by Veerendra

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 2 Exponents (Powers)

Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 2 Exponents (Powers)

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Mathematics Chapter 2 Exponents (Powers). You can download the Selina Concise Mathematics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Mathematics for Class 8 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

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Exponents Exercise 2A – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Evaluate:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 2 Exponents (Powers) image - 1
Solution:

Question 2.
If 1125 = 3^m x 5ⁿ; find m and n.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 2 Exponents (Powers) image - 8

Question 3.
Find x, if 9 × 3^x = (27)^2x-3
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 2 Exponents (Powers) image - 9

Exponents Exercise 2B – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Compute:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 2 Exponents (Powers) image - 10

Solution:

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 2 Exponents (Powers) image - 12

Question 2.
Simplify:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 2 Exponents (Powers) image - 48
Solution:

Question 3.
Evaluate:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 2 Exponents (Powers) image - 22
Solution:

Question 4.
Simplify:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 2 Exponents (Powers) image - 24
Solution:

Question 5.
Simplify:
(x^a+b)^a-b.(x^b+c)^b-c.(x^c+a)^c-a Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 2 Exponents (Powers) image - 28

Question 6.
Simplify:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 2 Exponents (Powers) image - 49
Solution:

Question 7.
(i) (a^-2)^-2. (ab)^-3
(ii) (xⁿy^-m)⁴× (x³y^-2)^-n

Solution:

Question 8.
Show that:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 2 Exponents (Powers) image - 34
Solution:

Question 9.
Evaluate:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 2 Exponents (Powers) image - 36
Solution:

Question 10.
Evaluate:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 2 Exponents (Powers) image - 38
Solution:

Question 11.
(m+n)^-1(m^-1+ n^-1) = (mn)^-1
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 2 Exponents (Powers) image - 40

Question 12.
Prove that:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 2 Exponents (Powers) image - 41
Solution:

Question 13.
Find the values of n, when:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 2 Exponents (Powers) image - 52
Solution:

Question 14.
Simplify:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 2 Exponents (Powers) image - 46
Solution:

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest

August 11, 2022January 27, 2019 by Prasanna

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest

Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Mathematics Chapter 9 Simple and Compound Interest. You can download the Selina Concise Mathematics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Mathematics for Class 8 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

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Simple and Compound Interest Exercise 9A – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Find the interest and the amount on:
(i) ₹ 750 in 3 years 4 months at 10% per annum.
(ii) ₹ 5,000 at 8% per year from 23rd December 2011 to 29th July 2012.
(iii) ₹ 2,600 in 2 years 3 months at 1% per month.
(iv) ₹ 4,000 in 1\(\frac { 1 }{ 3 }\) years at 2 paise per rupee per month.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -1

Question 2.
Rohit borrowed Rs. 24,000 at 7.5 percent per year. How much money will he pay at the end of 4th years to clear his debt ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -3

Question 3.
The interest on a certain sum of money is Rs. 1,480 in 2 years and at 10 per cent per year. Find the sum of money.
Solution:
Let P = Rs. x
Time (T) = 2 years
Rate (R) = 0%
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -4

Question 4.
On what principal will the simple interest be Rs. 7,008 in 6 years 3 months at 5% per year ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -5

Question 5.
Find the principal which will amount to Rs. 4,000 in 4 years at 6.25% Per annum.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -6

Question 6.
(i) At what rate per cent per annum will Rs. 630 produce an interest of Rs. 126 in 4 years ?
(ii) At what rate per cent per year will a sum double itself in 6\(\frac { 1 }{ 4 }\) years ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -8

Question 7.
(i) In how many years will Rs.950 produce Rs.399 as simple interest at 7% ?
(ii) Find the time in which Rs.1200 will amount to Rs.1536 at 3.5% per year.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -9

Question 8.
The simple interest on a certain sum of money is \(\frac { 3 }{ 8 }\) of the sum in 6\(\frac { 1 }{ 4 }\) years. Find the rate percent charged.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -11

Question 9.
What sum of money borrowed on 24th May will amount to Rs. 10210.20 on 17th October of the same year at 5 percent per annum simple interest.
Solution:
A = Rs. 10210.20
R = 5% P.A.
T=May + June + July + August + Sept.+ Oct.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -12

Question 10.
In what time will the interest on a certain sum of money at 6% be \(\frac { 5 }{ 8 }\) of itself ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -13

Question 11.
Ashok lent out Rs.7000 at 6% and Rs.9500 at 5%. Find his total income from the interest in 3 years.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -15

Question 12.
Raj borrows Rs.8,000; out of which Rs. 4500 at 5% and remainder at 6%. Find the total interest paid by him in 4 years.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -16

Question 13.
Mohan lends Rs.4800 to John for 4\(\frac { 1 }{ 2 }\) years and Rs.2500 to Shy am for 6 years and receives a total sum of Rs.2196 as interest. Find the rate percent per annum, it being the same in both the cases.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -18

Question 14.
John lent Rs. 2550 to Mohan at 7.5 per cent per annum. If Mohan discharges the debt after 8 months by giving an old black and white television and Rs. 1422.50; find the price of the television.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -20

Simple and Compound Interest Exercise 9B – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
The interest on a certain sum of money is 0.24 times of itself in 3 years. Find the rate of interest.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -21

Question 2.
If ₹ 3,750 amount to ₹ 4,620 in 3 years at simple interest. Find:
(i) the rate of interest
(ii) the amount of Rs. 7,500 in 5\(\frac { 1 }{ 2 }\) years at the same rate of interest
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -22

Question 3.
A sum of money, lent out at simple interst, doubles itself in 8 years. Find :
(i) the rate of interest
(ii) in how many years will the sum become triple (three times) of itself at the same rate per cent ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -24

Question 4.
Rupees 4000 amount to Rs.5000 in 8 years ; in what time will Rs.2100 amount to Rs.2800 at the same rate ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -25

Question 5.
What sum of money lent at 6.5% per annum will produce the same interest in 4 years as Rs.7500 produce in 6 years at 5% per annum ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -27

Question 6.
A certain sum amounts to Rs.3825 in 4 years and to Rs.4050 in 6 years. Find the rate percent and the sum.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -29

Question 7.
At what rafepercent of simple interest will the interest on Rs.3750 be one-fifth of itself in 4 years ? To what will it amount in 15 years ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -30

Question 8.
On what date will ₹ 1950 lent on 5th January, 2011 amount to ₹ 2125.50 at 5 percent per annum simple interest?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -32

Question 9.
If the interest on Rs.2400 be more than the interest on Rs.2000 by Rs.60 in 3 years at the same rate percent ; find the rate.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -34

Question 10.
Divide Rs. 15,600 into two parts such that the interest on one at 5 percent for 5 years may be equal to that on the other at 4\(\frac { 1 }{ 2 }\) per cent for 6 years.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -36

Simple and Compound Interest Exercise 9C – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
A sum of Rs. 8,000 is invested for 2 years at 10% per annum compound interest. Calculate:
(i) interest for the first year.
(ii) principal for the second year.
(iii) interest for the second year.
(iv) final amount at the end of second year
(v) compound interest earned in 2 years.
Solution:
(i) Here Principal (P) = Rs. 8,000
Rate of interest = 10%
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -37

Question 2.
A man borrowed Rs. 20,000 for 2 years at 8% per year compound interest. Calculate :
(i) the interest of the first year.
(ii) the interest of the second year.
(iii) the final amount at the end of second year.
(iv) the compound interest of two years.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -38

Question 3.
Calculate the amount and the compound interest on Rs. 12,000 in 2 years and at 10% per year.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -39

Question 4.
Calculate the amount and the compound interest on Rs. 10,000 in 3 years at 8% per annum.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -40

Question 5.
Calculate the compound interest on Rs. 5,000 in 2 years ; if the rates of interest for successive years be 10% and 12% respectively.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -42

Question 6.
Calculate the compound interest on Rs. 15,000 in 3 years ; if the rates of interest for successive years be 6%, 8% and 10% respectively.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -43

Question 7.
Mohan borrowed Rs. 16,000 for 3 years at 5% per annum compound interest. Calculate the amount that Mohan will pay at the end of 3 years.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -45

Question 8.
Rekha borrowed Rs. 40,000 for 3 years at 10% per annum compound interest. Calculate the interest paid by her for the second year.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -47

Question 9.
Calculate the compound interest for the second year on Rs. 15000 invested for 5 years at 6% per annum.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -48
= 159 x 6 = Rs. 954

Question 10.
A man invests Rs. 9600 at 10% per annum compound interest for 3 years. Calculate :
(i) the interest for the first year.
(ii) the amount at the end of the first year.
(iii) the interest for the second year.
(iv) the interest for the third year.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -49

Question 11.
A person invests Rs. 5,000 for two years at a certain rate of interest compounded annually. At the end of one year, this sum amounts to Rs. 5,600. Calculate :
(i) the rate of interest per year.
(ii) the amount at the end of the second year.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -50

Question 12.
Calculate the difference between the compound interest and the simple interest on ₹ 7,500 in two years and at 8% per annum.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -52

Question 13.
Calculate the difference between the compound interest and the simple interest on ₹ 8,000 in three years and at 10% per annum.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -53

Question 14.
Rohit borrowed ₹ 40,000 for 2 years at 10% per annum C.I. and Manish borrowed the same sum for the same time at 10.5% per annum simple interest. Which of these two gets less interest and by how much?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -54

