Practicing the AP Board Solutions Class 11 Maths and Chapter 5 Inter 1st Year Maths Linear Inequalities Solutions Exercise 5a Pdf Download will help students to clear their doubts quickly.
Intermediate 1st Year Maths Linear Inequalities Solutions Exercise 5a
Linear Inequalities Exercise 5a Solutions
Linear Inequalities Class 11 Exercise 5a Solutions – Linear Inequalities 5a Exercise Solutions
I.
Question 1.
Solve 24x < 100, when
(i) x is a natural number.
(ii) x is an integer.
Solution:
(i) Given that 24x <100
Now, we have to divide the inequality by 24, then we get x < \(\frac {25}{6}\)
Now, when x is a natural integer, then it is clear that the only natural numbers less than \(\frac {25}{6}\) are 1, 2, 3, 4.
Thus, 1, 2, 3, and 4 will be the solutions of the given inequality when x is a natural number.
Hence, {1, 2, 3, 4} is the solution set.
(ii) Given that 24x < 100
Now we have to divide the inequality by 24, then we get x < \(\frac {25}{6}\)
Now, when x is an integer, then.
It is clear that the integer numbers less than \(\frac {25}{6}\) are …… -1, 0, 1, 2, 3, 4.
Thus, solutions of 24x < 100 are …. -1, 0, 1, 2, 3, 4, when x is an integer.
Hence, {…., -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 that -12x > 30
Now, by dividing the inequality by -12 on both sides, we get, x < \(-\frac {5}{2}\)
When x is a natural integer, then.
It is clear that there is no natural number less than \(-\frac {2}{5}\) because \(-\frac {5}{2}\) is a negative number and natural numbers are positive numbers. Therefore, there would be no solution to the given inequality when x is a natural number.
(ii) Given that, -12x > 30
Now by dividing the inequality by -12 on both sides, we get, x < \(-\frac {5}{2}\)
When x is an integer, then.
It is clear that the integer numbers less than \(-\frac {5}{2}\) are…., -5, -4, -3
Thus, the solution of 12x > 30 is …., -5, -4, -3 when x is an integer.
Therefore the solution set is {…., -5, -4, -3}
Question 3.
Solve 5x – 3 < 7, when
(i) x is an integer.
(ii) x is a real number.
Solution:
(i) Given that, 5x – 3 < 7
Now by adding 3 on both sides, we get
5x – 3 + 3 < 7 + 3
The above inequality becomes 5x < 10
Again, by dividing both sides by 5, we get,
\(\frac{5 x}{5}<\frac{10}{5}\)
x < 2
When x is an integer, then.
The integer numbers less than 2 are …. -2, -1, 0, 1.
Thus, the solution of 5x – 3 < 7 is …., -2, -1, 0, 1, when x is an integer.
Therefore the solution set is {…. -2, -1, 0, 1}
(ii) Given that, 5x – 3 < 7
Now, by adding 3 on both sides, we get
5x – 3 + 3 < 7 + 3
The above inequality becomes
5x < 10
Again, by dividing both sides by 5, we get,
\(\frac{5 x}{5}<\frac{10}{5}\)
x < 2
When x is a real number, then
The solutions of 5x – 3 < 7 will be given by x < 2, which 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 that 3x + 8 > 2
Now, by subtracting 8 from both sides, we get
3x + 8 – 8 > 2 – 8
The above inequality becomes,
3x > -6
Again, by dividing both sides by 3, we get,
\(\frac{3 x}{3}<\frac{-6}{3}\) Hence x > -2
When x is an integer, then
It is clear that the integer numbers greater than -2 are -1, 0, 1, 2,….
Thus, the solution of 3x + 8 > 2 is -1, 0, 1, 2,… when x is an integer.
Hence, the solution set is {-1, 0, 1, 2,…}
(ii) Given that, 3x + 8 > 2
Now, by subtracting 8 from both sides, we get,
3x + 8 – 8 > 2 – 8
The above inequality becomes,
3x > -6
Again, by dividing both sides by 3, we get
\(\frac{3 x}{3}<\frac{-6}{3}\) Hence x > -2
When x is a real number.
The solutions of 3x + 8 > 2 will be given by x > -2, which means all the real numbers that are greater than -2.
Therefore the solution set x ∈ (-2, ∞)
Solve the inequalities in Exercises 5 to 12 for real x.
Question 5.
4x + 3 < 5x + 7
Solution:
Given that, 4x + 3 < 5x + 7
Now, by subtracting 7 from both sides, we get
4x + 3 – 7 < 5x + 7 – 7
The above inequality becomes,
4x – 4 < 5x
Again, by subtracting 4x from both sides.
4x – 4 – 4x < 5x – 4x x > -4
∴ The solutions of the given inequality are defined by all the real numbers greater than -4.
The required solution set is (-4, ∞)
Question 6.
