NCERT Exemplar Solutions for Class 11 Maths Chapter 6 Linear Inequalities are provided here for the students to refer. Linear inequality is a fundamental concept that forms the basis for much more advanced topics. Therefore, one of the guides to help you understand this concept is NCERT Exemplar Solutions for Class 11 Maths Chapter 6 â€“ Linear Inequalities. It is designed by knowledgeable teachers with years of relevant experience. NCERT Exemplar Solutions is one of the best guides you could adopt for your study needs. We mainly focus on providing answers, which are in easy to remember format, which further helps students to clearly understand and remember the formulae.

This chapter of NCERT Exemplar Solutions for Class 11 mainly focuses on the solutions of inequalities. A linear function with inequalities is called linear inequality. It contains one of the symbols of linear equality. It acts exactly like a linear equation except for the inequality sign, which replaces the equality sign. Some of the linear inequalities of real numbers are mentioned below:

- Two-dimensional linear equalities
- Linear inequalities in general dimensions
- Systems of linear inequalities

Linear programming, along with the polyhedra are some of the major applications of linear equalities. The PDF files of Chapter 6 exemplars can be downloaded easily and can also be used as a reference tool while preparing for final exams. Let us have a look at some of the important concepts that are discussed in this chapter.

- Solution of an inequality
- Representation of solution of linear inequality in one variable on a number line
- Graphical representation of the solution of a linear inequality
- Two important results on positive real numbers

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### Access Answers of NCERT Exemplar Class 10 Maths Chapter 6 Linear Inequalities

Exercise Page No: 107

**Short Answer Type**

**Solve for x, the inequalities in Exercises 1 to 12.**

**1. Solve for x, the inequalities in**

**Solution:**

According to the question,

** **

Multiplying each term by (x + 1)

â‡’Â 4 â‰¤ 3(x + 1) â‰¤ 6

â‡’Â 4 â‰¤ 3x + 3 â‰¤ 6

Subtracting each term by 3, we get,

â‡’Â 1 â‰¤ 3x â‰¤ 3

Dividing each term by 3, we get,

â‡’Â (1/3) â‰¤ x â‰¤ 1

**2. Solve for x, the inequalities in**

**Solution:**

According to the question,

** **

Hence,

1 â‰¤ y < 2

â‡’Â 1 â‰¤ |x â€“ 2| < 2

Here, there are two cases

â‡’Â 1 â‰¤ x â€“ 2 < 2

â‡’Â 3 â‰¤ x < 4

And

â‡’Â 1 â‰¤ -(x â€“ 2) < 2

â‡’Â 1 â‰¤ – x + 2 < 2

Multiplying each term by -1,

â‡’Â -2 â‰¤ x â€“ 2 < -1

Adding 2 to each term,

â‡’Â 0 â‰¤ x < 1

âˆ´Â Hence, solution is [0, 1)Â âˆªÂ [3, 4)

**3. Solve for x, the inequalities inÂ **

**Solution:**

According to the question,

â‡’Â 5 – |x| â‰¤ 0 and |x| – 3 > 0 or 5 – |x| â‰¥ 0 and |x| – 3 < 0

â‡’Â |x| â‰¥ 5 and |x| > 3 or |x| â‰¤ 5 and |x| < 3

â‡’Â |x| â‰¥ 5 or |x| < 3

â‡’Â xÂ âˆˆÂ (- âˆž , – 5] or [5, âˆž) or xÂ âˆˆÂ ( -3 , 3)

â‡’Â xÂ âˆˆÂ (- âˆž , – 5]Â âˆªÂ ( -3 , 3)Â âˆªÂ [5, âˆž)

**4. Solve for x, the inequalities in |x â€“ 1| â‰¤ 5, |x| â‰¥ 2**

**Solution:**

|x â€“ 1|â‰¤ 5

There are two cases,

1:-

x â€“ 1 â‰¤ 5

Adding 1 to LHS and RHS

â‡’Â x â‰¤ 6

2:-

â‡’Â -(x â€“ 1) â‰¤ 5

â‡’Â -x + 1 â‰¤ 5

Subtracting 1 from LHS and RHS,

â‡’Â -x â‰¤ 4

â‡’Â x â‰¥ -4

From cases 1 and 2, we have

â‡’Â -4 â‰¤ x â‰¤ 6 â€¦[i]

Also,

|x| â‰¥ 2

â‡’Â x â‰¥ 2 and

â‡’Â -x â‰¥ 2

â‡’Â x â‰¤ -2

â‡’Â xÂ âˆˆÂ (âˆž, -2]Â âˆªÂ [2, âˆž) â€¦[ii]

Combining equation [i] and [ii], we get

xÂ âˆˆÂ [-4, -2]Â âˆªÂ [2, 6]

**5. Solve for x, the inequalities inÂ **

**Solution:**

** **According to the question,

** **

Multiplying each term by 4, we get

â‡’Â -20 â‰¤ 2 â€“ 3x â‰¤ 36

Adding -2 each term, we get

â‡’Â -22 â‰¤ -3x â‰¤ 34

Dividing each term by 3, we get

â‡’Â -22/3 â‰¤ -x â‰¤ 34/3

We know that,

Multiplication by -1 inverts the inequality.

