NCERT Solutions Class 11 – Chapter 5 Linear Inequalities – Exercise 5.1

Question 1. Solve 24x < 100, when

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

Solution:

(i) when x is a natural number.

Clearly x>0 because from definition (N =1,2,3,4,5,6…..)
Now we have to divide the inequation by 24 we get x<25/6
But x is a natural number that is the solution will be {1,2,3,4} 
Which is less than 25/6 and greater than 0.
Hence, {…, -2,-1, 0, 1, 2, 3, 4} is the solution set.

x={1,2,3,4}

(ii) Given 24x<100
Now we will divide the equation by 24 we get x<25/6
but according to question x is an integer then the solution 
less than 25/6 are…-2,-1, 0, 1, 2, 3, 4
Hence, {…, -2,-1, 0, 1, 2, 3, 4} is the solution set.

Question 2. Solve – 12x > 30, when

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

Solution:

(i) Given, – 12x > 30
Now by dividing the equation by -12 on both sides we get, x < -5/2(as per the rule if we divide by a negative integer then the inequality sign changes)
As per question when x is a natural integer then
It is clear that there is no natural number less than -2/5 since, the result of -2/5 will be negative and x is smaller than the result but the natural number contains only positive number 
Therefore, there would be  no any solution of the given equation when x is a natural number(x>0).

(ii) Given that, – 12x > 30
Now by dividing the equation by -12 on both sides we get, x < -5/2 (As explained earlier why sign changes)
As per question now x is an integer then
It is clear that the integer number less than -5/2 are…, -6.-5, -4, – 3
Thus, the solution of – 12x > 30 is …,-6,-5, -4, -3, when x is an integer.
Therefore, the solution set is {…,-6, -5, -4, -3}

Question 3. Solve 5x – 3 < 7, when

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

Solution:

(i) Given 5x – 3 < 7
Now let us add 3 both side we get,
5x – 3 + 3 < 7 + 3
Above equation becomes
5x < 10
Again let us divide  both sides by 5 we get,
5x/5 < 10/5
x < 2
As per question x is an integer then
It is clear that the integer number less than 2 are…, -3, -2, -1, 0, 1.
Thus, the  solution of 5x – 3 < 7 is …, -3,-2, -1, 0, 1, when x is an integer.
Therefore, the solution set is {…, -3.-2, -1, 0, 1}

(ii) Given that, 5x – 3 < 7
Now let us add 3 both side we get,
5x – 3 + 3 < 7 + 3
Above equation becomes
5x < 10
Again let us divide both the sides by 5 we get,
5x/5 < 10/5
x < 2
As per question x is a real (x ∈  R) number then
It is clear that the solutions of 5x – 3 < 7 will be given by x < 2 i.e it states that all the real numbers that are less than 2.
Hence, the solution set is x ∈ (-∞, 2)

Question 4. Solve 3x + 8 > 2, when

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

Solution:

(i) Given, 3x + 8 > 2
Now let us subtract 8 from both the sides we get,
3x + 8 – 8 > 2 – 8
The above equation becomes,
3x > – 6
Again let us divide both the sides by 3 we get,
3x/3 > -6/3
Hence, x > -2
As per question x is an integer then
It is clear that the integer number greater than -2 are -1, 0, 1, 2, 3,…
Thus, the solution of 3x + 8 > 2is -1, 0, 1, 2, 3,… when x is an integer.
Hence, the solution set is {-1, 0, 1, 2, 3,…}

(ii) Given3x + 8 > 2
Now let us subtract 8 from both sides we get,
3x + 8 – 8 > 2 – 8
The above equation becomes,
3x > – 6
Again let us divide both the sides by 3 we get,
3x/3 > -6/3
Hence, x > -2
As per question x is a real number.
It is clear that the solutions of 3x + 8 >2 will be given by x > -2 which states that all the real numbers that are greater than -2.
Therefore, the solution set is x ∈ (-2, ∞) 

