RBSE Solutions for Class 11 Maths Chapter 6 Linear Inequalities Miscellaneous Exercise

Rajasthan Board RBSE Solutions for Class 11 Maths Chapter 6 Linear Inequalities Miscellaneous Exercise Textbook Exercise Questions and Answers.

Rajasthan Board RBSE Solutions for Class 11 Maths in Hindi Medium & English Medium are part of RBSE Solutions for Class 11. Students can also read RBSE Class 11 Maths Important Questions for exam preparation. Students can also go through RBSE Class 11 Maths Notes to understand and remember the concepts easily.

RBSE Class 11 Maths Solutions Chapter 6 Linear Inequalities Miscellaneous Exercise

Solve the inequalities in Exercises 1 to 6.

Question 1.
2 ≤ 3x - 4 ≤ 5
Answer:
From inequality 2 ≤ 3x - 4 ≤ 5 
2 + 4 ≤ 3x ≤ 5 + 4
(Transposing - 4 to left and right side)
=>    6 ≤ 3x ≤ 9
=>    2 ≤ x ≤ 3 (Dividing by 3)
Thus x e [2,3]

Second Method : We can write the given inequality 2 ≤ 3x - 4 ≤ 5 as :
2 ≤ 3x - 4    ...(1) 
and    3x - 4 ≤ 5    ........(2)
Adding 4 to both sides of equation (1),
2 + 4 ≤ 3x - 4 + 4 
⇒ 6 ≤ 3x
⇒ 2 ≤ 3 (Dividing    by 3)
Again, adding 4 to both sides of equation (2),
3x - 4 + 4 ≤ 5 + 4 
⇒ 3x ≤ 9
⇒ x ≤ 3 (Dividing    by 3)
Now 2 ≤ x and x ≤ 3
Then x lies between 2 and 3.
Thus, the solution set is [2, 3],

RBSE Solutions for Class 11 Maths Chapter 6 Linear Inequalities Miscellaneous Exercise 

Question 2.
6 ≤ - 3 (2x - 4) ≤ 12
Answer:
Inequality 6 ≤ - 3 (2x - 4) ≤ 12 can be written as:
65 - 3(2x - 4)    ...(1)
and    -3(2x - 4) ≤ 12    ...(2)
From inequality (1),
6    ≤ -6x + 12
Adding - 12 to both sides,
6 - 12 ≤ -6x + 12 - 12 
⇒ - 6 ≤ - 6x
⇒ -1 ≤ - x (Dividing both sides by 6)
⇒ 1 > x
⇒ x ≤ 1
(On multiplying by (- 1) to both sides sign < changes to >)
In inequality (2), dividing both sides by - 3,
\(\frac{-3(2 x-4)}{-3}>\frac{12}{-3}\)
⇒ 2x - 4 ≥ - 4
⇒ 2x - 4 + 4 ≥ -4 + 4
(Adding 4 to both sides)
⇒ 2x ≥ 0
⇒ x ≥ 0
Now, x lies from 0 to 1.
Thus, the solution is x ∈ (0,1]

Question 3.
-3 ≤ 4 - \(\frac{7 x}{2}\) ≤ 18
Answer:
Inequality - 3 ≤ 4 - \(\frac{7 x}{2}\) ≤ 18 can be written as:
-3 ≤ 4 - \(\frac{7 x}{2}\) .........(1)
and    4 - \(\frac{7 x}{2}\) ≤ 18 .........(2)
Adding - 4 to inequality (1), 
-3 - 4 ≤ 4 - \(\frac{7 x}{2}\) - 4
⇒ - 7 ≤ - \(\frac{7 x}{2}\)
 
Dividing both sides by - 7, we get
1 ≥ \(\frac{x}{2}\)
⇒ 2 ≥ x 
⇒ x ≤ 2
Adding - 4 to both sides of inequility (1),
-4 + 4 - \(\frac{7 x}{2}\) ≤ 18 - 4
\(\frac{7 x}{2}\) ≤ 14

Dividing both sides by - 7,
\(\frac{x}{2}\) ≥ -2
⇒ x ≥ -4
We see that x ≥ - 4 and 2 ≥ x 
⇒ 2 ≥ x ≥ -4
⇒ - 4 ≤ x ≤ 2
RBSE Solutions for Class 11 Maths Chapter 6 Linear Inequalities Miscellaneous Exercise 1
It means, the value of x is from - 4 to 2. Thus, the solution is x ∈ [- 4, 2]

