Chapter 6 of Class 11 NCERT Mathematics delves into the topic of the Linear Inequalities. Linear inequalities are mathematical expressions that involve inequalities rather than equalities representing a range of possible values for the variable rather than a single solution. In this chapter, students explore how to solve and graph linear inequalities in the one and two variables understanding how these inequalities can be applied in various real-life scenarios. The Miscellaneous Exercise at the end of the chapter serves as a comprehensive review challenging students with a variety of problems that test their understanding of the concepts covered.
Linear Inequalities
Linear inequalities are expressions that involve variables and inequality symbols like <, >, ≤ and ≥. Unlike linear equations, which have one specific solution linear inequalities represent a range of the values that satisfy the given condition. These inequalities can be solved using similar techniques to the linear equations with the additional step of considering the direction of the inequality when multiplying or dividing by the negative number. The solutions can be represented graphically on a number line or in a coordinate plane making it easier to visualize the set of all possible solutions.
Solve the inequalities in Exercises 1 to 6.
Question 1. 2 ≤ 3x – 4 ≤ 5
Solution:
In this case, we have two inequalities, 2 ≤ 3x – 4 and 3x – 4 ≤ 5, which we will solve simultaneously.
We have 2 ≤ 3x – 4 ≤ 5
or 2 ≤ 3x – 4 and 3x – 4 ≤ 5
⇒ 2 + 4 ≤ 3x and 3x ≤ 5 + 4
⇒ 6 ≤ 3x and 3x ≤ 9
⇒
\frac{6}{3} ≤ x and x ≤\frac{9}{3} ⇒ 2 ≤ x and x ≤ 3
⇒ 2 ≤ x ≤ 3
Hence, all real numbers x greater than or equal to 2 but less than or equal to 3 are solution of given 2 ≤ 3x – 4 ≤ 5 equality.
x ∈ [2, 3]
Question 2. 6 ≤ – 3 (2x – 4) < 12
Solution:
In this case, we have two inequalities, 6 ≤ – 3 (2x – 4) and – 3 (2x – 4) < 12, which we will solve simultaneously.
We have 6 ≤ – 3 (2x – 4) < 12
or 6 ≤ – 3 (2x – 4) and – 3 (2x – 4) < 12
⇒
\frac{6}{3} ≤ -(2x – 4) and -(2x – 4) <\frac{12}{3} ⇒ 2 ≤ -(2x – 4) and -(2x – 4) < 4
⇒ -2 ≥ (2x – 4) and (2x – 4) > -4 [multiplying the inequality with (-1) which changes the inequality sign]
⇒ -2+4 ≥ 2x and 2x > -4+4
⇒ 2 ≥ 2x and 2x > 0
⇒
\frac{2}{2} ≥ x >\frac{0}{2} ⇒ 1≥ x > 0
⇒ 0 < x ≤ 1
Hence, all real numbers x greater than 0 but less than or equal to 1 are solution of given 6 ≤ – 3 (2x – 4) < 12 equality.
