Selina Class 8 ICSE Solutions Archives

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 15 Linear Inequations

November 9, 2023May 27, 2022 by Raju

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 15 Linear Inequations (Including Number Lines)

Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 15 Linear Inequations (Including Number Lines)

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 15 Linear Inequations (Including Number Lines). 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 Inequations Exercise 15A – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
If the replacement set is the set of natural numbers, solve.
(i) x – 5 < 0
(ii) x + 1 < 7
(iii) 3x – 4 > 6
(iv) 4x + 1 > 17
Solution:
(i) x – 5 < 0
x – 5 + 5 <0 + 5 ………(Adding 5)
=> x < 5
Required answer = {1, 2, 3, 4}
(ii) x + 1 ≤ 7 => x + 1 – 1 ≤ 7 – 1 (Subtracting 1)
=> x ≤ 6
Required answer = {1, 2, 3, 4, 5, 6}
(iii) 3x – 4 > 6
3x – 4 + 4 > 6 + 4 (Adding 4)
=> 3x > 10
\(\frac { 3x }{ 3 }\) > \(\frac { 10 }{ 3 }\) …(Dividing by 3)
=> x > \(\frac { 10 }{ 3 }\)
=> x > \(3\frac { 1 }{ 3 }\)
Required answer = { 4, 5, 6, …}
(iv) 4x + 1 ≥ 17
=> 4x + 1 – 1 ≥ 17 – 1 (Subtracting)
=> 4x ≥ 16
=> \(\frac { 4x }{ 4 }\) ≥ \(\frac { 16 }{ 4 }\) (Dividing by 4)
=> x ≥ 4
Required answer = {4, 5, 6, …}

Question 2.
If the replacement set = {-6, -3, 0, 3, 6, 9}; find the truth set of the following:
(i) 2x – 1 > 9
(ii) 3x + 7 < 1
Solution:
(i) 2x – 1 > 9
⇒ 2x – 1 + 1 > 9 + 1 (Adding 1)
⇒ 2x > 10
⇒ x > 5 (Dividing by 2)
⇒ x > 5
Required answer = {6, 9}
(ii) 3x + 7 ≤ 1
⇒ 3x + 7 – 7 ≤ 1 – 7 (Subtracting 7)
⇒ 3x ≤ – 6
⇒ x ≤ – 2
Required Answer = {-6, -3}

Question 3.
Solve 7 > 3x – 8; x ∈ N
Solution:
7 > 3x – 8
=> 7 – 3x > 3x – 3x – 8 (Subtracting 3x)
=> 7 – 7 – 3x > 3x – 3x – 8 – 7 (Subtracting 7)
=> -3x > -15
=> x < 5 (Dividing by -3)
Required Answer = {1, 2, 3, 4}
Note : Division by negative number reverses the inequality.

Question 4.
-17 < 9y – 8 ; y ∈ Z
Solution:
-17 < 9y – 8
=> -17 + 8 < 9y – 8 + 8 (Adding 8)
=> -9 < 9y
=> -1 < y (Dividing by 9)
Required number = {0, 1, 2, 3, 4, …}

Question 5.
Solve 9x – 7 ≤ 28 + 4x; x ∈ W
Solution:
9x – 1 ≤ 28 + 4x
=> 9x – 4x – 7 ≤ 28 + 4x – 4x (Subtracting 4x)
=> 5x – 7 ≤ 28
=> 5x – 7 + 7 ≤ 28 + 7 (Adding 7)
=> 5x ≤ 35
=> x ≤ 7 (Dividing by 5)
Required answer = {0, 1, 2, 3, 4, 5, 6, 7}

Question 6.
Solve : \(\frac { 2 }{ 3 }\)x + 8 < 12 ; x ∈ W
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 15 Linear Inequations image -1

Question 7.
Solve -5 (x + 4) > 30 ; x ∈ Z
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 15 Linear Inequations image -2

Question 8.
Solve the inquation 8 – 2x > x – 5 ; x ∈ N.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 15 Linear Inequations image -3
x = 1, 2, 3, 4 (x ∈ N)
Solution set = {1, 2, 3, 4}

Question 9.
Solve the inequality 18 – 3 (2x – 5) > 12; x ∈ W.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 15 Linear Inequations image -4

Question 10.
Solve : \(\frac { 2x+1 }{ 3 }\) + 15 < 17; x ∈ W.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 15 Linear Inequations image -5

Question 11.
Solve : -3 + x < 2, x ∈ N
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 15 Linear Inequations image -6

Question 12.
Solve : 4x – 5 > 10 – x, x ∈ {0, 1, 2, 3, 4, 5, 6, 7}
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 15 Linear Inequations image -7
Solution set = {4, 5, 6, 7}
Question 13.
Solve : 15 – 2(2x – 1) < 15, x ∈ Z.
Solution:

Question 14.
Solve : \(\frac { 2x+3 }{ 5 }\) > \(\frac { 4x-1 }{ 2 }\) , x ∈ W.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 15 Linear Inequations image -9

Linear Inequations Exercise 15B – Selina Concise Mathematics Class 8 ICSE Solutions

Solve and graph the solution set on a number line :
Question 1.
x – 5 < -2 ; x ∈ N
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 15 Linear Inequations image -10

Question 2.
3x – 1 > 5 ; x ∈ W
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 15 Linear Inequations image -11

Question 3.
-3x + 12 < -15 ; x ∈ R.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 15 Linear Inequations image -12

Question 4.
7 > 3x – 8 ; x ∈ W
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 15 Linear Inequations image -13

Question 5.
8x – 8 < – 24 ; x ∈ Z
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 15 Linear Inequations image -14

Question 6.
8x – 9 > 35 – 3x ; x ∈ N
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 15 Linear Inequations image -16

Question 7.
5x + 4 > 8x – 11 ; x ∈ Z
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 15 Linear Inequations image -17

Question 8.
\(\frac { 2x }{ 5 }\) + 1 < -3 ; x ∈ R
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 15 Linear Inequations image -18

Question 9.
\(\frac { x }{ 2 }\) > -1 + \(\frac { 3x }{ 4 }\) ; x ∈ N
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 15 Linear Inequations image -20

Question 10.
\(\frac { 2 }{ 3 }\) x + 5 ≤ \(\frac { 1 }{ 2 }\) x + 6 ; x ∈ W
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 15 Linear Inequations image -21

Question 11.
Solve the inequation 5(x – 2) > 4 (x + 3) – 24 and represent its solution on a number line.
Given the replacement set is {-4, -3, -2, -1, 0, 1, 2, 3, 4}.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 15 Linear Inequations image -23

Question 12.
Solve \(\frac { 2 }{ 3 }\) (x – 1) + 4 < 10 and represent its solution on a number line.
Given replacement set is {-8, -6, -4, 3, 6, 8, 12}.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 15 Linear Inequations image -24

Question 13.
For each inequation, given below, represent the solution on a number line :
(i) \(\frac { 5 }{ 2 }\) – 2x ≥ \(\frac { 1 }{ 2 }\) ; x ∈ W
(ii) 3(2x – 1) ≥ 2(2x + 3), x ∈ Z
(iii) 2(4 – 3x) ≤ 4(x – 5), x ∈ W
(iv) 4(3x + 1) > 2(4x – 1), x is a negative integer
(v) \(\frac { 4 – x }{ 2 }\) < 3, x ∈ R
(vi) -2(x + 8) ≤ 8, x ∈ R
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 15 Linear Inequations image -26

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 5 Playing with Number

November 9, 2023May 27, 2022 by Raju

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 5 Playing with Number

Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 5 Playing with Number

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

APlusTopper.com Provides Selina Concise ICSE Solutions for Class 8 Mathematics Chapter 5 Playing with APIusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Mathematics Chapter 5 Playing with Number. 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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Playing with Number Exercise 5A – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Write the quotient when the sum of 73 and 37 is divided by
(i) 11
(ii) 10
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 5 Playing with Number image -1

Question 2.
Write the quotient when the sum of 94 and 49 is divided by
(i) 11
(ii) 13
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 5 Playing with Number image -2

Question 3.
Find the quotient when 73 – 37 is divided by
(i) 9
(ii) 4
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 5 Playing with Number image -4

Question 4.
Find the quotient when 94 – 49 is divided by
(i) 9
(ii) 5
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 5 Playing with Number image -5

Question 5.
Show that 527 + 752 + 275 is exactly divisible by 14.
Solution:
Property :
abc = 100a + 106 + c ………(i)
bca = 1006 + 10c + a ……..(ii)
and cab = 100c + 10a + b ……….(iii)
Adding, (i), (ii) and (iii), we get abc + bca + cab = 111a + 111b + 111c = 111(a + b + c) = 3 x 37(a + b + c)
Now, let us try this method on
527 + 752 + 275 to check is it exactly divisible by 14
Here, a = 5, 6 = 2, c = 7
527 + 752 + 275 = 3 x 37(5 + 2 + 7) = 3 x 37 x 14
Hence, it shows that 527 + 752 + 275 is exactly divisible by 14

Question 6.
If a = 6, show that abc = bac.
Solution:
Given : a = 6
To show : abc = bac
Proof: abc = 100a + 106 + c …….(i)
(By using property 3)
bac = 1006 + 10a + c —(ii)
(By using property 3)
Since, a = 6
Substitute the value of a = 6 in equation (i) and (ii), we get
abc = 1006 + 106 + c ………(iii)
bac = 1006 + 106 + c ………(iv)
Subtracting (iv) from (iii) abc – bac = 0
abc = bac
Hence proved.

