CBSE Class 11 Maths Important Questions - Chapter 5 Linear Inequalities

1 Mark Questions

1. Solve  $ \dfrac{3x-4}{2}\ge \dfrac{x+1}{4}-1 $.

Ans: Rewrite the given inequality.

$\dfrac{3x-4}{2}\ge \dfrac{x+1}{4}-\dfrac{1}{1}$ 

$\dfrac{3x-4}{2}\ge \dfrac{x+1-4}{4}$ 

$\dfrac{3x-4}{2}\ge \dfrac{x-3}{4}$ 

Multiply the left-hand side with 2 and the right-hand side by 4.

$2\left( 3x-4 \right)\ge \left( x-3 \right)$ 

$6x-8\ge x-3$ 

Subtract  $ x $  from both sides. Further, add 8 to both sides.

$6x-8-x+8\ge x-3-x+8$ 

$5x\ge 5$ 

Divide both sides by 5.

 $ x\ge 1 $ 

Hence, the solution set is  $ \left[ 1,\infty  \right) $ .

2. Solve  $ 3x+8>2 $  when  $ x $  is a real number.

Ans: Given,  $ 3x+8>2 $ . 

Subtract 2 from both sides.

$3x+8-8>2-8$ 

$3x>-6$  

Divide both parts of the inequality by 3.

$\dfrac{3x}{3}>\dfrac{-6}{3}$ 

$x>-2$  

Hence, the solution set is  $ \left( -2,\infty  \right) $ .

3. Solve the inequality  $ \dfrac{x}{4}<\dfrac{5x-2}{3}-\dfrac{7x-3}{5} $ .

Ans: Given,  $ \dfrac{x}{4}<\dfrac{5x-2}{3}-\dfrac{7x-3}{5} $ . 

Take LCM of the terms at the right-hand side of the inequality. 

$\dfrac{x}{4}<\dfrac{5\left( 5x-2 \right)-3\left( 7x-3 \right)}{15}$ 

$\dfrac{x}{4}<\dfrac{25x-10-21x+9}{15}$ 

$\dfrac{x}{4}<\dfrac{4x-1}{15}$ 

Cross-multiply the terms.

$15x<4\left( 4x-1 \right)$ 

$15x<16x-4$ 

Subtract  $ 16x $ from both sides.

$15x-16x<16x-4-16x$ 

$-x<-4$ 

$x>4$ 

Hence, the solution set is $ \left( 4,\infty  \right) $ .

4. If  $ 4x>-16 $ , then  $ x\,\,\,-4 $ .

Ans: Divide both sides of the inequality,  $ 4x>-16 $ by 4.

$\dfrac{4x}{4}>\dfrac{-16}{4}$ 

$x>-4$ 

Hence,  $ x\,\,-4 $ 

5. Solve the inequality $ \dfrac{1}{2}\left( \dfrac{3x}{5}+4 \right)\ge \dfrac{1}{3}\left( x-6 \right) $ .

Ans: Given,  $ \dfrac{1}{2}\left( \dfrac{3x}{5}+4 \right)\ge \dfrac{1}{3}\left( x-6 \right) $ .

Solve the parenthesis.

$ \dfrac{3x}{10}+2\ge \dfrac{x}{3}-2 $ 

Subtract 2 from both sides and add  $ \dfrac{x}{3} $  to both sides and take LCM. 

$\dfrac{3x}{10}-\dfrac{x}{3}\ge -4$ 

$\dfrac{9x-10x}{30}\ge -4$ 

$\dfrac{-x}{30}\ge -4$  

Multiply both sides by 30. 

$-x\ge -4\left( 30 \right)$ 

$-x\ge -120$ 

$x\le 120$

Hence, the solution set is $ \left( -\infty ,120 \right]$.

6. Solve the inequalities,   $ 2x-1\le 3 $  and  $ 3x+1\ge -5 $  is.

Ans: Given,  $ 2x-1\le 3 $ and  $ 3x+1\ge -5 $ .

Solve the equation,  $ 2x-1\le 3 $. 

