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## NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Exercise 6.1

Chapter 6 Linear Inequalities Exercise 6.1 Class 11 Maths NCERT Solutions that will be beneficial in understanding the basic concepts of the chapter. NCERT Solutions for Class 11 Maths provided here are free and accurate that can be helpful in solving your doubts and completing your homework on time. 1. Solve 24x < 100, when
(i) x is a natural number.
(ii) x is an integer.

We are given:
24x < 100
or 24x/24 < 100/24  or x< 100/24
(i) When x is natural number, the following values of x make the statement true
x = 1, 2, 3, 4.
The solution set = {1, 2, 3, 4}

(ii) When x is an integer, in this case the solutions of the given inequality are ......,
– 3, –2, –1, 0, 1, 2, 3, 4.
The solution set of the inequality is
{...., –3, –2, –1, 0, 1, 2, 3, 4}.

2. Solve –12x > 30, when
(i) x is a natural number
(ii) x is an integer.

–12x > 30, dividing by – 12 => x < -(30/12) = -5/2
(i) This inequality is not true for any natural number
(ii) Integers that satisfy this inequality are {........–5, –4, –3}

3. Solve 5x – 3 < 7, when
(i) x is an integer
(ii) x is a real number

5x – 3 < 7
Transposing 3 to R.H.S
5x < 7 + 3 or 5x < 10 (i) When x is an integer x = 1 satisfies this inequality solution set is {...., –2, –1, 0, 1}.
(ii) When x is real, the solution is (–∞, 2)

4. Solve 3x + 8 > 2, when
(i) x is an integer
(ii) x is a real number

Inequality is 3x + 8 > 2
Transposing 8 to RHS 3x > 2 – 8 = –6
Dividing by 3, x > –2
(i) When x is an integer the solution is {–1, 0, 1, 2, 3, ........}
(ii) When x is real, the solution is (–2, ∞).

5. Solve the following inequality for real x:
4x + 3 < 6 + 7

4x + 3 < 6x + 7
=>  4x + 3 – 3 < 6x + 7 – 3 => 4x < 6x + 4
=> 4x – 6x < 6x – 6x + 4 => – 2x < 4
=> -2x/-2 > 4/-2 =>  x > – 2
Solution set = (–2, ∞)

6. Solve the following inequality for real
x: 3x – 7 > 5x – 1

3x – 7 > 5x – 1
=>  3x – 7 + 7 > 5x – 1 + 7
=>  3x > 5x + 6
=> 3x – 5x > 5x – 5x + 6
=>  – 2x > 6
=>  -2x/-2 < 6/-2
=> x < - 3
Solution set = (– ∞, – 3)

7. Solve the inequality 3(x – 1) ≤ 2(x – 3) for real x.

The inequality is 3(x - 1) ≤ 2(x - 3)
=> 3x – 3 ≤ 2x – 6
Transposing 2x to L.H.S and –3 to RHS
=> 3x – 2x ≤ -6 + 3 => x ≤ -3
The solution is (–∞, –3]

8. Solve the inequality 3(2 – x) ≥ 2(1 – x) for real x.

The inequality is
3(2 – x) ≥ 2 (1 – x)
Simplifying
6 – 3x ≥ 2 – 2x
Transposing – 2x to L.H.S. and 6 to R.H.S.
– 3x + 2x ≥ 2 – 6
or – x ≥ – 4
Multiplying by – 1
x ≤ 4
The solution is (– ∞, 4]

9. Solve the inequality x + x/2 + x/3 < 11 for real x

The inequality is
x + x/2 + x/3 < 11
Simplifying
(6x + 3x + 2x)/6  < 11          or 11x/6  < 11
Dividing both sides by 6/11
x < 6
∴ The solution is (– ∞, 6)

10. Solve the inequality x/3 > x/2 + 1 for real x.

The inequality x/3 > x/2 + 1
Transposing x/2  to LHS.
Simplifying, x/3 – x/2 > 1 => (2x – 3x)/6  >  1
Or –x/6  > 1
Multiply by –6, x < –6
∴ The solution is (–∞, –6)