Question 5. 4x + 3 < 5x + 7

Solution:

Given, 4x + 3 < 5x + 7
Now let us subtract 7 from both the sides, we get
4x + 3 – 7 < 5x + 7 – 7
The above equation becomes,
4x – 4 < 5
Again let us subtract 4x from both the sides, we get
4x – 4 – 4x < 5x – 4x
x > – 4

Thus, solution of the given equation is defined by all the real numbers greater than -4.
Required solution set is (-4, ∞)

Question 6. 3x – 7 > 5x – 1

Solution:

Given, 3x – 7 > 5x – 1
Now let us add 7 to both the sides, we get
3x – 7 +7 > 5x – 1 + 7
3x > 5x + 6
Again let us subtract 5x from both the sides,
3x – 5x > 5x + 6 – 5x
-2x > 6
Now let us divide both the  sides by -2 to simplify we get
-2x/-2 < 6/-2
x < -3
Solutions of the given inequality are defined by all the real numbers less than -3.
Hence, the required solution set is (-∞, -3)

Question 7. 3(x – 1) ≤ 2 (x – 3)

Solution:

Given, 3(x – 1) ≤ 2 (x – 3)
After multiplying, the above inequation can be written as
3x – 3 ≤ 2x – 6
Now let us add 3 to both the sides, we get
3x – 3+ 3 ≤ 2x – 6+ 3
3x ≤ 2x – 3
Again let us subtract 2x from both the sides,
3x – 2x ≤ 2x – 3 – 2x
x ≤ -3
Therefore, the solutions of the given equation is defined by all the real numbers less than or equal to -3.
Hence, the required solution set is (-∞, -3]

Question 8. 3 (2 – x) ≥ 2 (1 – x)

Solution:

Given, 3 (2 – x) ≥ 2 (1 – x)
After multiplying, the above equation can be written as
6 – 3x ≥ 2 – 2x
Now let us add 2x to both the sides,
6 – 3x + 2x ≥ 2 – 2x + 2x
6 – x ≥ 2
Again let us subtract 6 from both the sides, we get
6 – x – 6 ≥ 2 – 6
– x ≥ – 4
Now multiplying the equation by negative sign we get
x ≤ 4
Thus, solutions of the given equation is defined by all the real numbers greater than or equal to 4.
Hence, the required solution set is (- ∞, 4]

Question 9. x + x/2 + x/3 < 11

Solution:

Given, x + x/2 + x/3 < 11
Now taking 6 as the lcm we will simplify the equation,

(6x+3x+2x)/6 <11
11x/6<11
Now let us multiply 6 at both the sides
11x<66
Now let us divide the equation by 11 at both the sides we get,
x<6

Thus, the solution of the given equation is defined by all the real numbers less than 6.
Hence, the solution set is (-∞, 6)

Question 10. x/3 > x/2 + 1

Solution:

Given x/3 > x/2 + 1
Firstly, we will move all the terms containing x to the left-hand side we get,
x/3-x/2>1
Now taking 6 as the lcm we get
(2x-3x)/6>1
-x/6>1
Now multiplying 6 at both the sides we get,
 -x>6
Now multiplying by -1 at both the ends
x<6
Thus, the solution of the given equation is defined by all the real numbers less than – 6.
Hence, the required solution set is (-∞, -6)

Question 11. 3(x – 2)/5 ≤ 5 (2 – x)/3

Solution:

Given, 3(x – 2)/5 ≤ 5 (2 – x)/3
Now by cross – multiplying the denominators, we get
9(x- 2) ≤ 25 (2 – x)
9x – 18 ≤ 50 – 25x
Now let us add 25x both the sides,
9x – 18 + 25x ≤ 50 – 25x + 25x
34x – 18 ≤ 50
Let us add 25x both the sides,
34x – 18 + 18 ≤ 50 + 18
34x ≤ 68
Dividing both the sides by 34,
34x/34 ≤ 68/34
x ≤ 2
Thus, the solution of the given equation is defined by all the real numbers less than or equal to 2.
Hence, solution set is (-∞, 2]

