Anand Classes provides precise and student-friendly Linear Inequalities Exercise 5.1 NCERT Solutions Class 11 Maths Chapter 5 (Set-3) to help learners master the core concepts of inequalities with ease. These step-by-step solutions are prepared according to the latest NCERT and CBSE syllabus, ensuring complete conceptual understanding and exam readiness. Ideal for Class 11 students, these solutions simplify complex inequality problems and make learning more efficient and result-oriented. Click the print button to download study material and notes.
NCERT Question 21 : 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 the marks obtained by Ravi in the third test be $x$.
According to the question:
$$\frac{70 + 75 + x}{3} \ge 60$$
$$145 + x \ge 180$$
$$x \ge 180 – 145$$
$$x \ge 35$$
Final Answer
$$\boxed{x \ge 35}$$
Therefore, Ravi must score at least 35 marks in the third test to maintain an average of 60 marks.
Thus, Ravi must obtain a minimum of 35 marks to get an average of at least 60 marks.
Note. A minimum of 35 marks.
⇒ Marks greater than or equal to 35.
NCERT 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 the first four examinations are 87, 92, 94 and 95, find minimum marks that Sunita must obtain in the fifth examination to get grade ‘A’.
Solution:
Let Sunita’s marks in the fifth exam be $x$.
According to the condition:
$$\frac{87 + 92 + 94 + 95 + x}{5} \ge 90$$
$$\frac{368 + x}{5} \ge 90$$
$$368 + x \ge 450$$
$$x \ge 450 – 368$$
$$x \ge 82$$
Final Answer
$$\boxed{x \ge 82}$$
Therefore, Sunita must score at least 82 marks in the fifth examination to get Grade ‘A’.
Thus, Sunita must obtain marks greater than or equal to 82, i.e., a minimum of 82 marks.
NCERT 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.
Then the other integer will be $x + 2$.
It is given that both the integers are smaller than $10$, i.e.,
$$x + 2 < 10$$
$$x < 8 \quad …(a)$$
Also, their sum is more than $11$,
$$x + (x + 2) > 11$$
$$2x + 2 > 11$$
$$2x > 9$$
$$x > \frac{9}{2} \quad …(b)$$
From $(a)$ and $(b)$, $x$ is an odd integer such that:
$$4.5 < x < 8$$
Thus, the possible values of $x$ are $5$ and $7$.
Therefore, the possible pairs are:
$$(5, 7) \quad \text{and} \quad (7, 9)$$
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NCERT 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 the smaller integer be $x$.
Then the next consecutive even integer will be $x + 2$.
Since both are larger than $5$:
$$x > 5 \quad …(a)$$
Given that their sum is less than $23$:
$$x + (x + 2) < 23$$
$$2x + 2 < 23$$
$$2x < 21$$
$$x < \frac{21}{2} = 10.5 \quad …(b)$$
From $(a)$ and $(b)$, $x$ is an even positive integer between $5$ and $10.5$:
$$x = 6, \ 8, \ 10$$
Therefore, the required pairs are:
$$(6, 8), \ (8, 10), \ (10, 12)$$
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NCERT 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,
$$x + 3x + (3x – 2) \ge 61$$
$$7x – 2 \ge 61$$
$$7x \ge 63$$
Dividing both sides by 7:
$$x \ge 9$$
Hence, the minimum length of the shortest side will be 9 cm.
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NCERT Question 26. A man wants to cut three lengths from a single piece of board of length $91\text{ cm}$. The second length is to be $3\text{ cm}$ longer than the shortest and the third length is twice as long as the shortest. What are the possible lengths of the shortest board if the third piece is to be at least $5\text{ cm}$ longer than the second?
Solution:
Let the shortest piece be $x\text{ cm}$.
Then the second piece is $x+3\text{ cm}$ and the third piece is $2x\text{ cm}$.
Since all three are cut from the $91\text{ cm}$ board:
$$x + (x+3) + 2x \le 91$$
$$4x + 3 \le 91$$
$$4x \le 88$$
$$x \le 22 \quad\text{…(1)}$$
The third piece is at least $5\text{ cm}$ longer than the second:
$$2x \ge (x+3) + 5$$
$$2x \ge x + 8$$
$$x \ge 8 \quad\text{…(2)}$$
Combining (1) and (2):
$$8 \le x \le 22$$
If $x$ may be any real number (continuous lengths), the possible shortest lengths are all real $x$ with
$$8 \le x \le 22.$$
If lengths are required to be whole centimetres (integers), the possible integer values are
$$x\in\{8,9,10,\dots,22\}.$$
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