NCERT Solutions for Miscellaneous Exercise Chapter 5 Class 11 - Linear Inequalities

Question 1: Solve the inequality $2\leq 3x-4\leq5$

Answer:

Given : $2\leq 3x-4\leq5$

$2\leq 3x-4\leq5$

$\Rightarrow\, \, 2+4\leq 3x\leq 5+4$

$\Rightarrow\, \, 6\leq 3x\leq 9$

$\Rightarrow\, \, \frac{6}{3}\leq x\leq \frac{9}{3}$

$\Rightarrow\, \, 2\leq x\leq 3$

Thus, all the real numbers greater than equal to 2 and less than equal to 3 are solutions to this inequality.

Solution set is $[2,3]$

Question 2: Solve the inequality $6 \leq -3(2x - 4) < 12$

Answer:

Given $6 \leq -3(2x - 4) < 12$

$6 \leq -3(2x - 4) < 12$

$\Rightarrow\, \ \frac{6}{3}\leq -(2x-4)< \frac{12}{3}$

$\Rightarrow\, \ -2\geq (2x-4)> -4$

$\Rightarrow\, \ -2+4\geq 2x> -4+4$

$\Rightarrow\, \ 2\geq 2x> 0$

$\Rightarrow\, \ 1\geq x> 0$

Solution set is $(01]$

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

Answer:

Given $-3 \leq 4 - \frac{7x}{2}\leq 18$

$\Rightarrow \, \, -3 \leq 4 - \frac{7x}{2}\leq 18$

$\Rightarrow \, \, -3-4 \leq - \frac{7x}{2}\leq 18-4$

$\Rightarrow \, \, -7 \leq - \frac{7x}{2}\leq 14$

$\Rightarrow \, \, 7 \geq \frac{7x}{2} \geq -14$

$\Rightarrow \, \, 7\times 2 \geq 7x\geq -14\times 2$

$\Rightarrow \, \, 14 \geq 7x \geq -28$

$\Rightarrow \, \, \frac{14}{7} \geq x \geq \frac{-28}{7}$

$\Rightarrow \, \, 2 \geq x \geq -4$

Solution set is $[-4,2]$

Question 4: Solve the inequality $-15 < \frac{3(x-2)}{5} \leq 0$

Answer:

Given The inequality

$-15 < \frac{3(x-2)}{5} \leq 0$

$-15 < \frac{3(x-2)}{5} \leq 0$

$\Rightarrow\, \ -15\times 5< 3(x-2)\leq 0\times 5$

$\Rightarrow\, \ -75< 3(x-2)\leq 0$

$\Rightarrow\, \ \frac{-75}{3}< (x-2)\leq \frac{0}{3}$

$\Rightarrow\, \ -25< (x-2)\leq 0$

$\Rightarrow\, \ -25+2< x\leq 0+2$

$\Rightarrow\, \ -23< x\leq 2$

The solution set is $(-23,2]$

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

Answer:

Given the inequality

$-12<4-\frac{3x}{-5} \leq 2$

$-12<4-\frac{3x}{-5} \leq 2$

$\Rightarrow\, \, -12-4< -\frac{3x}{-5}\leq 2-4$

$\Rightarrow\, \, -16< -\frac{3x}{-5}\leq -2$

$\Rightarrow\, \, -16< \frac{3x}{5}\leq -2$

$\Rightarrow\, \, -16\times 5< 3x\leq -2\times 5$

$\Rightarrow\, \, -80< 3x\leq -10$

$\Rightarrow\, \, \frac{-80}{3}< 3x\leq \frac{-10}{3}$

Solution set is $(\frac{-80}{3}, \frac{-10}{3}]$

Question 6: Solve the inequality $7 \leq \frac{(3x+ 11)}{2}\leq 11$

Answer:

Given the linear inequality

$7 \leq \frac{(3x+ 11)}{2}\leq 11$

$7 \leq \frac{(3x+ 11)}{2}\leq 11$

$\Rightarrow \, \, 7\times 2 \leq (3x+ 11)\leq 11\times 2$

$\Rightarrow \, \, 14 \leq (3x+ 11)\leq 22$

$\Rightarrow \, \, 14-11 \leq (3x)\leq 22-11$

$\Rightarrow \, \, 3 \leq 3x\leq 11$

$\Rightarrow \, \, 1 \leq x\leq \frac{11}{3}$

The solution set of the given inequality is $[1,\frac{11}{3}]$

Question 7: Solve the inequality and represent the solution graphically on number line. $5x + 1 > -24,\ 5x - 1 <24$