Question 7.
If a > c; show that abc – cba = 99(a – c).
Solution:
Given, a > c
To show : abc – cba = 99(a – c)
Proof: abc = 100a + 10b + c ……….(i)
(By using property 3)
cba = 100c + 10b + a ………..(ii)
(By using property 3)
Subtracting, equation (ii) from (i), we get
abc – cba = 100a + c – 100c – a
abc – cba = 99a – 99c
abc – cba = 99(a – c)
Hence proved.

Question 8.
If c > a; show that cba – abc = 99(c – a).
Solution:
Given : c > a
To show : cba – abc = 99(c – a)
Proof:
cba = 100c + 106 + a ……….(i)
(By using property 3)
abc = 100a + 106 + c ………(ii)
(By using property 3)
Subtracting (ii) from (i)
cba – abc= 100c+ 106 + a- 100a- 106-c
=> cba – abc = 99c – 99a
=> cba – abc = 99(c – a)
Hence proved.

Question 9.
If a = c, show that cba – abc = 0.
Solution:
Given : a = c
To show : cba – abc = 0
Proof:
cba = 100c + 106 + a …………(i)
(By using property 3)
abc = 100a + 106 + c …………(ii)
(By using property 3)
Since, a = c,
Substitute the value of a = c in equation (i) and (ii), we get
cba = 100c + 10b + c ……….(iii)
abc = 100c + 10b + c …………(iv)
Subtracting (iv) from (iii), we get
cba – abc – 100c + 106 + c – 100c – 106 – c
=> cba – abc = 0
=> cba = abc
Hence proved.

Question 10.
Show that 954 – 459 is exactly divisible by 99.
Solution:
To show : 954 – 459 is exactly divisible by 3 99, where a = 9, b = 5, c = 4
abc = 100a + 10b + c
=> 954 = 100 x 9 + 10 x 5 + 4
=> 954 = 900 + 50 + 4 ………(i)
and 459 = 100 x 4+ 10 x 5 + 9
=> 459 = 400 + 50 + 9 ……..(ii)
Subtracting (ii) from (i), we get
954 – 459 = 900 + 50 + 4 – 400 – 50 – 9
=> 954 – 459 = 500 – 5
=> 954 – 459 = 495
=> 954 – 459 = 99 x 5
Hence, 954 – 459 is exactly divisible by 99
Hence proved.

Playing with Number Exercise 5B – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 5 Playing with Number image -6
Solution:

Question 2.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 5 Playing with Number image -8
Solution:

Question 3.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 5 Playing with Number image -10
Solution:

Question 4.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 5 Playing with Number image -12
Solution:

Question 5.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 5 Playing with Number image -14
Solution:

Question 6.

Solution:
As we need A at unit place and 9 at ten’s place,
A = 6 as 6 x 6 = 36
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 5 Playing with Number image -17

Question 7.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 5 Playing with Number image -18
Solution:

Question 8.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 5 Playing with Number image -20
Solution:

Question 9.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 5 Playing with Number image -22
Solution:
As we need B at unit place and A at ten’s place,
B = 0 as 5 x 0 = 0
Now we want to find A, 5 x A = A (at unit’s place)
A = 5, as 5 x 5 = 25
C = 2

Question 10.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 5 Playing with Number image -24
Solution:

Question 11.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 5 Playing with Number image -26
Solution:

Playing with Number Exercise 5C – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Find which of the following nutpbers are divisible by 2:
(i) 192
(ii) 1660
(iii) 1101
(iv) 2079
Solution:
A number having its unit digit 2,4,6,8 or 0 is divisible by 2,
So, Number 192, 1660 are divisible by 2.

Question 2.
Find which of the following numbers are divisible by 3:
(i) 261
(ii) 111
(iii) 6657
(iv) 2574
Solution:
A number is divisible by 3 if the sum of its digits is divisible by 3,
So, 261, 111 are divisible by 3.

Question 3.
Find which of the following numbers are divisible by 4:
(i) 360
(ii) 3180
(iii) 5348
(iv) 7756
Solution:
A number is divisible by 4, if the number formed by the last two digits is divisible by 4.
So, Number 360, 5348, 7756 are divisible by 4.

Question 4.
Find which of the following numbers are divisible by 5 :
(i) 3250
(ii) 5557
(iii) 39255
(iv) 8204
Solution:
A number having its unit digit 5 or 0, is divisible by 5.
So, 3250, 39255 are all divisible by 5.

Question 5.
Find which of the following numbers are divisible by 10:
(i) 5100
(ii) 4612
(iii) 3400
(iv) 8399
Solution:
A number having its unit digit 0, is divisible by 10.
So, 5100, 3400 are all divisible by 10.

Question 6.
Which of the following numbers are divisible by 11 :
(i) 2563
(ii) 8307
(iii) 95635
Solution:
A number is divisible by 11 if the difference of the sum of digits at the odd places and sum of the digits at even places is zero or divisible by 11.
So, 2563 is divisible by 11.

Playing with Number Exercise 5D – Selina Concise Mathematics Class 8 ICSE Solutions

For what value of digit x, is :
Question 1.
1×5 divisible by 3 ?
Solution:
1×5 is divisible by 3
=> 1 + x + 5 is a multiple of 3
=> 6 + x = 0, 3, 6, 9,
=> x = -6, -3, 0, 3, 6, 9
Since, x is a digit
x = 0, 3, 6 or 9

Question 2.
31×5 divisible by 3 ?
Solution:
31×5 is divisible by 3
=> 3 + 1 + x + 5 is a multiple of 3
=> 9 + x = 0, 3, 6, 9,
=> x = -9, -6, -3, 0, 3, 6, 9,
Since, x is a digit
x = 0, 3, 6 or 9

Question 3.
28×6 a multiple of 3 ?
Solution:
28×6 is a multiple of 3
2 + 8+ x + 6 is a multiple of 3
=> 16 + x = 0, 3, 6, 9, 12, 15, 18
=> x = -18, -5, -2, 0, 2, 5, 8
Since, x is a digit = 2, 5, 8

Question 4.
24x divisible by 6 ?
Solution:
24x is divisible by 6
=> 2 + 4+ x is a multiple of 6
=> 6 + x = 0, 6, 12
=> x = -6, 0, 6
Since, x is a digit
x = 0, 6

Question 5.
3×26 a multiple of 6 ?
Solution:
3×26 is a multiple of 6
3 + x + 2 + 6 is a multiple of 3
=> 11 + x = 0, 3, 6, 9, 12, 15, 18,21,
=> x = -11, -8, -5, -2, 1, 4, 7, 10, ….
Since, x is a digit
x = 1, 4 or 7

Question 6.
42×8 divisible by 4 ?
Solution:
42×8 is divisible by 4
=> 4 + 2 + x + 8 is a multiple of 2
=> 14 + x = 0, 2, 4, 6, 8,
=> x = -8, -6, -4, -2, 2, 4, 6, 8,
Since, x is a digit 2, 4, 6, 8

Question 7.
9142x a multiple of 4 ?
Solution:
9142x is multiple of 4
=> 9 + 1 + 4 + 2 + x is a multiple of 4
=> 16 + x = 0, 4, 8, ………
x = -8, -4, 0, 4, 8
Since, x is a digit
4, 8

Question 8.
7×34 divisible by 9 ?
Solution:
7×34 is multiple of 9
=> 7 + x + 3+ 4 is a multiple of 9
=> 14 + x = 0, 9, 18, 27,
=> x = -1, 4, 13,
Since, x is a digit
x = 4

Question 9.
5×555 a multiple of 9 ?
Solution:
Sum of the digits of 5×555
=5 + x + 5 + 5 + 5 = 20 + x
It is multiple by 9
The sum should be divisible by 9
Value of x will be 7

Question 10.
3×2 divisible by 11 ?
Solution:
Sum of the digit in even place = x
and sum of the digits in odd place = 3 + 2 = 5
Difference of the sum of the digits in even places and in odd places = x – 5
3×2 is a multiple of 11
=> x – 5 = 0, 11, 22,
=> x = 5, 16, 27,
Since, x is a digit x = 5

Question 11.
5×2 a multiple of 11 ?
Solution:
Sum of a digit in even place = x
and sum of the digits in odd place = 5 + 2 = 7
Difference of the sum of the digits in even places and in odd places = x – 7
5×2 is a multiple of 11
=> x – 7 = 0, 11, 22,
=> x = 7, 18, 29,
Since, x is a digit
x = 7

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations

November 9, 2023May 27, 2022 by Raju

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations

Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations

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 10 Direct and Inverse Variations. 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

