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

Question 1: Solve the inequality $2\le 3x-4\le 5$

Answer:

Given : $2\le 3x-4\le 5$

$2\le 3x-4\le 5$

$\Rightarrow 2+4\le 3x\le 5+4$

$\Rightarrow 6\le 3x\le 9$

$\Rightarrow \frac{6}{3}\le x\le \frac{9}{3}$

$\Rightarrow 2\le x\le 3$

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

Answer:

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

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

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

$\Rightarrow -2\ge (2x-4)>-4$

$\Rightarrow -2+4\ge 2x>-4+4$

$\Rightarrow 2\ge 2x>0$

$\Rightarrow 1\ge x>0$

Solution set is $(01]$

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

Answer:

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

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

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

$\Rightarrow -7\le -\frac{7x}{2}\le 14$

$\Rightarrow 7\ge \frac{7x}{2}\ge -14$

$\Rightarrow 7\times 2\ge 7x\ge -14\times 2$

$\Rightarrow 14\ge 7x\ge -28$

$\Rightarrow \frac{14}{7}\ge x\ge \frac{-28}{7}$

$\Rightarrow 2\ge x\ge -4$

Solution set is $[-4,2]$

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

Answer:

Given The inequality

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

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

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

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

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

$\Rightarrow -25<(x-2)\le 0$

$\Rightarrow -25+2<x\le 0+2$

$\Rightarrow -23<x\le 2$

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

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

Answer:

Given the inequality

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

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

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

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

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

$\Rightarrow -16\times 5<3x\le -2\times 5$

$\Rightarrow -80<3x\le -10$

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

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

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

Answer:

Given the linear inequality

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

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

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

$\Rightarrow 14\le (3x+11)\le 22$

$\Rightarrow 14-11\le (3x)\le 22-11$

$\Rightarrow 3\le 3x\le 11$

$\Rightarrow 1\le x\le \frac{11}{3}$

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

$5x+1>-24and5x-1<24$

$\Rightarrow 5x>-24-1and5x<24+1$

$\Rightarrow 5x>-25and5x<25$

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

$\Rightarrow x>-5andx<5$

$(-5,5)$

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

$2(x-1)<x+5and3(x+2)>2-x$

$\Rightarrow 2x-2<x+5and3x+6>2-x$

$\Rightarrow 2x-x<2+5and3x+x>2-6$

$\Rightarrow x<7and4x>-4$

$\Rightarrow x<7andx>-1$

$(-1,7)$

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

$3x-7>2(x-6)and6-x>11-2x$

$\Rightarrow 3x-7>2x-12and6-x>11-2x$

$\Rightarrow 3x-2x>7-12and2x-x>11-6$

$\Rightarrow x>-5andx>5$

$x\in (5,\mathrm{\infty })$

The solution graphically on the number line is as shown :

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

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

Answer:

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

$5(2x-7)-3(2x+3)\le 0and2x+19\le 6x+47$

$\Rightarrow 10x-35-6x-9\le 0and2x-6x\le 47-19$

$\Rightarrow 4x-44\le 0and-4x\le 28$

$\Rightarrow 4x\le 44and4x\ge -28$

$\Rightarrow x\le 11andx\ge -7$

$x\in [-7,11]$

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.

$\therefore 2\mathrm{\%}x+8\mathrm{\%}of640>4\mathrm{\%}of(640+x)$ and $2\mathrm{\%}x+8\mathrm{\%}of640<6\mathrm{\%}of(x+640)$

$\Rightarrow 2\mathrm{\%}x+8\mathrm{\%}of640>4\mathrm{\%}of(640+x)$ and $2\mathrm{\%}x+8\mathrm{\%}of640<6\mathrm{\%}of(x+640)$

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

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

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

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

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

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.

$\therefore 30\mathrm{\%}of(1125+x)>45\mathrm{\%}of(1125)$ and $25\mathrm{\%}of(1125+x)<45\mathrm{\%}of1125$

$\Rightarrow 30\mathrm{\%}of(1125+x)>45\mathrm{\%}of(1125)$ and $25\mathrm{\%}of(1125+x)<45\mathrm{\%}of1125$

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

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

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

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

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

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

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 $80\le IQ\le 140$ for a group of 12 years old children, find the range of their mental age.

Answer:

Given that group of 12 years old children.

$80\le IQ\le 140$

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

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

Putting the value of IQ, we obtain

$80\le IQ\le 140$

$\Rightarrow 80\le \frac{MA}{CA}\times 100\le 140$

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

$\Rightarrow 80\times 12\le MA\times 100\le 140\times 12$

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

$\Rightarrow 9.6\le MA\le 16.8$

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