Understanding ICSE Mathematics Class 10 ML Aggarwal Solutions

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations Chapter Test

More Exercises

• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations Ex 4
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations MCQS
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations Chapter Test

Question 1.
Solve the inequation : 5x – 2 ≤ 3(3 – x) where x ∈ { – 2, – 1, 0, 1, 2, 3, 4}. Also represent its solution on the number line.
Solution:
5x – 2 < 3(3 – x)
⇒ 5x – 2 ≤ 9 – 3x
⇒ 5x + 3x ≤ 9 + 2

Question 2.
Solve the inequations :
6x – 5 < 3x + 4, x ∈ I.
Solution:
6x – 5 < 3x + 4
6x – 3x < 4 + 5
⇒ 3x <9
⇒ x < 3
x ∈ I
Solution Set = { -1, -2, 2, 1, 0….. }

Question 3.
Find the solution set of the inequation
x + 5 < 2 x + 3 ; x ∈ R
Graph the solution set on the number line.
Solution:
x + 5 ≤ 2x + 3
x – 2x ≤ 3 – 5
⇒ -x ≤ -2
⇒ x ≥ 2

Question 4.
If x ∈ R (real numbers) and – 1 < 3 – 2x ≤ 7, find solution set and represent it on a number line.
Solution:
-1 < 3 – 2x ≤ 7
-1 < 3 – 2x and 3 – 2x ≤ 7
⇒ 2x < 3 + 1 and – 2x ≤ 7 – 3
⇒ 2x < 4 and -2x ≤ 4
⇒ x < 2 and -x ≤ 2
and x ≥ -2 or -2 ≤ x
x ∈ R
Solution set -2 ≤ x < 2
Solution set on number line

Question 5.
Solve the inequation :
\(\frac { 5x+1 }{ 7 } -4\left( \frac { x }{ 7 } +\frac { 2 }{ 5 } \right) \le 1\frac { 3 }{ 5 } +\frac { 3x-1 }{ 7 } ,x\in R\)
Solution:
\(\frac { 5x+1 }{ 7 } -4\left( \frac { x }{ 7 } +\frac { 2 }{ 5 } \right) \le 1\frac { 3 }{ 5 } +\frac { 3x-1 }{ 7 } \)
\(\frac { 5x+1 }{ 7 } -4\left( \frac { x }{ 7 } +\frac { 2 }{ 5 } \right) \le \frac { 8 }{ 5 } +\frac { 3x-1 }{ 7 } \)

Question 6.
Find the range of values of a, which satisfy 7 ≤ – 4x + 2 < 12, x ∈ R. Graph these values of a on the real number line.
Solution:
7 < – 4x + 2 < 12
7 < – 4x + 2 and – 4x + 2 < 12

Question 7.
If x∈R, solve \(2x-3\ge x+\frac { 1-x }{ 3 } >\frac { 2 }{ 5 } x\)
Solution:
\(2x-3\ge x+\frac { 1-x }{ 3 } >\frac { 2 }{ 5 } x\)
\(2x-3\ge x+\frac { 1-x }{ 3 } \) and \(x+\frac { 1-x }{ 3 } >\frac { 2 }{ 5 } x\)

Question 8.
Find positive integers which are such that if 6 is subtracted from five times the integer then the resulting number cannot be greater than four times the integer.
Solution:
Let the positive integer = x
According to the problem,
5a – 6 < 4x
⇒ 5a – 4x < 6
⇒ x < 6
Solution set = {x : x < 6}
= { 1, 2, 3, 4, 5, 6}

Question 9.
Find three smallest consecutive natural numbers such that the difference between one-third of the largest and one-fifth of the smallest is at least 3.
Solution:
Let first least natural number = x
then second number = x + 1
and third number = x + 2

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations Chapter Test are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations MCQS

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations MCQS

More Exercises

• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations Ex 4
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations MCQS
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations Chapter Test

Choose the correct answer from the given four options (1 to 5) :