Direct and Inverse Variations Exercise 10A – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
In which ofthe following tables, x and y vary directly:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 1
Solution:

Question 2.
If x and y vary directly, find the values of x, y and z:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 4
Solution:

Question 3.
A truck consumes 28 litres of diesel for moving through a distance of 448 km. How much distance will it cover in 64 litres of diesel?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 7

Question 4.
For 100 km, a taxi charges ₹ 1,800. How much will it charge for a journey of 120 km?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 8

Question 5.
If 27 identical articles cost ₹ 1,890, how many articles can be bought for ₹ 1,750?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 9

Question 6.
7 kg of rice costs ₹ 1,120. How much rice can be bought for ₹ 3,680?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 10

Question 7.
6 note-books cost ₹ 156, find the cost of 54 such note-books.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 11

Question 8.
22 men can dig a 27 m long trench in one day. How many men should be employed for digging 135 m long trench of the same type in one day?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 12

Question 9.
If the total weight of 11 identical articles is 77 kg, how many articles of the same type would weigh 224 kg?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 13

Question 10.
A train is moving with uniform speed of 120 km per hour.
(i) How far will it travel in 36 minutes?
(ii) In how much time will it cover 210 km?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 14

Direct and Inverse Variations Exercise 10B – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Check whether x and y vary inversely or not.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 15
Solution:
x and y are inversely proportional.
Then xy are equal.
(i) xy = 4 x 6 = 24
xy = 3 x 8 = 24
xy = 12 x 2 = 24
xy = 1 x 24 = 24
xy in each case is equal.
x and y are inversely proportional
(ii) xy = 30 x 60 = 1800
xy= 120 x 30 = 3600
xy = 60 x 30= 1800
xy = 24 x 75 = 1800
xy in each case is not equal.
x and y are not inversely proportional.
(iii) xy = 10 x 90 = 900
= 30 x 30 = 900
= 60 x 20= 1200
= 10 x 90 = 900
xy in each case is not equal.
x and y are not inversely proportional.

Question 2.
If x and y vary inversely, find the values of l, m and n :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 16
Solution:

Question 3.
36 men can do a piece of work in 7 days. How many men will do the same work in 42 days?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 19

Question 4.
12 pipes, all of the same size, fill a tank in 42 minutes. How long will it take to fill the same tank, if 21 pipes of the same size are used?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 20

Question 5.
In a fort 150 men had provisions for 45 days. After 10 days, 25 men left the fort. How long would the food last at the same rate?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 21

Question 6.
72 men do a piece of work in 25 days. In how many days will 30 men do the same work?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 22

Question 7.
If 56 workers can build a wall in 180 hours, how many workers will be required to do the same work in 70 hours?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 23

Question 8.
A car takes 6 hours to reach a destination by travelling at the speed of 50 km per hour. How long will it take when the car travels at the speed of 75 km per hour?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 24

Direct and Inverse Variations Exercise 10C – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Cost of 24 identical articles is Rs. 108, Find the cost of 40 similar articles.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 25

Question 2.
If 15 men can complete a piece of work in 30 days, in how many days will 18 men complete it?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 26

Question 3.
In order to complete a work in 28 days, 60 men are required. How many men will be required if the same work is to be completed in 40 days ?
Solution:
Let x be number of men required 60 men can do the work in = 28 days
1 man can do the work in = 28 x 60 days
x man can do the work in = \(\frac { 28\times 60 }{ x }\) day
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 27

Question 4.
A fort had provisions for 450 soldiers for 40 days. After 10 days, 90 more soldiers come to the fort. Find in how many days will the remaining provisions last at the same rate ?
Solution:
After 10 days :
For 450 soldiers, provision are sufficient for (40 – 10) days = 30 days
For 1 soldier, provision are sufficient for 30 x 450 days
For 540 soldiers, the provision are sufficient for
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 28

Question 5.
A garrison has sufficient provisions for 480 men for 12 days. If the number of men is reduced by 160; find how long will the provisions last.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 29

Question 6.
\(\frac { 3 }{ 5 }\) quintal of wheat costs Rs.210. Find the cost of :
(i) 1 quintal of wheat
(ii) 0.4 quintal of wheat
Solution:
(i) \(\frac { 3 }{ 5 }\) quintal of wheat costs = Rs.210
1 quintal of wheat costs = 210 x \(\frac { 3 }{ 5 }\) = 70 x 5 = Rs.350
(ii) 1 quintal of wheat costs = Rs.350
0.4 quintal of wheat costs = 350 x 0.4 = Rs. 140.0 = Rs.140

Question 7.
If \(\frac { 2 }{ 9 }\) of a property costs Rs.2,52,000; find the cost of \(\frac { 4 }{ 7 }\) of it.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 30

Question 8.
4 men or 6 women earn Rs. 360 in one day. Find, how much will:
(i) a man earn in one day ?
(ii) a woman earn in one day ?
(iii) 6 men and 4 women earn in one day ?
Solution:
4 men earn Rs. 360 in one day
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 31

Question 9.
16 boys went to canteen to have tea and snacks together. The bill amounted to Rs. 114.40. What will be the contribution of a boy who pays for himself and 5 others ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 32

Question 10.
50 labourers can dig a pond in 16 days. How many labourers will be required to dig an another pond, double in size in 20 days ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 33

Question 11.
If 12 men or 18 women can complete a piece of work in 7 days, in how many days can 4 men and 8 women complete the same work?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 34

Question 12.
If 3 men or 6 boys can finish a work in 20 days, how long will 4 men and 12 boys take to finish the same work ?
Solution:
3 men = 6 boys
4 men = \(\frac { 6 }{ 3 }\) x 4 = 8
Total boys in second case :
= 4 men + 12 boys = 8 + 12 = 20 boys
6 boys can do a piece of work in 20 days
Then let 20 boys will do the same work in x days
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 35

Question 13.
A particular work can be completed by 6 men and 6 women in 24 days; whereas the same work can be completed by 8 men and 12 women in 15 days. Find :
(i) according to the amount of work done, one man is equivalent to how many women.
(ii) the time taken by 4 men and 6 women to complete the same work.
Solution:
6 men + 6 women can finish the work in = 24 days
144 men + 144 women can finish it in = 1 day
8 m + 12 women can finish the work in = 15 days
120 men + 180 women can finish it in = 1 day
(i) 144 men + 144 women = 120 men + 180 women
=> 144 men – 120 men
= 180 – women – 144 women
=> 24 men = 36 women
1 man = \(\frac { 36 }{ 24 }\)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 36

Question 14.
If 12 men and 16 boys can do a piece of work in 5 days and, 13 men and 24 boys can do it in 4 days, how long will 7 men and 10 boys take to do it ?
Solution:
12 men + 16 boys can do a piece of work in = 5 days
60 men + 80 boys can do a piece of work in = 1 day ………. (i)
and 13 men + 24 boys can do the same work in = 4 days
52 men + 96 boys can do the same work in = 1 day ………….(ii)
From (i) and (ii)
60 men + 80 boys = 52 men + 96 boys
=> 60 men – 52 men = 96 boys – 80 boys
=> 8 men = 16 boys
1 men = \(\frac { 16 }{ 8 }\) = 2 boys
Now, in first case,
12 men + 16 boys = 12 x 2 + 16 = 24+16 = 40 boys
In the second case,
7 men + 10 boys = 7 x 2 + 10 = 14 + 10 = 24 boys
Now 40 boys can do a piece of work in = 5 days
1 boy can do the same work in = 5 x 40 days
and 24 boys will do the same work in
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 37

Direct and Inverse Variations Exercise 10D – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Eight oranges can be bought for Rs. 10.40. How many more can be bought for Rs. 16.90?
Solution:
Number of oranges bought for Rs. 10.40 = 8
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 38

Question 2.
Fifteen men can build a wall in 60 days. How many more men are required to build another wall of same size in 45 days ?
Solution:
In 60 days a wall can be built by = 15 men
In 1 day a wall can be built by = 15 x 60 men
In 45 days a wall can be built by = \(\frac { 15\times 60 }{ 45 } =\frac { 900 }{ 45 }\) = 20 men
No. of more men required to build the wall in 45 days = 20 – 15 = 5 men

Question 3.
Six taps can fill an empty cistern in 8 hours. How much more time will be taken, if two taps go out of order ? Assume, all the taps supply water at the same rate.
Solution:
Total no. of taps = 6
Out of order taps = 2
Taps in working condition =6 – 2 = 4
6 taps can fill an empty cistern in = 8 hours
1 tap can fill an empty cistern in = 6 x 8 hours
4 taps can fill an empty cistern in = \(\frac { 48 }{ 4 }\) = 12 hours
More time taken when 2 taps are out of order = 12 – 8 = 4 hour