$2x-1+1\le 3+1$ 

$2x\le 4$ 

$x\le 2$ 

Solve the equation, $ 3x+1\ge -5$.

$3x+1\ge -5$ 

$3x\ge -6$ 

$x\ge -2$ 

From both the solutions it is concluded that,  $ -2\le x\le 2 $ . Hence, the solution set is          $ \left[ -2,2 \right] $ .

7. Solve  $ 7x+3<5x+9 $. Show the graph of the solution on the number line.

Ans: Given,  $ 7x+3<5x+9 $ .

Subtract  $ 5x $  and 3 from both sides. 

$7x+3-5x-3<5x-5x+9-3$ 

$2x<6$

Divide both sides by 2.

$ x<3 $ 

The graph of the solution of the given inequality is represented by the red color in the number line shown below.

8. Solve the inequality, $ \dfrac{2x-1}{3}\ge \dfrac{3x-2}{4}-\dfrac{2-x}{5} $ .     

Ans: Given,  $ \dfrac{2x-1}{3}\ge \dfrac{3x-2}{4}-\dfrac{2-x}{5} $ .

Take the LCM of the terms at the right-hand side.

$\dfrac{2x-1}{3}\ge \dfrac{5\left( 3x-2 \right)-4\left( 2-x \right)}{20}$ 

$\dfrac{2x-1}{3}\ge \dfrac{15x-10-8+4x}{20}$ 

$\dfrac{2x-1}{3}\ge \dfrac{19x-18}{20}$ 

Cross-multiply. 

$20\left( 2x-1 \right)\ge 3\left( 19x-18 \right)$ 

$40x-20\ge 57x-54$ 

$40x-57x\ge -54-20$ 

$-17x\ge -34$

Divide both sides by  $ -17 $ .

$ x\le 2 $ 

Hence, the solution set is  $ \left( -\infty ,2 \right] $ .

9. Solve  $ 5x-3\le 3x+1 $  when  $ x $  is an intege 

Ans: Given,  $ 5x-3\le 3x+1 $ .

Subtract  $ 3x $  from both sides and add  $ 3 $  to both sides.

$5x-3-3x+3\le 3x+1+3-3x$ 

$2x\le 4$ 

$x\le 2$ 

Hence, the solution set is  $ \left\{ \ldots .,-3,-2,-1,0,1,2 \right\} $ .

10. Solve  $ 30x<200 $ when  $ x $  is a natural number

Ans: Given,  $ 30x<200 $ .

Divide both sides by 30.

$x<\dfrac{200}{30}$ 

$x<\dfrac{20}{3}$  

Hence, the solution set that satisfies the given inequality is  $ \left\{ 1,2,3,4,5,6 \right\}$ . 

11.  Solve the inequality  $ \dfrac{x}{2}\ge \dfrac{5x-2}{3}-\dfrac{7x-3}{5} $ .

Ans: Given,  $ \dfrac{x}{2}\ge \dfrac{5x-2}{3}-\dfrac{7x-3}{5} $ .

Take the LCM of the terms at the right-hand side. 

$\dfrac{x}{2}\ge \dfrac{5\left( 5x-2 \right)-3\left( 7x-3 \right)}{15}$ 

$\dfrac{x}{2}\ge \dfrac{25x-10-21x+9}{15}$ 

$\dfrac{x}{2}\ge \dfrac{4x-1}{15}$  

Cross-multiply the terms. 

$15x\ge 2\left( 4x-1 \right)$ 

$15x\ge 8x-2$ 

$15x-8x\ge -2$ 

$7x\ge -2$  

Divide both sides by 7.

$ x\ge \dfrac{-2}{7} $ 

Hence, the solution set that satisfies the given inequality is  $ \left( \dfrac{-2}{7},\infty  \right) $ .

12.  Solve  $ 5x-3<3x+1 $  where  $ x $  is an integer.

Ans: Given,  $ 5x-3<3x+1 $ .

Isolate the like terms to one side of the inequality. 

