5 . Linear Inequalities

Class 11 Maths Chapter 5 Linear Inequalities

Chapter 5. LINEAR INEQUALITIES

Class 11 Maths Chapter 5. Linear Inequalities Exercise 5.1 Solutions :

1. Solve , when (i) is a natural number. (ii) is an integer.

Solution : We have,

(i) When is a natural number, in this case the following values of x make the statement true.

1, 2, 3, 4 .

The solution set of the inequality is {1,2,3,4}.

(ii) When is an integer, the solutions of the given inequality are

...,– 4 , – 3, –2, –1, 0, 1, 2, 3, 4 .

The solution set of the inequality is {...,– 4 ,–3, –2,–1, 0, 1, 2, 3, 4 }

2. Solve , when (i) is a natural number. (ii) is an integer.

Solution : We have,

(i) When is a natural number , the given inequality has no solutions .

(ii) When is an integer, the solutions of the inequality are given by ,

i.e., all real numbers x which are less than . Therefore, the solution set of the inequality is .

3. Solve , when (i) is an integer. (ii) is a real number.

Solution : We have,

(i) When is an integer, the solutions of the given inequality are {..., – 3, –2, –1, 0, 1 }

The solution set of the inequality is {...,–3, –2,–1, 0, 1}

(ii) When is a real number, the solutions of the inequality are given by x < 2,

i.e., all real numbers x which are less than 2. Therefore, the solution set of the inequality is .

4. Solve , when (i) is an integer. (ii) is a real number.

Solution : We have,

(i) When is an integer, the solutions of the given inequality are {– 1, 0, 1 , 2 , 3 , ……… } .

(ii) When is a real number, the solutions of the inequality are given by ,

Therefore, the solution set of the inequality is .

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

Solution : We have,

i.e., all the real numbers which are greater than – 4, are the solutions of the given inequality. Hence, the solution set is .

Solution : We have,

i.e., all the real numbers which are less than –3, are the solutions of the given inequality. Hence, the solution set is .
7.

Solution : We have,

Thus, all real numbers x which are less than or equal to – 3 are the solutions of the given inequality, i.e., .

Solution : We have,

Thus, all real numbers x which are greater than or equal to 4 are the solutions of the given inequality, i.e., .

Solution : We have,

Thus, all real numbers x which are less than to 6 are the solutions of the given inequality, i.e., .

10.

Solution : We have,

Thus, all real numbers x which are less than to – 6 are the solutions of the given inequality, i.e., .

11.

Solution : We have,

Thus, all real numbers x which are less than or equal to 2 are the solutions of the given inequality, i.e., .

12.

Solution : We have,

Thus, all real numbers x which are greater than or equal to 120 are the solutions of the given inequality, i.e., .

13.

Solution : We have,

Thus, all real numbers x which are greater than 4 are the solutions of the given inequality, i.e., .

14.

Solution : We have,

Thus, all real numbers x which are less than or equal to 2 are the solutions of the given inequality, i.e., .

15.

Solution : We have,

Thus, all real numbers x which are greater than 4 are the solutions of the given inequality, i.e.,.
16.

Solution : We have,

Thus, all real numbers x which are less than or equal to 2 are the solutions of the given inequality, i.e., .

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

Solution : We have,

Thus, all real numbers x which are less than 3 are the solutions of the given inequality, i.e., .

18.

Solution : We have,

Thus, all real numbers which are greater than or equal to – 1 are the solutions of the given inequality, i.e., .

19.

Solution : We have,

Thus, all real numbers which are greater than – 1 are the solutions of the given inequality, i.e., .

20.

Solution : We have,

Thus, all real numbers x which are greater than or equal to are the solutions of the given inequality, i.e.,

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 be the marks of Ravi needs to score in the third test .

The average of the three tests should be at least 60.

Average

A/Q,

Therefore, Ravi should score at least 35 marks in the third test to have an average of at least 60 marks.

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 Sunita must obtain in the fifth examination to achieve an average of 90 marks or more in the five examinations.