Answer:

Given : $5x + 1 > -24,\ 5x - 1 <24$

$5x + 1 > -24\, \, \, \, \, \, \, and\, \, \, \, \, \, \ 5x - 1 <24$

$\Rightarrow 5x > -24-1\, \, \, \, \, \, \, and\, \, \, \, \, \, \ 5x <24+1$

$\Rightarrow 5x > -25\, \, \, \, \, \, \, and\, \, \, \, \, \, \ 5x <25$

$\Rightarrow x > \frac{-25}{5}\, \, \, \, \, \, \, and\, \, \, \, \, \, \ x <\frac{25}{5}$

$\Rightarrow x > -5\, \, \, \, \, \, \, and\, \, \, \, \, \, \ x <5$

$(-5,5)$

The solution graphically on the number line is as shown :

Question 8: Solve the inequality and represent the solution graphically on number line. $2(x-1)<x+5,\ 3(x+2)> 2 -x$

Answer:

Given : $2(x-1)<x+5,\ 3(x+2)> 2 -x$

$2(x-1)<x+5\, \, \, \, and\, \, \, \, \, \ 3(x+2)> 2 -x$

$\Rightarrow \, \, 2x-2<x+5\, \, \, \, and\, \, \, \, \, \ 3x+6> 2 -x$

$\Rightarrow \, \, 2x-x<2+5\, \, \, \, and\, \, \, \, \, \ 3x+x> 2 -6$

$\Rightarrow \, \, x<7\, \, \, \, and\, \, \, \, \, \ 4x> -4$

$\Rightarrow \, \, x<7\, \, \, \, and\, \, \, \, \, \ x> -1$

$(-1,7)$

The solution graphically on the number line is as shown :

Question 9: Solve the inequality and represent the solution graphically on number line. $3x - 7 > 2(x-6),\ 6-x > 11 - 2x$

Answer:

Given : $3x - 7 > 2(x-6),\ 6-x > 11 - 2x$

$3x - 7 > 2(x-6)\, \, \, \, and\, \, \, \, \, \ 6-x > 11 - 2x$

$\Rightarrow \, \, 3x - 7 > 2x-12\, \, \, \, and\, \, \, \, \, \ 6-x > 11 - 2x$

$\Rightarrow \, \, 3x - 2x >7-12\, \, \, \, and\, \, \, \, \, \ 2x-x > 11 - 6$

$\Rightarrow \, \, x >-5\, \, \, \, and\, \, \, \, \, \ x > 5$

$x\in (5,\infty )$

The solution graphically on the number line is as shown :

Question 10: Solve the inequality and represent the solution graphically on number line.

$5(2x-7)-3(2x+3)\leq 0,\quad 2x + 19 \leq 6x +47$

Answer:

Given : $5(2x-7)-3(2x+3)\leq 0,\quad 2x + 19 \leq 6x +47$

$5(2x-7)-3(2x+3)\leq 0\, \, \, \, \, and\, \, \, \, \, \, \, \quad 2x + 19 \leq 6x +47$

$\Rightarrow \, \, 10x-35-6x-9\leq 0\, \, \, \, \, and\, \, \, \, \, \, \, \quad 2x -6x\leq 47-19$

$\Rightarrow \, \, 4x-44\leq 0\, \, \, \, \, and\, \, \, \, \, \, \, \quad -4x\leq 28$

$\Rightarrow \, \, 4x\leq 44\, \, \, \, \, and\, \, \, \, \, \, \, \quad 4x\geq - 28$

$\Rightarrow \, \, x\leq 11\, \, \, \, \, and\, \, \, \, \, \, \, \quad x\geq - 7$

$x\in [-7,11]$

The solution graphically on the number line is as shown :

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:

Since the solution is to be kept between 68° F and 77° F.