Question 4.
-15 < \(\frac{3(x-2)}{5}\) ≤ 0
Answer:
The given inequality can be written as :
-15 < \(\frac{3(x-2)}{5}\) ............(1)
and \(\frac{3(x-2)}{5}\) ≤ 0 ........(2)
Multiplying by \(\frac{5}{3}\) to both sides of inequality (1),
-15 × \(\frac{5}{3}<\frac{3(x-2)}{5} \times \frac{5}{3}\)
⇒ -25 < x - 2
Adding 2 to both sides,
- 25 + 2 < x - 2 + 2
⇒ - 23 < x ⇒ x > - 23
And multiplying by \(\frac{5}{3}\) to both sides of inequality (2),
\(\frac{5}{3} \times \frac{3(x-2)}{5}<\frac{5}{3}\) × 0
⇒ x - 2 ≤ 0

Adding 2 to both the sides,
x - 2 + 2 ≤ 0 + 2
⇒ x ≤ 2
We see that - 23 ≤ x and x ≤ 2.
So, the value of x is from - 23 to 2.
RBSE Solutions for Class 11 Maths Chapter 6 Linear Inequalities Miscellaneous Exercise 2
Thus, the solution is (- 23, 2] 

Alternative Method
Given, -15 < \(\frac{3(x-2)}{5}\) ≤ 0
Multiplying by 5,
-75 < 3x - 6 ≤ 0
⇒ - 75 + 6 ≤ 3x - 6 + 6 ≤ 0 + 6
(Adding 6 to both sides)
⇒ - 69 < 3x ≤ 6
Dividing by 3,
- 23 < x ≤ 2
RBSE Solutions for Class 11 Maths Chapter 6 Linear Inequalities Miscellaneous Exercise 3
So, the value of x is greater than -23 and is equal to or less than 2.
Thus, the solution is x ∈ (- 23,2]

RBSE Solutions for Class 11 Maths Chapter 6 Linear Inequalities Miscellaneous Exercise

Question 5.
-12 < 4 - \(\frac{3 x}{-5}\) ≤ 2
Answer:
The given inequality can be written as :
-12 < 4 - \(\frac{3 x}{-5}\) ......(1)
and    4 - \(\frac{3 x}{-5}\) ≤ 2
Adding - 4 to inequality (1),
⇒ -4 - 12 < -4 + 4 - \(\frac{3 x}{-5}\)
⇒ -16 < \(\frac{3 x}{-5}\)
⇒ -16 < \(\frac{3 x}{5}\)

Multiplying both sides by 5.
⇒ - 80 < 3x
⇒ -\(\frac{80}{3}\) < x    
⇒ x > -\(\frac{80}{3}\)
Adding - 4 to both sides of inequality (2),
⇒ -4 + 4 - \(\frac{3 x}{-5}\) ≤ 2 - 4
⇒ - \(\frac{3 x}{-5}\) ≤ -2
⇒  \(\frac{3 x}{5}\) ≤ -2
\(\frac{3 x}{5}\) ≤ -2
⇒ 3x ≤ -10
⇒ x ≤ \(-\frac{10}{3}\)

We see that x > \(-\frac{80}{3}\) and x ≤ \(-\frac{10}{3}\)
So, \(-\frac{80}{3}\) and x ≤ \(-\frac{10}{3}\)

Then the value of x is from -\(\frac{80}{3}\) to \(-\frac{10}{3}\)
RBSE Solutions for Class 11 Maths Chapter 6 Linear Inequalities Miscellaneous Exercise 4
Thus, the solution is x ∈ \(\left(-\frac{80}{3},-\frac{10}{3}\right]\)

Alternative Method
Given, -12 < 4 - \(\frac{3 x}{5}\) ≤ 2
⇒ -12 < 4 + \(\frac{3 x}{-5}\) ≤ 2
Subtracting 4 from the inequality,
-12 - 4 < 4 + \(\frac{3 x}{5}\) - 4 ≤ 2 - 4
⇒ -16 < \(\frac{3 x}{5}\) ≤ -2

Multiplying by 5,
-16 × 5 < \(\frac{3 x}{5}\) × 5 ≤ -2 × 5
⇒ -80 < 3x ≤ -10
\(-\frac{80}{3}<\frac{3 x}{3} \leq-\frac{10}{3}\)
\(-\frac{80}{3}<x \leq-\frac{10}{3}    \)
So, the value of x is from \(-\frac{80}{3}\) to \(-\frac{10}{3}\)
RBSE Solutions for Class 11 Maths Chapter 6 Linear Inequalities Miscellaneous Exercise 5
Thus, the solution is x ∈ \(\left(-\frac{80}{3},-\frac{10}{3}\right]\)

