# NCERT SOLUTIONS FOR CLASS 11 MATHS CHAPTER 6 / LINEAR INEQUALITIES EX 6.1 /

Question 1. Solve 24x < 100, when

• (i) x is a natural number
• (ii) x is an integer.

Solution.

Given inequality is 24x < 100

Dividing both sides by 24, we get This inequality is true when

(i) x is a natural number, {1, 2, 3, 4} satisfies this inequality.

(ii) x is an integer, {…., -4, -3,-2, -1, 0, 1, 2, 3, 4} satisfies this inequality.

Question 2. Solve – 12x > 30, when

• (i) x is a natural number
• (ii) x is an integer

Solution.

Given inequality is -12x > 30 Dividing both sides by -12, we get (i) This inequality is not true for any natural number.

(ii) Integers that satisfy this inequality are (…, -5, -4, -3}.

Question 3. Solve 5x – 3 < 7, when

• (i) x is an integer
• (ii) x is a real number

Solution.

Given inequality is 5x – 3 < 7

Transposing 3 to R.H.S., we get

5x < 7 + 3 or 5x < 10

Dividing both sides by 5, we get

x < 2

(i) When x is an integer, {…. -2, -1, 0, 1} satisfies this inequality.

(ii) When x is real number, the solution is (-∞, 2).

Question 4. Solve 3x + 8 > 2, when

• (i) x is an integer
• (ii) x is a real number.

Solution.

Given inequality is 3x + 8 > 2

Transposing 8 to R.H.S., we get

3x > 2 – 8 = -6

Dividing both sides by 3, we get

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, ∞).

Solve the inequalities in Exercises 5 to 16 for real x.

Question 5. 4x + 3 < 5x + 7

Solution.

The inequality is 4x + 3 < 5x + 7

Transposing 5x to L.H.S. and 3 to R.H.S., we get

4x – 5x < 7 – 3 or -x < 4 Dividing both sides by -1, we get x > -4

∴ The solution is (- 4, ∞).

Question 6. 3x – 7 > 5x – 1

Solution.

The inequality is 3x – 7 > 5x -1

Transposing 5x to L.H.S. and -7 to R.H.S., we get

3x – 5x > -1 + 7 or -2x > 6

Dividing both sides by -2, we get

x < -3

∴ The solution is (-∞, -3).

Question 7. 3(x – 1) < 2(x – 3)

Solution.

The inequality is 3(x – 1) < 2(x – 3) or 3x – 3 < 2x – 6

Transposing 2x to L.H.S. and -3 to R.H.S., we get

3x – 2x < – 6 + 3

⇒ x<-3

∴ The solution is (- ∞, -3],

Question 8. 3(2 -x) > 2(1 -x)

Solution.

The inequality is 3(2 – x) > 2(1 – x) or 6 – 3x > 2 – Zx

Transposing -2x to L.H.S. and 6 to R.H.S., we get

-3x + 2x > 2 – 6 or -x > -4

Multiplying both sides by -1, we get

x ≤ 4

∴ The solution is (- ∞, 4],

Question 9. Solution.

The inequality is Simplifying, Multiplying both sides by $\frac { 6 }{ 11 }$, we get

x < 6

∴ The solution is (- ∞, 6),

Question 10. $\frac { x }{ 3 } >\frac { x }{ 2 } +1$

Solution.

The inequality is $\frac { x }{ 3 } >\frac { x }{ 2 } +1$

Transposing $\frac { x }{ 2 }$ to L.H.S., we get $\frac { x }{ 3 } -\frac { x }{ 2 } >1$

Simplifying, $\frac { 2x-3x }{ 6 } >1\quad or\quad -\frac { x }{ 6 } >1$

Multiplying both sides by -6, we get

x < -6

∴ The solution is (- ∞, – 6).

Question 11. $\frac { 3\left( x-2 \right) }{ 5 } \le \frac { 5\left( 2-x \right) }{ 3 }$

Solution.