$68< F< 77$

Putting the value of $F = \frac{9}{5}C + 32$ , we have

$\Rightarrow \, \, \, 68< \frac{9}{5}C + 32< 77$

$\Rightarrow \, \, \, 68-32< \frac{9}{5}C < 77-32$

$\Rightarrow \, \, \, 36< \frac{9}{5}C < 45$

$\Rightarrow \, \, \, 36\times 5< 9C < 45\times 5$

$\Rightarrow \, \, \, 180< 9C < 225$

$\Rightarrow \, \, \, \frac{180}{9}< C < \frac{225}{9}$

$\Rightarrow \, \, \, 20< C < 25$

the range in temperature in degree Celsius (C) is between 20 to 25.

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 is required to be added.

Total mixture = (x+640) litres

The resulting mixture is to be more than 4% but less than 6% boric acid.

$\therefore \, 2\%x+8\%\, of\, 640> 4\%\, of\, (640+x)$ and $2\%x+8\%\, of\, 640< 6\%\, of\, (x+640)$

$\Rightarrow \, 2\%x+8\%\, of\, 640> 4\%\, of\, (640+x)$ and $2\%x+8\%\, of\, 640< 6\%\, of\, (x+640)$

$\Rightarrow \, \frac{2}{100}x+(\frac{8}{100}) 640> \frac{4}{100} (640+x)$ $\Rightarrow \, \frac{2}{100}x+(\frac{8}{100}) 640< \frac{6}{100} (640+x)$

$\Rightarrow \, 2x+5120> 4x+2560$ $\Rightarrow \, 2x+5120< 6x+3840$

$\Rightarrow \, 5120-2560> 4x-2x$ $\Rightarrow \, 5120-3840< 6x-2x$

$\Rightarrow \, 2560> 2x$ $\Rightarrow \, 1280< 4x$

$\Rightarrow \, 1280> x$ $\Rightarrow \, 320< x$

Thus, the number of litres 2% of boric acid solution that is to be added will have to be more than 320 and 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?

Answer:

Let x litres of water is required to be added.

Total mixture = (x+1125) litres

It is evident that amount of acid contained in the resulting mixture is 45% of 1125 litres.

The resulting mixture contain more than 25 % but less than 30% acid.

$\therefore \, 30\%\, of\, (1125+x) > 45\%\, of\, (1125)$ and $25\%\, of\, (1125+x)< 45\%\, of\, 1125$

$\Rightarrow \, 30\%\, of\, (1125+x) > 45\%\, of\, (1125)$ and $25\%\, of\, (1125+x)< 45\%\, of\, 1125$

$\Rightarrow \, \frac{30}{100}(1125+x)> \frac{45}{100} (1125)$ $\Rightarrow \, (\frac{25}{100}) (1125+x)< \frac{45}{100} (1125)$

$\Rightarrow \, 30\times 1125+30x> 45\times (1125)$ $\Rightarrow \, 25 (1125+x)< 45(1125)$

$\Rightarrow \, 30x> (45-30)\times (1125)$ $\Rightarrow \, 25 x< (45-25)1125$

$\Rightarrow \, 30x> (15)\times (1125)$ $\Rightarrow \, 25 x< (20)1125$

$\Rightarrow \, x> \frac{15\times 1125}{30}$ $\Rightarrow \, x< \frac{20\times 1125}{25}$

$\Rightarrow \, x> 562.5$ $\Rightarrow \, x< 900$

Thus, the number of litres water that is to be added will have to be more than 562.5 and less than 900 litres.

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 $80\leq IQ\leq140$ for a group of 12 years old children, find the range of their mental age.

Answer:

Given that group of 12 years old children.

$80\leq IQ\leq140$

For a group of 12 years old children, CA =12 years

$IQ= \frac{MA}{CA}\times 100$

Putting the value of IQ, we obtain

$80\leq IQ\leq140$

$\Rightarrow \, \, 80\leq \frac{MA}{CA}\times 100\leq140$

$\Rightarrow \, \, 80\leq \frac{MA}{12}\times 100\leq140$

$\Rightarrow \, \, 80\times 12\leq MA\times 100\leq140\times 12$

$\Rightarrow \, \, \frac{80\times 12}{100}\leq MA\leq \frac{140\times 12}{100}$

$\Rightarrow \, \, 9.6\leq MA\leq 16.8$

Thus, the range of mental age of the group of 12 years old children is $\, \, 9.6\leq MA\leq 16.8$