Question 1.
If x ∈ { – 3, – 1, 0, 1, 3, 5}, then the solution set of the inequation 3x – 2 ≤ 8 is
(a) { – 3, – 1, 1, 3}
(b) { – 3, – 1, 0, 1, 3}
(c) { – 3, – 2, – 1, 0, 1, 2, 3}
(d) { – 3, – 2, – 1, 0, 1, 2}
Solution:
x ∈ { -3, -1, 0, 1, 3, 5}
3x – 2 ≤ 8
⇒ 3x ≤ 8 + 2
⇒ 3x ≤ 10
⇒ x ≤ \(\\ \frac { 10 }{ 3 } \)
⇒ x < \(3 \frac { 1 }{ 3 } \)
Solution set = { -3, -1, 0, 1, 3} (b)

Question 2.
If x ∈ W, then the solution set of the inequation 3x + 11 ≥ x + 8 is
(a) { – 2, – 1, 0, 1, 2, …}
(b) { – 1, 0, 1, 2, …}
(c) {0, 1, 2, 3, …}
(d) {x : x∈R,x≥\(– \frac { 3 }{ 2 } \)}
Solution:
x ∈ W
3x + 11 ≥ x + 8
⇒ 3x – x ≥ 8 – 11

Question 3.
If x ∈ W, then the solution set of the inequation 5 – 4x ≤ 2 – 3x is
(a) {…, – 2, – 1, 0, 1, 2, 3}
(b) {1, 2, 3}
(c) {0, 1, 2, 3}
(d) {x : x ∈ R, x ≤ 3}
Solution:
x ∈ W
5 – 4x < 2 – 3x
⇒ 5 – 2 ≤ 3x + 4x
⇒ 3 ≤ x
Solution set = {0, 1, 2, 3,} (c)

Question 4.
If x ∈ I, then the solution set of the inequation 1 < 3x + 5 ≤ 11 is
(a) { – 1, 0, 1, 2}
(b) { – 2, – 1, 0, 1}
(c) { – 1, 0, 1}
(d) {x : x ∈ R, \(– \frac { 4 }{ 3 } \) < x ≤ 2}
Solution:
x ∈ I
1 < 3x + 5 ≤ 11
⇒ 1 < 3x + 5
⇒ 1 – 5 < 3x

Question 5.
If x ∈ R, the solution set of 6 ≤ – 3 (2x – 4) < 12 is
(a) {x : x ∈ R, 0 < x ≤ 1}
(b) {x : x ∈ R, 0 ≤ x < 1}
(c) {0, 1}
(d) none of these
Solution:
x ∈ R
6 ≤ – 3(2x – 4) < 12
⇒ 6 ≤ – 3(2x – 4)
⇒ 6 ≤ – 6x + 12

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations MCQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations Ex 4

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations Ex 4

More Exercises

• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations Ex 4
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations MCQS
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations Chapter Test

Question 1.
Solve the inequation 3x -11 < 3 where x ∈ {1, 2, 3,……, 10}. Also represent its solution on a number line
Solution:
3x – 11 < 3 => 3x < 3 + 11 => 3x < 14 x < \(\\ \frac { 14 }{ 3 } \)
But x ∈ 6 {1, 2, 3, ……., 10}
Solution set is (1, 2, 3, 4}
Ans. Solution set on number line

Question 2.
Solve 2(x – 3)< 1, x ∈ {1, 2, 3, …. 10}
Solution:
2(x – 3) < 1 => x – 3 < \(\\ \frac { 1 }{ 2 } \) => x < \(\\ \frac { 1 }{ 2 } \) + 3 => x < \(3 \frac { 1 }{ 2 } \)
But x ∈ {1, 2, 3 …..10}
Solution set = {1, 2, 3} Ans.