Question 15.
Mr. Sharma borrowed ₹ 24,000 at 13% p.a. simple interest and an equal sum at 12% p.a. compound interest. Find the total interest earned by Mr. Sharma in 2 years.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -56

Question 16.
Peter borrows ₹ 12,000 for 2 years at 10% p.a. compound interest. He repays ₹ 8,000 at the end of first year. Find:
(i) the amount at the end of first year, before making the repayment.
(ii) the amount at the end of first year, after making the repayment.
(iii) the principal for the second year.
(iv) the amount to be paid at the end of second year, to clear the account.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -58

Question 17.
Gautam takes a loan of ₹ 16,000 for 2 years at 15% p.a. compound interest. He repays ₹ 9,000 at the end of first year. How mucH must he pay at the end of second year to clear the debt?
Solution:
Loan taken (P) = ₹ 16000
Rate (R) = 15% p.a.
Time (T) = 2 years
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -59

Question 18.
A certain sum of money, invested for 5 years at 8% p.a. simple interest, earns an interest of ₹ 12,000. Find:
(i) the sum of money.
(ii) the compound interest earned by this money in two years and at 10% p.a. compound interest.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -60

Question 19.
Find the amount and the C.I. on ₹ 12,000 at 10% per annum compounded half-yearly.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -62

Question 20.
Find the amount and the C.I. on ₹ 8,000 in 1\(\frac { 1 }{ 2 }\) years at 20% per year compounded half- yearly.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -63

Question 21.
Find the amount and the compound interest on ₹ 24,000 for 2 years at 10% per annum compounded yearly.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -65

Question 22.
Find the amount and the compound interest on ₹ 16,000 for 3 years at 5% per annum compounded annually.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -66

Question 23.
Find the amount and the compound interest on ₹ 20,000 for 1\(\frac { 1 }{ 2 }\) years at 10% per annum compounded half-yearly.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -68

Question 24.
Find the amount and the compound interest on ₹ 32,000 for 1 year at 20% per annum compounded half-yearly.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -69

Question 25.
Find the amount and the compound interest on ₹ 4,000 in 2 years, if the rate of interest for first year is 10% and for the second year is 15%.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -71

Question 26.
Find the amount and the compound interest on ₹ 10,000 in 3 years, if the rates of interest for the successive years are 10%, 15% and 20% respectively.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest image -72

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 4 Cubes and Cube-Roots

August 11, 2022January 27, 2019 by Prasanna

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 4 Cubes and Cube-Roots (Including use of tables for natural numbers)

Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 4 Cubes and Cube-Roots (Including use of tables for natural numbers)

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Mathematics Chapter 4 Cubes and Cube-Roots (Including use of tables for natural numbers). You can download the Selina Concise Mathematics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Mathematics for Class 8 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

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Cubes and Cube-Roots Exercise 4A – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Find the cube of :
(i) 7
(ii) 11
(iii) 16
(iv) 23
(v) 31
(vi) 42
(vii) 54
Solution:
(i) (7)³ = 7 x 7 x 7 = 343
(ii) (11)³ =11 x 11 x 11 = 1331
(iii) (16)³ = 16 x 16 x 16 = 4096
(iv) (23)³ = 23 x 23 x 23 = 12167
(v) (31)³ = 31 x 31 x 31 = 29791
(vi) (42)³ = 42 x 42 x 42 = 74088
(vii) (54)³ = 54 x 54 x 54 = 157464

Question 2.
Find which of the following are perfect cubes :
(i) 243
(ii) 588
(iii) 1331
(iv) 24000
(v) 1728
(vi) 1938
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 4 Cubes and Cube-Roots image -1

1938 = 2 x 3 x 17 x 19
1938 is not a perfect cube.

Question 3.
Find the cubes of :
(i) 2.1
(ii) 0.4
(iii) 1.6
(iv) 2.5
(v) 0.12
(vi) 0.02
(vii) 0.8
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 4 Cubes and Cube-Roots image -4

Question 4.
Find the cubes of :
(i) \(\frac { 3 }{ 7 }\)
(ii) \(\frac { 8 }{ 9 }\)
(iii) \(\frac { 10 }{ 13 }\)
(iv) \(1\frac { 2 }{ 7 }\)
(v) \(2\frac { 1 }{ 2 }\)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 4 Cubes and Cube-Roots image -7

Question 5.
Find the cubes of :
(i) -3
(ii) -7
(iii) -12
(iv) -18
(v) -25
(vi) -30
(vii) -50
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 4 Cubes and Cube-Roots image -8

Question 6.
Which of the following are cubes of:
(i) an even number
(ii) an odd number
216, 729, 3375, 8000, 125, 343, 4096 and 9261.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 4 Cubes and Cube-Roots image -10

(i) Cubes of an even number are 216, 8000, 4096.
(ii) Cubes of an odd number are 729, 3375, 125, 343, 9261.

Question 7.
Find the least number by which 1323 must be multiplied so that the product is a perfect cube.
Solution:
The prime factor of 1323 are =3 x 3 x 3 x 7 x 7
= (3 x 3 x 3) x 7 x 7
Clearly, 1323 must be multiplied by 7.

Question 8.
Find the smallest number by which 8768 must be divided so that the quotient is a perfect cube.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 4 Cubes and Cube-Roots image -13

Question 9.
Find the smallest number by which 27783 be multiplied to get a perfect square number.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 4 Cubes and Cube-Roots image -14

Question 10.
With what least number must 8640 be divided so that the quotient is a perfect cube?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 4 Cubes and Cube-Roots image -15

Question 11.
Which is the smallest number that must be multiplied to 77175 to make it a perfect cube?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 4 Cubes and Cube-Roots image -17

Cubes and Cube-Roots Exercise 4B – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Find the cube-roots of :
(i) 64
(ii) 343
(iii) 729
(iv) 1728
(v) 9261
(vi) 4096
(vii) 8000
(viii) 3375
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 4 Cubes and Cube-Roots image -18

Question 2.
Find the cube-roots of :
(i) \(\frac { 27 }{ 64 }\)
(ii) \(\frac { 125 }{ 216 }\)
(iii) \(\frac { 343 }{ 512 }\)
(iv) 64 x 729
(v) 64 x 27
(vi) 729 x 8000
(vii) 3375 x 512
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 4 Cubes and Cube-Roots image -23

Question 3.
Find the cube-roots of :
(i) -216
(ii) -512
(iii) -1331
(iv) \(\frac { -27 }{ 125 }\)
(v) \(\frac { -64 }{ 343 }\)
(vi) \(\frac { -512 }{ 343 }\)
(vii) -2197
(viii) -5832
(ix) -2744000
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 4 Cubes and Cube-Roots image -24

Question 4.
Find the cube-roots of :
(i) 2.744
(ii) 9.261
(iii) 0.000027
(iv) -0.512
(v) -15.625
(vi) -125 x 1000
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 4 Cubes and Cube-Roots image -28

Question 5.
Find the smallest number by which 26244 may be divided so that the quotient is a perfect cube.
Solution:
The prime factors of 26244 are
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 4 Cubes and Cube-Roots image -31
Clearly, 26244 must be divided by
3 x 3 x 2 x 2 = 36

Question 6.
What is the least number by which 30375 should be multiplied to get a perfect cube?
Solution:
The prime factors of 30375 are
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 4 Cubes and Cube-Roots image -32

Question 7.
Find the cube-roots of :
(i) 700 x 2 x 49 x 5
(ii) -216 x 1728
(iii) -64 x -125
(iv) \(\frac { -27 }{ 343 }\)
(v) \(\frac { 729 }{ -1331 }\)
(vi) 250.047
(vii) -175616
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 4 Cubes and Cube-Roots image -33

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes

August 11, 2022January 27, 2019 by Prasanna

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes (Including Polygons)

Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes (Including Polygons)

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Mathematics Chapter 16 Understanding Shapes (Including Polygons). You can download the Selina Concise Mathematics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Mathematics for Class 8 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

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Understanding Shapes Exercise 16A – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
State which of the following are polygons :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 1
If the given figure is a polygon, name it as convex or concave.
Solution:
Only Fig. (ii), (iii) and (v) are polygons.
Fig. (ii) and (iii) are concave polygons while
Fig. (v) is convex.