x ∈ (0, 1]
Question 3. -3 ≤ 4- \frac{7x}{2} ≤ 18
Solution:
In this case, we have two inequalities, -3 ≤ 4-
\frac{7x}{2} and 4-\frac{7x}{2} ≤ 18, which we will solve simultaneously.We have -3 ≤ 4-
\mathbf{\frac{7x}{2}} ≤ 18or -3 ≤ 4-
\frac{7x}{2} and 4-\frac{7x}{2} ≤ 18⇒ -3-4 ≤ -
\frac{7x}{2} and -\frac{7x}{2} } ≤ 18-4⇒ -7 ≤ -
\frac{7x}{2} and -\frac{7x}{2} ≤ 14⇒ 7 ≥
\frac{7x}{2} and\frac{7x}{2} ≥ -14 [multiplying the inequality with (-1) which changes the inequality sign]⇒ 7×2 ≥ 7x and 7x ≥ -14×2
⇒
\frac{14}{7} ≥ x and x ≥\frac{-28}{7} ⇒ 2 ≥ x ≥ -4
⇒ -4 ≤ x ≤ 2
Hence, all real numbers x greater than or equal to -4 but less than or equal to 2 are solution of given -3 ≤ 4-
\frac{7x}{2} ≤ 18 equality.x ∈ [-4, 2]
Question 4. -15 < \frac{3(x-2)}{5} ≤ 0
Solution:
In this case, we have two inequalities, -15 <
\frac{3(x-2)}{5} and\frac{3(x-2)}{5} } ≤ 0, which we will solve simultaneously.We have -15 <
\frac{3(x-2)}{5} ≤ 0or -15 <
\frac{3(x-2)}{5} and\frac{3(x-2)}{5} ≤ 0⇒ -15×
\frac{5}{3} < (x-2) and (x-2) ≤ 0×\frac{5}{3} ⇒ -25 < x-2 and x-2 ≤ 0
⇒ -25+2 < x and x ≤ 0+2
⇒ -23 < x ≤ 2
Hence, all real numbers x greater than -23 but less than or equal to 2 are solution of given -15 ≤
\frac{3(x-2)}{5} ≤ 0 equality.x ∈ (-23, 2]
Question 5. -12 < 4- (\frac{3x}{-5} ) ≤ 2
Solution:
In this case, we have two inequalities, -12 < 4-
\frac{3x}{-5} and 4-\frac{3x}{-5} ≤ 2, which we will solve simultaneously.We have -12 < 4-
\mathbf{\frac{3x}{-5}} ≤ 2or -12 < 4-
\frac{3x}{-5} and 4-\frac{3x}{-5} ≤ 2⇒ -12-4 <
- \frac{3x}{-5} and- \frac{3x}{-5} ≤ 2-4⇒ -16 <
-\frac{3x}{-5} and-\frac{3x}{-5} ≤ -2⇒ -16 <
\frac{3x}{5} and\frac{3x}{5} ≤ -2⇒ -16×5 < 3x and 3x ≤ -2×5
⇒
\frac{-80}{3} < x and x ≤\frac{-10}{3} ⇒
\mathbf{\frac{-80}{3} < x ≤ \frac{-10}{3}} Hence, all real numbers x greater than -80/3 but less than or equal to -10/3 are solution of given -12 < 4-
\frac{3x}{-5} ≤ 2 equality.x ∈ (-80/3, -10/3]
Question 6. 7 ≤ \frac{3x+11}{2} ≤ 11
Solution:
In this case, we have two inequalities, 7 ≤
\frac{3x+11}{2} and\frac{3x+11}{2} ≤ 11, which we will solve simultaneously.We have 7 ≤
\mathbf{\frac{3x+11}{2}} ≤ 11or 7 ≤
\frac{3x+11}{2} and\frac{3x+11}{2} ≤ 11⇒ 7×2 ≤ 3x+11 and 3x+11 ≤ 11×2
⇒ 14 ≤ 3x+11 and 3x+11 ≤ 22
⇒ 14-11 ≤ 3x and 3x ≤ 22-11
⇒ 3 ≤ 3x and 3x ≤ 11
⇒ 1 ≤ x and x ≤
\frac{11}{3} ⇒ 1 ≤ x ≤
\mathbf{\frac{11}{3}} Hence, all real numbers x greater than or equal to 1 but less than or equal to 11/3 are solution of given 7 ≤
\frac{3x+11}{2} ≤ 11 equality.x ∈ [1, 11/3]
Solve the inequalities in Exercises 7 to 10 and represent the solution graphically on number line.
