NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Ex 6.1 are part of NCERT Solutions for Class 11 Maths. Here we have given NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Ex 6.1.
| Board | CBSE |
| Textbook | NCERT |
| Class | Class 11 |
| Subject | Maths |
| Chapter | Chapter 6 |
| Chapter Name | Linear Inequalities |
| Exercise | Ex 6.1 |
| Number of Questions Solved | 26 |
| Category | NCERT Solutions |
NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Ex 6.1
Ex 6.1 Class 11 Maths 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
\(x<\frac { 100 }{ 24 } =\frac { 25 }{ 6 } \)
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.
Ex 6.1 Class 11 Maths 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
\(x<-\frac { 30 }{ 12 } =-\frac { 5 }{ 2 } \)
(i) This inequality is not true for any natural number.
(ii) Integers that satisfy this inequality are (…, -5, -4, -3}.
Ex 6.1 Class 11 Maths 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).
Ex 6.1 Class 11 Maths 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.
Ex 6.1 Class 11 Maths 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, ∞).
Ex 6.1 Class 11 Maths 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).
Ex 6.1 Class 11 Maths 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],
Ex 6.1 Class 11 Maths 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],
Ex 6.1 Class 11 Maths Question 9.
\(x+\frac { x }{ 2 } +\frac { x }{ 3 } <11\)
Solution.
The inequality is \(x+\frac { x }{ 2 } +\frac { x }{ 3 } <11\)
Simplifying, \(\frac { 6x+3x+2x }{ 6 } <11\quad or\quad \frac { 11x }{ 6 } <11\)
Multiplying both sides by \(\frac { 6 }{ 11 } \), we get
x < 6
∴ The solution is (- ∞, 6),
Ex 6.1 Class 11 Maths 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).
Ex 6.1 Class 11 Maths 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].
Ex 6.1 Class 11 Maths 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].
Ex 6.1 Class 11 Maths 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, ∞).
Ex 6.1 Class 11 Maths 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].
Ex 6.1 Class 11 Maths Question 15.
\(\frac { x }{ 4 } <\frac { \left( 5x-2 \right) }{ 3 } -\frac { \left( 7x-3 \right) }{ 5 } \)
Solution.
The inequality is \(\frac { x }{ 4 } <\frac { \left( 5x-2 \right) }{ 3 } -\frac { \left( 7x-3 \right) }{ 5 } \)
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, ∞).
Ex 6.1 Class 11 Maths 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.
Ex 6.1 Class 11 Maths 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
Ex 6.1 Class 11 Maths 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
Ex 6.1 Class 11 Maths 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
Ex 6.1 Class 11 Maths Question 20.
\(\frac { x }{ 2 } \ge \frac { \left( 5x-2 \right) }{ 3 } -\frac { \left( 7x-3 \right) }{ 5 } \)
Solution.
The inequality is \(\frac { x }{ 2 } \ge \frac { \left( 5x-2 \right) }{ 3 } -\frac { \left( 7x-3 \right) }{ 5 } \)
Ex 6.1 Class 11 Maths 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,
∴ \(\frac { 70+75+x }{ 3 } \ge 60\quad or,\quad \frac { 145+x }{ 3 } \ge 60
\)
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.
Ex 6.1 Class 11 Maths 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.
Ex 6.1 Class 11 Maths 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).
Ex 6.1 Class 11 Maths 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, \(x<\frac { 21 }{ 2 } \)
Thus, the value of x may be 6,8,10 (even integers) Hence, the pairs may be (6, 8), (8,10), (10,12).
Ex 6.1 Class 11 Maths 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.
Ex 6.1 Class 11 Maths 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.
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