11. Solve the inequality (3(x - 2))/5 ≤ (5(2 - x))/3 for real x.

The inequality is
(3(x - 2))/5 ≤ (5(2 - x))/3
Multiply both sides by L.C.M. of 5, 3 i.e., by 15
3 × 3 (x – 2) ≤ 5 × 5 (2 – x)
or 9(x – 2) ≤ 25 (2 – x)
Simplifying
9x – 18 ≤ 50 – 25x
Transposing – 25x to L.H.S. and – 18 to R.H.S
∴ 9x + 25x ≤ 50 + 18
or 34x ≤ 68
Dividing by 34.
x ≤ 2
∴ Solution is (– ∞, 2]

12. Solve the inequality
1/2 (3x/5 + 4)  ≥ 1/3 (x - 6) for real x.

The inequality is
1/2 (3x/5 + 4)  ≥ 1/3 (x - 6)
=> 1/2((3x + 20)/5) ≥ 1/3(x - 6)
Multiplying both sides by 30
3(3x + 20) ≥ 10(x - 6)
=> 9x + 60 ≥ 10x - 60
Transposing 10x to LHS and 60 to RHS
∴ 9x - 10x ≥ -60 - 60 => -x ≥ - 120
Multiplying by –1, x ≤ 120
∴ The solution is (–∞, 120]

13. Solve the inequality 2(2x + 3) – 10 < 6 (x – 2) for real x.

The inequality is
2(2x + 3) – 10 < 6 (x – 2)
Simplifying.
4x + 6 – 10 < 6x – 12
or 4x – 4 < 6x – 12
Transposing 6x to L.H.S. and – 4 to R.H.S.
∴ 4x – 6x < – 12 + 4
or – 2x < – 8
Dividing by – 2
x > 4
∴ The solution is (4, ∞)

14. Solve the inequality
37 - (3x + 5) ≥ 9x - 8(x - 3) for real x.

The inequality is 37 – (3x + 5) ≥ 9x - 8(x - 3)
Simplifying 37 - 3x - 5 ≥ 9x - 8x + 24
=> 32 - 3x ≥  x + 24
Transposing x to L.H.S and 32 to R.H.S,
-3x -x ≥ 24 - 32 or -4x ≥ -8
Dividing by –4, we get x ≤ 2
The solution is (–∞, 2]

15. Solve the inequality x/4 < (5x - 2)/3 - (7x - 3)/5  for real x.

The inequality is x/4 < (5x - 2)/3  - (7x - 3)/5
Multiplying each term by the L.C.M. of 4, 3, 5 i.e. 60
15x < 20 (5x – 2) – 12(7x – 3)
Simplifying
15x < 100x – 40 – 84x + 36
or 15x < 100x – 84x – 40 + 36
or 15x < 16x – 4
Transposing 16x to L.H.S.
15x – 16x < – 4
or – x < – 4
Multiplying by – 1
x > 4
The solution is (4, ∞)

16. Solve the inequality
(2x - 1)/3 ≥ (3x - 2)/4 - (2 - x)/5 for real x.

The inequality is (2x - 1)/3 ≥ (3x - 2)/4 - (2 - x)/5

Multiplying each terms by L.C.M of 3, 4, 5 i.e. 60
(2x – 1)/3 × 60 ≥ (3x – 2)/ 4 × 60 – (2 – x)/ 5 × 60
=> 20(2x – 1) ≥  (3x – 2) × (15) – ( 2 – x) × 12
=> 40x -20 ≥ 45x – 30 – 24 + 12x
=> 40x – 20 ≥ 57x – 54
Transposing 57x to LHS and -20 to RHS
40x – 57x ≥ – 54 + 20 => - 17x ≥ – 34
=>  x ≤ 2. (dividing by -17) ∴ The solution is (-∞ , 2)

17. Solve the following inequality and show the graph of the solution on number line 3x – 2 < 2x + 1

3x – 2 < 2x + 1
=> 3x – 2 + 2 < 2x + 1 + 2
=> 3x < 2x + 3
=> 3x – 2x < 2x + 3 – 2x
=>  x < 3
i.e., all real numbers x less than 3 as shown on the number line. 18. Solve the following inequality and show the graph of the solution on number line 5x – 3 ≥ 3x – 5

5x – 3 ≥ 3x – 5
=> 5x – 3 + 3 ≥ 3x – 5 + 3
=> 5x ≥ 3x – 2
=> 5x – 3x ≥ 3x – 2 – 3x
=> 2x ≥ – 2
=> 2x/2 ≥ -2/2 => x ≥-1
i.e., all real numbers x greater than or equal to – 1 as shown on the number line. 19. Solve the following inequality and show the graph of the solution on number line.
3(1 – x) < 2 (x + 4)

3(1 – x) < 2(x + 4)
=> 3 – 3x < 2x + 8
=> 3 – 3x – 3 < 2x + 8 – 3
=> – 3x < 2x + 5
=> – 3x – 2x < 2x + 5 – 2x
=> – 5x < 5
=> -5x/-5 > 5/-5
=> x > – 1 i.e., all real numbers x greater than –1 as shown on the number line.

