Important Questions Class 11 Maths Chapter 6

Important Questions Class 11 Mathematics Chapter 6

Important Questions for CBSE Class 11 Mathematics Chapter 6 – Linear Inequalities

In Chapter 6 of Class 11 Mathematics, students will study linear inequalities in one and two variables. Students should refer to the set of Important Questions Class 11 Mathematics Chapter 6, Linear Inequalities to get an idea of the type of questions that would be asked for the exams.

Linear inequalities are the expressions where any two values are compared by the inequality symbols such as, ‘<‘, ‘>’, ‘≤’, or ‘≥’. These values could be numerical or algebraic or a combination of both.

Students can go over these Mathematics Class 11 Chapter 6 Important Questions and Answers to help them prepare for and ace the final exams.

CBSE Class 11 Mathematics Chapter-6 Important Questions

Study Important Questions for CBSE Class 11 Mathematics Chapter 6 – Linear Inequalities

Following are some of the important questions asked from Linear Inequalities. Click the below link to get Important Questions Class 11 Mathematics Chapter 6

Very Short Questions and Answers: (1 Mark)

Q.1 Solve 3x-42 x+41-1

Ans: Rewrite the given inequality.

3x-42 x+14 – 11

3x-42 x+1-44

3x-42 x-34

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

2(3x−4)≥(x−3)  

6x−8≥ x−3 

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

6x−8−x+8≥x−3−x+8 

5x≥5 

Divide both sides by 5.

x ≥1 Hence, the solution set is [1,∞)

Q.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.

3xx > -63 

x>−2  

Hence, the solution set is  (−2,∞) .

 Q.3 If  4x>−16 , then  x−4 .

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

4x4> -164

x >−4 

Hence,  x−4 

Q.4 Solve the inequalities,   2x−1≤3  and  3x+1≥−5  is.

Ans: Given,  2x−1≤3 and  3x+1≥−5 .

Solve equation,  2x−1≤3. 

2x−1+1≤3+1 

2x≤4 

x≤2 

Solve equation, 3x+1≥−5.

3x+1≥−5 

3x≥−6 

x≥−2 

From both solutions, it is concluded that −2≤x≤2 . Hence, the solution set is          [−2,2] .

Q.5  Solve  7x+3<5x+9. 

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 

Q.6 Solve  5x−3≤3x+1  when  x  is an integer 

Ans: Given,  5x−3≤3x+1 .

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

5x−3−3x+3≤3x+1+3−3x

2x≤4 

x≤2 

Hence, the solution set is  {….,−3,−2,−1,0,1,2} .

Long Questions and Answers: (4 Marks)

Q.1 Anil 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 Anil be  x.

According to the question, 

70+75+x3 ≥ 60 

145+x3 ≥ 60 

Multiply the inequality obtained by 3.

145+x  ≥ 60(3) 

145+x  ≥ 180 

Subtract 145 from both sides. 

x ≥ 180−145 

x ≥ 35  

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

Q.2 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 a consecutive odd natural number

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

According to the question, 

x+(x+2)<40 

2x+2<40 

Subtract 2 from both sides.

2x<38 … (1)

Divide both sides by 2.

x<19 … (2) 

From (1) and (2). The pairs obtained are,

(11,13) ,  (13,15) , (15,17) , (17,19) 

Hence, the required pairs are (11,13) ,  (13,15) , (15,17) , and (17,19) .

Q.3 The longest side of a  Δ  is 3 times the shortest side and the third side is 2 cm shorter than the longest side. If the perimeter of the  Δ  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 being 3x  cm and the third side be  (3x−2)  cm.  

According to the question,

(x)+(3x)+(3x−2) ≥ 61 

7x−2 ≥ 61 

7x ≥ 61+2 

7x ≥ 63 

Divide both sides by 7.

x ≥ 9 

Hence, the shortest side is 9 cm.

Q.4 A plumber can be paid under two schemes as given below.

1. Rs 600 and Rs 50 per hr.
2. Rs 170 per hr.

If the job takes n hours for what values of n  does the scheme  I  give the plumber the better wages?

Ans: The total wage of the labour in the scheme  I is  600+50n  and the total wage in scheme  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.

Q5. 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 being normal.

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

7.2 < 7.48+7.85+x3 <7.8

 Multiply all the parts of the inequality by 3.

21.6<15.33+x<23.4

 6.27<x<7.07

 Very Long Questions and Answers : (6 Marks)

Q.1 Solve graphically  4x+3y≤60 ,  y≥2x ,  x≥3 ,  x,y≥0 .

Ans: The corresponding equality of  4x+3y≤60 is 4x+3y=60. The two coordinates through which the line passes are,

Put  (0,0) in the equation,  4x+3y≤60 . 

4(0)+3(0)≤60

 0≤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  (0,0) . 

The corresponding equality of  y≥2x is y−2x=0 .

The two coordinates through which the line passes are,

Put  (0,0)  in the equation,  y≥2x.

y−2x≥0

 0≥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  (0,0) . 

The corresponding equality of  x≥3 is x=3 .

Put  (0,0) in the equation,  x≥3.

 0≥3 

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

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

Q.2 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, 

2x100 + 8100 640>4100 (640+x)

2x+8×640>4(640+x)

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  (1)

Now, the second inequality is 12

2x100 + 8100 640>6100 (640+x)

2x+8×640>6(640+x)

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  (2)

From  (1)  and  (2)

  320<x<1280 

Hence,  320<x<1280