3x – 7 > 5x – 1
Solution:
Given that, 3x – 7 > 5x – 1
Now, by adding 7 to both sides, we get
3x – 7 + 7 > 5x – 1 + 7
3x > 5x + 6
Again, by subtracting 5x from both sides,
3x – 5x > 5x + 6 – 5x
-2x > 6
Dividing both sides by -2 to simplify, we get
\(\frac{-2 x}{-2}<\frac{6}{-2}\)
x < -3
The 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 that, 3(x -1) ≤ 2(x – 3)
By multiplying, the above inequality can be written as
-3x – 3 ≤ 2x – 6
Now, by adding 3 to both sides, we get
3x – 3 + 3 ≤ 2x – 6 + 3
3x ≤ 2x – 3
Again, by subtracting 2x from both sides,
3x – 2x ≤ 2x – 3 – 2x
x ≤ -3
Therefore, the solutions of the given inequality are 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 that, 3(2 – x) ≥ 2(1 – x)
By multiplying, we get
6 – 3x ≥ 2 – 2x
Now, by adding 2x to both sides,
6 – 3x + 2x ≥ 2 – 2x + 2x
6 – x ≥ 2
Again, by subtracting 6 from both sides, we get
6 – x – 6 ≥ 2 – 6
-x ≥ -4
Multiplying throughout the inequality by the negative sign, we get x ≤ 4.
∴ The solutions of the given inequality are defined by all the real numbers greater than or equal to 4.
Hence, the required solution set is (-∞, 4).
Question 9.
\(x+\frac{x}{2}+\frac{x}{3}<11\)
Solution:
Given that
\(x+\frac{x}{2}+\frac{x}{3}<11\)
By taking x as common, then we get
\(x\left(1+\frac{1}{2}+\frac{1}{3}\right)\) < 11
By taking LCM
\(x\left(\frac{6+3+2}{6}\right)\) < 11
\(\frac {11x}{6}\) < 11
Dividing by 11 on both sides,
\(\frac{11 \mathrm{x}}{6 \times 11}<\frac{11}{11}\)
\(\frac {x}{6}\) < 1
x < 6
The solutions of the given inequality are defined by all the real numbers less than 6.
Hence the solution is (-∞, 6).
Question 10.
\(\frac{x}{3}>\frac{x}{2}+1\)
Solution:
Given that \(\frac{x}{3}>\frac{x}{2}+1\)
On rearranging and by taking LCM, we get
\(\left(\frac{2 x-3 x}{6}\right)\) > 1
\(\frac {1}{2}\) > 1
-x > 6
x < -6
∴ The solutions of the given inequality are defined by all the real numbers less than -6.
Hence, the required solution set is (-∞, -6).
Question 11.
2(2x + 3) – 10 < 6(x – 2)
Solution:
Given that, 2(2x + 3) – 10 < 6(x – 2)
By multiplying, we get,
4x + 6 – 10 < 6x – 12
On simplifying, we get
4x – 4 < 6x – 12
4x – 6x < -12 + 4
-2x < -8
Dividing by 2, we get,
-x < -4 Multiply by “-1” and change the sign. x > 4
∴ The solutions of the given inequality are defined by all the real numbers greater than 4.
Hence, the required solution set is (4, ∞).
Question 12.
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 that are less than or equal to 2 are the solutions of the given inequality.
Hence, (-∞, 2) will be the solution for the given inequality.
Question 13.
Ravi obtained 70 and 75 marks in the 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 us assume that x is the marks obtained by Ravi in his third unit test.
According to the question, all the students should have an average of atleast 60 marks.
\(\frac{70+75+x}{3}\) ≥ 60
145 + x ≥ 180
x ≥ 180 – 145
x ≥ 35
Hence, all the students must obtain 35 marks in order to have an average of at least 60 marks.
II. Solve the inequalities in Exercises 1 to 4 for real x.
Question 1.
\(\frac{3(x-2)}{5} \leq \frac{5(2-x)}{3}\)
Solution:
Given that \(\frac{3(x-2)}{5} \leq \frac{5(2-x)}{3}\)
Now, by cross-multiplying the denominators, we get
9(x – 2) ≤ 25(2 – x)
9x – 18 ≤ 50 – 25x
Now adding 25x to both sides,
9x – 18 + 25x ≤ 50 – 25x + 25x
34x – 18 ≤ 50
Adding 18 to both sides,
34x – 18 + 18 ≤ 50 + 18
34x ≤ 68
Dividing both sides by 34,
\(\frac{34 x}{34} \leq \frac{68}{34}\)
x ≤ 2
∴ The solutions of the given inequality are defined by all real numbers less than or equal to 2.
Required solution set is (-∞, 2).
Question 2.
\(\frac{1}{2}\left(\frac{3 x}{5}+4\right) \geq \frac{1}{3}(x-6)\)
Solution:
Given that, \(\frac{1}{2}\left(\frac{3 x}{5}+4\right) \geq \frac{1}{3}(x-6)\)
Now, by cross-multiplying the denominators, we get,
\(3\left(\frac{3 x}{5}+4\right)\) ≥ 2(x – 6)
Multiplying by 3, we get
\(\left(\frac{9 x}{5}+12\right)\) ≥ 2(x – 6)
On rearranging, we get
12 + 12 ≥ 2x – \(\frac {9x}{5}\)
24 ≥ \(\frac{10 x-9 x}{5}\)
24 ≥ \(\frac {x}{5}\)
120 ≥ x
∴ The solutions of the given inequality are defined by all the real numbers less than or equal to 120.
