11th maths NCERT Linear Inequalities Exercise Miscellaneous detailed solution with explanation

Solve the following inequalities in Exercise 1 to 6

Question (1)

Solve the inequality 2 ≤ 3x – 4 ≤ 5

Solution

2 ≤ 3x – 4 ≤ 5
⇒ 2 + 4 ≤ 3x – 4 + 4 ≤ 5 + 4
⇒ 6 ≤ 3x ≤ 9
⇒ 2 ≤ x ≤ 3
Thus, all the real numbers, x, which are greater than or equal to 2 but less than or equal to 3, are the solutions of the given inequality. The solution set for the given inequalityis [2, 3].

Question (2)

Solve the inequality 6 ≤ –3(2x – 4) < 12

Solution

6 ≤ – 3(2x – 4) < 12
rArr; 2 ≤ –(2x – 4) < 4
rArr; –2 ≥ 2x – 4 > –4
rArr; 4 – 2 ≥ 2x > 4 – 4
rArr; 2 ≥ 2x > 0
rArr;1 ≥ x > 0
Thus, the solution set for the given inequalityis (0, 1].

Question (3)

Solve the inequality : $ - 3 \le 4 - \frac{{7x}}{2} \le 18$

Solution

\[ - 3 \le 4 - \frac{{7x}}{2} \le 18\] \[ \Rightarrow - 3 - 4 \le - \frac{{7x}}{2} \le 18 - 4\] \[ \Rightarrow - 7 \le - \frac{{7x}}{2} \le 14\] \[ \Rightarrow 7 \ge \frac{{7x}}{2} \ge - 14\] \[ \Rightarrow 1 \ge \frac{x}{2} \ge - 2\] \[ \Rightarrow 2 \ge - x \ge - 4\] Thus, the solution set for the given inequalityis [–4, 2].

Question (4)

Solve the inequality :$ - 15 \le \frac{{3\left( {x - 2} \right)}}{2} \le 0$

Solution

\[ - 15 < \frac{{3\left( {x - 2} \right)}}{5} \le 0\] ⇒ –75 < 3(x – 2) ≤ 0
⇒ –25 < x – 2 ≤ 0
⇒ – 25 + 2 < x ≤ 2
⇒ –23 < x ≤ 2
Thus, the solution set for the given inequalityis (–23, 2].

Question (5)

Solve the inequality : $ - 12 < 4 - \frac{{3x}}{{ - 5}} \le 2$

Solution

\[ - 12 < 4 - \frac{{3x}}{{ - 5}} \le 2\] \[ - 12 - 4 < - \frac{{3x}}{{ - 5}} \le 2 - 4\] \[ - 16 < \frac{{3x}}{5} \le - 2\] \[ - 80 < 3x \le - 10\] \[ \Rightarrow \frac{{ - 80}}{3} < x \le \frac{{ - 10}}{3}\] Thus, the solution set for the given inequality is $\left( {\frac{{ - 80}}{3},\frac{{ - 10}}{3}} \right]$

Question (6)

Solve the inequality : $7 \le \frac{{\left( {3x + 11} \right)}}{2} \le 11$

Solution

\[7 \le \frac{{\left( {3x + 11} \right)}}{2} \le 11\] \[ \Rightarrow 14 \le 3x + 11 \le 22\] \[ \Rightarrow 14 - 11 \le 3x \le 22 - 11\] \[ \Rightarrow 3 \le 3x \le 11\] \[ \Rightarrow 1 \le 3x \le \frac{{11}}{3}\] Thus, the solution set for the given inequality is $\left[ {1,\frac{{11}}{3}} \right]$

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

Question (7)

5x + 1 > –24, 5x – 1 < 24

Solution

5x + 1 > –24
⇒ 5x > –25
⇒ x > –5 ... (1)
5x – 1 < 24
⇒ 5x < 25
⇒ x < 5 ... (2) From (1) and (2), it can be concluded that the solution set for the given system of inequalities is (–5, 5). The solution of the given system of inequalities can be represented on number line as

Question (8)

2(x – 1) < x + 5, 3(x + 2) > 2 – x

Solution

2(x – 1) < x + 5
rArr; 2x – 2 < x + 5
rArr; 2x – x < 5 + 2
rArr; x < 7 … (1)
3(x + 2) > 2 – x
rArr; 3x + 6 > 2 – x
rArr; 3x + x > 2 – 6
rArr; 4x > – 4
rArr; x > – 1 … (2)
From (1) and (2), it can be concluded that the solution set for the given system of inequalities is (–1, 7).
The solution of the given system of inequalities can be represented on number line as

Question (9)

3x – 7 > 2(x – 6), 6 – x > 11 – 2x

Solution

3x – 7 > 2(x – 6)
⇒ 3x – 7 > 2x – 12
⇒ 3x – 2x > – 12 + 7
⇒ x > –5 … (1)
6 – x > 11 – 2x
⇒ –x + 2x > 11 – 6
⇒ x > 5 … (2)
From (1) and (2), it can be concluded that the solution set for the given system of inequalities is (5, ∞). The solution of the given system of inequalities can be represented on number line as

