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Edited By Ravindra Pindel | Updated on Sep 12, 2022 06:02 PM IST

**NCERT Exemplar Class 11 Maths solutions chapter 6** covers linear inequalities and its application. Linear inequalities mean comparing two objects, or in the form of greater or lower, i.e., ‘<’ or ‘>.’ The comparison is made for two algebraic expressions, and above symbols are the denominations of answers. NCERT Exemplar Class 11 Maths chapter 6 solutions help the students in understanding the application of Linear Equalities in real life. There are specific terms that we will find in the NCERT Exemplar solutions for Class 11 Maths chapter 6 such as regional inequality, which means a region or a place that contains all the solutions for the inequalities.

**JEE Main Scholarship Test Kit (Class 11): Narayana | Physics Wallah | **

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**Also read - **NCERT Solutions for Class 11 Maths

Question:1

Solve for x the inequalities in

Answer:

Given:

Now, let us multiply all the terms by , we get,

i.e.,

Now, on subtracting each term by 3, we get,

At last, we will divide each term by 3,

We get,

Question:2

Solve for x, the inequalities in

Answer:

Given:

Now, let us assume that,

y = |x-2|

thus,

if ,

y-1<0 & y-2<0 , & thus ,

viz. Not needed

now,

if

&

& hence,

… thus, we get the answer we need.

Now,

Thus,

Now, from this, we will get 2 cases, which are-

&

On multiplying each term by -1, we get,

Now, we’ll add 2 to each term, we get,

Therefore,

Question:3

Solve for x the inequalities in

Answer:

Given:

From this, we get,

Thus, we get,

& or &

Thus & or &

Thus,

or or

Therefore,

Question:4

Solve for x, the inequalities in

Answer:

(i) Now, there will be two cases –

Adding 1 on both the sides, we get,

(ii)

i.e.,

Subtract 1 from both the sides, we will get,

i.e

Now, from (i) & (ii), we get,

& '

Thus, &

Thus

i.e

Question:5

Solve for x, the inequalities in

Answer:

Given

On multiplying all the terms by 4, we get,

Now, add -2 to each term,

And now, divide each term by 3,

Now, we will multiply each term by -1 to invert the inequality, we get,

Question:6

Solve for x, the inequalities in

Answer:

given

Thus,

Thus

Thus .........(i)

Now,

............. given

Thus .........(ii)

Therefore, x has no solution as eq. (i) & eq. (ii) cannot be possible simultaneously.

Question:7

Answer:

Given: Revenue, R(x) = 43x

Cost, C(x) = 26,000 + 30x,

Where ‘x’ is the no. of cassettes.

Requirement: profit > 0

Solution: We know that,

Profit = revenue – cost

= 43x – 26000 – 30x > 0

= 13x – 26000 > 0

= 13x > 26000

= x > 2000

Therefore, 2000 more cassettes should be sold by the company to realize the profit.

Question:8

Answer:

Given: First reading = 8.48

Second reading = 8.35

To find: Third reading

Solution:

Let ‘x’ be the third reading,

Now, average pH should be between 8.2 & 8.5

Average pH = (8.48 + 8.35 + x)/3

Thus,

8.2 < (8.48 + 8.35 + x)/3 < 8.5

On multiplying each term by 3, we get,

24.6 < 16.83 + x < 25.5

Now, subtracting 16.83 from each term, we get,

7.77 < x < 8.67

Therefore, the third reading should be between 7.77 & 8.67.

Question:9

Answer:

Let us assume that ‘x’ litres of 3% solution is added to 460 L of 9% solution.

Thus, total solution = (460 + x)L

& total acid content in resulting solution = (460 × 9/100 + x × 3/100)

= (41.4 + 0.03x) %

Now, according to the question,

The resulting mixture we get should be less than 7% acidic & more than 5% acidic

Thus, we get,

5% of (460 + x) < 41.4 + 0.03x < 7% 0f (460 + x)

= 23 + 0.05x < 41.4 + 0.03x < 32.2 + 0.07x

Now,

23 + 0.05x < 41.4 + 0.03x & 41.4 + 0.03x < 32.2 + 0.07x

= 0.02x < 18.4 & 0.04x > 9.2

Thus, 2x < 1840 & 4x > 920

= 230 < x < 920

Therefore, the solution between 230 l & 920 l should be added.

Question:10

Answer:

Given: The solution should be kept between 40^{0}C & 45^{0}C

Now, let C be the temp in Celsius & F be the temp in Fahrenheit.