Question 7. 5x + 1 > – 24, 5x – 1 < 24
Solution:
So, from given data
5x + 1 > – 24 ........................(1)
5x – 1 < 24 .........................(2)
From inequality (1), we have
5x + 1 > – 24
5x > – 24-1
x > – 25/5
x > -5 ...............................(3)
Also, from inequality (2), we have
5x – 1 < 24
5x < 24+1
x < 25/5
x < 5 .................................(4)
So, from (3) and (4), we can conclude that,
-5 < x < 5 ...................(5)
If we draw the graph of inequalities (5) on the number line, we see that the values of x, which are common to both, are
x ∈ (-5,5)
Thus, solution of the system are real numbers x lying between -5 and 5 excluding -5 and 5.
Question 8. 2 (x – 1) < x + 5, 3 (x + 2) > 2 – x
Solution:
So, from given data
2 (x – 1) < x + 5 ........................(1)
3 (x + 2) > 2 – x .........................(2)
From inequality (1), we have
2 (x – 1) < x + 5
2x – 2 < x + 5
2x – x < 5+2
x < 7 ...............................(3)
Also, from inequality (2), we have
3 (x + 2) > 2 – x
3x + 6 > 2 – x
3x + x > 2 – 6
4x > -4
x > -1 .................................(4)
So, from (3) and (4), we can conclude that,
-1 < x < 7 ...................(5)
If we draw the graph of inequalities (5) on the number line, we see that the values of x, which are common to both, are
x ∈ (-1,7)
Thus, solution of the system are real numbers x lying between -1 and 7 excluding -1 and 7.
Question 9. 3x – 7 > 2 (x – 6) , 6 – x > 11 – 2x
Solution:
So, from given data
3x – 7 > 2 (x – 6) ........................(1)
6 – x > 11 – 2x .........................(2)
From inequality (1), we have
3x – 7 > 2 (x – 6)
3x – 7 > 2x – 12
3x – 2x > – 12+7
x > -5 ...............................(3)
Also, from inequality (2), we have
6 – x > 11 – 2x
– x+2x > 11 -6
x > 5 .................................(4)
So, from (3) and (4), we can conclude that,
5 < x ...................(5)
If we draw the graph of inequalities (5) on the number line, we see that the values of x, which are common to both, are
x ∈ (5,∞)
Thus, solution of the system are real numbers x lying between 5 and ∞ excluding 5 .
Question 10. 5 (2x – 7) – 3 (2x + 3) ≤ 0 , 2x + 19 ≤ 6x + 47
Solution:
So, from given data
5 (2x – 7) – 3 (2x + 3) ≤ 0 ........................(1)
2x + 19 ≤ 6x + 47 .........................(2)
From inequality (1), we have
5 (2x – 7) – 3 (2x + 3) ≤ 0
10x - 35 -6x - 9 ≤ 0
4x - 44 ≤ 0
4x ≤ 44
x ≤ 44/4
x ≤ 11 ...............................(3)
Also, from inequality (2), we have
2x + 19 ≤ 6x + 47
2x - 6x ≤ 47 - 19
-4x ≤ 28
4x ≥ -28 [multiplying the inequality with (-1) which changes the inequality sign]
x ≥ -28/4
x ≥ -7 .................................(4)
So, from (3) and (4), we can conclude that,
-7 ≤ x ≤ 11 ...................(5)
If we draw the graph of inequalities (5) on the number line, we see that the values of x, which are common to both, are
x ∈ [-7,11)
Thus, solution of the system are real numbers x lying between -7 and 11 including -7 and 11 .
Question 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 F = (\frac{9}{5} )C + 32
Solution:
According to the given data
The solution has to be kept between 68° F and 77° F
So, we have,
68° < F < 77°
Substituting, F =
\mathbf{\frac{9}{5}} C + 32⇒ 68° <
\frac{9}{5} C + 32 < 77°⇒ 68°- 32° <
\frac{9}{5} C < 77°- 32°⇒ 36° <
\frac{9}{5} C < 45°⇒ 36×
\frac{5}{9} < C < 45×\frac{5}{9} ⇒ 20° < C < 25°
Hence, here we get,
The range of temperature in degree Celsius is between 20° C to 25° C.