20. Solve the inequality and show the graph of the solution in each case on number line.
x/2 < (5x - 2)/3 - (7x - 3)/5

The inequality is x/2 < (5x – 2)/3  - (7x – 3)/5
Multiply each term by L.C.M of 2, 3, 5 i.e. by 30
x/2 × 30 < (5x – 2)/3 × 30 – (7x – 3)/5 × 30
=> 15x < 10(5x – 2) – 6(7x – 3)
=> 15x< 50x – 20 – 42x + 18
=> 15x < 8x – 2
Transposing 8x to LHS 15x – 8x< -2 => 7x < -2
Dividing by 7 => x < - 2/7
∴ The solution is (- ∞, -(2/7))

21. Ravi obtained 70 and 75 marks in first two unit test. Find the number if minimum marks he should get in the third test to have an average of at least 60 marks.

Let Ravi get x marks in third unit test.
∴ Average marks obtained by Ravi
= (70 + 75 + x)/3
To obtain average of atleast 60 marks, we have
(70 + 75 + x)/3 ≥ 60 => (145 + x)/3 ≥ 60
Multiplying by 3, 145 + x ≥ 60 × 3 = 180
Transposing 145 to RHS
=> x ≥ 180 - 145 = 35
∴ Ravi should got atleast 35 marks in the third unit test.

22. To receive grade ‘A’ in a course, one must obtain an average of 90 marks or more in five examinations (each of 100 marks). If Sunita’s marks in first four examinations are 87, 92, 94 and 95. Find minimum marks that Sunita must obtain in fifth examination to get grade ‘A’ in the course.

Let Sunita obtain x marks in the fifth examination.
Average of marks of five examinations
= (87 + 92 + 94 + 95 + x)/5 = (368 + x)/5
This average must be atleast equal to 90
=>  (368 + x)/5  ≥ 90
=> 368 + x ≥ 5 × 90 = 450
=> x ≥ 450 – 368 = 82
i.e. She should obtain atleast 82 marks in the fifth examination.

23. Find all pair of consecutive odd positive integers both of which are smaller than 10 such that their sum is more than 11.

Let x be the smaller of the two odd positive integers so that the other is x + 2. Then we should have x < 10
∴ x + (x + 2) > 11 => 2x + 2 > 11
=> 2x > 11 – 2
=> 2x > 9 => x > 9/2
Hence, if one number is 5 (odd number), then the other is seven, since 7 < 10. If the smaller number is seven, then the other is nine. Hence, possible are (5, 7) and (7, 9).

24. Find all pairs of consecutive even positive integers, both of which are larger than 5 such that their sum is less than 23.

Let x be the smaller of the two positive even integers then the other one is x + 2, then we should have
x > 5 and x + x + 2 < 23 => 2x + 2 < 23
Since, if the sum is less than 23 then it is also less than 24
2x + 2 < 24 => 2x < 22 => x < 11
Thus, the value of x may be 6, 8, 10 (even integer).
Hence, the pairs may be (6, 8), (8, 10), (10, 12)

25. The longest side of a triangle is 3 times the shortest side and the third side is 2 cm shorter than the longest side. If the perimeter of the triangle is at least 61 cm, find the minimum length of the shortest side.

Let shortest side measure x cm. Therefore the longest side will be 3x cm and third side will be (3x– 2) cm
According to the problem,
x + 3x + 3x – 2 ≥ 61
=> 7x – 2 ≥ 61 or 7x ≥ 63
=> x ≥ 9 cm
Hence, the minimum length of the shortest side is 9 cm and the other sides measure 27 cm and 25 cm.

26. A man wants to cut three lengths from a single piece of board of length 91 cm, the second length is to be 3 cm longer than the shortest and the third length is to be twice as long as the shortest. What are the possible lengths of the shortest board if the third piece is to be at least 5 cm longer than the second?