Question 4.
A contractor undertakes to dig a canal, 6 kilometres long, in 35 days and employed 90 men. He finds that after 20 days only 2 km of canal have been completed. How many more men must be employed to finish the work in time ?
Solution:
Length of canal = 6 km
In 20 days canal made = 2 km
Remaining length of canal = 6 – 2 = 4 km
Remaining time = 35 – 20 = 15 days
In 20 days 2 km canal is made by = 90 men
In 1 day 2 km canal is made by = 90 x 20 men
In 15 days 2 km canal is made by = \(\frac { 90\times 20 }{ 15 }\) men
In 15 days 1 km canal is made by = \(\frac { 90\times 20 }{ 15\times 2 }\) men
In 15 days 4 km canal is made by = \(\frac { 90\times 20\times 4 }{ 15\times 2 }\) men = 6 x 10 x 4 men = 240 men
Number of more men to be employed to finish the work in time = 240-90 = 150 men

Question 5.
If 10 horses consume 18 bushels in 36 days. How long will 24 bushels last for 30 horses ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 39

Question 6.
A family of 5 persons can be main¬tained for 20 days with Rs.2,480. Find, how long Rs.6944 maintain a family of 8 persons
Solution:
A family of 5 persons can be maintained with Rs.2480 for = 20 days
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 40

Question 7.
90 men can complete a work in 24 days working 8 hours a day. How many men are required to complete the same work in 18 days working 7\(\frac { 1 }{ 2 }\) hours a day ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 41

Question 8.
Twelve typists, all working with same speed, type a certain number of pages in 18 days working 8 hours a day. Find, how many hours per day must sixteen typists work in order to type the same number of pages in 9 days ?
Solution:
12 typists can type in 18 days with number of working hours in day = 8 hours
1 typist can type in 18 days = 8 x 12 hour
1 typist can type in 9 days = 2 (8 x 12) hour
16 typist can type in a day = \(\frac { 2(8\times 12) }{ 16 }\) = 12 hours

Question 9.
If 25 horses consume 18 quintal in 36 days, how long will 28 quintal last for 30 horses ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 42

Question 10.
If 70 men dig 15,000 sq. m of a field in 5 days, how many men will dig 22,500 sq. m field in 25 days ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 43

Question 11.
A contractor undertakes to build a wall 1000 m long in 50 days. He employs 56 men, but at the end of 27 days, he finds that only 448 m of wall is built. How many extra men must the contractor employ so that the wall is completed in time ?
Solution:
Number of men employed in the beginning = 56
Length of wall = 1000 m No. of days = 50
In the time of 27 days, only 448 m of wall was completed
Remaining period = 50 – 27 = 23 days
and length of wall to be completed = 1000 – 448 = 552
Now in 27 days, 448 m long wall was completed by = 1000 m
in 1 day, 448 m long was completed by = 56 x 27
inf 1 day, 1 m long wall will be completed by
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 44

Question 12.
A group of labourers promises to do a piece of work in 10 days, but five of them become absent. If the remaining labourers complete the work in 12 days, find their original number in the group.
Solution:
Total period = 10 days
But work completed in = 12 days
No. of men were absent = 5
Let the number of men in the beginning = x
Now x men can do a piece work in = 10 days
1 man will do it in = 10x x days
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 45

Question 13.
Ten men, working for 6 days of 10 hours each, finish \(\frac { 5 }{ 21 }\) of a piece of work. How many men working at the same rate and for the same number of hours each day, will be required to complete the remaining work in 8 days ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 46

Direct and Inverse Variations Exercise 10E – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
A can do a piece of work in 10 days and B in 15 days. How long will they take together to finish it ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 47

Question 2.
A and B together can do a piece of work in 6\(\frac { 2 }{ 3 }\) days ; but B alone can do it in 10 days. How long will A take to do it alone ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 48

Question 3.
A can do a work in 15 days and B in 20 days. If they together work on it for 4 days ; what fraction of the work will be left ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 49

Question 4.
A, B and C can do a piece of work in 6 days, 12 days and 24 days respectively. In what time will they all together do it ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 50

Question 5.
A and B working together can mow a field in 56 days and with the help of C, they could have mowed it in 42 days. How long would C take by himself ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 52

Question 6.
A can do a piece of work in 24 days, A and B can do it in 16 days and A, B and C in 10\(\frac { 2 }{ 3 }\) days. In how many days can A and C do it ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 53

Question 7.
A can do a piece of work in 20 days and B in 15 days. They worked together on it for 6 days and then A left. How long will B take to finish the remaining work ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 55

Question 8.
A can finish a piece of work in 15 days and B can do it in 10 days. They worked together for 2 days and then B goes away. In how many days will A finish the remaining work?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 57

Question 9.
A can do a piece of work in 10 days ; B in 18 days; and A, B and C together in 4 days. In what time would C alone do it ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 58

Question 10.
A can do \(\frac { 1 }{ 4 }\) of a work in 5 days and B can do \(\frac { 1 }{ 3 }\) of the same work in 10 days. Find the number of days in which both working together will complete the work.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 60

Question 11.
One tap can fill a cistern in 3 hours and the waste pipe can empty the full cistern in 5 hours. In what time will the empty cistern be full, if the tap and the waste pipe are kept open together ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 62

Question 12.
A and B can do a work in 8 days; B and C in 12 days, and A and C in 16 days. In what time could they do it, all working together ?
Solution:
A and B can do a work in = 8 days
B and C can do a work in = 12 days
A and C can do a work in = 16 days
(A+B)’s 1 day work = \(\frac { 1 }{ 8 }\)
(B+C)’s 1 day work = \(\frac { 1 }{ 12 }\)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 63

Question 13.
A and B complete a piece of work in 24 days. B and C do the same work in 36 days ; and A, B and C together finish it in 18 days. In how many days will (i) A alone,
(ii) C alone,
(iii) A and C together, complete the work ?
Solution:
A and B complete a piece of work in = 24 days
B and C complete a piece of work in = 36 days
(A+B+C) complete a piece of work in = 18 days
(A+B)’s 1 day work = \(\frac { 1 }{ 24 }\)
(B+C)’s 1 day work = \(\frac { 1 }{ 36 }\)
(A+B+C)’s 1 day work = \(\frac { 1 }{ 18 }\)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 64

Question 14.
A and B can do a piece of work in 40 days; B and C in 30 days; and C and A in 24 days.
(i) How long will it take them to do the work together ?
(ii) In what time can each finish it working alone ?
Solution:
A and B can do a piece of work in = 40 days
B and C can do a piece of work in = 30 days
C and A can do a piece of work in = 24 days
(A+B)’s 1 day work = \(\frac { 1 }{ 40 }\)
(B+C)’s 1 day work = \(\frac { 1 }{ 30 }\)
(C+A)’s 1 day work = \(\frac { 1 }{ 24 }\)
(i) [(A+B)+(B+C)+(C+A)]’s 1 day work = \(\frac { 1 }{ 40 }\) + \(\frac { 1 }{ 30 }\) + \(\frac { 1 }{ 24 }\)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 65

C can do the work in 40 days
Hence A can do the work in = 60 days
B can do the work in = 120 days
C can do the work in = 40 days

Question 15.
A can do a piece of work in 10 days, B in 12 days and C in 15 days. All begin together but A leaves the work after 2 days and B leaves 3 days before the work is finished. How long did the work last ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 67

Question 16.
Two pipes P and Q would fill an empty cistern in 24 minutes and 32 minutes respectively. Both the pipes being opened together, find when the first pipe must be turned off so that the empty cistern may be just filled in 16 minutes.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 69

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets

November 9, 2023May 27, 2022 by Raju

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets

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APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Mathematics Chapter 6 Sets. 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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Sets Exercise 6A – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Write the following sets in roster (Tabular) form :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 1
Solution:

Question 2.
Write the following sets in set-builder (Rule Method) form :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 6
Solution:

Question 3.
(i) Is {1, 2, 4, 16, 64} = {x : x is a factor of 32} ? Give reason.
(ii) Is {x : x is a factor of 27} ≠ {3, 9, 27, 54} ? Give reason.
(iii) Write the set of even factors of 124.
(iv) Write the set of odd factors of 72.
(v) Write the set of prime factors of 3234.
(vi) Is {x : x² – 7x + 12 = 0} = {3, 4} ?
(vii) Is {x : x² – 5x – 6 = 0} = {2, 3} ?
Solution:
(i) No, {1, 2, 4, 16, 64} ≠ {x : x is factor of 32}
Because 64 is not a factor of 32
(ii) Yes, {x : x is a factor of 27} + {3, 9, 27, 54}
Because 54 is not a factor of 27
(iii) 1 x 124 = 124
2 x 62 = 124
4 x 31 = 124
Factors of 124 = 1, 2, 4, 31, 62, 124
Set of even factors of 124 = {2, 4, 62, 124}
(iv) 1 x 72 = 72
2 x 36 = 72
3 x 24 = 72
4 x 18 = 72
6 x 12 = 72
8 x 9 = 72
Factors of 72 = 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72
Set of odd factors of 72 = {1, 3, 9}
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 8