Question 2.
Calculate the sum of angles of a polygon with :
(i) 10 sides
(ii) 12 sides
(iii) 20 sides
(iv) 25 sides
Solution:
(i) No. of sides n = 10
sum of angles of polygon = (n – 2) x 180°
= (10 – 2) x 180° = 1440°
(ii) no. of sides n = 12
sum of angles = (n – 2) x 180°
= (12 – 2) x 180° = 10 x 180° = 1800°
(iii) n = 20
Sum of angles of Polygon = (n – 2) x 180°
= (20 – 2) x 180° = 3240°
(iv) n = 25
Sum of angles of polygon = (n – 2) x 180°
= (25 – 2) x 180° = 4140°

Question 3.
Find the number of sides in a polygon if the sum of its interior angles is :
(i) 900°
(ii) 1620°
(iii) 16 right-angles
(iv) 32 right-angles.
Solution:
(i) Let no. of sides = n
Sum of angles of polygon = 900°
(n – 2) x 180° = 900°
n – 2 = \(\frac { 900 }{ 180 }\)
n – 2 = 5
n = 5 + 2
n = 7
(ii) Let no. of sides = n
Sum of angles of polygon = 1620°
(n – 2) x 180° = 1620°
n – 2 = \(\frac { 1620 }{ 180 }\)
n – 2 = 9
n = 9 + 2
n = 11
(iii) Let no. of sides = n
Sum of angles of polygon = 16 right angles = 16 x 90 = 1440°
(n – 2) x 180° = 1440°
n – 2 = \(\frac { 1440 }{ 180 }\)
n – 2 = 8
n = 8 + 2
n = 10
(iv) Let no. of sides = n
Sum of angles of polygon = 32 right angles = 32 x 90 = 2880°
(n – 2) x 180° = 2880
n – 2 = \(\frac { 2880 }{ 180 }\)
n – 2 = 16
n = 16 + 2
n = 18

Question 4.
Is it possible to have a polygon ; whose sum of interior angles is :
(i) 870°
(ii) 2340°
(iii) 7 right-angles
(iv) 4500°
Solution:
(i) Let no. of sides = n
Sum of angles = 870°
(n – 2) x 180° = 870°
n – 2 = \(\frac { 870 }{ 180 }\)
n – 2 = \(\frac { 29 }{ 6 }\)
n = \(\frac { 29 }{ 6 }\) + 2
n = \(\frac { 41 }{ 6 }\)
Which is not a whole number.
Hence it is not possible to have a polygon, the sum of whose interior angles is 870°
(ii) Let no. of sides = n
Sum of angles = 2340°
(n – 2) x 180° = 2340°
n – 2 = \(\frac { 2340 }{ 180 }\)
n – 2 = 13
n = 13 + 2 = 15
Which is a whole number.
Hence it is possible to have a polygon, the sum of whose interior angles is 2340°.
(iii) Let no. of sides = n
Sum of angles = 7 right angles = 7 x 90 = 630°
(n – 2) x 180° = 630°
n – 2 = \(\frac { 630 }{ 180 }\)
n – 2 = \(\frac { 7 }{ 2 }\)
n = \(\frac { 7 }{ 2 }\) + 2
n = \(\frac { 11 }{ 2 }\)
Which is not a whole number. Hence it is not possible to have a polygon, the sum of whose interior angles is 7 right-angles.
(iv) Let no. of sides = n
(n – 2) x 180° = 4500°
n – 2 = \(\frac { 4500 }{ 180 }\)
n – 2 = 25
n = 25 + 2
n = 27
Which is a whole number.
Hence it is possible to have a polygon, the sum of whose interior angles is 4500°.

Question 5.
(i) If all the angles of a hexagon are equal ; find the measure of each angle.
(ii) If all the angles of a 14-sided figure are equal ; find the measure of each angle.
Solution:
(i) No. of sides of hexagon, n = 6
Let each angle be = x°
Sum of angles = 6x°
(n – 2) x 180° = Sum of angles
(6 – 2) x 180° = 6x°
4 x 180 = 6x
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 2

Question 6.
Find the sum of exterior angles obtained on producing, in order, the sides of a polygon with :
(i) 7 sides
(ii) 10 sides
(iii) 250 sides.
Solution:
(i) No. of sides n = 7
Sum of interior & exterior angles at one vertex = 180°
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 3

Question 7.
The sides of a hexagon are produced in order. If the measures of exterior angles so obtained are (6x – 1)°, (10x + 2)°, (8x + 2)° (9x – 3)°, (5x + 4)° and (12x + 6)° ; find each exterior angle.
Solution:
Sum of exterior angles of hexagon formed by producing sides of order = 360°
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 5
i.e. 41° ; 72°, 58° ; 60° ; 39° and 90°

Question 8.
The interior angles of a pentagon are in the ratio 4 : 5 : 6 : 7 : 5. Find each angle of the pentagon.
Solution:
Let the interior angles of the pentagon be 4x, 5x, 6x, 7x, 5x.
Their sum = 4x + 5x + 6x + 7x + 5x = 21x
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 6

Question 9.
Two angles of a hexagon are 120° and 160°. If the remaining four angles are equal, find each equal angle.
Solution:
Two angles of a hexagon are 120°, 160°
Let remaining four angles be x, x, x and x.
Their sum = 4x + 280°
But sum of all the interior angles of a hexagon
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 7

Question 10.
The figure, given below, shows a pentagon ABCDE with sides AB and ED parallel to each other, and ∠B : ∠C : ∠D = 5 : 6 : 7.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 8
(i) Using formula, find the sum of interior angles of the pentagon.
(ii) Write the value of ∠A + ∠E
(iii) Find angles B, C and D.
Solution:
(i) Sum of interior angles of the pentagon

Question 11.
Two angles of a polygon are right angles and the remaining are 120° each. Find the number of sides in it.
Solution:
Let number of sides = n
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 10
n = \(\frac { 300 }{ 60 }\)
n = 5

Question 12.
In a hexagon ABCDEF, side AB is parallel to side FE and ∠B : ∠C : ∠D : ∠E = 6 : 4 : 2 : 3. Find ∠B and ∠D.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 11

Question 13.
the angles of a hexagon are x + 10°, 2x + 20°, 2x – 20°, 3x – 50°, x + 40° and x + 20°. Find x.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 12

Question 14.
In a pentagon, two angles are 40° and 60°, and the rest are in the ratio 1 : 3 : 7. Find the biggest angle of the pentagon.
Solution:
In a pentagon, two angles are 40° and 60° Sum of remaining 3 angles = 3 x 180°
= 540° – 40° – 60° = 540° – 100° = 440°
Ratio in these 3 angles =1 : 3 : 7
Sum of ratios =1 + 3 + 7 = 11
Biggest angle = \(\frac { 440\times 7 }{ 11 }\) = 280°

Understanding Shapes Exercise 16B – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Fill in the blanks :
In case of regular polygon, with :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 14
Solution:

Question 2.
Find the number of sides in a regular polygon, if its each interior angle is :
(i) 160°
(ii) 135°
(iii) \(1\frac { 1 }{ 5 }\) of a right-angle
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 17

Question 3.
Find the number of sides in a regular polygon, if its each exterior angle is :
(i) \(\frac { 1 }{ 3 }\) of a right angle
(ii) two-fifth of a right-angle.
Solution:
(i) Each exterior angle = \(\frac { 1 }{ 3 }\) of a right angle
= \(\frac { 1 }{ 3 }\) x 90
= 30°
Let number of sides = n
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 19

Question 4.
Is it possible to have a regular polygon whose each interior angle is :
(i) 170°
(ii) 138°
Solution:
(i) No. of sides = n
each interior angle = 170°
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 20

Which is not a whole number.
Hence it is not possible to have a regular polygon having interior angle of 138°.

Question 5.
Is it possible to have a regular polygon whose each exterior angle is :
(i) 80°
(ii) 40% of a right angle.
Solution:
(i) Let no. of sides = n each exterior angle = 80°
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 22
Which is not a whole number.
Hence it is not possible to have a regular polygon whose each exterior angle is of 80°.
(ii) Let number of sides = n
Each exterior angle = 40% of a right angle

Which is a whole number.
Hence it is possible to have a regular polygon whose each exterior angle is 40% of a right angle.

Question 6.
Find the number of sides in a regular polygon, if its interior angle is equal to its exterior angle.
Solution:
Let each exterior angle or interior angle be = x°
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 24

Question 7.
The exterior angle of a regular polygon is one-third of its interior angle. Find the number of sides in the polygon.
Solution:
Let interior angle = x°
Exterior angle = \(\frac { 1 }{ 3 }\) x°
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 25

Question 8.
The measure of each interior angle of a regular polygon is five times the measure of its exterior angle. Find :
(i) measure of each interior angle ;
(ii) measure of each exterior angle and
(iii) number of sides in the polygon.
Solution:
Let exterior angle = x°
Interior angle = 5x°
x + 5x = 180°
6x = 180°
x = 30°
Each exterior angle = 30°
Each interior angle = 5 x 30° = 150°
Let no. of sides = n
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 27

Question 9.
The ratio between the interior angle and the exterior angle of a regular polygon is 2 : 1. Find :
(i) each exterior angle of the polygon ;
(ii) number of sides in the polygon
Solution:
Interior angle : exterior angle = 2 : 1
Let interior angle = 2x° & exterior angle = x°
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 28

Question 10.
The ratio between the exterior angle and the interior angle of a regular polygon is 1 : 4. Find the number of sides in the polygon.
Solution:
Let exterior angle = x° & interior angle = 4x°
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 29

Question 11.
The sum of interior angles of a regular polygon is twice the sum of its exterior angles. Find the number of sides of the polygon.
Solution:
Let number of sides = n
Sum of exterior angles = 360°
Sum of interior angles = 360° x 2 = 720°
Sum of interior angles = (n – 2) x 180°
720° = (n – 2) x 180°
n – 2 = \(\frac { 720 }{ 180 }\)
n – 2 = 4
n = 4 + 2
n = 6

Question 12.
AB, BC and CD are three consecutive sides of a regular polygon. If angle BAC = 20° ; find :
(i) its each interior angle,
(ii) its each exterior angle
(iii) the number of sides in the polygon.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 30

Question 13.
Two alternate sides of a regular polygon, when produced, meet at the right angle. Calculate the number of sides in the polygon.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 31

Question 14.
In a regular pentagon ABCDE, draw a diagonal BE and then find the measure of:
(i) ∠BAE
(ii) ∠ABE
(iii) ∠BED
Solution:
(i) Since number of sides in the pentagon = 5
Each exterior angle = \(\frac { 360 }{ 5 }\) = 72°
∠BAE = 180° – 72°= 108°
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 32