The inequality is $\frac { 3\left( x-2 \right) }{ 5 } \le \frac { 5\left( 2-x \right) }{ 3 }$

Multiply both sides by the L.C.M. of 5, 3 i.e., by 15.

3 x 3(x – 2) ≤ 5 x 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 both sides by 34, we get

x < 2

∴ Solution is (- ∞, 2].

Question 12. $\frac { 1 }{ 2 } \left( \frac { 3x }{ 5 } +4 \right) \ge \frac { 1 }{ 3 } \left( x-6 \right)$

Solution.

The inequality is $\frac { 1 }{ 2 } \left( \frac { 3x }{ 5 } +4 \right) \ge \frac { 1 }{ 3 } \left( x-6 \right)$

or, $\frac { 1 }{ 2 } \left( \frac { 3x+20 }{ 5 } \right) \ge \frac { 1 }{ 3 } \left( x-6 \right)$

Multiplying both sides by 30,

3(3x + 20) ≥ 10(x – 6) or, 9x + 60 ≥ 10x-60

Transposing 10 x to L.H.S. and 60 to R.H.S., we get

∴ 9x-10x ≥ -60 – 60 or -x ≥-120

Multiplying both sides by -1, we get

x < 120

∴ The solution is (- ∞, 120].

Question 13. 2(2x + 3) – 10 < 6(x – 2)

Solution.

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., we get

∴ 4x – 6x < -12 + 4 or -2x < – 8 Dividing both sides by -2, we get x>4

∴ The solution is (4, ∞).

Question 14. 37 – (3x + 5) ≥ 9x – 8(x – 3)

Solution.

The inequality is 37 – (3x + 5) ≥ 9x – 8(x – 3)

Simplifying, 37 – 3x – 5 ≥ 9x – 8x + 24

or 32 – 3x ≥ x + 24

Transposing x to L.H.S. and 32 to R.H.S., We get

-3x – x ≥ 24 – 32 or -4x ≥ -8

Dividing both sides by – 4, we get

x < 2

∴ The solution is (- ∞, 2].

Question 15. Solution.

The inequality is Multiplying each term by the L.C.M. of 4, 3, 5, i.e., by 60, we get

15x < 100x – 40 – 84x + 36

or, 15x < 100x – 84x -40 + 36

or, 15x < 16x – 4

Tranposing 16x to L.H.S., we get

15x – 16x < -4 or -x < -4 Multiplying both sides by -1, we get x >4

∴ The solution is (4, ∞).

Question 16. $\frac { \left( 2x-1 \right) }{ 3 } \ge \frac { \left( 3x-2 \right) }{ 4 } -\frac { 2-x }{ 5 }$

Solution.

The inequality is $\frac { \left( 2x-1 \right) }{ 3 } \ge \frac { \left( 3x-2 \right) }{ 4 } -\frac { 2-x }{ 5 }$

Multiplying each term by L.C.M. of 3,4, 5, i.e., by 60 $\frac { \left( 2x-1 \right) }{ 3 } \times 60\ge \frac { \left( 3x-2 \right) }{ 4 } \times 60-\frac { 2-x }{ 5 } \times 60$

or, 20(2x – 1) ≥ (3x – 2) x 15 – (2 – x) x 12

or, 40x – 20 ≥ 45x – 30 – 24 + 12x

or, 40x – 20 ≥ 57x – 54

Transposing 57x to L.H.S. and -20 to R.H.S., we get

40x – 57x ≥ -54 + 20 or -17x ≥ -34

Dividing both sides by -17, we get

x <2

∴ The solution is (- ∞, 2].

Solve the inequalities in Exercises 17 to 20 and show the graph of the solution in each case on number line.

Question 17. 3x – 2 < 2x + 1

Solution.

The inequality is 3x – 2 < 2x + 1

Transposing 2x to L.H.S. and -2 to R.H.S, we get

3x- 2x < 1 + 2 or, x <3

Question 18. 5x – 3 ≥ 3x – 5

Solution.