$5x-3x<1+3$ 

$2x<4$ 

Divide both sides by 2.

$ x<2 $ 

Hence, when  $ x $  is an integer the solutions of the given inequality are,  $ \left\{ \ldots ,-4,-3,-2,-1,0,1 \right\} $.

4 Mark Questions 

Refer to Page 6-12 for 4 Mark Questions in the PDF

1. Solve  $ 3x-6\ge 0 $  graphically.

Ans: The corresponding equality of the given inequality is  $ 3x-6=0 $ .

Solve the equation,  $ 3x-6=0 $  for  $ x $ .

$3x=6$ 

$x=2$

Now, find the region portioned by the inequality by substituting  $ \left( 0,0 \right) $  in $ 3x-6\ge 0 $.

 $3\left( 0 \right)-6\ge 0$ 

$-6\ge 0$ 

Thus, the inequality is false.

Since the inequality is false then, the solution set will not include the portion of  $ x=2 $ which contains the origin. 

The graph of the inequality  $ 3x-6\ge 0 $  is,

2. Ravi obtained 70 and 75 marks in the first unit test. Find the minimum marks he should get in the third test to have an average of at least 60 marks.

Ans: Let the marks secured by Ravi be  $ x $.

According to the question, 

$\dfrac{70+75+x}{3}\ge 60$ 

$\dfrac{145+x}{3}\ge 60$ 

Multiply the inequality obtained by 3.

$145+x\ge 60\left( 3 \right)$ 

$145+x\ge 180$ 

Subtract 145 from both sides. 

$x\ge 180-145$ 

$x\ge 35$  

The minimum mark he should score in the third test is 35.

3. Find all pairs of consecutive odd natural numbers both of which are larger than 10 such that their sum is less than 40.

Ans: Let  $ x $  and  $ x+2 $  be consecutive odd natural number

Since the numbers are larger than 10, then,  $ x>10 $ . Let this equation be  $ \left( 1 \right) $ .

According to the question, 

$x+\left( x+2 \right)<40$ 

$2x+2<40$ 

Subtract 2 from both sides.

$ 2x<38 $ 

Divide both sides by 2.

$ x<19 $ 

Let this equation be  $ \left( 2 \right) $ .

From  $ \left( 1 \right) $  and  $ \left( 2 \right) $ . The pairs obtained are,

$ \left( 11,13 \right) $ ,  $ \left( 13,15 \right) $ , $ \left( 15,17 \right) $ , $ \left( 17,19 \right) $ 

Hence, the required pairs are $ \left( 11,13 \right) $ ,  $ \left( 13,15 \right) $ , $ \left( 15,17 \right) $ , and $ \left( 17,19 \right) $ .

4.  A company manufactures cassettes and its cost equation for a week  $ C=300+1.5x $  and its revenue equation is  $ R=2x $, where  $ x $  is the number of cassettes sold in a week. How many cassettes must be sold by the company to get some profit?

Ans: The profit equation is  $ \text{Profit=Revnue}-\text{Cost} $ .

To get the profit, revenue should be more than the cost. Thus,  $ R>C $ .

Substitute  $ R $  as  $ 2x $  and  $ C $  as  $ 300+1.5x $  in the inequality,  $ R>C $

$2x>300+1.5x$ 

$2x-1.5x>300$ 

$0.5x>300  $ 

Divide both sides by  $ 0.5 $.

 $x>\dfrac{300}{0.5}$ 

$x>600$ 

5. The longest side of a  $ \Delta  $  is 3 times the shortest side and the third side is 2 cm shorter than the longest side. If the perimeter of the  $ \Delta  $  is at least 61 cm, find the minimum length of the shortest side.

Ans: Let the length of the shortest side be  $ x $  cm, the longest side be  $ 3x $  cm and the third side be  $ \left( 3x-2 \right) $  cm.  

According to the question,

$\left( x \right)+\left( 3x \right)+\left( 3x-2 \right)\ge 61$ 

$7x-2\ge 61$ 

$7x\ge 61+2$ 

$7x\ge 63$ 

Divide both sides by 7.