Question 15.
The difference between the exterior angles of two regular polygons, having the sides equal to (n – 1) and (n + 1) is 9°. Find the value of n.
Solution:
We know that sum of exterior angles of a polynomial is 360°
(i) If sides of a regular polygon = n – 1
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 33

Question 16.
If the difference between the exterior angle of a n sided regular polygon and an (n + 1) sided regular polygon is 12°, find the value of n.
Solution:
We know that sum of exterior angles of a polygon = 360°
Each exterior angle of a regular polygon of 360°
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 35

Question 17.
The ratio between the number of sides of two regular polygons is 3 : 4 and the ratio between the sum of their interior angles is 2 : 3. Find the number of sides in each polygon.
Solution:
Ratio of sides of two regular polygons = 3 : 4
Let sides of first polygon = 3n
and sides of second polygon = 4n
Sum of interior angles of first polygon
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 37

Question 18.
Three of the exterior angles of a hexagon are 40°, 51 ° and 86°. If each of the remaining exterior angles is x°, find the value of x.
Solution:
Sum of exterior angles of a hexagon = 4 x 90° = 360°
Three angles are 40°, 51° and 86°
Sum of three angle = 40° + 51° + 86° = 177°
Sum of other three angles = 360° – 177° = 183°
Each angle is x°
3x = 183°
x = \(\frac { 183 }{ 3 }\)
Hence x = 61

Question 19.
Calculate the number of sides of a regular polygon, if:
(i) its interior angle is five times its exterior angle.
(ii) the ratio between its exterior angle and interior angle is 2 : 7.
(iii) its exterior angle exceeds its interior angle by 60°.
Solution:
Let number of sides of a regular polygon = n
(i) Let exterior angle = x
Then interior angle = 5x
x + 5x = 180°
=> 6x = 180°
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 38

Question 20.
The sum of interior angles of a regular polygon is thrice the sum of its exterior angles. Find the number of sides in the polygon.
Solution:
Sum of interior angles = 3 x Sum of exterior angles
Let exterior angle = x
The interior angle = 3x
x + 3x=180°
=> 4x = 180°
=> x = \(\frac { 180 }{ 4 }\)
=> x = 45°
Number of sides = \(\frac { 360 }{ 45 }\) = 8

Understanding Shapes Exercise 16C – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Two angles of a quadrilateral are 89° and 113°. If the other two angles are equal; find the equal angles.
Solution:
Let the other angle = x°
According to given,
89° + 113° + x° + x° = 360°
2x° = 360° – 202°
2x° = 158°
x° = \(\frac { 158 }{ 2 }\)
other two angles = 79° each

Question 2.
Two angles of a quadrilateral are 68° and 76°. If the other two angles are in the ratio 5 : 7; find the measure of each of them.
Solution:
Two angles are 68° and 76°
Let other two angles be 5x and 7x
68° + 76°+ 5x + 7x = 360°
12x + 144° = 360°
12x = 360° – 144°
12x = 216°
x = 18°
angles are 5x and 7x
i.e. 5 x 18° and 7 x 18° i.e. 90° and 126°

Question 3.
Angles of a quadrilateral are (4x)°, 5(x+2)°, (7x – 20)° and 6(x+3)°. Find :
(i) the value of x.
(ii) each angle of the quadrilateral.
Solution:
Angles of quadrilateral are,
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 40

Question 4.
Use the information given in the following figure to find :
(i) x
(ii) ∠B and ∠C
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 42
Solution:

Question 5.
In quadrilateral ABCD, side AB is parallel to side DC. If ∠A : ∠D = 1 : 2 and ∠C : ∠B = 4 : 5
(i) Calculate each angle of the quadrilateral.
(ii) Assign a special name to quadrilateral ABCD
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 44

Question 6.
From the following figure find ;
(i) x
(ii) ∠ABC
(iii) ∠ACD
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 45
(i) In Quadrilateral ABCD,
x + 4x + 3x + 4x + 48° = 360°
12x = 360° – 48°
12x = 312

Question 7.
Given : In quadrilateral ABCD ; ∠C = 64°, ∠D = ∠C – 8° ; ∠A = 5(a+2)° and ∠B = 2(2a+7)°.
Calculate ∠A.
Solution:
∠C = 64° (Given)
∠D = ∠C – 8° = 64°- 8° = 56°
∠A = 5(a+2)°
∠B = 2(2a+7)°
Now ∠A + ∠B + ∠C + ∠D = 360°
5(a+2)° + 2(2a+7)° + 64° + 56° = 360°
5a + 10 + 4a + 14° + 64° + 56° = 360°
9a + 144° = 360°
9a = 360° – 144°
9a = 216°
a = 24°
∠A = 5 (a + 2) = 5(24+2) = 130°

Question 8.
In the given figure : ∠b = 2a + 15 and ∠c = 3a + 5; find the values of b and c.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 46
Solution:
Stun of angles of quadrilateral = 360°
70° + a + 2a + 15 + 3a + 5 = 360°
6a + 90° = 360°
6a = 270°
a = 45°
b = 2a + 15 = 2 x 45 + 15 = 105°
c = 3a + 5 = 3 x 45 + 5 = 140°
Hence ∠b and ∠c are 105° and 140°

Question 9.
Three angles of a quadrilateral are equal. If the fourth angle is 69°; find the measure of equal angles.
Solution:
Let each equal angle be x°
x + x + x + 69° = 360°
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 47
3x = 360°- 69
3x = 291
x = 97°
Each, equal angle = 97°

Question 10.
In quadrilateral PQRS, ∠P : ∠Q : ∠R : ∠S = 3 : 4 : 6 : 7.
Calculate each angle of the quadrilateral and then prove that PQ and SR are parallel to each other
(i) Is PS also parallel to QR ?
(ii) Assign a special name to quadrilateral PQRS.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 48

Question 11.
Use the informations given in the following figure to find the value of x.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 49
Solution:
Take A, B, C, D as the vertices of Quadrilateral and BA is produced to E (say).
Since ∠EAD = 70°
∠DAB = 180° – 70°= 110°
[EAB is a straight line and AD stands on it ∠EAD+ ∠DAB = 180°]
110° + 80° + 56° + 3x – 6° = 360°
[sum of interior angles of a quadrilateral = 360°]
3x = 360° – 110° – 80° – 56° + 6°
3x = 360° – 240° = 120°
x = 40°

Question 12.
The following figure shows a quadrilateral in which sides AB and DC are parallel. If ∠A : ∠D = 4 : 5, ∠B = (3x – 15)° and ∠C = (4x + 20)°, find each angle of the quadrilateral ABCD.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 50
Solution:
Let ∠A = 4x
∠D = 5x
Since ∠A + ∠D = 180° [AB||DC]
4x + 5x = 180°
=> 9x = 180°
=> x = 20°
∠A = 4 (20) = 80°,
∠D = 5 (20) = 100°
Again ∠B + ∠C = 180° [ AB||DC]
3x – 15° + 4x + 20° = 180°
7x = 180° – 5°
=> 7x = 175°
=> x = 25°
∠B = 75° – 15° = 60°
and ∠C = 4 (25) + 20 = 100°+ 20°= 120°

Question 13.
Use the following figure to find the value of x
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 51
Solution:
The sum of exterior angles of a quadrilateral

=> y + 80° + 60° + 90° = 360°
=> y + 230° = 360°
=> y = 360° – 230° = 130°
At vertex A,
∠y + ∠x = 180° (Linear pair)
x = 180° – 130°
=> x = 50°

Question 14.
ABCDE is a regular pentagon. The bisector of angle A of the pentagon meets the side CD in point M. Show that ∠AMC = 90°.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 53
Given : ABCDE is a regular pentagon.
The bisector ∠A of the pentagon meets the side CD at point M.
To prove : ∠AMC = 90°
Proof: We know that, the measure of each interior angle of a regular pentagon is 108°.
∠BAM = \(\frac { 1 }{ 2 }\) x 108° = 54°
Since, we know that the sum of a quadrilateral is 360°
In quadrilateral ABCM, we have
∠BAM + ∠ABC + ∠BCM + ∠AMC = 360°
54° + 108° + 108° + ∠AMC = 360°
∠AMC = 360° – 270°
∠AMC = 90°

Question 15.
In a quadrilateral ABCD, AO and BO are bisectors of angle A and angle B respectively. Show that:
∠AOB = \(\frac { 1 }{ 2 }\) (∠C + ∠D)
Solution:
Given : AO and BO are the bisectors of ∠A and ∠B respectively.
∠1 = ∠4 and ∠3 = ∠5 ……..(i)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes image - 54
To prove : ∠AOB = \(\frac { 1 }{ 2 }\) (∠C + ∠D)
Proof: In quadrilateral ABCD
∠A + ∠B + ∠C + ∠D = 360°
\(\frac { 1 }{ 2 }\) (∠A + ∠B + ∠C + ∠D) = 180° …………(ii)
Now in ∆AOB
∠1 + ∠2 + ∠3 = 180° ………(iii)
Equating equation (ii) and equation (iii), we get
∠1 + ∠2 + ∠3 = ∠A + ∠B + \(\frac { 1 }{ 2 }\) (∠C + ∠D)
∠1 + ∠2 + ∠3 = ∠1 + ∠3 + \(\frac { 1 }{ 2 }\) (∠C + ∠D)
∠2 = \(\frac { 1 }{ 2 }\) (∠C + ∠D)
∠AOB = \(\frac { 1 }{ 2 }\) (∠C + ∠D)
Hence proved.