Question 3.
Solve : 5 – 4x > 2 – 3x, x ∈ W. Also represent its solution on the number line.
Solution:
5 – 4x > 2 – 3x
– 4x + 3x > 2 – 5
=> – x > – 3
=> x < 3
x ∈ w,
solution set {0, 1, 2}
Solution set on Number Line :

Question 4.
List the solution set of 30 – 4 (2.x – 1) < 30, given that x is a positive integer.
Solution:
30 – 4 (2x – 1) < 30
30 – 8x + 4 < 30
– 8x < 30 – 30 – 4
– 8x < – 4 x > \(\\ \frac { -4 }{ -8 } \)
=> x > \(\\ \frac { 1 }{ 2 } \)
x is a positive integer
x = {1, 2, 3, 4…..} Ans.

Question 5.
Solve : 2 (x – 2) < 3x – 2, x ∈ { – 3, – 2, – 1, 0, 1, 2, 3} .
Solution:
2(x – 2) < 3x – 2
=> 2x – 4 < 3x – 2
=> 2x – 3x < – 2 + 4
=> – x < 2
=> x > – 2
Solution set = { – 1, 0, 1, 2, 3} Ans.

Question 6.
If x is a negative integer, find the solution set of \(\\ \frac { 2 }{ 3 } \)+\(\\ \frac { 1 }{ 3 } \) (x + 1) > 0.
Solution:
\(\\ \frac { 2 }{ 3 } \)+\(\\ \frac { 1 }{ 3 } \) x + \(\\ \frac { 1 }{ 3 } \) > 0
=> \(\\ \frac { 1 }{ 3 } \) x + 1 > 0
=> \(\\ \frac { 1 }{ 3 } \) x > – 1

x is a negative integer
Solution set = {- 2, – 1} Ans.

Question 7.
Solve: \(\\ \frac { 2x-3 }{ 4 } \)≥\(\\ \frac { 1 }{ 2 } \), x ∈ {0, 1, 2,…,8}
Solution:
\(\\ \frac { 2x-3 }{ 4 } \)≥\(\\ \frac { 1 }{ 2 } \)
=> 2x – 3 ≥ \(\\ \frac { 4 }{ 2 } \)

Question 8.
Solve x – 3 (2 + x) > 2 (3x – 1), x ∈ { – 3, – 2, – 1, 0, 1, 2, 3}. Also represent its solution on the number line.
Solution:
x – 3 (2 + x) > 2 (3x – 1)
=> x – 6 – 3x > 6x – 2
=> x – 3x – 6x > – 2 + 6
=> – 8x > 4
=> x < \(\\ \frac { -4 }{ 8 } \) => x < \(– \frac { 1 }{ 2 } \)
x ∈ { – 3, – 2, – 1, 0, 1, 2}
.’. Solution set = { – 3, – 2, – 1}
Solution set on Number Line :

Question 9.
Given x ∈ {1, 2, 3, 4, 5, 6, 7, 9} solve x – 3 < 2x – 1.
Solution:
x – 3 < 2x – 1
x – 2x < – 1 + 3 => – x < 2 x > – 2
But x ∈ {1, 2, 3, 4, 5, 6, 7, 9}
Solution set = {1, 2, 3, 4, 5, 6, 7, 9} Ans.

Question 10.
Given A = {x : x ∈ I, – 4 ≤ x ≤ 4}, solve 2x – 3 < 3 where x has the domain A Graph the solution set on the number line.
Solution:
2x – 3 < 3 => 2x < 3 + 3 => 2x < 6 => x < 3
But x has the domain A = {x : x ∈ I – 4 ≤ x ≤ 4}
Solution set = { – 4, – 3, – 2, – 1, 0, 1, 2}
Solution set on Number line :

Question 11.
List the solution set of the inequation
\(\\ \frac { 1 }{ 2 } \) + 8x > 5x \(– \frac { 3 }{ 2 } \), x ∈ Z
Solution:
\(\\ \frac { 1 }{ 2 } \) +8x > 5x \(– \frac { 3 }{ 2 } \)

Question 12.
List the solution set of \(\\ \frac { 11-2x }{ 5 } \) ≥ \(\\ \frac { 9-3x }{ 8 } \) + \(\\ \frac { 3 }{ 4 } \),
x ∈ N
Solution:
\(\\ \frac { 11-2x }{ 5 } \) ≥ \(\\ \frac { 9-3x }{ 8 } \) + \(\\ \frac { 3 }{ 4 } \)
=> 88 – 16x ≥ 45 – 15x + 30
(L.C.M. of 8, 5, 4 = 40}
=> – 16x + 15x ≥ 45 + 30 – 88
=> – x ≥ – 13
=>x ≤ 13
x ≤ N.
Solution set = {1, 2, 3, 4, 5, .. , 13} Ans.