The inequality is 5x – 3 ≥ 3x – 5

Transposing 3x to L.H.S. and -3 to R.H.S., we get

∴ 5x – 3x ≥ -5 + 3 or, 2x ≥ -2

Dividing both sides by 2, we get

Question 19. 3(1 – x) < 2(x + 4)

Solution.

3(1-x) < 2(x + 4)

Simplifying 3 – 3x < 2x + 8

Transposing 2x to L.H.S. and 3 to R.H.S., we get

-3x – 2x < 8 – 3 or -5x < 5

Dividing both sides by -5, we get

Question 20. $\frac { x }{ 2 } \ge \frac { \left( 5x-2 \right) }{ 3 } -\frac { \left( 7x-3 \right) }{ 5 }$

Solution.

The inequality is $\frac { x }{ 2 } \ge \frac { \left( 5x-2 \right) }{ 3 } -\frac { \left( 7x-3 \right) }{ 5 }$

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

Solution.

Let Ravi gets x marks in third unit test.

∴ Average marks obtained by Ravi $=\frac { 70+75+x }{ 3 }$

He has to obtain atleast 60 marks,

∴ Multiplying both sides by 3,

145 + x ≥ 60 x 3 = 180

Transposing 145 to R.H.S., we get

x ≥ 180 – 145 = 35

∴ Ravi should get atleast 35 marks in the third unit test.

Question 22. To receive Grade ‘A’ in the 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.

Solution.

Let Sunita obtained x marks in the fifth examination.

∴ Average marks of 5 examinations $=\frac { 87+92+94+95+x }{ 5 } =\frac { 368+x }{ 5 }$

This average must be atleast 90

∴ $\frac { 368+x }{ 5 } \ge 90$

Multiplying both sides by 5

368 + x ≥ 5 x 90 = 450

Transposing 368 to R.H.S., we get

x ≥ 450 – 368 = 82

∴ Sunita should obtain atleast 82 marks in the fifth examination.

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

Solution.

Let x be the smaller of the two odd positive integers. Then the other integer is x + 2. We should havex + 2< 10 and x + (x + 2) > 11 or, 2x + 2 > 11

or, 2x > 11 – 2 or, 2x > 9 or, $x>\frac { 9 }{ 2 }$

Hence, if one number is 5 (odd number), then the other is 7. If the smaller number is 7, then the other is 9. Hence, possible pairs are (5, 7) and (7, 9).

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

Solution.

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 or, 2x + 2 < 23

or, 2x < 21 or, Thus, the value of x may be 6,8,10 (even integers) Hence, the pairs may be (6, 8), (8,10), (10,12).

Question 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.

Solution.

Let the shortest side measures x cm

The longest side will be 3x cm.

Third side will be (3x – 2) cm.

According to the problem, x + 3x + 3x – 2 ≥ 61

or, 7x – 2 ≥ 61 or, 7x ≥ 63 or, x ≥ 9

Hence, the minimum length of the shortest side is 9 cm.

Question 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 for the shortest board if the third piece is to be atleast 5 cm longer than the second?

[Hint: If x is the length of the shortest board, then x, (x + 3) and 2x are the lengths of the second and third piece, respectively. Thus, x + (x + 3) + 2x ≤ 91 and 2x ≥ (x + 3) + 5].

Solution.

Let x be the length of the shortest board, then x + 3 is the second length and 2x is the third length. Thus, x + (x + 3) + 2x ≤ 91

or 4x + 3 ≤ 91 or 4x ≤ 88 or x ≤ 22

According to the problem, 2x ≥ (x + 3) + 5 or x ≥ 8

∴ Atleast 8 cm but not more than 22 cm are the possible lengths for the shortest board.

#### NCERT SOLUTIONS FOR CLASS 11 MATHS CHAPTER 6 / LINEAR INEQUALITIES EX 6.2 /

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