6. Linear Inequalities Mathematics Exercise - 6.1 class 11 Maths in English - CBSE Study
NCERT Solutions for Class 11 Mathematics are carefully prepared according to the latest CBSE syllabus and NCERT textbooks to help students understand every concept clearly. These solutions cover all important 6. Linear Inequalities with detailed explanations and step-by-step answers for better exam preparation. Each Exercise 6.1 is explained in simple language so that students can easily grasp the fundamentals and improve their academic performance. The study material is designed to support daily homework, revision practice, and final exam preparation for Class 11 students. With accurate answers, concept clarity, and structured content, these NCERT solutions help learners build confidence and score higher marks in their examinations. Whether you are revising a specific topic or preparing an entire chapter, this resource provides reliable and syllabus-based guidance for complete success in Mathematics.
Class 11 English Medium Mathematics All Chapters:
6. Linear Inequalities
1. Exercise 6.1
Exercise 6.1
Q1. Solve 24x < 100, when
(i) x is a natural number
(ii) x is an integer
Solution:
The given inequality is 24x < 100.
(i) It is evident that 1, 2, 3, and 4 are the only natural numbers less than
∴ when x is a natural number, the solutions of the given inequality are 1, 2, 3, and 4.
Hence, in this case, the solution set is {1, 2, 3, 4}.
(i) It is evident that 1, 2, 3, and 4 are the only natural numbers less than
Thus, when x is a natural number, the solutions of the given inequality are 1, 2, 3, and 4.
Hence, in this case, the solution set is {1, 2, 3, 4}.
Q2. Solve –12x > 30, when
(i) x is a natural number
(ii) x is an integer
Solution:
The given inequality is –12x > 30.
(i) There is no natural number less than
Thus, when x is an integer, the solutions of the given inequality are …, –5, –4, –3.
Hence, in this case, the solution set is {…, –5, –4, –3}.
Q3. Solve 5x– 3 < 7, when
(i) x is an integer
(ii) x is a real number
Soluution:
The given inequality is 5x– 3 < 7.
Q5. Solve the given inequality for real x: 4x + 3 < 5x + 7
Solution :
4x + 3 < 5x + 7
⇒ 4x + 3 – 7 < 5x + 7 – 7
⇒ 4x – 4 < 5x
⇒ 4x – 4 – 4x < 5x – 4x
⇒ –4 < x
Thus, all real numbers x, which are greater than –4, are the solutions of the given inequality.
Hence, the solution set of the given inequality is (–4, ∞).
Q23. 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 consecutive odd positive integers.
Then, the other integer is x + 2.
Since both the integers are smaller than 10, x + 2 < 10
⇒ x < 10 – 2
⇒ x < 8 … (1)
Also, the sum of the two integers is more than 11.
∴x + (x + 2) > 11
⇒ 2x + 2 > 11
⇒ 2x > 11 – 2
⇒ 2x > 9
⇒ x > 9/2
⇒ x > 4.5 ....... (2)
From (1) and (2), we get .
Since x is an odd number, x can take the values, 5 and 7.
Therefore, the required possible pairs are (5, 7) and (7, 9).
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