Question 13.
Find the values of x, which satisfy the inequation : \(-2\le \frac { 1 }{ 2 } -\frac { 2x }{ 3 } \le 1\frac { 5 }{ 6 } \), x ∈ N.
Graph the solution set on the number line. (2001)
Solution:
\(-2\le \frac { 1 }{ 2 } -\frac { 2x }{ 3 } \le 1\frac { 5 }{ 6 } \), x ∈ N

Question 14.
If x ∈ W, find the solution set of
\(\frac { 3 }{ 5 } x-\frac { 2x-1 }{ 3 } >1\)
Also graph the solution set on the number line, if possible.
Solution:
\(\frac { 3 }{ 5 } x-\frac { 2x-1 }{ 3 } >1\)
9x – (10x – 5) > 15 (L.C.M. of 5, 3 = 15)
=> 9x – 10x + 5 > 15
=> – x > 15 – 5
=> – x > 10
=> x < – 10
But x ∈ W
Solution set = Φ
Hence it can’t be represented on number line.

Question 15.
Solve:
(i)\(\frac { x }{ 2 } +5\le \frac { x }{ 3 } +6\) where x is a positive odd integer.
(ii)\(\frac { 2x+3 }{ 3 } \ge \frac { 3x-1 }{ 4 } \) where x is positive even integer.
Solution:
(i) \(\frac { x }{ 2 } +5\le \frac { x }{ 3 } +6\)

Question 16.
Given that x ∈ I, solve the inequation and graph the solution on the number line :
\(3\ge \frac { x-4 }{ 2 } +\frac { x }{ 3 } \ge 2 \) (2004)
Solution:
\(3\ge \frac { x-4 }{ 2 } +\frac { x }{ 3 } \) and \(3\ge \frac { x-4 }{ 2 } +\frac { x }{ 3 } \ge 2 \)

Question 17.
Given x ∈ {1, 2, 3, 4, 5, 6, 7, 9}, find the values of x for which -3 < 2x – 1 < x + 4.
Solution:
-3 < 2x – 1 < x + 4.
=> – 3 < 2x – 1 and 2x – 1 < x + 4
=> – 2x < – 1 + 3 and 2x – x < 4 + 1
=> – 2x < 2 and x < 5
=> – x < 1
=> x > – 1
– 1 < x < 5
x ∈ {1, 2, 3, 4, 5, 6, 7, 9}
Solution set = {1, 2, 3, 4} Ans.

Question 18.
Solve : 1 ≥ 15 – 7x > 2x – 27, x ∈ N
Solution:
1 ≥ 15 – 7x > 2x – 27
1 ≥ 15 – 7x and 15 – 7x > 2x – 27

Question 19.
If x ∈ Z, solve 2 + 4x < 2x – 5 ≤ 3x. Also represent its solution on the number line.
Solution:
2 + 4x < 2x – 5 ≤ 3x
2 + 4x < 2x – 5 and 2x – 5 ≤ 3x => 4x – 2x < – 5 – 2 ,and 2x – 3x ≤ 5

Question 20.
Solve the inequation = 12 + \(1 \frac { 5 }{ 6 } x\) ≤ 5 + 3x, x ∈ R. Represent the solution on a number line. (1999)
Solution:
12 + \(1 \frac { 5 }{ 6 } x\) ≤ 5 + 3x

Question 21.
Solve : \(\\ \frac { 4x-10 }{ 3 } \)≤\(\\ \frac { 5x-7 }{ 2 } \) x ∈ R and represent the solution set on the number line.
Solution:
\(\\ \frac { 4x-10 }{ 3 } \)≤\(\\ \frac { 5x-7 }{ 2 } \)
=> 8x – 20 ≤ 15x – 21
(L.C.M. of 3, 2 = 6)

