ML Aggarwal Solutions for Class 10 Maths Chapter 4 Linear Inequations can be utilised by the students to comprehend the concepts and solve problems in an effective way. The solutions are clearly briefed in a step-by-step manner, in order to help students understand easily. Further, these solutions are in accordance with the latest ICSE board guidelines. To learn more about these concepts, students can access the ML Aggarwal Solutions for Class 10 Maths Chapter 4 Linear Inequations PDF, from the link provided below.
The main topics covered in this chapter are inequalities among real numbers, linear inequations and solving linear inequations in one-variable using the two permissible rules. Students can start solving the problems on a regular basis, so that their conceptual knowledge becomes stronger and also problem-solving skills will be improved, which is important from the examination perspective. Moreover, the ML Aggarwal Solutions can be used as a reference to answers and to learn the other methods of solving problems effortlessly.
ML Aggarwal Solutions for Class 10 Maths Chapter 4 Linear Inequations :-
Access ML Aggarwal Solutions for Class 10 Maths Chapter 4 Linear Inequations
Exercise 4
1. Solve the inequation, 3x – 11 < 3 where x ∈ {1, 2, 3, ……, 10}. Also, represent its solution on a number line.
Solution:
Given inequation, 3x – 11 < 3
3x < 3 + 11
3x < 14
⇒ x < 14/3
But, x ∈ {1, 2, 3,……, 10}
Hence, the solution set is {1, 2, 3, 4}.
Representing the solution on a number line:
ML Aggarawal Solutions for Class 10 Chapter 4 - 1
2. Solve 2(x – 3) < 1, x ∈ {1, 2, 3, …. 10}
Solution:
Given inequation, 2(x – 3) < 1
2x – 6 < 1
2x < 7
⇒ x < 7/2
But, x ∈ {1, 2, 3, …. 10}
Hence, the solution set is {1, 2, 3}
3. Solve 5 – 4x > 2 – 3x, x ∈ W. Also represent its solution on the number line.
Solution:
Given inequation, 5 – 4x > 2 – 3x
– 4x + 3x > 2 – 5
-x > -3
On multiplying both sides by -1, the inequality reverses
⇒ x < 3
Since, x ∈ W
The solution set is {0, 1, 2}
Representing the solution on a number line:.
4. List the solution set of 30 – 4 (2x – 1) < 30, given that x is a positive integer.
Solution:
Given inequation, 30 – 4 (2x – 1) < 30
30 – 8x + 4 < 30
34 – 8x < 30
-8x < 30 – 34
-8x < -4 [On multiplying both sides by -1, the inequality reverses]
8x > 4
x > 4/8
⇒ x > 1/2
As x is a positive integer
The solution set is {1, 2, 3, … }
5. Solve: 2 (x – 2) < 3x – 2, x ∈ {– 3, – 2, – 1, 0, 1, 2, 3}.
Solution:
Given inequation, 2 (x – 2) < 3x – 2
2x – 4 < 3x – 2
2x – 3x < -2 + 4
-x < 2
⇒ x > -2
But, x ∈ {– 3, – 2, – 1, 0, 1, 2, 3}
Hence, the solution set is {– 1, 0, 1, 2, 3}.
6. If x is a negative integer, find the solution set of 2/3 + 1/3 (x + 1) > 0.
Solution:
Given inequation, 2/3 + 1/3 (x + 1) > 0.
2/3 + x/3 + 1/3 > 0
x/3 + 1 > 0
x/3 > -1
⇒ x > -3
As x is a negative integer
The solution set is {-1, -2}.
7. 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:
Given inequation, x – 3 (2 + x) > 2 (3x – 1)
x – 6 – 3x > 6x – 2
-2x – 6 > 6x – 2
-6x – 2x > -2 + 6
-8x > 4
x < -4/8
⇒ x < -1/2
But, x ∈ { – 3, – 2, – 1, 0, 1, 2, 3}
Hence, the solution set is {-3, -2, -1}
8. Given x ∈ {1, 2, 3, 4, 5, 6, 7, 9} solve x – 3 < 2x – 1.
Solution:
Given inequation, x – 3 < 2x – 1
x – 2x < – 1 + 3
-x < 2
⇒ x > -2
But, x ∈ {1, 2, 3, 4, 5, 6, 7, 9}
Hence, the solution set is {1, 2, 3, 4, 5, 6, 7, 9}.
9. List the solution set of the inequation ½ + 8x > 5x – 3/2, x ∈ Z
Solution:
Given inequation, ½ + 8x > 5x – 3/2
8x – 5x > -3/2 – ½
3x > -4/2
⇒ x > -2/3
As x ∈ Z
The solution set is {0, 1, 2, 3, 4, 5, …}
10. List the solution set of (11 – 2x)/5 ≥ (9 – 3x)/8 + 3/4, x ∈ N
Solution:
Given inequation, (11 – 2x)/5 ≥ (9 – 3x)/8 + ¾
(11 – 2x)/5 ≥ (9 – 3x + 6)/8
8 (11 – 2x) ≥ 5 (15 – 3x)
88 – 16x ≥ 75 – 15x
15x – 16x ≥ 75 – 88
-x ≥ -13
⇒ x ≤ 13
As x ∈ N
Hence, the solution set is {1, 2, 3, 4, …, 13}.
10. List the solution set of (11 – 2x)/5 ≥ (9 – 3x)/8 + 3/4, x ∈ N
Solution:
Given inequation, (11 – 2x)/5 ≥ (9 – 3x)/8 + ¾
(11 – 2x)/5 ≥ (9 – 3x + 6)/8
8 (11 – 2x) ≥ 5 (15 – 3x)
88 – 16x ≥ 75 – 15x
15x – 16x ≥ 75 – 88
-x ≥ -13
⇒ x ≤ 13
As x ∈ N
Hence, the solution set is {1, 2, 3, 4, …, 13}.