Question (10)

5(2x – 7) – 3(2x + 3) ≤ 0, 2x + 19 ≤ 6x + 47

Solution

5(2x – 7) – 3(2x + 3) ≤ 0
rArr; 10x – 35 – 6x – 9 ≤ 0
rArr; 4x – 44 ≤ 0
rArr; 4x ≤ 44
rArr; x ≤ 11 ... (1)
2x + 19 ≤ 6x + 47
rArr; 19 – 47 ≤ 6x – 2x
rArr; –28 ≤ 4x
rArr; –7 ≤ x ... (2)
From (1) and (2), it can be concluded that the solution set for the given system of inequalities is [–7, 11].
The solution of the given system of inequalities can be represented on number line as

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

Since the solution is to be kept between 68°F and 77°F, 68 < F < 77
Putting $F = \frac{9}{5}C + 32$ we obtained
\[68 < \frac{9}{5}C + 32 < 77\] \[68 - 32 < \frac{9}{5}C < 77 - 32\] \[36 < \frac{9}{5}C < 45\] \[36 \times \frac{5}{9} < C < 45 \times \frac{5}{9}\] \[ \Rightarrow 20 < C < 25\] Thus, the required range of temperature in degree Celsius is between 20°C and 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

Let x litres of 2% boric acid solution is required to be added.
Then, total mixture = (x + 640) litres
This resulting mixture is to be more than 4% but less than 6% boric acid.
∴ 2%x + 8% of 640 > 4% of (x + 640)
And, 2% x + 8% of 640 < 6% of (x + 640)
2%x + 8% of 640 > 4% of (x + 640)
\[ \Rightarrow \frac{2}{{100}}x + \frac{8}{{100}}\left( {640} \right) > \frac{4}{{100}}\left( {x + 640} \right)\] ⇒ 2x + 5120 > 4x + 2560
⇒ 5120 – 2560 > 4x – 2x
⇒ 5120 – 2560 > 2x
⇒ 2560 > 2x
⇒ 1280 > x
2% x + 8% of 640 < 6% of (x + 640) \[ \frac{2}{{100}}x + \frac{8}{{100}}\left( {640} \right) < \frac{6}{{100}}\left( {x + 640} \right)\] ⇒ 2x + 5120 < 6x + 3840
⇒ 5120 – 3840 < 6x – 2x
⇒ 1280 < 4x
⇒ 320 < x
∴ 320 < x < 1280
Thus, the number of litres of 2% of boric acid solution that is to be added will have to 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

Let x litres of water is required to be added. Then, total mixture = (x + 1125) litres It is evident that the amount of acid contained in the resulting mixture is 45% of 1125 litres. This resulting mixture will contain more than 25% but less than 30% acid content. ∴30% of (1125 + x) > 45% of 1125 And, 25% of (1125 + x) < 45% of 1125 30% of (1125 + x) > 45% of 1125 \[ \Rightarrow \frac{{30}}{{100}}\left( {1125 + x} \right) > \frac{{45}}{{100}} \times 1125\] \[ \Rightarrow 30\left( {1125 + x} \right) > 45 \times 1125\] \[ \Rightarrow 30 \times 1125 + 30x > 45 \times 1125\] \[ \Rightarrow 30x > 45 \times 1125 - 30 \times 1125\] \[ \Rightarrow 30x > \left( {45 - 30} \right) \times 1125\] \[ \Rightarrow x > \frac{{15 \times 1125}}{{30}} = 562.5\] \[25\% of\left( {1125 + x} \right) < 45\% of1125\] \[\frac{{25}}{{100}}\left( {1125 + x} \right) < \frac{{45}}{{100}}\left( {1125} \right)\] \[1125 \times 25 + 25x < 1125 \times 45\] \[25x < 1125 \times 45 - 1125 \times 25\] \[25x < 1125\left( {45 - 25} \right)\] \[x < \frac{{1125 \times 20}}{{25}}\] \[x < 45 \times 20\] \[x < 900\] ∴ 562.5 < x < 900
Thus, the required number of litres of water that is to be added will have to be more than 562.5 but less than 900.

Question (14)

IQ of a person is given by the formula $IQ = \frac{{MA}}{{CA}} \times 100$
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

It is given that for a group of 12 years old children, 80 ≤ IQ ≤ 140 ... (i)
For a group of 12 years old children, CA = 12 years
\[IQ = \frac{{MA}}{{12}} \times 100\] Putting this value of IQ in (i), we obtain \[80 \le \frac{{MA}}{{12}} \times 100 \le 140\] \[ \Rightarrow 80 \times \frac{{100}}{{12}} \le MA \le 140 \times \frac{{100}}{{12}}\] \[ \Rightarrow 9.6 \le MA \le 16.8\] Thus, the range of mental age of the group of 12 years old children is 9.6 ≤ MA ≤ 16. 8