Question 22.
Solve \(\frac { 3x }{ 5 } -\frac { 2x-1 }{ 3 } \) > 1, x ∈ R and represent the solution set on the number line.
Solution:
\(\frac { 3x }{ 5 } -\frac { 2x-1 }{ 3 } \) > 1
=> 9x – (10x – 5) > 15
=> 9x – 10x + 5 > 15
=> – x > 15 – 5
=> – x > 10
=> x < – 10
x ∈ R.
.’. Solution set = {x : x ∈R, x < – 10}
Solution set on the number line

Question 23.
Solve the inequation – 3 ≤ 3 – 2x < 9, x ∈ R. Represent your solution on a number line. (2000)
Solution:
– 3 ≤ 3 – 2x < 9
– 3 ≤ 3 – 2x and 3 – 2x < 9
2x ≤ 3 + 3 and – 2x < 9 – 3
2x ≤ 6 and – 2x < 6 => x ≤ 3 and – x < 3 => x ≤ – 3 and – 3 < x
– 3 < x ≤ 3.
Solution set= {x : x ∈ R, – 3 < x ≤ 3)
Solution on number line

Question 24.
Solve 2 ≤ 2x – 3 ≤ 5, x ∈ R and mark it on number line. (2003)
Solution:
2 ≤ 2x – 3 ≤ 5 .
2 ≤ 2x – 3 and 2x – 3 ≤ 5
2 + 3 ≤ 2x and 2x ≤ 5 + 3
5 ≤ 2x and 2x ≤ 8.

Question 25.
Given that x ∈ R, solve the following inequation and graph the solution on the number line: – 1 ≤ 3 + 4x < 23. (2006)
Solution:
We have
– 1 ≤ 3 + 4x < 23 => – 1 – 3 ≤ 4x < 23 – 3 => – 4 ≤ 4x < 20 => – 1 ≤ x < 5, x ∈ R
Solution Set = { – 1 ≤ x < 5; x ∈ R}

Question 26.
Solve tlie following inequation and graph the solution on the number line. (2007)
\(-2\frac { 2 }{ 3 } \le x+\frac { 1 }{ 3 } <3+\frac { 1 }{ 3 } \) x∈R
Solution:
Given \(-2\frac { 2 }{ 3 } \le x+\frac { 1 }{ 3 } <3+\frac { 1 }{ 3 } \) x∈R
\(-\frac { 8 }{ 3 } \le x+\frac { 1 }{ 3 } <\frac { 10 }{ 3 } \)
Multiplying by 3, L.C.M. of fractions, we get

Question 27.
Solve the following inequation and represent the solution set on the number line :
\(-3<-\frac { 1 }{ 2 } -\frac { 2x }{ 3 } \le \frac { 5 }{ 6 } ,x\in R\)
Solution:
\(-3<-\frac { 1 }{ 2 } -\frac { 2x }{ 3 } \le \frac { 5 }{ 6 } ,x\in R\)

Question 28.
Solve \(\frac { 2x+1 }{ 2 } +2(3-x)\ge 7,x\in R\). Also graph the solution set on the number line
Solution:
\(\frac { 2x+1 }{ 2 } +2(3-x)\ge 7,x\in R\)

Question 29.
Solving the following inequation, write the solution set and represent it on the number line. – 3(x – 7)≥15 – 7x > \(\\ \frac { x+1 }{ 3 } \), n ∈R
Solution:
– 3(x – 7)≥15 – 7x > \(\\ \frac { x+1 }{ 3 } \), n ∈R

Question 30.
Solve the inequation :
\(-2\frac { 1 }{ 2 } +2x\le \frac { 4x }{ 3 } \le \frac { 4 }{ 3 } +2x,\quad x\in W\). Graph the solution set on the number line.
Solution:
\(-2\frac { 1 }{ 2 } +2x\le \frac { 4x }{ 3 } \le \frac { 4 }{ 3 } +2x,\quad x\in W\)

