NCERT Solutions for Class 11 Maths Chapter 5 Ex 5.1

# NCERT Solutions for Class 11 Maths Chapter 5 Ex 5.1

NCERT Solutions for Class 11 Maths Chapter 5 Linear Inequalities Ex 5.1 are the part of NCERT Solutions for Class 11 Maths. In this post, you can find the NCERT Solutions for Class 11 Maths chapter 5 Linear Inequalities Ex 5.1.

## NCERT Solutions for Class 11 Maths Chapter 5 Linear Inequalities Ex 5.1

### Ex 5.1 Class 11 Maths Question 1.

Solve 24x < 100, when
(i) x is a natural number.
(ii) x is an integer.

Solution.
We have, 24x < 100
Dividing both sides by 24, we get
x < 100/24

x < 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 5.1 Class 11 Maths Question 2.

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

Solution.
We have, –12x > 30

Dividing both sides by –12, we get
< −30/12

< −5/2
(i) This inequality is not true for any natural number.
(ii) Integers that satisfy this inequality are {…,  -5,  -4,  -3}.

### Ex 5.1 Class 11 Maths Question 3.

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

Solution.
We have, 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 a real number, the solution is (-∞, 2).

### Ex 5.1 Class 11 Maths Question 4.

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

Solution.
We have, 3x + 8 > 2
Transposing 8 to R.H.S., we get
3x > 2 – 8

3x > –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 a real number, the solution is (-2, ∞).

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

### Ex 5.1 Class 11 Maths Question 5.

4x + 3 < 5x + 7

Solution.
We have, 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 5.1 Class 11 Maths Question 6.

3x – 7 > 5x – 1

Solution.
We have, 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 5.1 Class 11 Maths Question 7.

3(x – 1) ≤ 2(x – 3)

Solution.
We have, 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 5.1 Class 11 Maths Question 8.

3(2 – x) ≥ 2(1 – x)

Solution.
We have, 3(2 – x) ≥ 2(1 – x) or 6 – 3x ≥ 2 – 2x
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 5.1 Class 11 Maths Question 9.

x/x/11

Solution.
We have, x/x/11
Simplifying, (632x)/11 or 11x/11
Multiplying both sides by 6/11, we get
x < 6
The solution is (-∞, 6).

### Ex 5.1 Class 11 Maths Question 10.

x/x/1

Solution.
We have, x/x/1
Transposing x/2 to L.H.S., we get
x/– x/1
Simplifying, (2− 3x)/or x/1
Multiplying both sides by -6, we get
x < -6
The solution is (-∞, –6).

### Ex 5.1 Class 11 Maths Question 11.

3(− 2)/≤ 5(− x)/3

Solution.
We have, 3(− 2)/≤ 5(− x)/3
Multiplying both sides by the L.C.M. of 5 and 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 both sides by 34, we get
x ≤ 2
The solution is (-∞, 2].

### Ex 5.1 Class 11 Maths Question 12.

½ (3x/4) ≥ 1/3 (− 6)

Solution.
We have, ½ (3x/4) ≥ 1/3 (− 6)

or, ½ (320)/5 ≥ 1/3 (− 6)
Multiplying both sides by 30, we get
3(3x + 20) ≥ 10(x – 6)

or, 9x + 60 ≥ 10x – 60
Transposing 10x 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 5.1 Class 11 Maths Question 13.

2(2x + 3) – 10 < 6(x – 2)

Solution.
We have, 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 5.1 Class 11 Maths Question 14.

37 – (3x + 5) ≥ 9x – 8(x – 3)

Solution.
We have, 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 5.1 Class 11 Maths Question 15.

x/< (5− 2)/− (7− 3)/5

Solution.
We have, x/< (5− 2)/− (7− 3)/5
Multiplying each term by the L.C.M. of 4, 3 and 5, i.e., by 60, we get
15x < 100x – 40 – 84x + 36
or, 15x < 100x – 84x – 40 + 36
or, 15x < 16x – 4
Transposing 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 5.1 Class 11 Maths Question 16.

(2− 1)/≥ (3− 2)/– (− x)/5

Solution.
We have, (2− 1)/≥ (3− 2)/– (− x)/5
Multiplying each term by L.C.M. of 3,4 and 5, i.e., by 60.
(2− 1)/× 60 ≥ (3− 2)/× 60 – (– x)/× 60
or, 20(2x – 1) ≥ (3x – 2) × 15 – (2 – 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 5.1 Class 11 Maths Question 17.

3x – 2 < 2x + 1

Solution.
We have, 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

The solution is (-∞, 3).

### Ex 5.1 Class 11 Maths Question 18.

5x – 3 ≥ 3x – 5

Solution.
We have, 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 x ≥ -1.

The solution is [-1, ∞).

### Ex 5.1 Class 11 Maths Question 19.

3(1 – x) < 2(x + 4)

Solution.
We have, 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 x > -1.

The solution is (-1, ∞).

### Ex 5.1 Class 11 Maths Question 20.

x/≥ (5− 2)/− (7− 3)/5

Solution.
We have, x/≥ (5− 2)/− (7− 3)/5

### Ex 5.1 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 obtains x marks in third unit test.
Average marks obtained by Ravi = (70 75 x)/3
He has to obtain at least 60 marks,
(70 75 x)/≥ 60 or, (145 x)/≥ 60
Multiplying both sides by 3, we get
145 + x ≥ 180
Transposing 145 to R.H.S., we get
x ≥ 180 – 145

x ≥ 35
Ravi should get at least 35 marks in the third unit test to have an average of at least 60 marks.

### Ex 5.1 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
= (87 92 94 95 x)/= (368 x)/5
This average must be at least 90.
(368 x)/≥ 90
Multiplying both sides by 5, we get
368 + x ≥ 450
Transposing 368 to R.H.S., we get
x ≥ 450 – 368

x ≥ 82
Sunita must obtain at least 82 marks in the fifth examination to get Grade ‘A’ in the course.

### Ex 5.1 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 have x + 2 < 10 and x + (x + 2) > 11 or, 2x + 2 > 11
or, 2x > 11 – 2 or, 2x > 9 or, 9/2
Hence, if one number is 5 (odd number), then the other number is 7. If the smaller number is 7, then the other number is 9.

Hence, possible pairs are (5, 7) and (7, 9).

### Ex 5.1 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 number is x + 2.

Then, we should have, x > 5 and x + x + 2 < 23 or, 2x + 2 < 23
or, 2x < 21 or, 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 5.1 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 length of the shortest side be x cm.
Then, the longest side will be 3x cm.
The third side will be (3x – 2) cm.
According to the question, 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 5.1 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 at least 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 piece, then x + 3 is the length of the second piece and 2x is the length of the third piece.

Thus, x + (x + 3) + 2x ≤ 91 or 4x + 3 ≤ 91 or 4x ≤ 88 or x ≤ 22
According to the question, 2x ≥ (x + 3) + 5 or x ≥ 8
At least 8 cm but not more than 22 cm is the possible length of the shortest board.