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NCERT Class 11 Mathematics Chapter 5 Linear Inequalities
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Linear Inequalities
Chapter – 5
| Exercise 5.1 |
1. Solve 24x < 100 when
(i) x is a natural number.
Ans: It is evident that 1, 2, 3, and 4 are the only natural numbers less than 25/6
Thus, when x is a natural number, the solutions of the given inequality are 1, 2, 3, and 4.
Hence, in this case, the solution set is {1, 2, 3, 4} .
(ii) x is an integer.
Ans: The integers less than 25/6 are …-3 , -2, -1 , 0, 1, 2, 3, 4.
Thus, when x is an integer, the solutions of the given inequality are
…-3 , -2, -1, 0,
1, 2, 3, 4.
Hence, in this case, the solution set is { … -3, -2 , – 1 , 0, 1, 2, 3, 4}.
2. Solve – 12x > 30 when
(i) x is a natural number.
Ans:There is no natural number less than (- 5/2)
Thus, when xis a natural number, there is no solution of the given inequality.
(ii) x is an integer.
Ans: The integers less than (- 5/2) are … – 5 , – 4, – 3 .
Thus, when xis an integer, the solutions of the given inequality are
…, – 5, – 4, -3.
Hence, in this case, the solution set is {…, – 5 , – 4, -3}.
3. Solve 5x – 3 < 7 when.
(i) x is an integer.
Ans: The integers less than 2are…, -4, -3, -2, -1, 0, 1.
Thus, when xis an integer, the solutions of the given inequality are
…, -4, – 3, – 2, – 1 , 0, 1.
Hence, in this case, the solution set is { …, – 4, – 3,- 2 , – 1 ,0,1}.
(ii) x is a real number.
Ans: When xis a real number, the solutions of the given inequality are given by x< 2, that is, all real numbers xwhich are less than 2.
Thus, the solution set of the given inequality isx ∈(-∞, 2).
4. Solve 3x + 8 > 2 when.
(i) x is an integer.
Ans: The integers greater than -2are -1, 0, 1, 2, …
Thus, when xis an integer, the solutions of the given inequality are
– 1,0,1,2…
Hence, in this case, the solution set is {- 1,0,1,2,…}
(ii) x is a real number.
Ans: When xis a real number, the solutions of the given inequality are all the real numbers, which are greater than – 2.
Thus, in this case, the solution set is (- 2, ∞).
Solve the inequalities in Exercises 5 to 16 for real x.
5. 4x + 3 < 5x + 7.
Ans: 4x + 3 < 5x + 7
⇒4x + 3 – 7 < 5x + 7 – 7
⇒ 4x – 4 < 5x
⇒ 4x – 4 – 4x < 5x – 4x
⇒ – 4 < x
Thus, all real numbers x, which are greater than -4, are the solutions of the given inequality.
Hence, the solution set of the given inequality is (-4, ∠A³/₄ ).
6. 3x – 7 > 5x – 1.
Ans: 3x – 7 > 5x – 1
⇒3x – 7 + 7 > 5x – 1 + 7
⇒ 3x > 5x + 6
⇒ 3x – 5x>5x+6 – 5x
⇒ – 2x > 6
⇒x < – 3
Thus, all real numbers x, which are less than -3, are the solutions of the given inequality.
Hence, the solution set of the given inequality is ( – ∞, – 3).
7. 3(x – 1) ≤ 2(x – 3).
Ans: Given that, 3(x – 1) ≤ 2 (x – 3)
By multiplying, the above inequality can be written as
3x – 3 ≤ 2x – 6
Now, by adding 3 to both the sides, we get
3x – 3+ 3 ≤ 2x – 6+ 3
3x ≤ 2x – 3
Again, by subtracting 2x from both the sides,
3x – 2x ≤ 2x – 3 – 2x
x ≤ -3
Therefore, the solutions of the given inequality are defined by all the real numbers less than or equal to -3.
Hence, the required solution set is (-∞, -3]
8. 3(2 – x) ≥ 2(1 – x).
Ans: Given that, 3 (2 – x) ≥ 2 (1 – x)
By multiplying, we get
6 – 3x ≥ 2 – 2x
Now, by adding 2x to both the sides,
6 – 3x + 2x ≥ 2 – 2x + 2x
6 – x ≥ 2
Again, by subtracting 6 from both the sides, we get
6 – x – 6 ≥ 2 – 6
– x ≥ – 4
Multiplying throughout inequality by negative sign, we get
x ≤ 4
∴ The solutions of the given inequality are defined by all the real numbers greater than or equal to 4.
Hence the required solution set is (- ∞, 4]
Ans:
Thus, all real numbers x, which are less than 6, are the solutions of the given inequality.
Hence, the solution set of the given inequality is ( – ∞, 6) .
Ans:
⇒ – x > 6
⇒ x < – 6
Thus, all real numbers x, which are less than -6, are the solutions of the given inequality.
Hence, the solution set of the given inequality is ( – ∞, – ∞ 6).
Ans:
Thus, all real numbers x, which are less than or equal to 2, are the solutions of the given inequality.
Hence, the solution set of the given inequality is ( – ∞ ,2].
Ans:
Thus, all real numbers x, which are less than or equal to 120, are the solutions of the given inequality.
Hence, the solution set of the given inequality is ( – ∞ ,120] .
13. 2(2x + 3) – 10 < 6(x – 2).
Ans: 2(2x+3)-10<6(x-2)
⇒ 4x+6-10<6x-12
⇒ 4x-4<6x-12
⇒ -4+12<6x-4x
⇒ 8<2.x
⇒ 4<x
Thus, all real numbers x, which are greater than or equal to 4, are the solutions of the given inequality.