Question 4.
Write the following sets in Roster form :
(i) The set of letters in the word ‘MEERUT’.
(ii) The set of letters in the word ‘UNIVERSAL’.
(iii) A = {x : x = y + 3, y ∈N and y > 3}
(iv) B = {p : p ∈ W and p² < 20}
(v) C = {x : x is composite number and 5 < x < 21}
Solution:
(i) Roster form of the set of letters in the word “MEERUT” = {m, e, r, u, t}
(ii) Roster form of the set of letters in the word “UNIVERSAL” = {u, n, i, v, e, r, s, a, l}
(iii) A = {x : x = y + 3, y ∈ N and y > 3}
x = y + 3
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 10

Question 5.
List the elements of the following sets :
(i) {x : x² – 2x – 3 = 0}
(ii) {x : x = 2y + 5; y ∈ N and 2 ≤ y < 6}
(iii) {x : x is a factor of 24}
(iv) {x : x ∈ Z and x² ≤ 4}
(v) {x : 3x – 2 ≤ 10, x ∈ N}
(vi) {x : 4 – 2x > -6, x ∈ Z}
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 12

Sets Exercise 6B – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Find the cardinal number of the following sets :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 16
Solution:

Question 2.
If P = {P : P is a letter in the word “PERMANENT”}. Find n (P).
Solution:
P = (P : P is a letter in the word “PERMANENT”}
or P = {p, e, r, m, a, n, t)
n (P) = 7

Question 3.
State, which of the following sets are finite and which are infinite :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 19
Solution:

Question 4.
Find, which of the following sets are singleton sets :
(i) The set of points of intersection of two non-parallel st. lines in the same plane
(ii) A = {x : 7x – 3 = 11}
(iii) B = {y : 2y + 1 < 3 and y ∈ W}
Note : A set, which has only one element in it, is called a SINGLETON or unit set.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 24

Question 5.
Find, which of the following sets are empty :
(i) The set of points of intersection of two parallel lines.
(ii) A = {x : x ∈ N and 5 < x < 6}
(iii) B = {x : x² + 4 = 0, x ∈ N}
(iv) C = {even numbers between 6 & 10}
(v) D = {prime numbers between 7 & 11}
Note : The set, which has no element in it, is called the empty or null set.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 25

Question 6.
(i) Are the sets A = {4, 5, 6} and B = {x : x² – 5x – 6 = 0} disjoint ?
(ii) Are the sets A = {b, c, d, e} and B = {x : x is a letter in the word ‘MASTER’} joint ?
Note :
(i) Two sets are said to be joint sets, if they have atleast one element in common.
(ii) Two sets are said to be disjoint, if they have no element in common.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 26

Question 7.
State, whether the following pairs of sets are equivalent or not :
(i) A = {x : x ∈ N and 11 ≥ 2x – 1} and B = {y : y ∈ W and 3 ≤ y ≤ 9}
(ii) Set of integers and set of natural numbers.
(iii) Set of whole numbers and set of multiples of 3.
(iv) P = {5, 6, 7, 8} and M = {x : x ∈ W and x < 4}
Note : Two sets are said to be equivalent, if they contain the same number of elements.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 28
B = {3,4,5,6,7,8,9}
n (B) = 7
Cardinal number of set A = 6 and cardinal number of set B = 7
Set A and set B are not equivalent.
(ii) Set of integers has infinite number of elements. Set of natural numbers has infinite number of elements.
Set of integers and set of natural numbers are equivalent because both these sets have infinite number of elements.
(iii) Set of whole numbers, has infinite number of elements. Set of multiples of 3, has infinite number of element.
Set of whole numbers and set of multiples of 3 are equivalent because both these sets have infinite number of elements.
(iv) P = {5,6,7,8}
n (P) = 4
M = {x : x ∈ W and x ≤ 4}
M = {0, 1, 2, 3, 4}
n (M) = 5
Now Cardinal number of set P = 4 and
Cardinal number of set M = 5
These sets are not equivalent.

Question 8.
State, whether the following pairs of sets are equal or not :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 29
Solution:

Question 9.
State whether each of the following sets is a finite set or an infinite set:
(i) The set of multiples of 8.
(ii) The set of integers less than 10.
(iii) The set of whole numbers less than 12.
(iv) {x : x = 3n – 2, n ∈ W, n ≤ 8}
(v) {x : x = 3n – 2,n ∈ Z, n ≤ 8}
(vi) {x : x = \(\frac { n-2 }{ n+1 }\) , n ∈ w)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 33

Question 10.
Answer, whether the following statements are true or false. Give reasons.
(i) The set of even natural numbers less than 21 and the set of odd natural numbers less than 21 are equivalent sets.
(ii) If E = {factors of 16} and F = {factors of 20}, then E = F.
(iii) The set A = {integers less than 20} is a finite set.
(iv) If A = {x : x is an even prime number}, then set A is empty.
(v) The set of odd prime numbers is the empty set.
(vi) The set of squares of integers and the set of whole numbers are equal sets.
(vii) In n(P) = n(M), then P → M.
(viii) If set P = set M, then n(P) = n(M).
(ix) n(A) = n(B) => A = B.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 34

Sets Exercise 6C – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Find all the subsets of each of the following sets :
(i) A = {5, 7}
(ii) B = {a, b, c}
(iii) C = {x : x ∈ W, x ≤ 2}
(iv) {p : p is a letter in the word ‘poor’}
Solution:
(i) A = {5,7}
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 38

Question 2.
If C is the set of letters in the word “cooler”, find :
(i) Set C
(ii) n(C)
(iii) Number of its subsets
(iv) Number of its proper subsets.
Note : (i) If a set has n elements, the number of its subsets = 2ⁿ
(ii) If a set has n elements, the number of its proper subsets = 2ⁿ – 1
Solution:
(i) C = {c, o, l, e, r}
(ii) n(C) = 5
(iii) Number of its subsets : 2⁵= 2 x 2 x 2 x 2 x 2 = 32
(iv) Number of its proper subsets = 2⁵ – 1 = 32 – 1 = 31

Question 3.
If T = {x : x is a letter in the word ‘TEETH’}, find all its subsets.
Solution:
T = {t,e,h}
Subsets of set T = φ, {r}, {e}, {h}, {t,e}, {t,h}, {e,h}, {t,e,h}

Question 4.
Given the universal set = {-7,-3, -1, 0, 5, 6, 8, 9}, find :
(i) A = {x : x < 2}
(ii) B = {x : -4 < x < 6}
Solution:
Universal set = {-7, -3, -1, 0, 5, 6, 8, 9},
(i) A = {x : x < 2} = {-7, -3, -1, 0}
(ii) B = {x : -4 < x < 6} = {-3, -1, 0, 5}

Question 5.
Given the universal set = {x : x ∈ N and x < 20}, find :
(i) A = {x : x = 3p ; p ∈ N}
(ii) B = {y : y – 2n + 3, n ∈ N}
(iii) C = {x : x is divisible by 4}
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 39

Question 6.
Find the proper subsets of {x : x² – 9x – 10 = 0}
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 40
x = 10
=> x = -1
Given set = {-1, 10}
Proper subsets of this set = φ, {-1}, {10}

Question 7.
Given, A = {Triangles}, B = {Isosceles triangles}, C = {Equilateral triangles}. State whether the following are true or false. Give reasons.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 41
Solution:

Question 8.
Given, A = {Quadrilaterals}, B = {Rectangles}, C = {Squares}, D= {Rhombuses}. State, giving reasons, whether the following are true or false.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 43
Solution:

Question 9.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 46
Solution:

Question 10.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 48
Solution:

Question 11.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 50
Solution:

Question 12.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 52
Solution:

Sets Exercise 6D – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 54
Solution:

Question 2.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 56
Solution:

Question 3.
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Solution:

Question 4.
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Solution:

Question 5.
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Solution:

Question 6.
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Solution:

Question 7.
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Solution:

Question 8.
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Solution:

Question 9.
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Solution:

Sets Exercise 6E – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 77
Solution:

Question 2.
From the given diagram, find :
(i) A’
(ii) B’
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 79
Solution:

Question 3.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 81
Solution:

Question 4.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 84
Solution:

Question 5.
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Solution:

Question 6.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 89
Solution:

Question 7.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 91
Solution:

Question 8.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 93
Solution:

Question 9.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 96
Solution:

Question 10.
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Solution:

Question 11.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 100
Solution:

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions

November 9, 2023May 27, 2022 by Raju

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions (Using ruler and compass only)

Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions (Using ruler and compass only)

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Constructions Exercise 18A – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Given below are the angles x and y.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 1
Without measuring these angles, construct :
(i) ∠ABC = x + y
(ii) ∠ABC = 2x + y
(iii) ∠ABC = x + 2y
Solution:

(i) Steps of Construction :

Draw a line segment BC of any suitable length.
With B as centre, draw an arc of any suitable radius. With the same radius, draw arcs with the vertices of given angles as centres. Let these arcs cut arms of the arc x at points P and Q and arms of angle y at points R and S.
From the arc, with centre B, cut DE = PQ arc of x and EF = RS arc of y
Join BF and produce upto point A.
Thus ∠ABC = x + y