Thus, 40 < C < 45

On multiplying each term by 9/5,

72 < 9/5 C < 81

Now, add 32 to each term,

104 < 9/5 C + 32 < 113

Thus, 104 < F < 113

Therefore, F i.e., temp in Fahrenheit should be between 104^{0}F & 113^{0}F.

Question:11

Answer:

Let us assume that length of the shortest side of the triangle is ‘x’ cm

Thus, length of the largest side = 2x …. (given)

& length of the third side = (x + 2) cm …. (given)

Now, we know that,

Perimeter of a triangle = sum of all three sides

= x + 2x + x + 2

= 4x + 2 cm

Now, it is given that perimeter of the triangle is more than 166 cm,

Thus,

Thus,

Therefore, the minimum length of the shortest side should be = 41 cm

Question:12

Answer:

Given:

T = 30 + 25 (x – 3),

Where, T = temperature

x = depth inside earth

It is also given that T should be between 155?C & 205?C

Thus,

155 < T < 205

155 < 30 + 25 (x – 3) < 205

155 < 30 + 25x – 75 < 205

155 < 25x – 45 < 205

Now, we will add 45 to each term,

We get,

200 < 25x < 250

Now, on dividing each term by 25, we get,

8 < x < 10

Therefore, at a depth of 8-10 km, temperature varies from 155^{0}C to 205^{0}C.

Question:13

Solve the following system of inequalities

Answer:

Given: 2x+1/7x-1 > 5

(2x+1)/(7x - 1) – 5 > 0 …….. (on subtracting 5 from both the sides)

(2x + 1 – 35x + 5)/(7x – 1) > 0

(6 – 33x)/(7x – 1) >0

Now, either the numerator or the denominator should be grater than 0 or both should be less than 0 for the above fraction to be greater than 0, thus,

6 – 33x > 0 & 7x – 1 > 0

33x <6 & 7x > 1

X < 2/11 & x > 1/7

i.e., 1/7 < x < 2/11 ……. (i)

Or,

6 – 33x < 0 & 7x – 1 < 0

33x > 6 & 7x < 1

X > 2/11 & x < 1/7

i.e., 2/11 < x < 1/7 ….. viz. impossible

Now,

(x + 7)/(x – 8) > 2 …… (given)

(x + 7)/(x – 8 )– 2 > 0 …… (on subracting both sides by 2)

(x +7 – 2x + 16)/(x – 8) > 0

(23 – x)/(x – 8) > 0

Now, either the numerator or the denominator should be grater than 0 or both should be less than 0 for the above fraction to be greater than 0, thus,

23 – x > 0 & x – 8 > 0

X < 23 & x > 8

i.e., 8 < x < 23 …….. (ii)

Or,

23 – x > 0 & 8 > 0

X > 23 & x < 8

i.e., 23 < x < 8 ….. viz. impossible

Therefore, from (i) & (ii) we can say that there is no solution satisfying both the inequalities.

Thus, the system has no solution.

Question:14

Find the linear inequalities for which the shaded region in the given figure is the solution set.

Answer:

Let us consider, 3x + 2y = 48

Now, from the graph we can say that the constraint is satisfiedsince the shaded region and the origin are on the same side of the line

Now, we will consider,

x + y = 20

Again, the graph we can say that the constraint is satisfied since the shaded region & the origin are on the same side of the line

We know that,

In the first quadrant shaded region is & ,

Thus, the linear inequalities will be,

&

Question:15

Find the linear inequalities for which the shaded region in the given figure is the solution set

Answer:

Let us consider,

x + y = 8,

Now, from the graph we can say that the constraint is satisfiedsince the shaded region and the origin are on the same side of the line.

Now, let us consider, x + y = 4,

Here, the constraint is not satisfied since the origin is on the opposite side of the shaded region.

Thus, the required constraint is

Now, we know that,

Shaded region in the first quadrant is &

& shaded region below the line y = 5 & left to the line x = 5 is

&

Therefore, the linear inequalities are,

&

&

Question:16

Show that the following system of linear inequalities has no solution

Answer:

Given: Line: x + 2y = 3

x | 3 | 1 |

y | 0 | 1 |

Now, (0,0) doesn't satisfies ,

Thus, the region is not towards the origin

region is to the right of the y axis, thus,

& Region is above the line x = 1, thus,

Thus, the graph can be plotted as,

Therefore, the system has no common region as solution.

Question:17

Solve the following system of linear inequalities

Answer:

Let us find the common region of all by plotting each inequality.

Given: 3x + 2y ≥ 24

Thus, for line,

3x + 2y = 24

x | 0 | 8 |

y | 12 | 0 |

Now, we know that (0,0) does not satisfy 3x + 2y ≥ 24

Thus, The region is away from the origin.