Question 31.
Solve the inequation 2x – 5 ≤ 5x + 4 < 11, where x ∈ I. Also represent the solution set on the number line. (2011)
Solution:
2x – 5 ≤ 5x + 4 < 11 2x – 5 ≤ 5x + 4
=> 2x – 5 – 4 ≤ 5x and 5x + 4 < 11
=> 2x – 9 ≤ 5x and 5x < 11 – 4
and 5x < 7
=> 2x – 5x ≤ 9 and x < \(\\ \frac { 7 }{ 5 } \)
=> 3x > – 9 and x< 1.4
=> x > – 3

Question 32.
If x ∈ I, A is the solution set of 2 (x – 1) < 3 x – 1 and B is the solution set of 4x – 3 ≤ 8 + x, find A ∩B.
Solution:
2 (x – 1) < 3 x – 1
2x – 2 < 3x – 1
2x – 3x < – 1 + 2 => – x < 1 x > – 1
Solution set A = {0, 1, 2, 3, ..,.}
4x – 3 ≤ 8 + x
4x – x ≤ 8 + 3
=> 3x ≤ 11
=> x ≤ \(\\ \frac { 11 }{ 3 } \)
Solution set B = {3, 2, 1, 0, – 1…}
A ∩ B = {0, 1, 2, 3} Ans.

Question 33.
If P is the solution set of – 3x + 4 < 2x – 3, x ∈ N and Q is the solution set of 4x – 5 < 12, x ∈ W, find
(i) P ∩ Q
(ii) Q – P.
Solution:
(i) – 3 x + 4 < 2 x – 3
– 3x – 2x < – 3 – 4 => – 5x < – 7

Question 34.
A = {x : 11x – 5 > 7x + 3, x ∈R} and B = {x : 18x – 9 ≥ 15 + 12x, x ∈R}
Find the range of set A ∩ B and represent it on a number line
Solution:
A = {x : 11x – 5 > 7x + 3, x ∈R}
B = {x : 18x – 9 ≥ 15 + 12x, x ∈R}
Now, A = 11x – 5 > 7x + 3
=> 11x – 7x > 3 + 5
=> 4x > 8
=>x > 2, x ∈ R

Question 35.
Given: P {x : 5 < 2x – 1 ≤ 11, x∈R)
Q{x : – 1 ≤ 3 + 4x < 23, x∈I) where
R = (real numbers), I = (integers)
Represent P and Q on number line. Write down the elements of P ∩ Q. (1996)
Solution:
P= {x : 5 < 2x – 1 ≤ 11}
5 < 2x – 1 ≤ 11

Question 36.
If x ∈ I, find the smallest value of x which satisfies the inequation \(2x+\frac { 5 }{ 2 } >\frac { 5x }{ 3 } +2\)
Solution:
\(2x+\frac { 5 }{ 2 } >\frac { 5x }{ 3 } +2\)
=>\(2x-\frac { 5x }{ 3 } >2-\frac { 5 }{ 2 } \)
=>12x – 10x > 12 – 15
=> 2x > – 3
=>\(x>-\frac { 3 }{ 2 } \)
Smallest value of x = – 1 Ans.

Question 37.
Given 20 – 5 x < 5 (x + 8), find the smallest value of x, when
(i) x ∈ I
(ii) x ∈ W
(iii) x ∈ N.
Solution:
20 – 5 x < 5 (x + 8)
⇒ 20 – 5x < 5x + 40
⇒ – 5x – 5x < 40 – 20
⇒ – 10x < 20
⇒ – x < 2
⇒ x > – 2
(i) When x ∈ I, then smallest value = – 1.
(ii) When x ∈ W, then smallest value = 0.
(iii) When x ∈ N, then smallest value = 1. Ans.

Question 38.
Solve the following inequation and represent the solution set on the number line :
\(4x-19<\frac { 3x }{ 5 } -2\le -\frac { 2 }{ 5 } +x,x\in R\)
Solution:
We have
\(4x-19<\frac { 3x }{ 5 } -2\le -\frac { 2 }{ 5 } +x,x\in R\)
Hence, solution set is {x : -4 < x < 5, x ∈ R}
The solution set is represented on the number line as below.

