# Exercise 6.1 (Linear Inequalities)

Ex 6.1 Linear Inequalities ncert math solution class 11

Question 1: Solve , when

(i) is a natural number.

(ii) is an integer.

Solution: Given that:

(i) is a natural number.

Hence, in this case, the solution set .

(ii) is an integer.

The integers less than are …-3, .

Hence, in this case, the solution set .

Question 2: Solve , when

(i) is a natural number.

(ii) is an integer.

Solution : Given that:

(i) is a natural number.

We know that there is no natural number less than .

Thus, when is a natural number, there is no solution of the given inequality.

(ii) is an integer.

The integers less than are .

Hence, in this case, the solution set .

Question 3: Solve , when

(i) is an integer.

(ii) is a real number.

Solution : Given that:

(i) is an integer.

The integers less than 2 are .

Hence, in this case, the solution set .

(ii) is a real number.

Thus, the solution set of the given inequality is .

Question 4: Solve , when

(i) is an integer.

(ii) is a real number.

Solution : Given that:

(i) is an integer.

The integers greater than -2 are

Hence, in this case, the solution set .

(ii) is a real number.

Thus, in this case, the solution set is .

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

Question 5: Solve:

Solution : Given that

Hence, the solution set of the given inequality is .

Question 6:

Solution: Given that:Â

Hence, the solution set of the given inequality is .

Question 7:

Solution : Given that

Hence, the solution set of the given inequality is .

Question 8:

Solution: Given that

Hence, the solution set of the given inequality is .

Question 9:

Solution: Given that

Hence, the solution set of the given inequality is .

Question 10:

Solution : Given that

Hence, the solution set of the given inequality is .

Question 11:

Solution : Given that

Hence, the solution set of the given inequality is .

Question 12:

Solution : Given that

Hence, the solution set of the given inequality is .

Question 13:

Solution : Given that

Hence, the solution set of the given inequality is .

Question 14:

Solution :Given that:

Hence, the solution set of the given inequality is .

Question 15:

Solution : Given that:

Hence, the solution set of the given inequality is .

Question 16:

Solution : Given that :

Hence, the solution set of the given inequality is .

### Solve the inequalities in Exercises 17 to 20 and show the graph of the solution in each case on number line

Question 17:

Â

The graphical representation of the solutions of the given inequality is as follows:

Question 18:

Solution : Given that

The graphical representation of the solutions of the given inequality is as follows:

Question 19:

Solution: Given that:

The graphical representation of the solutions of the given inequality is as follows:

Question 20:

Solution :Given that:

The graphical representation of the solutions of the given inequality is as follows:

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

Solution: Let be the marks obtained by Ravi in the third unit test.

Since the student should have an average of at least 60 marks, so

Hence, the student must obtain a minimum of 35 marks 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 be the marks obtained by Sunita in the fifth examination.

In order to receive grade ‘ ‘ in the course, she must obtain an average of 90 marks or more in five examinations. Therefore,

Hence, Sunita must obtain greater than or equal to 82 marks in the 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 be the smaller of the two consecutive odd positive integers. Then, the other integer is .

Since both the integers are smaller than 10 , therefore

Also, the sum of the two integers is more than 11.

From (i) and (ii), we obtain that the value of can be 4, 5, 6 or 7 . .

Since is an odd number, can take the values, 5 and 7 .

Hence, the required possible pairs of numbers are and .

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 be the smaller of the two consecutive even positive integers. Then, the other integer is .

Since both the integers are larger than 5 ,

Also, the sum of the two integers is less than 23.

From (1) and (2), we obtain .

Since is an even number, so can take the values, 6,8 and 10 .

Hence, the required possible pairs are and .

Question 25: The longest side of a triangle is 3 times the shortest side and the third side is shorter than the longest side. If the perimeter of the triangle is at least , find the minimum length of the shortest side.

Solution: Let the length of the shortest side of the triangle=.

Then, length of the longest side and the length of the third side

Since the perimeter of the triangle is at least ,

Hence, the minimum length of the shortest side is .

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

[Hint: If is the length of the shortest board, then and are the lengths of the second and third piece, respectively. Thus, and .

Solution: Let the length of the shortest piece = . Then, length of the second piece and the third piece are and respectively.

Since the three lengths are to be cut from a single piece of board of length ,

Also, the third piece is at least longer than the second piece.

From (1) and (2), we obtain,

Thus, the possible length of the shortest board is greater than or equal to but less than or equal to .

Exercise 5.1 complex no. ncert math solution class 11

Exercise 5.2 complex no. ncert math solution class 11

Exercise 5.3 complex no. ncert math solution class 11

Chapter 5 Miscellaneous ncert math solution class 11

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