Question 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:
According to the given data,
Here, 8% of solution of boric acid = 640 litres
So, we can take the amount of 2% boric acid solution added as x litres
Hence, Total mixture = (x + 640) litres
As it is given,
The resulting mixture has to be more than 4% but less than 6% boric acid
⇒ (2% of x + 8% of 640) > (4% of (x + 640)) and (2% of x + 8% of 640) < (6% of (x + 640))
⇒ (
\frac{4}{100} ) × (x + 640) < (\frac{2}{100} ) × x + (\frac{8}{100} ) × 640) < (\frac{6}{100} ) × (x + 640)⇒ 4(x + 640) < (2×x + 8× 640) < 6(x + 640)
⇒ 4x + 2560 < 2x +5120 < 6x+3840
In this case, we have two inequalities,
⇒ 4x + 2560 < 2x +5120 and 2x +5120 < 6x+3840
⇒ 4x - 2x < 5120 - 2560 and 5120-3840 < 6x-2x
⇒ 2x < 2560 and 1280 < 4x
⇒ x < 2560/2 and 1280/4 < x
⇒ x < 1280 and 320 < x
⇒ 320 < x < 1280
Therefore, the number of litres of 2% of boric acid solution that has to be added will be more than 320 litres but less than 1280 litres.
Question 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:
According to the given data,
Here, 45% of solution of acid = 1125 litres
Let the amount of water added in the solution = x litres
Resulting mixture = (x + 1125) litres
As it is given,
The resulting mixture has to be more than 25% but less than 30% acid content
Amount of acid in resulting mixture = 45% of 1125 litres.
⇒ 45% of 1125 < 30% of (x + 1125) and 45% of 1125 > 25% of (x + 1125)
⇒ 25% of (x + 1125) < 45% of 1125 < 30% of (x + 1125)
In this case, we have two inequalities,
⇒ (
\frac{25}{100} × (x + 1125)) < (\frac{45}{100} × 1125) and (\frac{30}{100} × (x + 1125)) > (\frac{45}{100} × 1125)⇒ (25(x + 1125)) < (45×1125) and (30(x + 1125)) > (45×1125)
⇒ (x + 1125) < (45×1125)/25 and (x + 1125) > (45×1125)/30
⇒ (x + 1125) < 2025 and (x + 1125) > 3375/2
⇒ 3375/2 < (x + 1125) < 2025
⇒ (3375/2)-1125 < x < 2025-1125
⇒ 1125/2 < x < 900
⇒ 562.5 < x < 900
Therefore, the number of litres of water that has to be added will have to be more than 562.5 litres but less than 900 litres.
Question 14. IQ of a person is given by the formula, IQ = (\frac{MA}{CA} ) × 100, where MA is mental age and CA is chronological age. If 80 ≤ IQ ≤ 140 for a group of 12 years old children, find the range of their mental age.
Solution:
According to the given data, we have
Chronological age = CA = 12 years
IQ for age group of 12 is in the range,
80 ≤ IQ ≤ 140
Substituting, IQ = (
\mathbf{\frac{MA}{CA}} ) × 100⇒ 80 ≤
\frac{MA}{CA} × 100 ≤ 140⇒ 80 ≤
\frac{MA}{12} × 100 ≤ 140⇒ 80×12/100 ≤ MA ≤ 140×12/100
⇒ 96/10 ≤ MA ≤ 168/10
⇒ 9.6 ≤ MA ≤ 16.8
Hence, Range of mental age (MA) of the group of 12 years old children is 9.6 ≤ MA ≤ 16.8
Conclusion
The Linear inequalities play a significant role in mathematical modeling and problem-solving providing the foundation for the understanding more complex mathematical concepts. The Miscellaneous Exercise in Chapter 6 of Class 11 NCERT is designed to the reinforce students' comprehension of the linear inequalities by the presenting diverse problems that require critical thinking and application of the concepts learned. Mastery of these exercises is crucial for the students as they progress to the more advanced topics in the mathematics.