$ x\ge 9 $ 

Hence, the shortest side is 9 cm.

6. In drilling the world's deepest hole, it was found that the temperature T in degree Celsius,  $ x $  km below the surface of the earth was given by  $T=30+25\left( x-3 \right) $ ,  $3 < x < 15$ . At what depth will the temperature be between  $ 200{}^\circ  $  C and  $ 300{}^\circ  $  C.

Ans: Let  $ x $  km be the depth where the temperature lies between  $ 200{}^\circ$  C and  $ 300{}^\circ  $ C. Here,  $200^o C<30+25\left( x-3 \right)<300^o C$ .

$200<30+25\left( x-3 \right)<300$ 

$200<30+25x-75<300$ 

$200<25x-45<300$ 

Solve the inequality.

$\dfrac{49}{5} < x < \dfrac{69}{5}$ 

$9.8 < x < 13.8$ 

7. A man wants to cut three lengths from a single piece of board of length 91 cm. The second length is to be 3 cm longer than the shortest and the third length is to be twice as long as the shortest. What are the possible lengths of the shortest board if the third piece is to be at least 5 cm longer than the second? 

Ans: Let the length of the shortest piece be  $ x $  cm, second piece be  $ \left( x+3 \right) $  cm and the length of the third piece be  $ 2x $  cm.

The total length of the piece is 91 cm. According to the question, 

$x+\left( x+3 \right)+2x\le 91$ 

$4x+3\le 91$ 

$4x\le 91-3$ 

$4x\le 88$ 

Divide both sides by 4.

$x\le \dfrac{88}{4}$ 

$x\le 22$ 

Now, the third piece should be at least 5 cm longer than the second. 

Hence, 

$2x\ge 5+\left( x+3 \right)$ 

$2x-x\ge 5+3$ 

$x\ge 8$  

From  $ x\le 22 $  and  $ x\ge 8 $ ,

$ 8\le x\le 22 $ 

Hence,  $ x\in \left[ 8,22 \right] $ .

8. The water acidity in a pool is considered normal when the average pH reading of three daily measurements is between  $ 7.2 $  and  $ 7.8 $ . If the first pH reading is  $ 7.48 $  and  $ 7.85 $ , find the range of pH value for the third reading that will result in the acidity level is normal.

Ans: Let the third reading of the pH level be  $ x $ . Then, 

$ 7.2<\dfrac{7.48+7.85+x}{3}<7.8 $ 

Multiply all the parts of the inequality by 3.

$21.6 < 15.33+ x < 23.4$ 

$6.27 < x < 7.07$ 

9. A plumber can be paid under two schemes as given below.

 $ \text{I}\text{.} $ Rs 600 and Rs 50 per hr.

 $ \text{II}\text{.} $ Rs 170 per hr.

If the job takes  $ n $  hour for what values of  $ n $  does the scheme  $ \text{I} $  gives the plumber the better wages.

Ans: The total wage of the labor in scheme  $ \text{I} $ is  $ 600+50n $  and the total wage in scheme  $ \text{II} $  is  $ 170n $ . According to the question,

$600+50n>170n$ 

$50n-170n>-600$ 

$-120n>-600$ 

$n<5$

Thus, for better wages, the working hours should be less than 5 hours.

10.  Solve the inequality $ 3x+4y\le 12 $  graphically.

Ans: Given,  $ 3x+4y\le 12 $ .

The corresponding equality is  $ 3x+4y=12 $ .

The two coordinates through which the line passes are,

Put  $ \left( 0,0 \right) $  in the equation,  $ 3x+4y\le 12$. 

$3\left( 0 \right)+4\left( 0 \right)\le 12$ 

$0\le 12$ 

The inequality is true. Thus, the solution region of the given inequality will be that portion of the graph of  $ 3x+4y=12 $ that contains the origin.

11. Solve graphically  $ x-y\le 0 $ .

 Ans: Given,  $ x-y\le 0 $ .