So, multiplying each term by -1, we get

â‡’Â -34/3â‰¤ x â‰¤ 22/3

**6. Solve for x, the inequalities in 4x + 3 â‰¥ 2x + 17, 3x â€“ 5 < â€“ 2.**

**Solution:**

According to the question,

4x + 3 â‰¥ 2x + 17

â‡’Â 4x â€“ 2x â‰¥ 17 â€“ 3

â‡’Â 2x â‰¥ 14

â‡’Â x â‰¥ 7 â€¦(i)

Also,

3x â€“ 5 < – 2

â‡’Â 3x < 3

â‡’Â x < 1 â€¦(2)

Since, equations [i] and [ii] cannot be possible simultaneously,

We conclude that x has no solution.

**7. A company manufactures cassettes. Its cost and revenue functions are C(x) = 26,000 + 30x and R(x) = 43x, respectively, where x is the number of cassettes produced and sold in a week. How many cassettes must be sold by the company to realise some profit?**

**Solution:**

We know that,

Profit = Revenue â€“ cost

Requirement is, profit > 0

According to the question,

Revenue, R(x) = 43 x

Cost, C(x) = 26,000 + 30 x; where x is number of cassettes

â‡’Â Profit = 43x â€“ (26,000 + 30x) > 0

â‡’Â 13x â€“ 26,000 > 0

â‡’Â 13x > 26000

â‡’Â x > 2000

Therefore, the company should sell more than 2000 cassettes to realise profit.

**8. The water acidity in a pool is considered normal when the average pH reading of three daily measurements is between 8.2 and 8.5. If the first two pH readings are 8.48 and 8.35, find the range of pH value for the third reading that will result in the acidity level being normal.**

**Solution:**

According to the question,

First reading = 8.48

Second reading = 8.35

Now, let the third reading be â€˜xâ€™

Average pH should be between 8.2 and 8.5

Average pHÂ = (8.48 + 8.35 + x)/3

Multiplying each term by 3, we get

â‡’Â 24.6 < 16.83 + x < 25.5

Subtracting 16.83 from each term, we get

â‡’Â 7.77 < x < 8.67

Therefore, from the above equation,

We get that,

The third pH reading should be between 7.77 and 8.67

**9. A solution of 9% acid is to be diluted by adding 3% acid solution to it. The resulting mixture is to be more than 5% but less than 7% acid. If there is 460 litres of the 9% solution, how many litres of 3% solution will have to be added?**

**Solution:**

According to the question,

Let x litres of 3% solution is to be added to 460 liters of the 9% of solution

Then, we get,

Total solution = (460 + x) litres

Total acid content in resulting solution

= (460 Ã— 9/100 + x Ã— 3/100)

= (41.4 + 0.03x)%

According to the question, we have,

Resulting mixture should be more than 5% acidic but less than 7% acidic

So we get,

â‡’Â 5 % of (460 + x) < 41.4 + 0.03x < 7% of (460 + x)

â‡’Â 5/100 Ã— (460 + x) < 41.4 + 0.03x < 7/100 Ã— (460 + x)

â‡’Â 23 + 0.05 x < 41.4 + 0.03x < 32.2 + 0.07x

Now, we have

â‡’Â 23 + 0.05x < 41.4 + 0.03x and 41.4 + 0.03x < 32.2 + 0.07x

i.e., 0.02x < 18.4 and 0.04x > 9.2

â‡’Â 2x < 1840 and 4x > 920

â‡’Â x < 920 and x > 230

â‡’Â 230 < x < 920

Hence, solution between 230 *l* and 920 *l* should be added**.**

**10. A solution is to be kept between 40Â°C and 45Â°C. What is the range of temperature in degree Fahrenheit, if the conversion formula is F = 9/5 C + 32?**

**Solution:**

** **Let temperature in Celsius be C

Let temperature in Fahrenheit be F

According to the question,

Solution should be kept between 40Â° C and 45Â°C

â‡’Â 40 < C < 45

Multiplying each term by 9/5, we get

â‡’Â 72 < 9/5 C < 81

Adding 32 to each term, we get

â‡’Â 104 < 9/5 C + 32 < 113

â‡’Â 104 < F < 113

Hence, the range of temperature in Fahrenheit should be between 104Â° F and 113Â° F.