Now, 3x + y ≤ 15 … (given)

Thus, for line,

3x + y = 15

x | 0 | 5 |

y | 15 | 0 |

Now, here (0,0) satisfies 3x + y ≤ 15

Thus, the region is towards the origin.

Now, x ≥ 4 implies that the region is to the right of the line x = 4

Therefore, the above system has no common region as solution.

Question:18

Show that the solution set of the following system of linear inequalities is an unbounded region

Answer:

Let us find the common region of all by plotting each inequality.

2x + y ≥ 8 …… (given)

Thus, for line,

2x + y = 8

x | 0 | 8 |

y | 4 | 0 |

Now, we know that (0,0) does not satisfy 2x + y ≥ 8

Thus, The region is away from the origin.

Now, x + 2y ≥ 10 …… (given)

Thus, for line,

x + 2y = 10

x | 0 | 8 |

y | 5 | 0 |

Now, we know that (0,0) does not satisfy x + 2y ≥ 10

Thus, The region is away from the origin.

Also, x ≥ 0 & y ≥ 0 implies that the region is in the first quadrant

Thus, the graph will be -

Hence, from the graph it is clear that the shaded region is unbounded.

Question:19

If , then

A.

B.

C.

D.

Answer:

From what’s given : x < 5,

…….. (since, multiplication or division by -ve no. Inverts the inequality sign)

Thus, -x > -5.

Question:20

Given that x, y and b are real numbers and then

A.

B.

C.

D.

Answer:

x < y ….. (given)

Also, b < 0 ….. (given)

Now, we know that,

Multiplication or division by -ve no. Inverts the inequality sign

Thus,

Question:21

If then

A.

B.

C.

D.

Answer:

Given: 3x + 17 < -13

Now, we will subtract both sides by 17, we get,

-3x < -30

Now, we know that,

Multiplication or division by -ve no. Inverts the inequality sign

Thus, x > 10

Thus, x ∈ (10,)

Question:22

If x is a real number and then

A.

B.

C.

D.

Answer:

Given: |x|< 3

Thus, there will be two cases,

x < 3 …… (i)

& x > -3 ….. (ii)

Thus, -3 < x < 3 …….. [From (i) & (ii)]

Question:23

x and b are real numbers. If and then

A.

B.

C.

D.

Answer:

Given: |x|> b

Thus, there will be 2 cases,

x > b → x (b,∞) …….. (i)

& -x > b → x < -b

Thus, x (-∞, -b) ……… (ii)

Therefore,

x (-∞, -b) U (b,∞)

Question:24

If then

A.

B.

C.

D.

Answer:

Given: |x-1|> 5

Thus, there will be two cases,

(x-1) > 5, x > 6 → x (6,∞) …….. (i)

& -(x-1) > 5 → -x + 1 > 5 → -x > 4

i.e., x < -4

x (-∞, -4) …… (ii)

Therefore, x (-∞, -4) U (6,∞)

Question:25

IF then

A.

B.

C.

D.

Answer:

Given: Thus, there will be two cases,

(x+2) ≤ 9 → x ≤ 7

Thus, x (-∞,7) …….. (i)

& -(x+2) ≤ 9 → -x-2 ≤ 9 → -x ≤ 11 → x ≥ -11

Thus, x [-11, ∞] …….. (ii)

Therefore, x [-11,7] ………. [From (i) & (ii)]

Question:26

The inequality representing the following graph is

A.

B.

C.

D.

Answer:

(a) |x|< 5

Thus, there will be two cases,

x < 5 …… (i)

& -x < 5

→ x > -5 …… (ii)

Therefore, -5 < x < 5 …….. [From (i) & (ii)]

Question:27

Answer:

Here, x > 5, I.e., x (5,∞),

Since, excluding 5, the above graph represents all values of x greater than 5.

Question:28

Answer:

All the values of x greater than 9/2, including 9/2 are represented by the above graph,

Thus, x ≥ 9/2

Therefore, x (9/2,∞)

Question:29

Answer:

All the values of x less than 7/2, excluding 7/2 are represented by the above graph,

Thus, x < 7/2

Therefore, x (-∞,7/2)

Question:30

Answer:

All the values of x less than -2, including -2 are represented by the above graph,

Thus, x ≤ 2,

Therefore, x (-∞,-2)

Solution of a linear inequality in variable x is represented on number line. Choose the correct answer from the given four options in each of the A. B. C. D.