(ii) Steps of Construction :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 3
Proceed in exactly the same way as in part(i)
takes DE = PQ = arc of x.
EF = PQ = arc of x and FG = RS = arc of y.
Join BG and produce it upto A.
Thus ∠ABC = x + x+ y = 2x + y
(iii) Steps of Construction :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 4
Proceed in exactly the same way as in (ii)
taking DE = PQ = arc of x. and EF = RS = arc of y and FG = RS = arc of y.
4. Join BF and produce upto point A.
Thus ∠ABC = x + y + y = x + 2y

Question 2.
Given below are the angles x, y and z.
Without measuring these angles construct :
(i) ∠ABC = x + y + z
(ii) ∠ABC = 2x + y + z
(iii) ∠ABC = x + 2y + z
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 5
Solution:
(i) Steps of Construction :

Draw line segment BC of any suitable length.
With B as centre, draw an arc of any suitable radius. With the same radius, draw arcs with the vertices of given angles as centres. Let these arcs cut arms of the anlge x at the points P and Q and arms of the angle y at points R and S and arms of the angle z at the points L and M.
From the arc, with centre B, cut
DE = PQ = arc of x, EF = RS = arc of y and FG = LM = arc of z.
Join BG and produce it upto A.
Then ∠ABC = x + y + z

(ii) Proceed as in part (i) upto step 2.
3. From the arc, with centre B, cut
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 8
DE = 2PQ = 2 arc of x
EF = RS = arc of y
FG = LM = arc of z
4. Join BG and produce it upto point A
Then ∠ABC = 2x + y + z
(iii) proceed as in (i) upto step 2

3. Here cut arc DE = arc PQ = arc of x arc EF = 2 arc RS = 2 arc of y arc FG = arc LM = arc of z.
4. Join BG and produce it upto A
5. Then ∠ABC = x + 2y + z

Question 3.
Draw a line segment BC = 4 cm. Construct angle ABC = 60°.
Solution:
Steps of Construction :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 10

Draw a line segment BC = 4 cm
With B as centre, draw an arc of any suitable radius which cuts BC at the point D.
With D as centre, and the same radius as in step 2, draw one more arc which cuts the previous arc at the point E.
Join BE and produce it to the point A.
Thus ∠ABC = 60°

Question 4.
Construct angle ABC = 45° in which BC = 5 cm and AB = 4.6 cm.
Solution:
Steps of Construction :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 11

Draw a line segment BC = 5 cm
Taking B as centre, draw an arc of any suitable radius, which cuts BC at the point D.
With D as centre and the same radius, as taken in step 2, draw an arc which cuts the previous arc at point E.
With E as centre and the same radius, draw one more arc which cuts the first arc at point F.
With E and F as centres and radii equal to more than half the distance between E at F, draw arc which cut each other at point P.
Join BP to meet EF at L and produce to point O. Then ∠OBC = 90°
Draw BA, the bisector of angle OBC. [With D, L as centres and suitable radius draw two arc meeting each other at Q produced it to R]
=> ∠ABC = 45° [∴ BA is bisector of ∠OBC ∴ ∠ABC = = 45°]
From BR cut arc AB = 4.6 cm

Question 5.
Construct angle ABC = 90°. Draw BP, the bisector of angle ABC. State the measure of angle PBC.
Solution:

Draw ∠ABC = 90° (as in Ques. 4)
Draw bisector of ∠ABC
Then ∠PBC = \(\frac { 1 }{ 2 }\) (90°) = 45°

Question 6.
6. Draw angle ABC of any suitable measure.
(i) Draw BP, the bisector of angle ABC.
(ii) Draw BR, the bisector of angle PBC and draw BQ, the bisector of angle ABP.
(iii) Are the angles ABQ, QBP, PBR and RBC equal?
(iv) Are the angles ABR and QBC equal ?
Solution:
Steps of Construction :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 13

Construct any angle ABC
With B as centre, draw an arc EF meeting BC at E and AB at F.
With E, F as centres draw two arc of equal radii meeting each other at the point P.
Join BP. Then BP is the bisector of ∠ABC
∠ABP = ∠PBC = \(\frac { 1 }{ 2 }\) ∠ABC
Similarly draw BR, the bisector of ∠PBC and draw BQ as the bisector of ∠ABP [With the same method as in steps 2, 3]
Then ∠ABQ = ∠QBP = ∠PBR = ∠RBC
∠ABR = \(\frac { 3 }{ 4 }\) ∠ABC and ∠QBC = \(\frac { 3 }{ 4 }\) ∠ABC
∠ABR = ∠QBC.

Constructions Exercise 18B – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Draw a line segment AB of length 5.3 cm. Using two different methods bisect AB.
Solution:
Steps of Construction :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 14

Draw a line segment AB = 5.3 cm
With A as centre and radius equal to more than half of AB, draw arcs on both sides of AB.
With B as centre and with the same radius as taken in step 2, draw arcs on both the sides of AB.
Let the arcs intersect each other at points P and Q.
Join P and Q.
The line PQ cuts the given line segment AB at the point O.
Thus, PQ is a bisector of AB such that
OA = OB = \(\frac { 1 }{ 2 }\) AB

Second Method
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 15
Steps of Construction :

Draw the given line segment AB = 5.3 cm.
At A, construct ∠PAB of any suitable measure. Then ∠PAB = 60° construct ∠QBA = 60°
3.From AP, cut AR of any suitable length and from BQ ; cut BS = AR.
Join R and S
Let RS cut the given line segment AB at the point O.
Thus RS is a bisector of AB such that OA = OB = \(\frac { 1 }{ 2 }\) AB

Question 2.
Draw a line segment PQ = 4.8 cm. Construct the perpendicular bisector of PQ.
Solution:
Steps of Construction :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 16

Draw a line segment PQ = 4.8 cm.
With P as centre and radius equal than half of PQ, draw arc on both the PQ.
With Q as centre and the same radius as taken in step 2, draw arcs on both sides of PQ.
Let the arcs intersect each other at point A and B
Join A and B.
The line AB cuts the line segment PQ at the point O. Here OP = OQ and ∠AOQ = 90°. Then the line AB is perpendicular bisector of PQ.

Question 3.
In each of the following, draw perpendicular through point P to the line segment AB :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 17
Solution:
(i) Steps of Construction :

With P as centre, draw an arc of a suitable radius which cuts AB at points C and D.
With C and D as centres, draw arcs of equal radii and let these arcs intersect each other at the point Q.
[The radius of these arcs must be more than half of CD and both the arcs must be drawn on the other side]
Join P and Q
Let PQ cut AB at the point O.
Thus, OP is the required perpendicular clearly, ∠AOP = ∠BOP = 90°

(ii) Steps of Construction :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 19

With P as centre, draw an arc of any suitable radius which cuts AB at points C and D.
With C and D as centres, draw arcs of equal radii. Which intersect each other at point A.
[This radius must be more than half of CD and let these arc intersect each other at the point 0]
Join P and O. Then OP is the required perpendicular.
∠OPA = ∠OPB = 90°

(iii) Steps of Construction :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 20

With P as centre, draw an arc of any suitable radius which cuts AB at points C and D.
With C and D as centre, draw arcs of equal radii
[The radius of these arcs must be more than half of CD and both the arcs must be drawn on the other side.]
and let these arcs intersect each other at the point Q.
Join Q and P. Let QP cut AB at the point O. Then OP is the required perpendicular.
Clearly, ∠AOP = ∠BOP = 90°

Question 4.
Draw a line segment AB = 5.5 cm. Mark a point P, such that PA = 6 cm and PB = 4.8 cm. From the point P, draw perpendicular to AB.
Solution:
Step of Construction :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 21

Draw a line segment AB = 5.5 cm
With A as centre and radius = 6 cm, draw an arc.
With B as centre and radius = 4.8 cm draw another arc.
Let these arcs meet each other at the point P.
PA = 6 cm, PB = 4.8
With P as centre and some suitable radius draw an arc meeting AB at the points C and D.
With C as centre and radius more than half of CD, draw an arc.
With D as centre and same radius as in step 6, draw an arc.
Let these arcs meet each other at the point Q.
Join PQ.
The PQ meet AB at point O.
Then PO ⊥ AB i.e; ∠AOP = 90° = ∠POB.

Question 5.
Draw a line segment AB = 6.2 cm. Mark a point P in AB such that BP = 4 cm. Through point P draw perpendicular to AB.
Solution:
Steps of Construction :

Draw a line segment AB = 6.2 cm
Cut off BP = 4 cm
With P as centre and some radius draw arc meeting AB at the points C, D.
With C, D as centres and equal radii [each is more than half of CD] draw two arcs, meeting each other at the point O.
Join OP. Then OP is perpendicular for AB.

Constructions Exercise 18C – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Draw a line AB = 6 cm. Mark a point P any where outside the line AB. Through the point P, construct a line parallel to AB.
Solution:
Steps of construction :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 23

Draw a line AB = 6 cm
Take any point Q on the line AB and join it with the given point P.
At point P, construct ∠CPQ = ∠PQB
Produce CP upto any point D.
Thus, CPD is the required parallel line.