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities

August 11, 2022January 27, 2019 by Prasanna

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities

Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Mathematics Chapter 12 Algebraic Identities. You can download the Selina Concise Mathematics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Mathematics for Class 8 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

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Algebraic Identities Exercise 12A – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Use direct method to evaluate the following products :
(i) (x + 8)(x + 3)
(ii) (y + 5)(y – 3)
(iii) (a – 8)(a + 2)
(iv) (b – 3)(b – 5)
(v) (3x – 2y)(2x + y)
(vi) (5a + 16)(3a – 7)
(vii) (8 – b) (3 + b)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -1

Question 2.
Use direct method to evaluate :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -3
Solution:

Question 3.
Evaluate :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -7
Solution:

Question 4.
Use the product (a + b) (a – b) = a² – b² to evaluate:
(i) 21 x 19
(ii) 33 x 27
(iii) 103 x 97
(iv) 9.8 x 10.2
(v) 7.7 x 8.3
(vi) 4.6 x 5.4
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -10

Question 5.
Evaluate :
(i) (6 – xy) (6 + xy)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -11
Solution:

Algebraic Identities Exercise 12B – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Expand :
(i) (2a + b)²
(ii) (a – 2b)²
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -15
Solution:

Question 2.
Find the square of :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -19
Solution:

Question 3.
Evaluate:
Using expansion of (a + b)² or (a – b)²
(i) (208)²
(ii) (92)²
(iii)(415)²
(iv) (188)²
(v) (9.4)²
(vi) (20.7)²
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -23

Question 4.
Expand :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -25
Solution:

Question 5.
Find the cube of :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -28
Solution:

Algebraic Identities Exercise 12C – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
If a+b=5 and ab = 6; find a² + b²
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -31

Question 2.
If a – b = 6 and ab = 16; find a² + b²
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -32

Question 3.
If a² + b² = 29 and ab = 10 ; find :
(i) a + b
(ii) a – b
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -33

Question 4.
If a² + b²= 10 and ab = 3; find :
(i) a – b
(ii) a + b
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -34

Question 5.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -35
Solution:

Question 6.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -38
Solution:

Question 7.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -41
Solution:

Question 8.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -42
Solution:

Question 9.
If a + b + c = 10 and a² + b² + c² = 38; find ab + bc + ca
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -46

Question 10.
Find a² + b² + c² ; if a + b + c = 9 and ab + bc + ca = 24
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -47

Question 11.
Find a + b + c; if a² + b² + c² = 83 and ab + bc + ca = 71
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -49

Question 12.
If a + b = 6 and ab=8; find a³ + b³
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -50

Question 13.
If a – b=3 and ab = 10; find a³ – b³
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -51

Question 14.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -53
Solution:

Question 15.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -55
Solution:

Question 16.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -58
Solution:

Question 17.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -62
Solution:

Question 18.
The sum of the squares of two numbers is 13 and their product is 6. Find:
(i) the sum of the two numbers.
(ii) the difference between them.
Solution:
Let x and y be the two numbers, then
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -65

Algebraic Identities Exercise 12D – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Evaluate:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -66
Solution:

Question 2.
Evaluate:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -69
Solution:

Question 3.
Find the square of :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -75
Solution:

Question 4.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -78
Solution:

Question 5.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -80
Solution:

Question 6.
If a² + b² = 41 and ab = 4, find :
(i) a – b
(ii) a + b
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -82

Question 7.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -83
Solution:

Question 8.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -85.
Solution:

Question 9.
Expand :
(i) (3x – 4y + 5z)²
(ii) (2a – 5b – 4c)²
(iii) (5x + 3y)³
(iv) (6a – 7b)³
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -87

Question 10.
If a + b + c = 9 and ab + bc + ca = 15, find: a² + b² + c².
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -89

Question 11.
If a + b + c = 11 and a² + b² + c² = 81, find ab + bc + ca.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -90

Question 12.
If 3x – 4y = 5 and xy = 3, find : 27x³ – 64y³.
Solution:
27x³ – 64x³ = (3x)³ – (4y)³
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -91

Question 13.
If a + b = 8 and ab = 15, find : a³ + b³.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -92

Question 14.
If 3x + 2y = 9 and xy = 3, find : 27x³ + 8y³
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -93

Question 15.
If 5x – 4y = 7 and xy = 8, find : 125x³ – 64y³
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -94

Question 16.
The difference between two numbers is 5 and their products is 14. Find the difference between their cubes.
Solution:
Let x and y be two numbers, then x – y = 5 and xy = 14
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 12 Algebraic Identities image -95

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots

August 11, 2022January 27, 2019 by Prasanna

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots

Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Mathematics Chapter 3 Squares and Square Roots. You can download the Selina Concise Mathematics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Mathematics for Class 8 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

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Squares and Square Roots Exercise 3A – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Find the square of :
(i) 59
(ii) 63
(iii) 15
Solution:
(i) Square of 59= 59 x 59 = 3481
(ii) Square of 6.3 = 6.3 x 6.3 = 39.69
(iii) Square of 15 = 15 x 15 = 225

Question 2.
By splitting into prime factors, find the square root of :
(i) 11025
(if) 396900
(iii) 194481
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -1

Question 3.
(i) Find the smallest number by which 2592 be multiplied so that the product is a perfect square.
(ii) Find the smallest number by which 12748 be mutliplied so that the product is a perfect square?
Solution:

On grouping the prime factors of 2592 as shown; on factor i.e. 2 is left which cannot be paired with equal factor.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -4
The given number should be multiplied by 2 to make the given number a perfect square.

On grouping the prime factors of 12748 as shown; one factor i.e. 3187 is left which cannot be paired with equal factor.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -6
The given number should be multiplied by 3187.

Question 4.
Find the smallest number by which 10368 be divided, so that the result is a perfect square. Also, find the square root of the resulting numbers.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -7

Question 5.
Find the square root of :
(i) 0.1764
(ii) \(96\frac { 1 }{ 25 }\)
(iii) 0.0169
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -8

Question 6.
Evaluate
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -9

Solution:

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -11

Question 7.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -14
Solution:

Question 8.
A man, after a tour, finds that he had spent every day as many rupees as the number of days he had been on tour. How long did his tour last, if he had spent in all ₹ 1,296
Solution:
Let the number of days he had spent = x
Number of rupees spent in each day = x
Total money spent = x x x = x² = 1,296 (given)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -18

Question 9.
Out of 745 students, maximum are to be arranged in the school field for a P.T. display, such that the number of rows is equal to the number of columns. Find the number of rows if 16 students were left out after the arrangement.
Solution:
Total number of students = 745
Students left after standing in arrangement = 16
No. of students who were to be arranged = 745 – 16 = 729
The number of rows = no. of students in each row
No. of rows = √729
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -19

Question 10.
13 and 31 is a strange pair of numbers such that their squares 169 and 961 are also mirror images of each other. Find two more such pairs.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -20

Question 11.
Find the smallest perfect square divisible by 3, 4, 5 and 6.
Solution:
L.C.M. of 3, 4, 5, 6 = 2 x 2 x 3 x 5 = 60
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -21
in which 3 and 5 are not in pairs L.C.M. = 2 x 3 x 2 x 5 = 60
We should multiple it by 3 x 5 i.e. by 15
Required perfect square = 60 x 15 = 900

Question 12.
If √784 = 28, find the value of:
(i) √7.84 + √78400
(ii) √0.0784 + √0.000784
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -22

Squares and Square Roots Exercise 3B – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Find the square root of:
(i) 4761
(ii) 7744
(iii) 15129
(iv) 0.2916
(v) 0.001225
(vi) 0.023104
(vii) 27.3529
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -23

Question 2.
Find the square root of:
(i) 4.2025
(ii) 531.7636
(iii) 0.007225
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -26

Question 3.
Find the square root of:
(i) 245 correct to two places of decimal.
(ii) 496 correct to three places of decimal.
(iii) 82.6 correct to two places of decimal.
(iv) 0.065 correct to three places of decimal.
(v) 5.2005 correct to two places of decimal.
(vi) 0.602 correct to two places of decimal
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -28

Required square root = 0.78 upto two places of decimals.