(11) Find the values of x, which satisfy the inequation
x ∈ N. Graph the solution set on the number line.
Solution:
Given inequation
-2 ≤ (3 – 4x)/ 6 ≤ 11/6
-12 ≤ 3 – 4x ≤ 11
-12 – 3 ≤ -4x ≤ 11 – 3
-15 ≤ -4x ≤ 8
-15/4 ≤ -x ≤ 8/4
⇒ 15/4 ≥ x ≥ -2
As x ∈ N,
The solution set is {1, 2, 3}.
ML Aggarawal Solutions for Class 10 Chapter 4 - 6
12. If x ∈ W, find the solution set of 3/5 x – (2x – 1)/3 > 1. Also graph the solution set on the number line, if possible.
Solution:
Given inequation, 3/5 x – (2x – 1)/3 > 1
9/15 x – 5(2x – 1)/15 > 1 [Taking L.C.M]
9x – 5(2x – 1) > 15 [Multiplying by 15 on both sides]
9x – 10x + 5 > 15
-x > 15 – 5
-x > 10
⇒ x < -10
But, x ∈ W
Hence, the solution set is a null set.
Thus, it can’t be represented on number line.
13. Solve:
(i) x/2 + 5 ≤ x/3 + 6, where x is a positive odd integer.
(ii) (2x + 3)/3 ≥ (3x – 1)/4, where x is positive even integer.
Solution:
(i) Given inequation, x/2 + 5 ≤ x/3 + 6
(x + 10)/2 ≤ (x + 18)/3 [Taking L.C.M on both sides]
3 (x + 10) ≤ 2 (x + 18) [On cross-multiplying]
3x + 30 ≤ 2x + 36
3x – 2x ≤ 36 – 30
⇒ x ≤ 6
As x is a positive odd integer.
Hence, the solution set is {1, 3, 5}.
(ii) Given inequation, (2x + 3)/3 ≥ (3x – 1)/4
4 (2x + 3) ≥ 3 (3x – 1) [On cross-multiplying]
8x + 12 ≥ 9x – 3
-9x + 8x ≥ -12 – 3
-x ≥ -15
⇒ x ≤ 15
As x is positive even integer.
Hence, the solution set is {2, 4, 6, 8, 10, 12, 14}.
14. Given that x ∈ I, solve the inequation and graph the solution on the number line:
3 ≥ (x – 4)/2 + x/3 ≥ 2
Solution:
Given inequation, 3 ≥ (x – 4)/2 + x/3 ≥ 2
Now, let’s take
3 ≥ (x – 4)/2 + x/3, we have
3 ≥ (3x – 12 + 2x)/6 [Taking L.C.M]
18 ≥ 5x – 12
30 ≥ 5x
⇒ x ≤ 6 …. (i)
Next,
(x – 4)/2 + x/3 ≥ 2
(3x – 12 + 2x)/6 ≥ 2
5x – 12 ≥ 12
5x ≥ 24
x ≥ 24/5 ⇒ x ≥ 4.8 … (ii)
Hence, from (i) and (ii) we have
Solution of x = {5, 6}
Representing the solution on a number line
Hence, from (i) and (ii) we have
Hence, from (i) and (ii) we have
Solution of x = {5, 6}
Representing the solution on a number line:
ML Aggarawal Solutions for Class 10 Chapter 4 - 7
15. Solve: 1 ≥ 15 – 7x > 2x – 27, x ∈ N
Solution:
Given inequation, 1 ≥ 15 – 7x > 2x – 27,
So, we have
1 ≥ 15 – 7x and 15 – 7x > 2x – 27
7x ≥ 15 – 1 and -2x – 7x > -27 – 15
7x ≥ 14 and -9x > -42
x ≥ 2 and -x > -42/9
x ≥ 2 and x < 14/3
⇒ 2 ≤ x < 14/3
But as x ∈ N
The solution set is {2, 3, 4}.
16. If x ∈ Z, solve 2 + 4x < 2x – 5 ≤ 3x. Also represent its solution on the number line.
Solution
Given inequation, 2 + 4x < 2x – 5 ≤ 3x
So, we have
2 + 4x < 2x – 5 and 2x – 5 ≤ 3x
4x – 2x < -5 – 2 and 2x – 3x ≤ 5
2x < -7 and -x ≤ 5
x < -7/2 and x ≥ -5
⇒ -5 ≤ x < -7/2
As x ∈ Z
The solution set is {-5, -4}.
Representing the solution on a number line:
17. Solve: (4x – 10)/3 ≤ (5x – 7)/2, x ∈ R and represent the solution set on the number line.
Solution:
Given inequation, (4x – 10)/3 ≤ (5x – 7)/2
2 (4x – 10) ≤ 3 (5x – 7) [On cross-multiplying]
8x – 20 ≤ 15x – 21
8x – 15x ≤ -21 + 20
-7x ≤ -1
-x ≤ -1/7
x ≥ 1/7
As x ∈ R
Hence, the solution set is {x: x ∈ R, x ≥ 1/7}
Representing the solution on a number line:
Solution:
Given inequation, 3x/5 – (2x – 1)/3 > 1
(9x – 10x + 5)/15 > 1 [Taking L.C.M]
-x + 5 > 15
-x > 15 – 5
-x > 10
x < -10
As x ∈ R
Hence, the solution set is {x: x ∈ R, x < -10}
Representing the solution on a number line
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