Question 2.
Draw a line MN = 5.8 cm. Locate a point A which is 4.5 cm from M and 5 cm from N. Through A draw a line parallel to line MN.
Solution:
Steps of construction :

Draw a line MN = 5.8 cm
With M as centre and radius = 4.5 cm, draw an arc.
With N as centre draw another arc of radius 5 cm. These arcs intersect each other at A.
Join AM and AN.
At point A, draw ∠DAN = ∠ANM
Produce DA to any point C.
Thus CAD is the required parallel line.

Question 3.
Draw a straight line AB = 6.5 cm. Draw another line which is parallel to AB at a distance of 2.8 cm from it.
Solution:
Steps of construction :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 25

Draw a straight line AB = 6.5 cm
Taking point A as centre, draw an arc of radius 2.8 cm.
Taking B as centre, drawn another arc of radius 2.8 cm.
Draw a line CD which touches the two arcs drawn.
Thus CD is the required parallel line.

Question 4.
Construct an angle PQR = 80°. Draw a line parallel to PQ at a distance of 3 cm from it and another line parallel to QR at a distance of 3.5 cm from it. Mark the point of intersection of these parallel lines as A.
Solution:
Steps of construction :

Draw ∠PQR = 80°
With P as centre draw an arc of radius 2 cm.
Again with Q as centre, draw another arc of radius 2 cm. Then BM is a line which touches the two arcs. Then BM is a line parallel to PQ.
With Q as centre, draw an arc of radius 3.5 cm. With R as centre draw another arc of radius 3.5 cm. Draw a line HC which touches these two arcs. Let these two parallel line intersect at A.

Question 5.
Draw an angle ABC = 60°. Draw the bisector of it. Also draw a line parallel to BC a distance of 2.5 cm from it.
Let this parallel line meet AB at point P and angle bisector at point Q. Measure the length of BP and PQ. Is BP = PQ ?
Solution:
Steps of construction :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 27

Draw, ∠ABC = 60°
Draw BD, the bisector of ∠ABC.
Taking B as centre, draw an arc of radius 2.5 cm.
Taking C as centre, draw another arc of radius 2.5 cm.
Draw a line MN which touches these two arcs drawn. Then MN is the required line parallel to BC.
Let this line MN meets AB at P and bisector BD at Q.
Measure BP and PQ.
By measurement we see BP = PQ.

Question 6.
Construct an angle ABC = 90°. Locate a point P which is 2.5 cm from AB and 3.2 cm from BC.
Solution:
Steps of construction :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 28

Draw ∠ABC = 90°
From AB, cut BD = 3.2 cm.
Through point C, draw CH⊥BC. From CH, cut CE = 3.2. Join DE. Now DE is a line parallel to BC and at a distance of 3.2 cm from BC.
From BC cut BM = 2.5 cm.
Through point A, draw AK ⊥ AB. From AK cut AN = 2.5 cm. Join NM. Therefore NM is parallel to AB and at a distance of 2.5 cm from AB.
DE and MN intersect each other at P. Thus P is the required point which is 2.5 cm from AB and 3.2 cm from BC.

Constructions Exercise 18D – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Construct a quadrilateral ABCD; if:
(i) AB = 4.3 cm, BC = 5.4, CD = 5 cm, DA = 4.8 cm and angle ABC = 75°.
(ii) AB = 6 cm, CD = 4.5 cm, BC = AD = 5 cm and ∠BCD = 60°.
(iii) AB = 8 cm, BC = 5.4 cm, AD = 6 cm, ∠A = 60° and ∠B = 75°.
(iv) AB = 5 cm, BC = 6.5 cm, CD =4.8 cm, ∠B = 75° and ∠C = 120°.
(v) AB = 6 cm = AC, BC = 4 cm, CD = 5 cm and AD = 4.5 cm.
(vi) AB = AD = 5cm, BD = 7 cm and BC = DC = 5.5 cm
Solution:
(i) Rough figure is as follow :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 29
Steps :

Draw AB = 4.3 cm.
At B, draw ∠PBA = 75°
Cut BC = 5.4 cm.
From C & A, draw arcs of radii 5 cm and 4.8 cm respectively which intersect at D.
Join AD and DC.
ABCD is the required quadrilateral.

(ii) Rough figure is as follow :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 30
Steps :

Draw BC = 5 cm.
Draw ∠PCB = 60° and cut CD = 4.5 cm.
From B and D, draw arcs of radii 6 cm and 5 cm respectively which intersect at A.
Join AB and AD.
Thus ABCD is the required quadrilateral.

(iii) Rough figure is as follow :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 31
Actual quadrilateral is constructed with the help of above rough figure.

Steps :

Draw AB = 8 cm.
At A, draw ∠PAB = 60° and cut DA = 6 cm.
At B, draw ∠QBA = 75° and cut BC = 5.4 cm.
Join DC.
Thus ABCD is the required quadrilateral.

(iv) Rough figure is as shown below.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 33
Steps :

Draw BC = 6-5 cm.
Draw ∠B = 75° and cut BA = 5 cm.
Draw ∠C = 120° and cut CD = 4.8 cm.
Join AD.
Thus ABCD is the required quadrilateral.

(v) Rough figure is as shown below.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 34
Steps :

Draw AB = 6 cm.
From A and B, draw arcs of radii 6 cm and 4 cm which cut at C.
From A and C, draw arcs of radii 4.5 cm and 5 cm respectively which intersect at D.
Join BC, CD and DA. Thus ABCD is the required quadrilateral.

(vi) Rough figure is as follow :

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 35
Actual construction is as follow (using above rough fig.)

Steps :

Draw AB = 5 cm.
From A & B draw arcs of radii 5 cm and 7cm which intersect at D.
From B & D draw arcs of radii 5.5 cm each which intersect at C.
Join AD, BD, DC and BC.
Thus ABCD is the required quadrilateral.

Question 2.
Construct a parallelogram ABCD, if :
(i) AB = 3.6 cm, BC = 4.5 cm and ∠ABC = 120°.
(ii) BC = 4.5 cm, CD = 5.2 cm and ∠ADC = 75°.
(iii) AD = 4 cm, DC = 5 cm and diagonal BD = 7 cm.
(iv) AB = 5.8 cm, AD = 4.6 cm and diagonal AC = 7.5 cm.
(v) diagonal AC = 6.4 cm, diagonal BD = 5.6 cm and angle between the diagonals is 75°.
(vi) lengths of diagonals AC and BD are 6.3 cm and 7.0 cm respectively, and the angle between them is 45°.
(vii) lengths of diagonals AC and BD are 5.4 cm and 6.7 cm respectively and the angle between them is 60°.
Solution:
(i) Rough figure is as follow :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 37
The above rough figure is used to construct the actual ||gm as follow :

Steps :

Draw AB = 3.6 cm.
Draw BP such that ∠B = 120°.
Cut BC = 4.5 cm.
From A, draw arc of radius 4.5 cm.
From C, draw arc of radius 3.6 cm. Which interescts first arc at D.
Join AD and CD.
Hence ABCD is the required ||gm.

(ii) Rough figure is as follow :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 39
Steps :

Draw CD = 5.2 cm.
Draw ZCDP = 75°
Cut DA = 4.5 cm.
From A draw arc of radius 5.2 cm.
From C, draw arc of radius 4.5 cm which meets first arc at B.
Join AB and CB.
Thus ABCD is the required ||gm.

(iii) Rough figure is as follow :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 40
Steps :

Draw AD = 4 cm.
From A, draw an arc of radius 5 cm.
From B, draw an arc of radius 4 cm.
From D, draw an arc of ardius 5 cm which intersect first arc at C.
Join AB, BD, BC and CD.
Thus ABCD is the required || gm.

(iv) Rough figure is as follow :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 41
opposite sides of ||gm are equal
BC = AD = 4.6 cm.
Actual figure is constructed as follow :

Steps:

Draw AB = 5.8 cm.
Draw an arc of radius 4.6 cm with centre B.
Draw an arc of radius 7.5 cm from A which intersects first arc at C.
From A, draw an arc of radius 4.6 cm.
From C, draw an arc of radius 5.8 cm which intersects first arc at D.
Join AD, CD, BC and AC.
Thus ABCD is the required //gm.

(v) Rough figure is as follow.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 43
Steps :

Draw AC = 6.4 cm.
Bisect AC at O.
Draw ∠XOC = 75° and produce XO to Y.
Cut OB = OD = 2 8 cm.
Join AB, BC, AD and CD.
Thus ABCD is the required ||gm.

(vi)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 44
Steps :

Draw AC = 6.3 cm.
Bisect AC at O.
At O, draw ∠XOC = 45° and produce XO to Y.
Cut OB = OD = 3.5 cm (half the diagonal 7 cm.)
Join AB, CB, AD and CD. Thus ABCD is the required || gm.

(vii)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 45
Steps :

Draw BD 6.7 cm.
Bisect BD at O.
At O, draw ∠XOD = 60° and produce XO to Y.
Cut OA = OC = 2.7 cm (half the diagonals 5.4 cm)
Join AB, AD, BC and CD.
Thus ABCD is the required ||gm.