The corresponding equality is  $ x-y=0 $ . Thus,  $ x=y $ . It will be the straight line passing through the origin.

Put  $ \left( 1,0 \right) $  in inequality,  $ x-y\le 0 $. 

$1-0\le 0$ 

$1\le 0$ 

The inequality is false. Thus, the solution region of the given inequality will be that portion of the graph of  $ x=y $  that is away from the point  $ \left( 1,0 \right) $ .

12. Solve the inequality  $ 3x+2y>6 $  graphically.

Ans: Given,  $ 3x+2y>6 $ .

The corresponding equality is  $ 3x+2y=6 $ .

The two coordinates through which the line passes are,

Put  $ \left( 0,0 \right) $  in the equation,  $ 3x+2y>6 $.

$3\left( 0 \right)+2\left( 0 \right)>6$ 

$0>6$ 

The inequality is false. Thus, the solution region of the given inequality will be that portion of the graph of  $ 3x+2y=6 $  that does not contain the point  $ \left( 0,0 \right) $ . 

6 Mark Questions

Refer to Page 13-18 for 6 Mark Questions in the PDF

1. The IQ of a person is given by the formula  $ \text{IQ=}\dfrac{\text{MA}}{\text{CA}}\times 100 $  where MA is mental age and CA is chronological age. If  $ 80\le \text{IQ}\le 140 $  for a group of 12 yr old children, find the range of their mental age.

Ans: Given,  $ 80\le \text{IQ}\le 140 $ .

Substitute  $ \text{IQ} $ as  $ \dfrac{\text{MA}}{\text{CA}}\times 100 $ .

$ 80\le \dfrac{\text{MA}}{\text{CA}}\times 100\le 140 $ 

The chronological age,  $ \text{CA} $ is 12 yr. Thus, substitute  $ \text{CA} $ as 12.

$ 80\le \dfrac{\text{MA}}{\text{12}}\times 100\le 140 $ 

Multiply the inequality obtained by  $ \dfrac{12}{100} $ .

$80\times \dfrac{12}{100}\le \dfrac{\text{MA}}{{12}}\times {100}\times \dfrac{{12}}{{100}}\le 140\times \dfrac{12}{100}$ 

$80\times \dfrac{12}{100}\le \text{MA}\le 140\times \dfrac{12}{100}$ 

$\dfrac{96}{100}\le \text{MA}\le \dfrac{168}{10}$ 

$9.6\le \text{MA}\le 16.8$  

Hence, the required range is  $ 9.6\le \text{MA}\le 16.8 $ .

2. Solve graphically  $ 4x+3y\le 60 $ ,  $ y\ge 2x $ ,  $ x\ge 3 $ ,  $ x,y\ge 0 $ .

Ans: The corresponding equality of  $ 4x+3y\le 60 $ is $ 4x+3y=60 $ .

The two coordinates through which the line passes are,

Put  $ \left( 0,0 \right) $  in the equation,  $ 4x+3y\le 60 $ . 

$4\left( 0 \right)+3\left( 0 \right)\le 60$ 

$0\le 60$

The inequality is true. Thus, the solution region of the given inequality will be that portion of the graph of  $ 3x+2y=6 $  that contains the point  $ \left( 0,0 \right) $ . 

The corresponding equality of  $ y\ge 2x $ is $ y-2x=0 $ .

The two coordinates through which the line passes are,

Put  $ \left( 0,0 \right) $  in the equation,  $ y\ge 2x $.

$y-2x\ge 0$ 

$0\ge 0$ 

The inequality is true. Thus, the solution region of the given inequality will be that portion of the graph of  $ y-2x=0 $  that contains the point  $ \left( 0,0 \right) $ . 

The corresponding equality of  $ x\ge 3 $ is $ x=3 $ .

Put  $ \left( 0,0 \right) $  in the equation,  $ x\ge 3 $ .

 $ 0\ge 3 $ 

The inequality is false. Thus, the solution region of the given inequality will be that portion of the graph of  $ x\ge 3 $  that does not contain the point  $ \left( 0,0 \right) $ . 