Question 4.
Find the square root of each of the following correct to two decimal places:
(i) \(3\frac { 4 }{ 5 }\)
(ii) \(6\frac { 7 }{ 8 }\)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -31

Question 5.
For each of the following, find the least number that must be subtracted so that the resulting number is a perfect square.
(i) 796
(ii) 1886
(iii) 23497
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -33

Question 6.
For each of the following, find the least number that must be added so that the resulting number is a perfect square.
(i) 511
(ii) 7172
(iii) 55078
Solution:
(i) 511
Taking square root of 511, we find that 27 has been left We see that 511 is greater than (22)²
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -35
On adding the required number to 511, we get (23)² i.e., 529
So, the required number = 529 – 511 = 18
(ii) 7172
Taking square root of 7172, we find that 116 has been left
We see that 7172 is greater than (84)²
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -36
Taking square root of 55078, we find that 322 has been left
We see that 55078 is greater than (234)²
On adding the required number to 55078, we get (235)² i.e., 55225
Required number = 55225 – 55078 = 147

Question 7.
Find the square root of 7 correct to two decimal places; then use it to find the value of \(\sqrt { \frac { 4+\sqrt { 7 } }{ 4-\sqrt { 7 } } }\) correct to three significant digits.
Solution:
√7 = 2.645 = 2.65
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -37

Question 8.
Find the value of √5 correct to 2 decimal places; then use it to find the square root of \(\sqrt { \frac { 3-\sqrt { 5 } }{ 3+\sqrt { 5 } } }\) correct to 2 significant digits.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -38

Question 9.
Find the square root of:
(i) \(\frac { 1764 }{ 2809 }\)
(ii) \(\frac { 507 }{ 4107 }\)
(iii) \(\sqrt { 108\times 2028 }\)
(iv) 0.01 + √0.0064
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -40

Question 10.
Find the square root of 7.832 correct to :
(i) 2 decimal places
(ii) 2 significant digits.
Solution:
Square root of 7.832
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -42
√7.832 = 2.80 upto two decimal places
= 2.8 upto two significant places

Question 11.
Find the least number which must be subtracted from 1205 so that the resulting number is a perfect square.
Solution:
Clearly, if 49 is subtracted from 1205, the number will be a perfect square.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -43

Question 12.
Find the least number which must be added to 1205 so that the resulting number is a perfect square.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -44

Question 13.
Find the least number which must be subtracted from 2037 so that the resulting number is a perfect square.
Solution:
Clearly; if 12 is subtracted from 2037, the remainder will be a perfect square.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -45

Question 14.
Find the least number which must be added to 5483 so that the resulting number is a perfect square.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -46

Squares and Square Roots Exercise 3A – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Seeing the value of the digit at unit’s place, state which of the following can be square of a number :
(i) 3051
(ii) 2332
(iii) 5684
(iv) 6908
(v) 50699
Solution:
We know that the ending digit (the digit at units place) of the square of a number is 0, 1, 4, 5, 6, or 9
So, the following numbers can be squares : 3051, 5684, and 50699 i.e., (i), (iii), and (v)

Question 2.
Squares of which of the following numbers will have 1 (one) at their unit’s place :
(i) 57
(ii) 81
(iii) 139
(iv) 73
(v) 64
Solution:
The square of the following numbers will have 1 at their units place as (1)² = 1, (9)² = 81
81 and 139 i.e., (i) and (iii)

Question 3.
Which of the following numbers will not have 1 (one) at their unit’s place :
(i) 322
(ii) 572
(iii) 692
(iv) 3212
(v) 2652
Solution:
The square of the following numbers will not have 1 at their units place : as only (1)² = 1, (9)² = 81 have 1 at then units place
322, 572, 2652 i.e., (i), (ii) and (v)

Question 4.
Square of which of the following numbers will not have 6 at their unit’s place :
(i) 35
(ii) 23
(iii) 64
(iv) 76
(v) 98
Solution:
The squares of the following numbers, Will not have 6 at their units place as only (4)² = 16, (6)² = 36 has but its units place 35, 23 and 98 i.e., (i), (ii), and (v)

Question 5.
Which of the following numbers will have 6 at their unit’s place :
(i) 262
(ii) 492
(iii) 342
(iv) 432
(v) 2442
Solution:
The following numbers have 6 at their units place as (4)² = 16, (6)² = 36 has 6 at their units place 262, 342, 2442 i.e., (i), (iii) and (v)

Question 6.
If a number ends with 3 zeroes, how many zeroes will its square have ?
Solution:
We know that if a number ends with n zeros, then its square will have 2n zeroes at their ends
A number ends with 3 zeroes, then its square will have 3 x 2 = 6 zeroes

Question 7.
If the square of a number ends with 10 zeroes, how many zeroes will the number have ?
Solution:
We know that if a number ends with n zeros Then its square will have 2n zeroes Conversely, if square of a number have 2n zeros at their ends then the number will have n zeroes
The square of a number ends 10 zeroes, then the number will have \(\frac { 10 }{ 2 }\) = 5 zeroes

Question 8.
Is it possible for the square of a number to end with 5 zeroes ? Give reason.
Solution:
No, it is not possible for the square of a number, to have 5 zeroes which is odd because the number of zeros of the square must be 2n zeroes i.e., even number of zeroes.

Question 9.
Give reason to show that none of the numbers, given below, is a perfect square.
(i) 2162
(ii) 6843
(iii) 9637
(iv) 6598
Solution:
A number having 2,3,7 or 8 at the unit place is never a perfect square.

Question 10.
State, whether the square of the following numbers is even or odd?
(i) 23
(ii) 54
(iii) 76
(iv) 75
Solution:
(i) 23 – odd
(ii) 54 – even
(iii) 76 – odd
(iv) 75 – even

Question 11.
Give reason to show that none of the numbers 640, 81000 and 3600000 is a perfect square.
Solution:
No, number has an even number of zeroes.

Question 12.
Evaluate:
(i) 37² – 36²
(ii) 85² – 84²
(iii) 101² – 100²
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -47

Question 13.
Without doing the actual addition, find the sum of:
(i) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23
(ii) 1 + 3 + 5 + 7 + 9 + ……………… + 39 + 41
(iii) 1 + 3 + 5 + 7 + 9 + ………………… + 51 + 53
Solution:
(i) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 +23
= Sum of first 12 odd natural numbers = 122 = 144
(ii) 1+3 + 5 + 7 + 9 + ……….. + 39 + 41
= Sum of first 21 odd natural numbers = 212 = 441
(iii) 1 + 3 + 5 + 7 + 9 + ……………. + 51 + 53
= Sum of first 27 odd natural number = 272 = 729

Question 14.
Write three sets of Pythagorean triplets such that each set has numbers less than 30.
Solution:
The three sets of Pythagorean triplets such that each set has numbers less than 30 are 3, 4 and 5; 6, 8 and 10; 5, 12 and 13
Proof:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots image -49

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable

August 11, 2022January 27, 2019 by Prasanna

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable (With Problems Based on Linear equations)

Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable (With Problems Based on Linear equations)

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APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Mathematics Chapter 14 Linear Equations in one Variable (With Problems Based on Linear equations). You can download the Selina Concise Mathematics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Mathematics for Class 8 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

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Linear Equations in one Variable Exercise 14A – Selina Concise Mathematics Class 8 ICSE Solutions

Solve the following equations:
Question 1.
20 = 6 + 2x
Solution:
20 = 6 + 2x
20 – 6 = 2x
14 = 2x
7 = x
x = 7

Question 2.
15 + x = 5x + 3
Solution:
15 – 3 = 5x – x
12 = 4x
3 = x
x = 3

Question 3.
\(\frac { 3x+2 }{ x-6 }\) = -7
Solution:
3x + 2 = -7 (x – 6) (by cross multiplying)
3x + 2 = -7x + 42
3x + 7x = 42 – 2
10x = 40
x = 4

Question 4.
3a – 4 = 2(4 – a)
Solution:
3a – 4 = 8 – 2a
3a + 2a = 8+4
5a = 12
a = 2.4

Question 5.
3(b – 4) = 2(4 – b)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 1

Question 6.
\(\frac { x+2 }{ 9 } =\frac { x+4 }{ 11 }\)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 2

Question 7.
\(\frac { x-8 }{ 5 } =\frac { x-12 }{ 9 }\)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 3

Question 8.
5(8x + 3) = 9(4x + 7)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 4

Question 9.
3(x +1) = 12 + 4(x – 1)
Solution:
3(x + 1) = 12 + 4(x – 1)
3x + 3 = 12 + 4x – 4
3x – 4x = 12 – 4 – 3
-x = 5
x = -5

Question 10.
\(\frac { 3x }{ 4 } -\frac { 1 }{ 4 } \left( x-20 \right) =\frac { x }{ 4 } +32\)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 5

Question 11.
\(3a-\frac { 1 }{ 5 } =\frac { a }{ 5 } +5\frac { 2 }{ 5 }\)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 6

Question 12.
\(\frac { x }{ 3 } -2\frac { 1 }{ 2 } =\frac { 4x }{ 9 } -\frac { 2x }{ 3 }\)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 7

Question 13.
\(\frac { 4\left( y+2 \right) }{ 5 } =7+\frac { 5y }{ 13 }\)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 9

Question 14.
\(\frac { a+5 }{ 6 } -\frac { a+1 }{ 9 } =\frac { a+3 }{ 4 }\)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 8

Question 15.
\(\frac { 2x-13 }{ 5 } -\frac { x-3 }{ 11 } =\frac { x-9 }{ 5 } +1\)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 13

Question 16.
6(6x – 5) – 5 (7x – 8) = 12 (4 – x) + 1
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 14

Question 17.
(x – 5) (x + 3) = (x – 7) (x + 4)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 15

Question 18.
(x – 5)2 – (x + 2)2 = -2
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 16

Question 19.
(x – 1) (x + 6) – (x – 2) (x – 3) = 3
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 18

Question 20.
\(\frac { 3x }{ x+6 } -\frac { x }{ x+5 } =2\)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 19

Question 21.
\(\frac { 1 }{ x-1 } +\frac { 2 }{ x-2 } =\frac { 3 }{ x-3 }\)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 20

Question 22.
\(\frac { x-1 }{ 7x-14 } =\frac { x-3 }{ 7x-26 }\)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 22

Question 23.
\(\frac { 1 }{ x-1 } -\frac { 1 }{ x } =\frac { 1 }{ x+3 } -\frac { 1 }{ x+4 }\)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 23

Question 24.
Solve: \(\frac { 2x }{ 3 } -\frac { x-1 }{ 6 } +\frac { 7x-1 }{ 4 } =2\frac { 1 }{ 6 }\)
Hence, find the value of ‘a’, if \(\frac { 1 }{ a }\) + 5x = 8.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 24