Question 6.
7 ≤ \(\left(\frac{3 x+11}{2}\right)\) ≤ 11
Answer:
Given, 7 ≤ \(\left(\frac{3 x+11}{2}\right)\) ≤ 11
Multiplying both sides by 2,
⇒ 14 ≤ 3x + 11 ≤ 22
We can write it as :
14 ≤ 3x + 11    ...(1)
and    3x + 11 < 22    ......(2)
Adding - 11 to both sides of inequality (1),
14 - 11 ≤ 3x + 11 - 11
⇒ 3 < 3x
⇒ 1 < x
Again, adding -11 to both sides of inequality (2),
3x+ 11 - 11 ≤ 22 - 11 
3x ≤ 11
x ≤ \(\frac{11}{3}\)
We see that 1 ≤ x ≤ \(\frac{11}{3}\)
RBSE Solutions for Class 11 Maths Chapter 6 Linear Inequalities Miscellaneous Exercise 6
So, the value of x lies from 1 to \(\frac{11}{3}\)
Thus, the solution is x ∈ [1, \(\frac{11}{3}\)]

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

Question 7.
5x + 1 > - 24, 5x - 1 < 24
Answer:
5x + 1 > - 24    ...(1)
and    5x - 1 < 24 ..........(2)
From inequality (1),
5x + 1 - 1 > -24 - 1
(Adding - 1 to both sides)
⇒ 5x > - 25
Dividing both sides by 5
x > -5
From inequation (2)
5x - 1 < 24 
Adding 1 to both the sides
5x - 1 + 1 < 24 + 1 
⇒ 5x < 25
Dividing both sides by 5,
x < 5
Now,    x > - 5 and x < 5
We see that - 5 < x < 5
i.e., the value of x is between - 5 and 5.
The solution is x ∈ (- 5, 5).
The graph of the solution is shown by the thick portion of the number line. Here, - 5 and 5 are not included.
RBSE Solutions for Class 11 Maths Chapter 6 Linear Inequalities Miscellaneous Exercise 7

Question 8.
2(x -1) < x + 5,3(x + 2) > 2 - x 
Answer:
2(x - 1) < x + 5
3(x + 2) >2 - x 
Now, from inequality (1),
2(x - 1) < x + 5 
2x - 2 < x + 5
Adding 2 to both sides,
2x - 2 + 2 < x + 5 + 2
2x < x + 7
Subtracting x from both sides
2x - x < x - x + 7 
⇒ x    <7
Now, from inequality (2),
3(x + 2) > 2 - x 
⇒ 3x + 6 > 2 - x
Subtracting 6 from both sides
3x + 6 - 6 > 2 - x- 6
⇒ 3x > -x - 4
Adding x to both sides,
3x + x > -x + x - 4
⇒ 4x > - 4
⇒ x > -1.
⇒ x < - 1
We see that -1 < x and x < 7
So,    -1 < x < 7
i.e., the value ofx is in between -1 to 7.
The solution is x ∈ (-1, 7).
The graph of the solution is shown by the thick portion of the number line. Here, - 1 and 7 are not included.
RBSE Solutions for Class 11 Maths Chapter 6 Linear Inequalities Miscellaneous Exercise 8

RBSE Solutions for Class 11 Maths Chapter 6 Linear Inequalities Miscellaneous Exercise

Question 9.
3x - 7 > 2(x - 6),(6 - x) > 11 - 2x
Answer:
3x - 7 > 2(x - 6)
⇒ 3x - 7 > 2x - 12
Subtracting 2x from both sides,
3x - 7 - 2x > 2x - 12 - 2x
⇒ x - 7 > -12
Adding 7 to both sides,
x - 7 + 7 > -12 + 7
⇒ x > -5 
Again, from (6 - x) > 11 - 2x
⇒ 6 - x > 11 - 2x
Adding 2x to both sides,
6 - x + 2x >11 - 2x + 2x 
⇒ 6 + x > 11
Subtracting 6 from both sides,
6 + x - 6 >11 - 6 ⇒  x > 5 
We see that x > - 5 and x > 5. 
i. e., the common value of x is x > 5.
The solution is x ∈ (5, ∞)
The graph of the solution set is shown by the thick portion of the number line.
RBSE Solutions for Class 11 Maths Chapter 6 Linear Inequalities Miscellaneous Exercise 9

Question 10.
5(2x - 7) - 3(2x + 3) ≤ 0, 2x + 19 ≤ 6x + 47
Answer:
From 5(2x - 7)    - 3(2x + 3) ≤ 0
⇒ 10x -    35 - 6x - 9 ≤ 0
⇒ 4x - 44 ≤ 0
⇒ 4x ≤ 44
⇒ x ≤ 11
Now, from 2x + 19 ≤ 6x + 47
Adding -19 to both the sides,
2x + 19 - 19 ≤ 6x + 47-19 
⇒ 2x ≤ 6x + 28
Subtracting 6x from both sides,
2x - 6x < 6x + 28 - 6x
⇒ - 4x < 28
Dividing both sides by - 4, 
\(\frac{-4 x}{-4} \geq \frac{28}{-4}\)
⇒ x ≥ -7
From x ≤ 11 and x ≥ -7, we see that
-7 ≤ x ≤ 11
i. e, the values of x are from - 7 and 11.
The graph of the solution is shown by the thick postion of the number line. Here, - 7 and 11 are included.
RBSE Solutions for Class 11 Maths Chapter 6 Linear Inequalities Miscellaneous Exercise 10