Thus, (-∞, 120) is the required solution set.
Question 3.
\(\frac{x}{4}<\frac{(5 x-2)}{3}-\frac{(7 x-3)}{5}\)
Solution:
Given,
15x < 4(4x – 1)
15x < 16x – 4
4 < x
∴ All the real numbers of x that are greater than 4 are the solutions of the given inequality.
Hence, (4, ∞) will be the solution for the given inequality.
Question 4.
\(\frac{(2 x-1)}{3} \geq \frac{(3 x-2)}{4}-\frac{(2-x)}{5}\)
Solution:
Given,
20(2x – 1) ≥ 3(19x – 18)
40x – 20 ≥ 57x – 54
-20 + 54 ≥ 57x – 40x
34 ≥ 17x
2 ≥ x
∴ All the real numbers of x that are less than or equal to 2 are the solutions of the given inequality.
Hence, (-∞, 2) will be the solution for the given inequality.
Solve the inequalities in Exercises 5 to 8 and show the graph of the solution in each case on the number line.
Question 5.
3x – 2 < 2x + 1
Solution:
Given, 3x – 2 < 2x + 1
Solving the given inequality, we get
3x – 2 < 2x + 1
3x – 2x < 1 + 2
x < 3
Solution set is (-∞, 3)
Now, the graphical representation of the solution is as follows.
Question 6.
5x – 3 ≥ 3x – 5
Solution:
Given inequality is 5x – 3 ≥ 3x – 5
5x – 3 ≥ 3x – 5
2x ≥ -2
x ≥ -1
Solution set is [-1, ∞)
The graphical representation of the solution of the given inequality is as follows.
Question 7.
3(1 – x) < 2(x + 4)
Solution:
Given inequality is 3(1 – x) < 2(x + 4)
3 – 3x < 2x + 8
3 – 8 < 2x + 3x
-5 < 5x
-1 < x
Solution set is (-1, ∞)
The graphical representation of the solution of the given inequality is as follows.
Question 8.
\(\frac{x}{2} \geq \frac{(5 x-2)}{3}-\frac{(7 x-3)}{5}\)
Solution:
Given,
15x ≥ 2(4x – 1)
15x ≥ 8x – 2
15x – 8x > 8x – 2 – 8x
7x ≥ -2
x ≥ \(-\frac {2}{7}\)
Now, the graphical representation of the solution is as follows.
Question 9.
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 the first four examinations are 87, 92, 94, and 95, find the minimum marks that Sunita must obtain in the fifth examination to get a grade ‘A’ in the course.
Solution:
Let us assume Sunita scored x marks in her fifth examination.
Now, according to the question, to receive.
To obtain an A grade in the course, she must obtain an average of 90 marks or more in her five examinations.
\(\frac{87+92+94+95+x}{5}\) ≥ 90
\(\frac{368+x}{5}\) ≥ 90
368 + x ≥ 450
x ≥ 450 – 368
x ≥ 82
Hence, she must obtain 82 or more marks in her fifth examination.
Question 10.
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 to be the smaller of the two consecutive odd positive integers.
∴ The other integer is x + 2
It is also given in the question that both the integers are greater than 10
∴ x + 2 < 10
x < 8 ………(i)
Additionally, it is stated in the question that the sum of two integers is greater than 11.
∴ x + (x + 2) > 11
2x + 2 > 11
x > \(\frac {9}{2}\)
x > 4.5 ………(ii)
Thus, from (i) & (ii), we have x is an odd integer and it can take values 5 & 7.
Hence, possible pairs are (5, 7) and (7, 9).
Question 11.
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 is the smaller of the two consecutive even positive integers.
∴ The other integer = x + 2
It is also given in the question that both the integers are larger than 5.
∴ x > 5 ……..(i)
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 < \(\frac {21}{2}\)
x < 10.5 ……….(ii)
Thus, from (i) and (ii) we have x is an even number and it can take values 6, 8, and 10.
Hence, possible pairs are (6, 8), (8, 10), and (10, 12).
Question 12.
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 the length of the shortest side of the triangle to be x cm.
∴ According to the question,
The length of the longest side = 3x cm
And, the 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 dividing by 7, we get x ≥ 9.
Hence, the minimum length of the shortest side will be 9 cm.
III.
Question 1.
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?
[Hint: If x is the length of the shortest board, then x, (x + 3), and 2x are the lengths of the first, the second, and the third pieces, respectively.
Thus, x + (x + 3) + 2x ≤ 91 and 2x ≥ (x + 3) + 5]
Solution:
Let us assume the length of the shortest piece to be x cm
According to the question,
length of the second piece = (x + 3) cm
And, length of the third piece = 2x cm
As all 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
\(\frac{4 x}{4} \leq \frac{88}{4}\)
x ≤ 22 ……….(i)
Also, it is given in the question that the second piece.
∴ 2x ≥ (x + 3) + 5
2x ≥ x + 8
x ≥ 8 ……….(ii)
Thus, from equations (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.