Question 3.
Construct a rectangle ABCD ; if :
(i) AB = 4.5 cm and BC = 5.5 cm.
(ii) BC = 61 cm and CD = 6.8 cm.
(iii) AB = 5.0 cm and diagonal AC = 6.7 cm.
(iv) AD = 4.8 cm and diagonal AC = 6.4 cm.
(v) each diagonal is 6 cm and the angle between them is 45°.
(vi) each diagonal is 5.5 cm and the angle between them is 60°.
Solution:
(i)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 46
Steps :

Draw BC = 5.5 cm.
At B, draw ∠XBC = 90°
Cut BA = 4.5 cm.
From A, draw an arc of radius 5.5 cm.
From C, draw an arc of radius 4 5 cm which meets first arc at D.
Join AD and CD.
Thus ABCD is the required rectangle.

(ii)
na Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 47
Steps:

Draw BC = 6.1 cm.
At C, draw ∠PCB = 90°.
Cut CD = 6.8 cm.
Draw an arc of radius 6.8 cm from B.
From D, draw an arc of radius 6.1 cm which meets the first arc at A.
Join AB and AD.
Thus ABCD is the required rectangle.

(iii)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 48
Steps :

Draw AB = 5 cm.
At B, draw ∠XBA = 90°.
From A, draw an arc of radius 6.7 cm which meets XB at C.
From C, draw an arc of a radius 5 cm.
From A, draw an arc of radius equal to BC which meets first arc at D.
Join AD and CD. Thus ABCD is the required rectangle.

(iv)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 49
Steps :

Draw AD = 4.8 cm.
At D, draw ∠XDA = 90°.
From A, draw an arc of radius 6-4 cm which meets DX at C.
From A, draw an arc of radius equal to DC.
From C, draw an arc of radius 4.8 cm which meets first arc at B.
Join AB and CB. Thus ABCD is the required rectangle.

(v)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 50
Steps :

Draw AC = 6 cm.
Bisect AC at O.
At O, draw ∠XOC = 45° and produce XO to Y.
Cut OB = OD = 3 cm (half the diagonal 6 cm)
Join AB, CB, AD and CD.
Thus ABCD is the required rectangle.

(vi)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 51
Steps :

Draw AC = 5.5 cm.
Bisect AC at O.
At O, draw ∠XOC = 60° and produce XO to Y.
Cut OB = OA and OD = OA (half the diagonal AC).
Join AB, BC, AD and CD.
Thus ABCD is the required rectangle.

Question 4.
Construct a rhombus ABCD, if ;
(i) AB = 4 cm and ∠B = 120°.
(ii) BC = 4.7 cm and ∠B = 75°.
(iii) CD = 5 cm and diagonal BD = 8.5 cm.
(iv) BC = 4.8cm, and diagonal AC = 7cm.
(v) diagonal AC = 6 cm and diagonal BD = 5.8 cm.
(vi) diagonal AC = 4.9 cm and diagonal BD = 6 cm.
(vii) diagonal AC = 6.6 cm and diagonal BD = 5.3 cm.
Solution:
(i)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 52
Steps :

Draw AB = 4 cm.
At B, draw ∠XBA = 120°
Cut BC = 4 cm.
Draw arcs of radii 4 cm each from A and C which intersect at D.
Join CD and AD.
Thus ABCD is the required rhombus.

(ii)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 53
Steps :

Draw BC = 4.7 cm.
At B, draw ∠XBC = 75°
Cut BA = 4.7 cm.
From A and C, draw arcs of radii 4.7 cm each which intersect at D.
Join AD and CD.
Thus ABCD is the rhombus.

(iii)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 54
Steps :

Draw CD = 5 cm.
From C & D draw arcs of radii 5 cm and 8.5 cm respectively which intersect at B.
From B and D, draw arcs of radii 5 cm each which intersect at A.
Join AB and AD.
Thus ABCD is the required rhombus.

(iv)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 55
Steps :

Draw AC = 7 cm.
Draw arcs of radii 4.8 cm each from A and C which intersect at B.
From A & C again draw arcs of radii 4.8 cm each which intersect at D.
Join AB, BC, AD and CD.
Thus ABCD is the required rhombus.

(v)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 56
Steps :

Draw BD = 5.8 cm.
Draw perpendicular bisector XY of BD.
Cut OA = OC = 3 cm (half the diagonal 6 cm)
Join AB, AD, BC and CD.
Thus ABCD is the required rhombus.

(vi)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 57
Steps :

Draw AC = 4.9 cm.
Draw perpendicular bisector XY of AC.
Cut OB = OD = 3 cm (half the diagonal 6 cm)
Join AB, BC, AD and CD.
Thus ABCD is the required rhombus.

(vii)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 58
Steps :

Draw BD = 5.3 cm.
Draw perpendicular bisector XY of BD.
Cut OA = OC = 3.3 cm (half the diagonal 6.6 cm)
Join AB, AD, BC and CD.
Thus ABCD is the required rhombus.

Question 5.
Construct a square, if :
(i) its one side is 3.8 cm.
(ii) its each side is 4.3 cm.
(iii) one diagonal is 6.2 cm.
(iv) each diagonal is 5.7 cm.
Solution:
(i)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 59
Steps :

Draw AB = 3.8 cm.
At B, draw ∠PBA = 90°.
Cut BC = 3.8 cm.
From A and C, draw arcs of radii 3.8 cm each which intersect at D.
Join AD and CD.
Thus ABCD is the required square.

(ii)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 60
Steps :

Draw AB = 4.3 cm.
Draw ∠PAB = 90° at A.
Cut AD = 4.3 cm.
From B and D, draw arcs of radii 4.3 cm each which intersect at C.
Join AD, BC and CD.
Hence ABCD is the required square.

(iii)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 61
Steps :

Draw BD = 6.2 cm.
Draw perpendicular bisector XY of BD.
Cut OA = OC = 3.1 cm (half the diagonal)
Join AB, AD, BC and CD.
Thus ABCD is the required square.

(iv)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 62
Steps :

Draw BD = 5.7 cm.
Draw perpendicular bisector XY of BD.
From 0, draw arcs of radii equal to OB which cuts XY at A and C.
Join AB, AD, BC and CD.
Thus ABCD is the required square.

Question 6.
Construct a quadrilateral ABCD in which ; ∠A = 120°, ∠B = 60°, AB = 4 cm, BC = 4.5 cm and CD = 5 cm.
Solution:
Rough figure is as follow :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 63
Steps :

Draw AB = 4 cm.
At A, draw ∠PAB = 120°.
At B, draw ∠QBA = 60°.
From BQ, cut BC = 4.5 cm.
From C, draw an arc of radius 5 cm which meets AP at D.
Join CD.
Thus ABCD is the required quadrilateral.

Question 7.
Construct a quadrilateral ABCD, such that AB = BC = CD = 4.4 cm, ∠B = 90° and ∠C = 120°.
Solution:
Rough figure is as follow
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 64
Steps :

Draw BC = 4.4 cm.
At B, draw ∠PBC = 90°.
Cut BA = 4.4 cm.
At C, draw ∠QCB = 120°.
Cut CD = 4.4 cm.
Join AD.
Thus ABCD is the required quadrilateral.

Question 8.
Using ruler and compasses only, construct a parallelogram ABCD, in which : AB = 6 cm, AD = 3 cm and ∠DAB = 60°. In the same figure draw the bisector of angle DAB and let it meet DC at point P. Measure angle APB.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 65
Steps :

Draw AB = 6 cm.
At A draw ∠QAB = 60°.
From AQ cut AD = 3 cm.
From D, draw an arc of radius 6 cm.
From B, draw an arc of radius 3 cm which meets first arc at C.
Join CD and BC.
Thus ABCD is the required ||gm.
Bisect ∠DAB, so that bisector meets CD at P.
Join PB and measure ZAPB.
∴ ∠APB = 90°.

Question 9.
Draw a parallelogram ABCD, with AB = 6 cm, AD = 4.8 cm and ∠DAB = 45°. Draw the perpendicular bisector of side AD and let it meet AD at point P. Also draw the diagonals AC and BD ; and let they intersect at point O. Join O and P. Measure OP.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 66
Steps :

Draw AB = 6 cm.
Draw ∠PAB = 45°.
Cut AD = 4.8 cm.
From D, draw an arc of radius 6 cm.
From B, draw an arc of radius 4.8 cm which meets first arc at C.
Join BC, CD, AD.
Thus ABCD is the required ||gm.
Draw perpendicular bisector XY of AD which cuts AD at P.
Join AC and BD which intersect at O.
Join OP and measure it.
OP = 3 cm.

Question 10.
Using ruler and compasses only, construct a rhombus whose diagonals are 8 cm and 6 cm. Measure the length of its one side.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 67
Steps :

Draw BD = 8 cm.
Draw perpendicular bisector PQ of BD.
Cut OA = OC = 3 cm [half the diagonal 6 cm]
Join AB, AD, BC and CD.
Measure side AB which is 5 cm.
Thus ABCD is the required rhombus.