Question 25.
Solve: \(\frac { 4-3x }{ 5 } +\frac { 7-x }{ 3 } +4\frac { 1 }{ 3 } =0\)
Hence find the value of ‘p’ if 2p – 2x + 1 = 0
Solution:

Hence x = 8
Now, 3p – 2x + 1=0
⇒ 3p – 2 x 8 + 1 = 0
⇒ 3p – 16 + 1 =0
⇒ 3p – 15 = 0.
⇒ 3p=15
⇒ p = 5

Question 26.
Solve: \(0.25+\frac { 1.95 }{ x } =0.9\)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 28

Question 27.
Solve: \(5x-\left( 4x+\frac { 5x-4 }{ 7 } \right) =\frac { 4x-14 }{ 3 }\)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 29

Linear Equations in one Variable Exercise 14B – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Fifteen less than 4 times a number is 9. Find the number.
Solution:
Let the required number be x
4 times the number = 4x
15 less than 4 times the number = 4x-15
According to the statement :
4x – 15 = 9
⇒ 4x = 9 + 15
⇒ 4x = 24
⇒ x = 6

Question 2.
If Megha’s age is increased by three times her age, the result is 60 years. Find her age
Solution:
Let Megha’s age = x years
Three times Megha’s age = 3x years
According to the statement :
x + 3x = 60
=> 4x = 60
=> x = 15
Megha’s age = 15 years

Question 3.
28 is 12 less than 4 times a number. Find the number.
Solution:
Let the required number be x
4 times the number = 4x
12 less than 4 times the number = 4x – 12
According to the statement
4x – 12 = 28
=> 4x = 28 + 12
=> 4x = 40
x = 10
Required number = 10

Question 4.
Five less than 3 times a number is -20. Find the number.
Solution:
Let the required number = x
3 times the number = 3x
5 less than 3 times the number = 3x – 5
According to statement :
3x – 5 = -20
=> 3x = -20 + 5
=> 3x = -15
=> x = -5
Required number = -5

Question 5.
Fifteen more than 3 times Neetu’s age is the same as 4 times her age. How old is she ?
Solution:
Let Neetu’s age = x years
3 times Neetu’s age = 3x years
Fifteen more than 3 times Neetu’s age = (3x + 15) years
4 times Neetu’s age = 4x
According to the statement :
4x = 3x + 15
=> 4x – 3x = 15
=> x = 15
Neetu’s age = 15 years

Question 6.
A number decreased by 30 is the same as 14 decreased by 3 times the number; Find the number.
Solution:
Let the required number = x
The number decreased by 30 = x – 30
14 decreased by 3 times the number = 14 – 3x
According to the statement :
x – 30 = 14 – 3x
=> x + 3x = 14 + 30
=> 4x = 44
x = 11
Required number =11

Question 7.
A’s salary is same as 4 times B’s salary. If together they earn Rs.3,750 a month, find the salary of each.
Solution:
Let B’s salary = Rs. x
A’s salary = Rs. 4x
According to the statement :
x + 4x = 3750
=> 5x = 3750
=> x = 750
4x = 750 x 4 = 3000
A’s salary = Rs. 3000
B’s salary = Rs. 750

Question 8.
Separate 178 into two parts so that the first part is 8 less than twice the second part.
Solution:
Let first part = x
Second part = 178 – x
According to the problem :
First Part = 8 less than twice the second part
x = 2(178 – x) – 8
=> x = 356 – 2x – 8
=> x+2x = 356 – 8
=> 3x = 348
=> x = 116
First Part = 116
=> Second Part = 178 – x = 178 – 116 = 62
First Part = 116
=> Second Part = 62

Alternative Method :
Let Second part = x
First part = 2x – 8
According to the problem :
x + 2x – 8 = 178
=> x + 2x = 178 + 8
=> 3x = 186
=> x = 62
First part = 2x – 8 = 2 x 62 – 8 = 124 – 8 = 116
First part = 116
Second part = 62

Question 9.
Six more than one-fourth of a number is two-fifth of the number. Find the number.
Solution:
Let the required number = x
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 30
x = 40
Required number = 40

Question 10.
The length of a rectangle is twice its width. If its perimeter is 54 cm; find its length.
Solution:
Let width of the rectangle = x cm
Length of the rectangle = 2x cm
Perimeter of the rectangle = 2 [Length + Width] = 2 [2x + x] = 2 x 3x = 6x cm
Given perimeter = 54 cm
6x = 54
=> x = 9
Length = 2x = 2 x 9 = 18 cm

Question 11.
A rectangle’s length is 5 cm less than twice its width. If the length is decreased by 5 cm and width is increased by 2 cm; the perimeter of the resulting rectangle will be 74 cm. Find the length and the width of the origi¬nal rectangle.
Solution:
Let width of the original rectangle = x cm
Length of the original rectangle = (2x – 5)cm
Now, new length of the rectangle = 2x – 5 – 5 = (2x – 10) cm
New width of the rectangle = (x + 2) cm
New perimeter = 2[Length+Width] = 2[2x – 10 + x + 2] = 2[3x – 8] = (6x – 16) cm
Given; new perimeter = 74 cm
6x – 16 = 74
=> 6x = 74 + 16
=> 6x = 90
=>x = 15
Length of the original rectangle = 2x – 5 = 2 x 15 – 5 = 30 – 5 = 25 cm
Width of the original rectangle = x = 15 cm

Question 12.
The sum of three consecutive odd numbers is 57. Find the numbers.
Solution:
Let the three consecutive odd numbers be x, x+2, x+4.
According to the statement :
x + x + 2 + x + 4 = 57
=> x + x + x = 57 – 2 – 4
=> 3x = 51
=> x = 17
Three consecutive odd numbers are 17, 19, 21

Question 13.
A man’s age is three times that of his son, and in twelve years he will be twice as old as his son would be. What are their present ages.
Solution:
Let present age of the son = x years
present age of the man = 3x years
In 12 years :
Son’s age will be = (x + 12) years
The man’s age will be = (3x + 12) years
According to the statement :
3x + 12 = 2(x + 12)
=> 3x + 12 = 2x + 24
=> 3x – 2x = 24 – 12
=> x = 12
3x = 3 x 12 = 36
Hence, present age of the man = 36 years
Present age of the son = 12 years.

Question 14.
A man is 42 years old and his son is 12 years old. In how many years will the age of the son be half the age of the man at that time?
Solution:
Man’s age = 42 years
Son’s age = 12 years
Let after x years the age of the son will be half the age of the man.
Man’s age after x years = (42 + x) years
Son’s age after x years = (12 + x) years
According to the statement :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 31
Hence after 18 years, the age of the son will be half the age of the man

Question 15.
A man completed a trip of 136 km in 8 hours. Some part of the trip was covered at 15 km/hr and the remaining at 18 km/hr. Find the part of the trip covered at 18 km/hr.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 32

Question 16.
The difference of two numbers is 3 and the difference of their squares is 69. Find the numbers.
Solution:
Let one number = x
Second number = x + 3 [Difference of two numbers is 3]
According to the statement :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 34
One number = 10
Second number = x + 3 = 10 + 3 = 13

Question 17.
Two consecutive natural numbers are such that one-fourth of the smaller exceeds one-fifth of the greater by 1. Find the numbers.
Solution:
Let two consecutive natural numbers = x, x+1
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 35
Two consecutive numbers are 24 and 25

Question 18.
Three consecutive whole numbers are such that if they be divided by 5, 3 and 4 respectively; the sum of the quotients is 40. Find the numbers.
Solution:
Let the three consecutive whole numbers be x, x + 1 and x + 2
According to the statement:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 36
x = 50
x + 1 = 50+1 = 51
x + 2 = 50 + 2 = 52
Three consecutive whole numbers are 50, 51 and 52

Question 19.
If the same number be added to the numbers 5, 11, 15 and 31, the resulting numbers are in proportion. Find the number.
Solution:
Let x be added to each number, then the numbers will be 5 + x, 11 + x, 15 + x and 31 + x
According to the condition
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 37
1 should be added

Question 20.
The present age of a man is twice that of his son. Eight years hence, their ages will be in the ratio 7 : 4. Find their present ages.
Solution:
Let present age of son = x year
Then age of his father = 2x
8 years hence,
Age of son = (x + 8) years and age of father = (2x + 8) years
According to the condition,
\(\frac { 2x+8 }{ x+8 } =\frac { 7 }{ 4 }\)
=> 8x + 32 = 7x + 56
=> 8x – 7x = 56 – 32
=> x = 24
Present age of son = 24 years
and age of father = 2x = 2 x 24 = 48 years
Hence age of man = 48 years and age of his son = 24 years