The graph of the inequalities is given below. The solution region is labelled as a feasible region.

3. A manufacturer has 600 litre of a  $ 12% $  sol. of acid. How many litres of a $ 30% $  acid sol. must be added to it so that the acid content in the resulting mixture will be more than  $ 15% $  but less than  $ 18% $ .

Ans: Let  $ x $ litres of  $ 30% $  acid sol. is required to be added in the mixture. The inequality that shows that the acid content will be more than  $ 15% $  is,

$ 30%\text{ of }x+12%\text{ of 60015 }\!\!%\!\!\text{  of }\left( x+600 \right) $ Solve the inequality.

$\dfrac{30x}{100}+\dfrac{12}{100}\times \text{600}\,>\,\dfrac{15}{100}\times \left( x+600 \right)$ 

$0.30x+12\times \text{6}>\,\text{0}\text{.15}\times \left( x+600 \right)$ 

$0.30x+12\times \text{6}>\text{0}\text{.15}x+0.15\times 600$ 

$0.30x+72>\text{0}\text{.15}x+90$ 

$0.15x>18$

Divide both sides by  $ 0.15 $ .

$ x>120 $ 

The other inequality that shows the acid content will be less than  $ 18% $  is, \[30%\text{ of }x+12%\text{ of 600}<18%\text{ of }\left( x+600 \right)\]. Solve the inequality.

$\dfrac{30x}{100}+\dfrac{12}{100}\times \text{600}<\dfrac{18}{100}\times \left( x+600 \right)$ 

$0.30x+12\times \text{6}<\text{0}\text{.18}\times \left( x+600 \right)$ 

$0.30x+12\times \text{6}<\text{0}\text{.18}x+0.18\times 600$ 

$0.30x+72<\text{0}\text{.18}x+108$ 

$0.12x<36$

Divide both sides by  $ 0.12 $ .

 $ x<300 $ 

Hence,  $120 < x < 300$. 

4. Solve graphically  $ x-2y\le 3 $ ,  $ 3x+4y\ge 12 $ ,  $ x\ge 0 $ ,  $ y\ge 1 $ .

Ans: The corresponding equality of  $ x-2y\le 3 $ is $ x-2y=3 $ .

The two coordinates through which the line passes are,

Substitute $ \left( 0,0 \right) $  in the equation,  $ x-2y\le 3 $.

$0-2\left( 0 \right)\le 3$ 

$0\le 3$ 

The inequality is true. Thus, the solution region of the given inequality will be that portion of the graph of  $ x-2y\le 3 $  that contains the point  $ \left( 0,0 \right) $ . 

The corresponding equality of  $ 3x+4y\ge 12 $ is $ 3x+4y=12 $ .

The two coordinates through which the line passes are,

Put  $ \left( 0,0 \right) $  in the equation,  $ 3x+4y\ge 12 $. 

$3\left( 0 \right)+4\left( 0 \right)\ge 12$ 

$0\ge 12$ 

The inequality is false. Thus, the solution region of the given inequality will be that portion of the graph of  $ 3x+4y=12 $  that does not contain the point  $ \left( 0,0 \right) $ . 

The corresponding equality of  $ y\ge 1 $ is $ y=1 $ .

Put  $ \left( 0,0 \right) $  in the equation,  $ y\ge 1 $ .

$ 0\ge 1 $ 

The inequality is false. Thus, the solution region of the given inequality will be that portion of the graph of  $ y=1 $  that does not contain the point  $ \left( 0,0 \right) $ . 

The graph of the inequalities is given below. The solution region is labelled as feasible region.

5. A solution of  $ 8% $   boric acid is to be diluted by adding a  $ 2% $  boric acid sol. 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% $  sol. will be added.

Ans: Let  $ x $ litres of the  $ 2% $  solution be added. 