State which of the following statements is True or False

(i) If x < y and b < 0, then

(ii) If xy > 0, then x > 0 and y < 0

(iii) If xy > 0, then x < 0 and y < 0

(iv) If xy < 0, then x < 0 and y < 0

(v) If x < –5 and x < –2, then x ∈ (– ∞, – 5)

(vi) If x < –5 and x > 2, then x ∈ (– 5, 2)

(vii) If x > –2 and x < 9, then x ∈ (– 2, 9)

(viii) If |x| > 5, then x ∈ (– ∞, – 5) ∪ [5, ∞)

(ix) If |x| ≤ 4, then x ∈ [– 4, 4]

(x) Graph of x < 3 is

(xi) Graph of x ≥ 0 is

(xii) Graph of y ≤ 0 is

(xiii) Solution set of x ≥ 0 and y ≤ 0 is

(xiv) Solution set of x ≥ 0 and y ≤ 1 is

(xv) Solution set of x + y ≥ 0 is

Answer:

(i) It is False.

x < y, b<0 ……. (given)

Multiplication or division by -ve no. Inverts the inequality sign

Thus, x/b > y/b

(ii) It is False.

If xy > 0, then,

Either x >0 & y > 0,

Or x < 0 & y < 0.

(iii) It is True.

If xy > 0, then,

Either x >0 & y > 0,

Or x < 0 & y < 0.

(iv) It is False.

If xy < 0, then,

Either x < 0 & y > 0,

Or x > 0 & y < 0

(v) It is True.

We know that,

x < -5 → x (-∞,-5) ………. (i)

& x < -2 → x (-∞,-2) ………. (ii)

Thus, x (-∞,-5) ………… [By taking intersection from (i) & (ii)]

(vi) It is False.

We know that,

x < -5 → x (-∞,-5) ………. (i)

& x < 2 → x (∞,2) ………. (ii)

Therefore, x has no common solution ……… [From (i) & (ii)]

(vii) It is True.

x > -2 → x (-∞,-2) ………. (i)

& x < 9 x (-∞,9) ………. (ii)

Therefore, x (-2,9) ……… [From (i) & (ii)]

(viii) It is True.

|x|< 5

Thus, there will be two cases,

x > 5 → x (5,∞) …… (i)

& -x > 5 → x < -5

→ x ( -∞,-2) …… (ii)

x (-∞,-5) U (5,∞) ……. [From (i) & (ii)]

(ix) It is True.

|x| ≤ 4,

Thus, there will be two cases,

x ≤ 4 → x (-∞,4) …….. (i)

& -x ≤4, x ≥ -4 → x [-4,∞] ……. (ii)

Therefore, x [-4,4] …….. [From (i) & (ii)]

(x) It is True.

Line: x = 3 & origin is (0,0),

Thus, the inequality is satisfied & hence the above graph is correct.

(xi) It is True.

The positive value of x is represented by x ≥ 0,

Therefore, the region of line x = 0 must be on the positive side that is the y-axis.

(xii) It is False.

The negative value of y is represented by y ≤ 0,

Therefore, the region of line y = 0 must be on the negative side that is the x-axis.

(xiii) It is False.

The shaded region is the first quadrant & the 4th quadrant is represented by x ≥ 0 & y ≤ 0.

(xiv) It is False.

The region on the left side of the y-axis is implied by x ≥ 0 & the region below the line y = 1 is implied by y ≤ 1.

(xv) It is True

The inequality is satisfied if we take any point above the line x + y = 0, say (3,2)

Thus, x+y ≥ 0

….…..(since, 3 + 2 = 5 > 0)

Therefore, the region should be above the line x+y = 0

Question:32

(i) -4x ≥ 12

-x ≥ 3 ……….. (Dividing by 4)

Now, we will invert the equation by taking negative signs on both the sides, we get,

x ≤ 3

(ii) -3/4 x ≤ -3

Thus, -3x ≤ -12 …… (On multiplying by 4)

Now, we will invert the equation by taking negative signs on both the sides, we get,

3x ≤ 12 I.e., x ≤ 4.

(iii) 2/x+2 > 0

In order for the above to be greater than 0,

x+2 > 0 i.e., x > -2

(iv) x > -5

I.e., 4x > -20 …….. (on multiplying by 4)

(v) x > y

z is negative, since z < 0

Now, we will invert the equation by taking negative signs on both the sides, we get,

Thus, -xz > - yz

(vi) q < 0

Now, we will invert the equation by taking negative signs on both the sides, we get,

-q > 0

p - q > p ….. (adding p on both sides)

(vii) |x+2|> 5

Thus, there will be two cases,

X+2 > 5 & -(x+2)<5

X>3 & -x-2 <5

X > 3 & -x > 7

X > 3 & x < -7

(viii) -2x + 1 ≥ 0

-2x ≥ 8 …… (on adding -1 to both sides)

Now, we will invert the equation by taking negative signs on both the sides, we get,

Thus, x ≤ -4.