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?
Answer:
Temperature is from 68°F to 77°F.
Then by F = \(\frac{9}{5}\)C + 32
\(\frac{9}{5}\)C + 32 > 68 .....(1)
and \(\frac{9}{5}\)C + 32 < 77 ..........(2)
From in inequality (1),
\(\frac{9}{5}\)C + 32 >68
\(\frac{9}{5}\)C > 68 - 32
(Adding 32 to both the sides) 9
\(\frac{9}{5}\)C > 36
(Multiplying both sides by \(\frac{9}{5}\))
C > 20

From inequality (2),
\(\frac{9}{5}\)C + 32 < 77
\(\frac{9}{5}\)C < 77 - 32
(Subtracting 32 from both sides)
\(\frac{9}{5}\)C < 45

Multiplying both side by 5,
⇒ 9C < 45 × 5
⇒ C < \(\frac{45 \times 5}{9}\)
⇒ C < 25
Thus, on the celsius scale, the range of the temperature is between 20°C and 25 °C.

RBSE Solutions for Class 11 Maths Chapter 6 Linear Inequalities Miscellaneous Exercise

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?
Answer:
Let x litres of 2% boric acid solution be added to 640 litres of 8% boric acid solution.
Total mixture = (x + 640) litres According to given,
2% of x + 8% of 640 > 4% of (x + 640) 
and 2% of x + 8% of 640 < 6% of (x + 640)
\(\frac{2}{100}\)x + \frac{8}{100} × 640 > \(\frac{4}{100}\)(x + 640) and \(\frac{2}{100}\)x + \(\frac{8}{100}\) × 640 < \(\frac{6}{100}\)(x + 640)
⇒ 2x + 8 × 640 > 4x + 4 × 640 
and 2x + 8 × 640 < 6x + 6 × 640 
⇒ 4 × 640 > 2x and 2 × 640 < 4x 
⇒ 128 > x and 320 < x 
⇒ x < 1280 and 320 <x 
⇒ 320 < x < 1280
Thus, the number of litres of 2% boric acid solution must be more than 320 but less than 1280.

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?
Answer:
Let x litres of water be added to 1125 litres of 45% acid solution, then 25% of (x+ 1125) <45% of 1125 < 30% of (x +1125)
We have 25% of (x + 1125) < 45% of 1125
\(\frac{25}{100}\) × (x + 1125) < \(\frac{45}{100}\) × 1125
Multiplying both sides by 100, 
⇒ 25(x + 1125) <45 × 1125
Dividing both sides by 25,
x+ 1125 < 45 × 45 
x+ 1125 < 2025

Subtracting 1125 from both sides
x + 1125 < 2025 - 1125 
⇒ x < 900
Again, we have
45% of 1125 < 30% of (x +1125)
\(\frac{45}{100}\) × 1125 < \(\frac{30}{100}\) × (x + 1125)
Multiplying both sides by 100,
45 × 1125 < 30(x + 1125)
Dividing both sides by 5,
9 × 1125 < 6(x +1125)
⇒ 10125 < 6x + 6750
Subtracting 6750 from both sides
10125 - 6750 < 6x + 6750 - 6750
⇒ 3375 < 6x
Dividing both sides by 6,
5625 < x ⇒ x > 5625
We see that x < 900 and x > 562.5
i.e.,    562.5 < x <900
The value of x lies between 562.5 and 900.
Thus, the quantity of water in the mixture must be from 562.5 litres to 900 litres.

RBSE Solutions for Class 11 Maths Chapter 6 Linear Inequalities Miscellaneous Exercise

Question 14.
IQ of a person is given by the formula IQ = \(\frac{\text { MA }}{\mathrm{C A}}\) × 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.
Answer:
Given, IQ = \(\frac{\mathrm{MA}}{12}\) × 100 and CA = 12 years
∴ IQ = \(\frac{\mathrm{MA}}{12}\) × 100 = \(\frac{25}{5}\) MA
Given, 80 ≤ IQ ≤ 140
⇒ 80 ≤ \(\frac{25}{3}\) MA ≤ 140
\(\frac{3}{25}\) × 80 ≤ MA ≤ \(\frac{3}{25}\) × 140 (Multiplying by \(\frac{3}{25}\)
\(\frac{48}{5}\) ≤ MA ≤ \(\frac{84}{5}\)
⇒ 9.6 ≤ MA ≤ 16.8
Thus, the mental age lies between 9.6 years and 16.8 years i.e., the mental age is alteast 9.6 years but not more than 16.8 years.

Prasanna
Last Updated on Nov. 3, 2023, 5:07 p.m.
Published Nov. 2, 2023