According to the question, 

$2%\text{ of }x+8%\text{ of 640}>\text{4 }\!\!%\!\!\text{  of }\left( 640+x \right)$ 

$\dfrac{2x}{100}+\dfrac{8}{100}\times \text{640}>\dfrac{4}{100}\times \left( 640+x \right)$ 

$2x+8\times 640 > 4\left( 640+x \right)$ 

$2x+5120 > 2560+4x$ 

Subtract  $ 4x $  from both sides and then subtract 5120 from both sides. 

$2x-4x>2560-5120$ 

$-2x>-2560 $ 

Divide both sides by  $ -2 $ .

$ x>1280 $ 

Let this equation be  $ \left( 1 \right) $ .

Now, the second inequality is \[12%\text{ of }x+8%\text{ of }640<6%\text{ of }\left( 640+x \right)\].

$\dfrac{2x}{100}+\dfrac{8}{100}\times \text{640}>\dfrac{6}{100}\times \left( 640+x \right)$ 

$2x+8\times 640>6\left( 640+x \right)$ 

$2x+5120>3840+6x$ 

Subtract  $ 6x $  from both sides and then subtract 5120 from both sides.

\[-4x>-1280\]

Divide both sides by  $ -4 $ .

$ x<320 $ 

Let this be inequality  $ \left( 2 \right) $ .

From  $ \left( 1 \right) $  and  $ \left( 2 \right) $ .

 $ 320 < x < 1280 $ 

Hence,  $ 320 < x < 1280 $ .

6. Solve graphically  $ x+2y\le 10 $ ,  $ x+y\ge 1 $ ,  $ x-y\le 0 $ ,  $ x\ge 0 $ ,  $ y\ge 0 $ .

Ans: The corresponding equality of  $ x+2y\le 10 $ is $ x+2y=10 $ .

The two coordinates through which the line passes are,

Substitute  $ \left( 0,0 \right) $  in the equation,  $ x+2y\le 10 $.

$0+2\left( 0 \right)\le 10$ 

$0\le 10 $ 

The inequality is true. Thus, the solution region of the given inequality will be that portion of the graph of  $ x+2y\le 10 $  that contains the point  $ \left( 0,0 \right) $ . 

The corresponding equality of  $ x+y\ge 1 $ is $ x+y=1 $ .

The two coordinates through which the line passes are,

Put  $ \left( 0,0 \right) $  in the equation,  $ x+y\ge 1 $. 

$\left( 0 \right)+\left( 0 \right)\ge 1$ 

$0\ge 1$ 

The inequality is false. Thus, the solution region of the given inequality will be that portion of the graph of  $ x+y\ge 1 $  that does not contain the point  $ \left( 0,0 \right) $ . 

The corresponding equality of  $ x-y\le 0 $ is $ x-y=0 $ . Thus,  $ x=y $ . From the equation, it can be seen that it is a straight line passing through the origin.

Put  $ \left( 1,0 \right) $  in inequality,  $ x-y\le 0 $ .

$1-0\le 0$ 

$1\le 0$ 

The inequality is false. Thus, the solution region of the given inequality will be that portion of the graph of  $ x=y $  that is away from the point  $ \left( 1,0 \right) $ .

7. How many litres of water will have to be added to 1125 litres of the  $ 45% $  sol. of acid so that the resulting mixture will contain more than  $ 25% $  but less than  $ 30% $  acid content.

Ans: Let  $ x $ litres of water be added to 1125 litres of 45 acid sol. in the resulting mixture.

According to the question,  $ 45%\text{ of }1125>25%\text{ of }\left( x+1125 \right) $  and  $ 45%\text{ of }1125<30%\text{ of }\left( x+1125 \right) $ .

Solve the inequality,  $ 45%\text{ of }1125>25%\text{ of }\left( x+1125 \right) $.

$\dfrac{45}{100}\times 1125>\dfrac{25}{100}\times \left( x+1125 \right)$ 

$0.45\times 1125>0.25\times \left( x+1125 \right)$ 

$506.25>0.25x+281.25$

Subtract \[281.25\] from both sides.

$506.25-281.25>0.25x$ 

$225>0.25x$

Divide both sides by  $ 0.25 $.