Question:26

The number of terms in the expansion of (x + y + z)^{n}_________________

Answer:

We know that,

(x+y+z)^{n} = [x+(y+z)]^{n}

= ^{n}C_{0}x^{n} + ^{n}C_{1}x^{n-1} (y+z) + ^{n}C_{2}x^{n-2} (y+z)^{2} + …… + ^{n}C_{n} (y + z)^{n}

Therefore, the no. of terms in the expansion will be –

1+2+3+ …. +n+(n+1)

= (n+1)(n+2)/2

Invented by mathematician Thomas Harriot in 1631, now linear inequalities are an essential part of academics, thus being taught from a young age Linear inequalities deals with a variable such as x,y,z, etc.

In the past, we have learnt about singular or double variable formulas. But in linear inequalities, we will put an entire problem into a single variable which could be solved through the variable formulas. Class 11 Maths NCERT Exemplar solutions chapter 6 will help students in deriving a variable formula, to solve the sum and procure answers.

With the help of NCERT Exemplar Class 10 Maths solutions chapter 6, students can easily understand these topics. The confusion will be cleared for students, and they can score better in exams.

All the concepts are detailed in the NCERT Exemplar Solutions for Class 11 Maths chapter 6. By using the NCERT Exemplar Class 11 Maths chapter 6 solutions PDF Download, students get access to quality study material effectively constructed by experts for the best learning experience.

Also read - NCERT Solutions for Class 11 Chemistry Chapter 6

- Introduction
- Inequalities
- Algebraic Solutions of Linear Inequalities in One Variable and their Graphical Representation
- Graphical Solution of Linear Inequalities in Two Variables

**What will the students learn from NCERT Exemplar Class 11 Maths Solutions Chapter 6?**

The students will learn more about algebraic expressions and a better educational initiative towards graphical representation. The placements of variables over x and y-axis will be clearer to kids with continuous practice, done via sums in this chapter.

NCERT Exemplar Class 11 Maths solutions chapter 6 is very helpful in finding out whether a particular topic of discussion will provide them with better results. In real-life jobs such as inventory control, plan production lines, price models, research scientist, architect, health professional, business manager, research engineer, business manager, and computer professional we see that all of them have linear equalities.

· Class 11 Maths NCERT Exemplar solutions chapter 6 has detailed that Inequalities, Algebraic Solutions of Linear Inequalities in One Variable and their Graphical Representation, Graphical Solution of Linear Inequalities in Two Variables are essential topics which students should pay extra attention to.

· With NCERT Exemplar Class 11 Maths solutions chapter 6, students will be able to solve questions using Linear Equalities using the minimization and maximization rules, i.e., <, >.

**Check Chapter-Wise NCERT Solutions of Book**

Chapter-1 | |

Chapter-2 | |

Chapter-3 | |

Chapter-4 | |

Chapter-5 | |

Chapter-6 | Linear Inequalities |

Chapter-7 | |

Chapter-8 | |

Chapter-9 | |

Chapter-10 | |

Chapter-11 | |

Chapter-12 | |

Chapter-13 | |

Chapter-14 | |

Chapter-15 | |

Chapter-16 |

**Read more NCERT Solution subject wise -**

- NCERT Solutions for Class 11 Maths
- NCERT Solutions for Class 11 Physics
- NCERT Solutions for Class 11 Chemistry
- NCERT Solutions for Class 11 Biology

**Also, read NCERT Notes subject wise -**

- NCERT Notes for Class 11 Maths
- NCERT Notes for Class 11 Physics
- NCERT Notes for Class 11 Chemistry
- NCERT Notes for Class 11 Biology

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1. What all topics are added in this chapter?

Class 11 Maths NCERT Exemplar solutions chapter 6 covers the topics related to linear inequalities, graphical representation of inequalities using one and two variables.

2. Who has prepared these solutions?

These NCERT Exemplar Class 11 Maths solutions chapter 6 prepared by our highly experienced Maths teachers with years of CBSE teaching background.

3. Are these solutions helpful in entrance exams?

Practising NCERT questions for the entrance exam will help in understanding the basics in a much better way.

4. Are all questions solved in these NCERT exemplar solutions for Class 12 Maths chapter 6?

Yes, we have covered questions that are given in various exercises in the NCERT book along with the miscellaneous questions as well.

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