Average

A/Q,

Therefore, Sunita must obtain at least 82 marks in the fifth examination to get a grade 'A' in the course.

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 and are two consecutive odd positive integers .

A/Q,

and

And

Since x must be a positive integer, the smallest positive integer value for x that satisfies this inequality is 2 and 3 [ Since ] .

For

The consecutive odd positive integers is (2×2+1,2×2+3) = (5 , 7)

For

The consecutive odd integer is (2×3+1 ,2×3+3) = (7,9)

So, the pair of consecutive odd positive integers smaller than 10 such that their sum is more than 11 is (5, 7) and (7,9).

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 and are the two consecutive even positive integers respectively .

A/Q,

And

Again,

We have,

Since is an even positive integer , can take the values 3, 4 and 5 . So, the required possible pairs will be (6,8),(8,10) and (10,12) .

25. The longestside 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 be the length of the shortest side .

The longest side cm.

The third side cm.

A/Q,

So, the minimum length of the shortest side is 9 cm.

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?
[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 is the length of the shortest board.

The lengths of the second

The lengths of third

A/Q,

and

We have,

So, the possible lengths of the shortest board are all values greater than or equal to 8 cm.

Class 11 Maths Chapter 5. Linear Inequalities Miscellaneous Exercise

Solve the inequalities in Exercises 1 to 6.
1.

Solution : We have,

The solution = [2,3]

Solution : We have,

The solution = (0,1] or ]0,1]

Solution : We have,

The solution = [– 4 ,2]

Solution : We have,

The solution = (– 23,2] or ]– 23 ,2]

Solution : We have,

The solution or

Solution : We have,

The solution

Solve the inequalities in Exercises 7 to 10 and represent the solution graphically on number line.
7.

Solution : We have,

And

Thus, solution of the system are real numbers x lying between – 5 and 5 excluding – 5 and 5 , i.e.,

Solution : We have,

and

Thus, solution of the system are real numbers x lying between – 1 and 7 excluding – 1 and 7 , i.e.,

Solution : We have,

and

Thus, solution of the system are real numbers x lying between – 5 and 5 excluding – 5 and 5 , i.e.,

10. .

Solution : We have,

and

Thus, solution of the system are real numbers x lying between – 7 and 11 including – 7 and 11 , i.e.,

11. A solution is to be kept between 68°F and 77°F. What is the range in temperature in degree Celsius (C) if the Celsius / Fahrenheit (F) conversion formula is given by ?

Solution : We have ,

Therefore, the range in temperature in degree Celsius lies between 20° and 25° .

12. A solution of 8% boric acid is to be diluted by adding a 2% boric acid solution to it. The resulting mixture is to be more than 4% but less than 6% boric acid. If we have 640 litres of the 8% solution, how many litres of the 2% solution will have to be added?

Solution : Let be the volume of the 2% boric acid solution (in liters).

The total volume of the resulting mixture liters .

The amount of boric acid in the 8% solution liters

The amount of boric acid in the 2% solution to be added liters.

For more than 4% boric acid:

For less than 6% boric acid :

Therefore, the volume of the 2% boric acid solution to be added should be between 320 and 1280 liters.
13. How many litres of water will have to be added to 1125 litres of the 45% solution of acid so that the resulting mixture will contain more than 25% but less than 30% acid content?

Solution : Let be the amount of water to be added in the acid solution (in litres) .

The total volume of the resulting mixture liters.

The acid content in the original solution is 45%, so the amount of acid in the original solutionliters.

The acid content in the resulting mixture should be more than 25% but less than 30%.

Lower limit (25% acid content):

A/Q,

Upper limit (30% acid content):

So, the solution to the inequalities is

Therefore, the amount of water to be added is between 562.5 and 900 liters to the 1125 liters of the 45% acid solution, resulting in a mixture with more than 25% but less than 30% acid content.

14. IQ of a person is given by the formula , where is mental age and is chronological age. If for a group of 12 years old children, find the range of their mental age.

Solution : We have, and years

Therefore, the range of their mental age between 9.6 years and 16.8 years .