Question 12. 1/2(3x/5+4)>=1/3(x-6)

Solution:

Given, 1/2(3x/5+4)>=1/3(x-6)
Now let us cross multiply,
3(3x/5+4)>=2(x-6)
Now multiply the respective terms at both the sides
9x/5+12>=2x-12
Now subtracting 12 at both the sides
9x/5+12-12 >= 2x-12-12
9x/5>=2x-24
Now multiplying by 5 both the sides,
9x>=10x-120
Now subtracting 10x both the sides.
-x>=-120
Now multiplying with -1 both the sides
x <= 120
Thus, the solutions of the given equation is defined by all the real numbers less than or equal to 120.
Thus, (-∞, 120] is the required solution set.

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

Solution:

Given, 2(2x + 3) – 10 < 6 (x – 2)
By multiplying we get
4x + 6 – 10 < 6x – 12
On simplifying we get
4x – 4 < 6x – 12
– 4 + 12 < 6x – 4x
8 < 2x
4 < x
Thus, the solutions of the given equation are defined by all the real numbers greater than or equal to 4.
Hence, the required solution set is [4, ∞)

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

Solution:

Given that, 37 – (3x + 5) ≥ 9x – 8 (x – 3)
On simplifying we get 
37 – 3x – 5 ≥ 9x – 8x + 24 
32 – 3x ≥ x + 24
On rearranging 
32 – 24 ≥ x + 3x 
8 ≥ 4x 
2 ≥ x
All the real numbers of x which are less than or equal to 2 are the solutions of the given equation
Hence, (-∞, 2] will be the solution for the given equation

Question 15. x/4<(5x-2)/3-(7x-3)/5

Solution:

Given, x/4<(5x-2)/3-(7x-3)/5
Taking 15 as the lcm on the right side
x/4<(5(5x-2)-3(7x-3))/15
Further, simplifying we get,
x/4<((25x-10)-(21x-9))/15
Multiplying 15 at both the sides
15x/4<25x-10-21x+9
15x/4<4x-1
Multiplying by 4 we get
15x < 4 (4x – 1)
15x < 16x – 4
4 < x
All the real numbers of x which are greater than 4 are the solutions of the given equation
Hence, (4, ∞) will be the solution for the given equation

Question 16. (2x-1)/3 >= (3x-2)/4-(2-x)/5

Solution:

Let us simplify the inequation
(2x-1)/3>=(5(3x-2)-4(2-x))/20
(2x-1)/3>=((15x-10)-(8-4x))/20
(2x-1)/3>=(15x-10-8+4x)/20
(2x-1)/3>=(19x-18)/20
Now cross multiplying at both the sides we get
20 (2x – 1) ≥ 3 (19x – 18) 
40x – 20 ≥ 57x – 54
– 20 + 54 ≥57x – 40x
34 ≥ 17x 
2 ≥ x
∴ All the real numbers of x which are less than or equal to 2 are the solutions of the given equation
Hence, (-∞, 2] will be the solution for the given equation

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:

Given, 3x – 2 < 2x + 1Solving the given inequality, we get
3x – 2 < 2x + 1 
3x – 2x < 1 + 2 
x < 3
Now, the graphical representation of the solution is as follows:
 

Question 18. 5x – 3 ≥ 3x – 5

Solution:

We have,5x – 3 ≥ 3x – 5
Solving the given inequality, we get
5x – 3 ≥ 3x – 5
On rearranging we get
 5x – 3x ≥ -5 + 3
On simplifying 2x ≥ -2
Now divide by 2 on both sides we get
 x ≥ -1
The graphical representation of the solution is as follows:

 

Question 19. 3 (1 – x) < 2 (x + 4)

Solution:

Given,3 (1 – x) < 2 (x + 4)
Solving the given inequality, we get
3 (1 – x) < 2 (x + 4)
Multiplying we get 3 – 3x < 2x + 8
On rearranging we get
 3 – 8 < 2x + 3x
 – 5 < 5x
Now by dividing 5 on both sides we get
-5/5 < 5x/5
 – 1 < x
Now, the graphical representation of the solution is as follows:

 

Question 20. x/2>=(5x-2)/3-(7x-3)/5

Solution:

Let us solve the right side of the inequation
x/2>=(5(5x-2)-(7x-3)3)/15
x/2>=(25x-10-21x+9)/15
x/2>=(4x+1)/15
15x ≥ 2 (4x – 1)
15x ≥ 8x -2
15x -8x ≥ 8x -2 -8x
7x ≥ -2
x ≥ -2/7

Now, the graphical representation of the solution is as follows:

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

Solution:

Let us assume that the marks obtained by Ravi in his third unit test be x,
According to question, 
the entire students should have an average of at least 60 marks i.e,
(70 + 75 + x)/3 ≥ 60
145 + x ≥ 180 
x ≥ 180 – 145
x ≥ 35

Thus, all the students must obtain 35 marks in order to have an average of at least 60 marks

Question 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.

Solution:

Let us assume Sunita scored x marks in her 5th examination
Now, according to the question
 In order to receive A grade in the course she must have to obtain average 90 marks or more in her five examinations that is,
(87 + 92 + 94 + 95 + x)/5 ≥ 90
 (368 + x)/5 ≥ 90
 368 + x ≥ 450
 x ≥ 450 – 368 
x ≥ 82
Thus, she must have to obtain 82 or more marks in her 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 us assume x be the smaller of the two consecutive odd positive integers
Let other integer = x + 2
It is also given in the question that, 
both the integers are smaller than 10 that is, 
x + 2 < 10

x< 8 … (a)
Also, it is given in the question that sum of two integers is more than 11
∴ x + (x + 2) > 11
2x + 2 > 11
x > 9/2
x > 4.5 … (b)
Thus, from (a) and (b) we have x is an odd integer and it can take values 5 and 7
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 us assume x be the smaller of the two consecutive even positive integers
Let other integer= x + 2
It is also given in the question that, both the integers are larger than 5∴
 x > 5   ….(a)
Also, it is given in the question that the sum of two integers is less than 23.
∴ x + (x + 2) < 23
2x + 2 < 23
x < 21/2
x < 10.5   …. (b)
Thus, from (a and (b) we have x is an even number and it can take values 6, 8 and 10

Thus, the possible pairs are (6, 8), (8, 10) and (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 us assume that the length of the shortest side of the triangle be x cm
 According to the question, 
length of the longest side = 3x cm
And, length of third side = (3x – 2) cm
As, the least perimeter of the triangle = 61 cm
Thus, x + 3x + (3x – 2) cm ≥ 61 cm 
7x – 2 ≥ 61 
7x ≥ 63
Now let us divide by 7 we get=
7x/7 ≥ 63/7=
x ≥ 9

Hence, the minimum length of the shortest side will be 9 cm

Question 26. A man wants to cut three lengths from a single piece of board of length 91cm. The second length is to be 3cm 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 5cm longer than the second?

Solution:

Let us assume that the length of the shortest piece be x cm
∴ According to the question,
Length of the second piece = (x + 3) cm
And, length of third piece = 2x cm
As all the three lengths are to be cut from a single piece of board having a length of 91 cm
x + (x + 3) + 2x ≤ 91 cm
4x + 3 ≤ 91
4x ≤ 88
4x/4 ≤ 88/4
x ≤ 22 … (i)
Also, it is given in the question that, the third piece is at least 5 cm longer than the second piece
2x ≥ (x+3) + 5
2x ≥ x + 8
x ≥ 8 … (ii)
Thus, from equation (i) and (ii) we have:
8 ≤ x ≤ 22
Hence, it is clear that the length of the shortest board is greater than or equal to 8 cm and less than or equal to 22 cm