**11. The longest side of a triangle is twice the shortest side and the third side is 2 cm longer than the shortest side. If the perimeter of the triangle is more than 166 cm then find the minimum length of the shortest side.**

**Solution:**

Let the length of shortest side = â€˜xâ€™ cm

According to the question,

The longest side of a triangle is twice the shortest side

â‡’Â Length of largest side = 2x

Also, the third side is 2 cm longer than the shortest side

â‡’Â Length of third side = (x + 2) cm

Perimeter of triangle = sum of three sides

= x + 2x + x + 2

= 4x + 2 cm

Now, we know that,

Perimeter is more than 166 cm

â‡’Â 4x + 2 â‰¥ 166

â‡’Â 4x â‰¥ 164

â‡’Â x â‰¥ 41

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

**12. In drilling worldâ€™s deepest hole it was found that the temperature T in degree Celsius, x km below the earthâ€™s surface was given by T = 30 + 25 (x â€“ 3), 3 â‰¤ x â‰¤ 15. At what depth will the temperature be between 155Â°C and 205Â°C?**

**Solution:**

** **According to the question,

T = 30 + 25(x â€“ 3), 3 â‰¤ x â‰¤ 15; where, T = temperature and x = depth inside the earth

The Temperature should be between 155Â°C and 205Â°C,

So, we get,

â‡’Â 155 < T < 205

â‡’Â 155 < 30 + 25(x â€“ 3) < 205

â‡’Â 155 < 30 + 25x â€“ 75 < 205

â‡’Â 155 < 25x â€“ 45 < 205

Adding 45 to each term, we get

â‡’Â 200 < 25x < 250

Dividing each term by 25, we get

â‡’Â 8 < x < 10

Hence, temperature varies from 155Â° C to 205Â° C at a depth of 8 km to 10 km.

**Long Answer Type**

**13. Solve the following system of inequalities**

**Solution:**

According to the question,

For above fraction be greater than 0, either both denominator and numerator should be greater than 0 or both should be less than 0.

â‡’Â 6 â€“ 33x > 0 and 7x â€“ 1 > 0

â‡’Â 33x < 6 and 7x > 1

â‡’Â x < 2/11 and x > 1/7

â‡’Â 1/7 < x < 2/11 â€¦(i)

Or

â‡’Â 6 â€“ 33x < 0 and 7x â€“ 1 < 0

â‡’Â 33x > 6 and 7x < 1

â‡’Â x > 2/11 and x < 1/7

â‡’Â 2/11< x < 1/7 â€¦(which is not possible since 1/7 > 2/11)

Also,

For above fraction be greater than 0, either both denominator and numerator should be greater than 0 or both should be less than 0.

â‡’Â 23 â€“ x > 0 and x â€“ 8 > 0

â‡’Â x < 23 and x > 8

â‡’Â 8 < x < 23 â€¦(ii)

Or

23 â€“ x < 0 and x â€“ 8 < 0

â‡’Â x > 23 and x < 8

â‡’Â 23 < x < 8 â€¦(which is not possible, as 23 > 8]

Therefore, from equations (i) and (ii), we infer that there is no solution satisfying both inequalities.

Hence, the given system has no solution.

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

**Solution:**

According to the question,

Considering 3x + 2y = 48,

The shaded region and the origin both are on the same side of the graph of the line and (0, 0) satisfy the constraint 3x + 2y â‰¤ 48.

Considering x + y = 20,

The shaded region and the origin both are on the same side of the graph of the line and (0, 0) satisfy the constraint x + y â‰¤ 20.

We also know that,

Shaded region is in the first quadrant i.e. x â‰¥ 0 and y â‰¥ 0,

Hence, the linear inequalities are

3x + 2y â‰¤ 48

x + y â‰¤ 20

x â‰¥ 0

y â‰¥ 0

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

**Solution:**

According to the question,

Considering x + y = 8,

The shaded region and the origin both are on the same side of the graph of the line and (0, 0) satisfy the constraint x + y â‰¤ 8.

Considering x + y = 4,

The origin is on the opposite side of the shaded region and (0, 0), hence, doesnâ€™t satisfy the constraint x + y â‰¥ 4, therefore required constraint is x + y â‰¥ 4

We see that,

The shaded region is in the first quadrant i.e. x â‰¥ 0 and y â‰¥ 0,

Also, shades region is below the line y = 5 and left to the line x = 5

â‡’Â y â‰¤ 5 and x â‰¥ 5

Hence, the linear inequalities are

x + y â‰¤ 8

x + y â‰¥ 4

x â‰¥ 0

y â‰¥ 0

x â‰¤ 5

y â‰¤ 5

**16. Show that the following system of linear inequalities has no solution x + 2y â‰¤ 3, 3x + 4y â‰¥ 12, x â‰¥ 0, y â‰¥ 1**

**Solution:**

According to the question,

x + 2y â‰¤ 3

Line: x + 2y = 3

x |
3 |
1 |

y |
0 |
1 |

Also, (0, 0) satisfies the x + 2y â‰¤ 3, hence region is towards the origin

3x + 4y â‰¤ 12

Line: 3x + 4y = 12

x |
0 |
4 |

y |
3 |
0 |

Also, (0, 0) satisfies the 3x + 4y â‰¤ 3, hence region is towards the origin

x â‰¥ 0 â‡’ region is to the right of the y-axis

And, y â‰¥ 1 â‡’ region is up above the line x = 1,

Therefore graph can be plotted as,

Hence, we can conclude from the graph that the above system has no common region as solution