$\dfrac{225}{0.25}>x$ 

$900>x$ 

Solve the inequality,  $ 45%\text{ of }1125<30%\text{ of }\left( x+1125 \right) $.

$\dfrac{45}{100}\times 1125<\dfrac{30}{100}\times \left( x+1125 \right)$ 

$0.45\times 1125<0.3\left( x+1125 \right)$ 

$506.25<0.3x+337.5$ 

Subtract  $ 337.5 $  from both sides.

 $506.25-337.5<0.3x$ 

$168.75<0.3x$

Divide both sides by  $ 0.25 $.

$\dfrac{168.75}{0.3}>x$ 

$562.3>x$ 

From both the inequalities  $ 900>x>562.5 $ .

Hence, the required amount of water to be added is  $ 900>x>562.5 $ .

8. Solve graphically  $ 3x+2y\le 150 $ ,  $ x+4y\le 80 $ ,  $ x\le 15 $ ,  $ y\ge 0 $ ,  $ x\ge 0 $ .

Ans: The corresponding equality of  $ 3x+2y\le 150 $ is $ 3x+2y=150 $ .

The two coordinates through which the line passes are,

Substitute  $ \left( 0,0 \right) $  in the equation,  $ 3x+2y\le 150 $.

$3\left( 0 \right)+2\left( 0 \right)\le 150$ 

$0\le 150$ 

The inequality is true. Thus, the solution region of the given inequality will be that portion of the graph of  $ 3x+2y\le 150 $  that contains the point  $ \left( 0,0 \right) $ . 

The corresponding equality of  $ x+4y\le 80 $ is $ x+4y=80 $ .

The two coordinates through which the line passes are,

Put  $ \left( 0,0 \right) $  in the equation,  $ x+4y\le 80 $.

$0+4\left( 0 \right)\le 80$ 

$0\le 80$ 

The inequality is true. Thus, the solution region of the given inequality will be that portion of the graph of  $ x+4y\le 80 $  that contains the point  $ \left( 0,0 \right) $ . 

The corresponding equality of  $ x\le 15 $ is $ x=15 $ .

Put  $ \left( 0,0 \right) $  in the equation,  $ x\le 15 $ .

 $ 0\le 15 $ 

The inequality is true. Thus, the solution region of the given inequality will be that portion of the graph of  $ x\le 15 $  that contains the point  $ \left( 0,0 \right) $ . 

The graph of the inequalities is given below. The solution region is labeled as a feasible region.

Benefits of Referring to Important Questions for Class 11 Maths Chapter 5 Linear Inequalities

• Focused Exam Preparation: Important questions are designed to target the key concepts and problem types that frequently appear in exams, helping you concentrate on what matters most.

• Improved Problem-Solving Skills: By practising these questions, you develop a better understanding of solving linear inequalities, graphing solutions, and interpreting results.

• Time Management: Working on curated important questions helps you learn how to solve problems quickly and accurately, improving your efficiency during exams.

• Clarity in Concepts: These questions simplify challenging topics like solving inequalities in one or two variables and representing solutions on a number line or graph, ensuring clear understanding.

• Boosts Confidence: Regular practise of important questions builds confidence, as you become familiar with the types of questions likely to appear in exams.

• Exam Readiness: Referring to important questions ensures that you are well-prepared to understand both straightforward and complex problems, reducing stress during the board exams.

Conclusion

Practising important questions from Chapter 5: Linear Inequalities is essential for mastering this topic and performing well in CBSE exams. These questions cover all key areas, including solving inequalities, representing solutions graphically, and interpreting results. By focusing on important questions, students can strengthen their conceptual understanding, improve problem-solving skills, and boost confidence for exams.

Important Study Materials for Class 11 Maths Chapter 5 Linear Inequalities

CBSE Class 11 Maths Chapter-wise Important Questions

CBSE Class 11 Maths Chapter-wise Important Questions and Answers cover topics from all 14 chapters, helping students prepare thoroughly by focusing on key topics for easier revision.

Important Related Links for CBSE Class 11 Maths