Linear Equations in one Variable Exercise 14C – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Solve:
(i) \(\frac { 1 }{ 3 } x-6=\frac { 5 }{ 2 }\)
(ii) \(\frac { 2x }{ 3 } -\frac { 3x }{ 8 } =\frac { 7 }{ 12 }\)
(iii) (x + 2)(x + 3) + (x – 3)(x – 2) – 2x(x + 1) = 0
(iv) \(\frac { 1 }{ 10 } -\frac { 7 }{ x } =35\)
(v) 13(x – 4) – 3(x – 9) – 5(x + 4) = 0
(vi) x + 7 – \(\frac { 8x }{ 3 } =\frac { 17x }{ 6 } -\frac { 5x }{ 8 }\)
(vii) \(\frac { 3x-2 }{ 4 } -\frac { 2x+3 }{ 3 } =\frac { 2 }{ 3 } -x\)
(viii) \(\frac { x+2 }{ 6 } -\left( \frac { 11-x }{ 3 } -\frac { 1 }{ 4 } \right) =\frac { 3x-4 }{ 12 }\)
(ix) \(\frac { 2 }{ 5x } -\frac { 5 }{ 3x } =\frac { 1 }{ 15 }\)
(x) \(\frac { x+2 }{ 3 } -\frac { x+1 }{ 5 } =\frac { x-3 }{ 4 } -1\)
(xi) \(\frac { 3x-2 }{ 3 } +\frac { 2x+3 }{ 2 } =x+\frac { 7 }{ 6 }\)
(xii) \(x-\frac { x-1 }{ 2 } =1-\frac { x-2 }{ 3 }\)
(xiii) \(\frac { 9x+7 }{ 2 } -\left( x-\frac { x-2 }{ 7 } \right) =36\)
(xiv) \(\frac { 6x+1 }{ 2 } +1=\frac { 7x-3 }{ 3 }\)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 38

Question 2.
After 12 years, I shall be 3 times as old as 1 was 4 years ago. Find my present age.
Solution:
Let present age = x years
According to question,
(x + 12) = 3(x – 4)
x + 12 = 3x – 12
2x = 24
=> x = 12 years
Present age = 12 years

Question 3.
A man sold an article for 7396 and gained 10% on it. Find the cost price of the article
Solution:
S.P. of article = ₹ 396
Gain = 10%
Let cost price = ₹ x
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 52
Cost price of an article = ₹ 360

Question 4.
The sum of two numbers is 4500. If 10% of one number is 12.5% of the other, find the numbers.
Solution:
Let the first number = x
and the second number = y
According to question,
x + y = 4500 ……(i)
and 10% x = 12.5% y
i.e. 10x = 12.5y
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 53
x = 2500
Hence, the numbers are 2500 and 2000

Question 5.
The sum of two numbers is 405 and their ratio is 8 : 7. Find the numbers.
Solution:
Let the first number = x
and the second number = 7
According to the question, x + y = 405 ……..(i)
and the numbers are in the ratio 8 : 7
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 54

x = 189
Hence, the numbers are 189 and 216

Question 6.
The ages of A and B are in the ratio 7 : 5. Ten years hence, the ratio of their ages will be 9 : 7. Find their present ages.
Solution:
Ratio in the present ages of A and B = 7 : 5
Let age of A = 7x years
Let age of B = 5x years
10 years hence,
Then age of A = 7x + 10 years
and age of B = 5x + 10 years
According to the condition,
\(\frac { 7x+10 }{ 5x+10 } =\frac { 9 }{ 7 }\)
By crossing multiplication
7(7x + 10) = 9(5x + 10)
=> 49x + 70 = 45x + 90
=> 49x – 45x = 90 – 70
=> 4x = 20
=> x = 5
Present age of A = 7x = 7 x 5 = 35 years
and present age of B = 5x = 5 x 5 = 25 years

Question 7.
Find the number whose double is 45 greater than its half.
Solution:
Let the required number = x
Double of it = 2x
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 56
Required number = 30

Question 8.
The difference between the squares of two consecutive numbers is 31. Find the numbers.
Solution:
Let first number = x
and The second number = x + 1
According to the condition,
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 57
First number = 15
and second number = 15 + 1 = 16
Hence, the numbers are 15, 16

Question 9.
Find a number such that when 5 is subtracted from 5 times the number, the result is 4 more than twice the number.
Solution:
Let the required number = x
5 times of it = 5x
Twice of it = 2x
According to the condition,
5x – 5 = 2x + 4
=> 5x – 2x = 4 + 5
=> 3x = 9
=> x = 3
Required number = 3

Question 10.
The numerator of a fraction is 5 less than its denominator. If 3 is added to the numerator, and denominator both, the fraction becomes \(\frac { 2 }{ 3 }\). Find the original fraction.
Solution:
Let denominator of the original fraction = x
Then numerator = x – 5
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable image - 58 .

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation

August 11, 2022January 27, 2019 by Prasanna

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation

Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Mathematics Chapter 13 Factorisation. You can download the Selina Concise Mathematics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Mathematics for Class 8 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

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Factorisation Exercise 13A – Selina Concise Mathematics Class 8 ICSE Solutions

Factorise :
Question 1.
15x + 5
Solution:
15x + 5 = 5(3x + 1)

Question 2.

Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.

Solution:

Question 6.

Solution:

Question 7.

Solution:

Question 8.

Solution:

Question 9.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 16

Question 10.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 18

Question 11.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 20

Question 12.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 22

Question 13.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 25

Question 14.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 27

Question 15.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 29

Question 16.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 30
Solution:

Question 17.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 33

Factorisation Exercise 13B – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Factorise : a² + ax + ab + bx
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 34

Question 2.
Factorise : a² – ab – ca + bc
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 35

Question 3.
Factorise : ab – 2b + a² – 2a
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 36

Question 4.
Factorise : a³ – a² + a – 1
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 37

Question 5.
Factorise : 2a – 4b – xa + 2bx
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 38

Question 6.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 40

Question 7.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 42

Question 8.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 44

Question 9.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 46

Question 10.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 48

Question 11.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 50

Question 12.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 52

Question 13.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 54

Question 14.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 56

Question 15.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 58

Question 16.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 60

Question 17.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 62

Question 18.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 64

Question 19.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 66

Question 20.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 68

Factorisation Exercise 13C – Selina Concise Mathematics Class 8 ICSE Solutions

Note : a² – b² = (a + b) (a – b)
Question 1.
Factorise : 16 – 9x²
Solution:
16 – 9x² = (4)² – (3x)² = (4 + 3x) (4 – 3x)

Question 2.
Factorise : 1 – 100a²
Solution:
1 – 100a² = (1)² – (10a)² = (1 + 10a) (1 – 10a)

Question 3.
Factorise : 4x² – 81y²
Solution:
4x² – 81y² = (2x)² – (9y)² = (2x + 9y) (2x – 9y)

Question 4.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 69
Solution:

Question 5.
Factorise : (a+2b)² – a²
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 71

Question 6.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 73

Question 7.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 75

Question 8.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 77

Question 9.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 79

Question 10.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 81

Question 11.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 83

Question 12.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 85

Question 13.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 86
Solution:

Question 14.
Evaluate :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 88
Solution:

Question 15.
Evaluate :
(0.7)² – (0.3)²
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 90

Question 16.
Evaluate :
(4.5)² – (1.5)²
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 91

Question 17.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 93

Question 18.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 95

Question 19.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 97

Question 20.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 99

Factorisation Exercise 13D – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 100
= (x+4) (x+2)

Question 2.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 102

Question 3.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 104

Question 4.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 106

Question 5.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 108

Question 6.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 110

Question 7.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 112

Question 8.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 114

Question 9.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 116

Question 10.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 118

Question 11.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 119
= (a – 1)(3a – 2)

Question 12.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 121

Question 13.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 123

Question 14.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 125

Question 15.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 127

Question 16.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 129

Question 17.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 132

Question 18.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 134

Question 19.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 136

Question 20.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 138

Question 21.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 140

Question 22.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 142

Question 23.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 144

Question 24.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 146

Question 25.

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 149

Factorisation Exercise 13E – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
In each case find whether the trinomial is a perfect square or not:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 150
Solution:

Question 2.
Factorise completely 2 – 8x².
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 153

Question 3.
Factorise completely : 8x²y – 18y³
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 154

Question 4.
Factorise completely : ax² – ay²
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 155

Question 5.
Factorise completely : 25x³ – x
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 156

Question 6.
Factorise completely : a⁴ – b⁴
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 157

Question 7.
Factorise completely : 16x⁴ – 81y⁴
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 158

Question 8.
Factorise completely : 625 – x⁴
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 159

Question 9.
Factorise completely : x² – y² – 3x – 3y
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 160

Question 10.
Factorise completely : x² – y² – 2x + 2y
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 162

Question 11.
Factorise completely : 3x² + 15x – 72
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 163

Question 12.
Factorise completely : 2a² – 8a – 64
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 164

Question 13.
Factorise completely : 5b² + 45b + 90
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 165

Question 14.
Factorise completely : 3x²y + 11xy + 6y
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 166

Question 15.
Factorise completely : 5ap² + 11ap + 2a
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 167

Question 16.
Factorise completely : a² + 2ab + b² – c²
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 168

Question 17.
Factorise completely : x² + 6xy + 9y² + x + 3y
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 169

Question 18.
Factorise completely : 4a² – 12ab + 9b² + 4a – 6b
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 170

Question 19.
Factorise completely : 2a²b² – 98b⁴
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 171

Question 20.
Factorise completely : a² – 16b² – 2a – 8b
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 172

Factorisation Exercise 13F – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Factorise :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 173
Solution:

Question 2.
Factorise :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 175
Solution:

Question 3.
Factorise :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 178
Solution:

Question 4.
Factorise :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 182
Solution:

Question 5.
Factorise :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 186
Solution:

Question 6.
Factorise :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 188

Solution:

Question 7.
Factorise xy² – xz², Hence, find the value of:

Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 192

Question 8.
Factorise :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 193
Solution:

Question 9.
Factorise a²b – b³ Using this result, find the value of 101² x 100 – 100³.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 197

Question 10.
Evaluate (using factors): 301² x 300 – 300³.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 13 Factorisation image - 198