Question 39.
Solve the given inequation and graph the solution on the number line :
2y – 3 < y + 1 ≤ 4y + 7; y ∈ R.
Solution:
2y – 3 < y + 1 ≤ 4y + 7; y ∈ R.
(a) 2y – 3 < y + 1
⇒ 2y – y < 1 + 3
⇒ y < 4
⇒ 4 > y ….(i)
(b) y + 1 ≤ 4y + 7

Question 40.
Solve the inequation and represent the solution set on the number line.
\(-3+x\le \frac { 8x }{ 3 } +2\le \frac { 14 }{ 3 } +2x,Where\quad x\in I\)
Solution:
Given : \(-3+x\le \frac { 8x }{ 3 } +2\le \frac { 14 }{ 3 } +2x,Where\quad x\in I\)

Question 41.
Find the greatest integer which is such that if 7 is added to its double, the resulting number becomes greater than three times the integer.
Solution:
Let the greatest integer = x
According to the condition,
2x + 7 > 3x
⇒ 2x – 3x > – 7
⇒ – x > – 7
⇒ x < 7
Value of x which is greatest = 6 Ans.

Question 42.
One-third of a bamboo pole is buried in mud, one-sixth of it is in water and the part above the water is greater than or equal to 3 metres. Find the length of the shortest pole.
Solution:
Let the length of the shortest pole = x metre
Length of pole which is burried in mud = \(\\ \frac { x }{ 3 } \)
Length of pole which is in the water = \(\\ \frac { x }{ 6 } \)

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations Ex 4 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends Chapter Test

More Exercises

• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends EX 3
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends MCQS
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends Chapter Test

Question 1.
If a man received ₹1080 as dividend from 9% ₹20 shares, find the number of shares purchased by him.
Solution:
Income on one share = \(\\ \frac { 9 }{ 100 } \) x 20
= Rs \(\\ \frac { 9 }{ 5 } \)
.’. No. of shares = 1080 x \(\\ \frac { 5 }{ 9 } \)
= 120 x 5 = 600 Ans.

Question 2.
Find the percentage interest on capital invested in 18% shares when a Rs 10 share costs Rs 12.
Solution:
Dividend on one share = 18% of Rs 10
= \(\\ \frac { 18\times 10 }{ 100 } \)
= Rs \(\\ \frac { 9 }{ 5 } \)

Question 3.
Rohit Kulkami invests Rs 10000 in 10% Rs 100 shares of a company. If his annual dividend is Rs 800, find :
(i) The market value of each share.
(ii) The rate percent which he earns on his investment.
Solution:
Investment = Rs 10000
Face value of each share = Rs 100
Rate of dividend = 10%
Annual dividend = Rs 800

Question 4.
At what price should a 9% Rs 100 share be quoted when the money is worth 6% ?
Solution:
If interest is 6 then investment = Rs 100
and if interest is 9, then investment
= Rs \(\\ \frac { 100\times 9 }{ 6 } \)
= Rs 150
Market value of each share = Rs 150 Ans

Question 5.
By selling at Rs 92, some 2.5% Rs 100 shares and investing the proceeds in 5% Rs 100 shares at Rs 115, a person increased his annual income by Rs 90. Find:
(i) the number of shares sold.
(ii) the number of shares purchased.
(iii) the new income.
(iv) the rate percent which he earns on his investment.
Solution:
Rate of dividend = 2.5% and market price = Rs 92
Let number of shares purchased = x.
Selling price of x shares = 92 x
Income from investing

Question 6.
A man has some shares of Rs. 100 par value paying 6% dividend. He sells half of these at a discount of 10% and invests the proceeds in 7% Rs. 50 shares at a premium of Rs. 10. This transaction decreases his income from dividends by Rs. 120. Calculate:
(i) the number of shares before the transaction.
(ii) the number of shares he sold.
(iii) his initial annual income from shares.
Solution:
Let no. of shares = x
Value of x shares = x × 100 = 100 x

Question 7.
Divide Rs. 101520 into two parts such that if one part is invested in 8% Rs. 100 shares at 8% discount and the other in 9% Rs. 50 shares at 8% premium, the annual incomes are equal.
Solution:
Total investment = Rs. 101520
Let investment in first part = x
and in second part = (101520 – x)
Market value of first kind of shares = Rs. 100 – Rs. 8
= Rs. 92
and rate of dividend = 8%

Question 8.
A man buys Rs. 40 shares of a company which pays 10% dividend. He buys the shares at such a price that his profit is 16% on his investment. At what price did he buy each share ?
Solution:
Face value of each share = Rs. 40
Dividend = 10%
Gain on investment = 10%

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends Chapter Test are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends MCQS

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends MCQS

More Exercises

• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends EX 3
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends MCQS
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends Chapter Test

Question 1.
If Jagbeer invest ₹10320 on ₹100 shares at a discount of ₹ 14, then the number of shares he buys is
(a) 110
(b) 120
(c) 130
(d) 150
Solution:
Investment = ₹ 10320
Face value of each share = ₹100
M.V. of each share = ₹100 – 14 = ₹86
No. of shares = \(\\ \frac { 10320 }{ 86 } \) = 120 (b)

Question 2.
If Nisha invests ₹19200 on ₹50 shares at a premium of 20%, then the number of shares she buys is
(a) 640
(b) 384
(c) 320
(d) 160
Solution:
Investment = ₹19200
Face value of each share = ₹50
M.V. = ₹50 x \(\\ \frac { 120 }{ 100 } \) = ₹60
Number of shares = \(\\ \frac { 19200 }{ 60 } \)
= 320 (c)

Question 3.
₹40 shares of a company are selling at 25% premium. If Mr. Jacob wants to buy 280 shares of the company, then the investment required by him is
(a) ₹11200
(b) ₹14000
(c) ₹16800
(d) ₹8400
Solution:
Face value of each share = ₹40
M.V. = 40 x \(\\ \frac { 125 }{ 100 } \)= ₹50
Number of shares = 280
Total investment = ₹280 x 50 = ₹ 14000 (d)

Question 4.
Arun possesses 600 shares of ₹25 of a company. If the company announces a dividend of 8%, then Arun’s annual income is
(a) ₹48
(b) ₹480
(c) ₹600
(d) ₹1200
Solution:
Number of shares = 600
F.V. of each share = ₹25
Rate of dividend = 8%
Annual income = 600 x 25 x \(\\ \frac { 8 }{ 100 } \)
= ₹1200 (d)

Question 5.
A man invests ₹24000 on ₹60 shares at a discount of 20%. if the dividend declared by the company is 10%, then his annual income is
(a) ₹3000
(b) ₹2880
(c) ₹ 1500
(d) 1440
Solution:
Investment = ₹24000
F.V. of each share = ₹60
M.V. at discount of 20% = 60 x \(\\ \frac { 80 }{ 100 } \)= ₹48
Rate of dividend = 10%

Question 6.
Salman has some shares of ₹50 of a company paying 15% dividend. If his annual income is ₹3000, then the number of shares he possesses is
(a) 80
(b) 400
(c) 600
(d) 800
Solution:
F.V. of each share = ₹50
Dividend = 15%
Annual income = ₹3000
Let x be the share, then
F.V. of shares = x × 50 = ₹50x

Question 7.
₹25 shares of a company are selling at ₹20. If the company is paying a dividend of 12%, then the rate of return is
(a) 10%
(b) 12%
(c) 15%
(d) 18%
Solution:
F.V. of each share = ₹25 ,
M.V. = ₹20
Rate of dividend = 12%
Dividend on each share = \(\\ \frac { 12 }{ 100 } \) x 25 = ₹3
Return on ₹20 = ₹3
and on ₹100 = ₹ \(\\ \frac { 3 }{ 20 } \) x \(\\ \frac { 5 }{ 100 } \) = 15% (c)

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends MCQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.