Hence, the solution set of the given inequality is (4,∞).
14. 37 – (3x + 5) ≥ 9x – 8(x – 3).
Ans: 37 – (3x + 5) ≥ 9x – 8(x – 3)
⇒ 37 – 3x – 5 ≥9x – 8x + 24
⇒ 32 – 3x ≥ x + 24
⇒ 32 – 24 > x + 3x
⇒ 8 ≥ 4x
⇒ 2 ≥x
Thus, all real numbers x, which are less than or equal to 2, are the solutions of the given inequality.
Hence, the solution set of the given inequality is (- ∞, 2].
Ans:
⇒ 15x < 4(4x – 1)
⇒ 15x < 16x – 4
⇒ 4 < 16x – 15x
⇒ 4 < x
Thus, all real numbers x, which are greater than 4, are the solutions of the given inequality.
Hence, the solution set of the given inequality is (4, ∞).
Ans:
⇒ 20(2x – 1) ≥ 3(19x – 18)
⇒ 40x – 20 ≥ 57x – 54
⇒-20 + 54 ≥ 57x – 40x
⇒ 34 ≥ 17x
⇒ 2 ≥ x
Thus, all real numbers x, which are less than or equal to 2, are the solutions of the given inequality.
Hence, the solution set of the given inequality is ( – ∞,2].
Solve the inequalities in Exercises 17 to 20 and show the graph of the solution in each case on number line.
17. 3x – 2 < 2x + 1.
Ans: 3x – 2 < 2x + 1
⇒ 3x – 2x < 1 + 2
⇒ x < 3
The graphical representation of the solutions of the given inequality is as follows.
18. 5x – 3 ≥ 3x – 5.
Ans: 5x – 3 ≥ 3x – 5
⇒ 5x – 3x ≥ – 5 + 3
⇒ 2x ≥ – 2
⇒ 2x/2 ≥ – 2/2
⇒ x ≥ – 1
The graphical representation of the solutions of the given inequality is as follows.
19. 3(1 – x) < 2(x + 4).
Ans: 3(1-x)<2(x+4)
⇒ 3 – 3x <2x + 8
⇒ 3 – 8 < 2x + 3x
⇒ -5 <5x
⇒ -1<x
The graphical representation of the solutions of the given inequality is as follows.
Ans:
UJ
21. Ravi obtained 70 and 75 marks in first two unit test. Find the minimum marks he should get in the third test to have an average of at least 60 marks.
Ans: Let xbe the marks obtained by Ravi in the third unit test.
Since the student should have an average of at least 60 marks,
⇒145 + x ≥ 180
⇒x ≥ 180 – 145
⇒x ≥ 35
Thus, the student must obtain a minimum of 35 marks to have an average of at least 60 marks.
22. To receive Grade ‘A’ in a course, one must obtain an average of 90 marks or more in five examinations (each of 100 marks). If Sunita’s marks in first four examinations are 87, 92, 94 and 95, find minimum marks that Sunita must obtain in fifth examination to get grade ‘A’ in the course.
Ans: Let xbe the marks obtained by Sunita in the fifth examination.
In order to receive grade ‘A’ in the course, she must obtain an average of 90 marks or more in five examinations.
Therefore,
⇒ 368 + x ≥= 450
⇒ x ≥ 450 – 368
⇒ x ≥ 82
Thus, Sunita must obtain greater than or equal to 82 marks in the fifth examination.
23. Find all pairs of consecutive odd positive integers both of which are smaller than 10 such that their sum is more than 11.
Ans: Let x be the smaller of the two consecutive odd positive integers. Then, the other integer is x + 2.
Since both the integers are smaller than 10,
x + 2 < 10
⇒ x< 10 – 2
⇒ x<8… (i)
Also, the sum of the two integers is more than 11.
∴ x + (x + 2) > 11
⇒ 2x + 2 > 11
⇒ 2x >11 – 2
⇒ 2x > 9
⇒ x > 9/2
⇒ x > 4.5 …(ii)
From (i) and (ii), we obtain
Since x is an odd number, xcan take the values, 5 and 7.
Thus, the required possible pairs are (5, 7) and (7, 9).
24. Find all pairs of consecutive even positive integers, both of which are larger than 5 such that their sum is less than 23.
Ans: Let the required consecutive even integers be x and x + 2 Then,
x > 5 and x + (x + 2) < 23
⇒ x > 5 and 2x < 21
⇒ 5 < x and x < 10.5
⇒ 5 < x < 10.5
∴ x can take the even integral value 6, 8 and 10.
Hence, the required pairs of even integers are (6, 8), (8, 10) and (10, 12).
25. The longest side of a triangle is 3 times the shortest side and the third side is 2 cm shorter than the longest side. If the perimeter of the triangle is at least 61 cm, find the minimum length of the shortest side.
Ans: Let the length of the shortest side of the triangle be x cm.
Then, length of the longest side = 3xcm
Length of the third side = (3x- 2) cm
Since the perimeter of the triangle is at least 61 cm,
x cm + 3x cm + (3x – 2) cm ≥ 61cm
⇒7x – 2 ≥ 61
⇒ 7x ≥ 61 + 2
⇒ 7x ≥ 63
⇒ 7x/7 ≥ 63/7
⇒ x ≥ 9
Thus, the minimum length of the shortest side is 9 cm.
26. A man wants to cut three lengths from a single piece of board of length 91cm. 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?[Hint: If x is the length of the shortest board, then x, (x + 3) and 2x are the lengths of the second and third piece, respectively. Thus, x+(x+3)+2 x≤91 and 2x ≥ (x+3) +5].
Ans: Let the length of the shortest piece be xcm. Then, length of the second piece and the third piece are (x + 3) cm and 2xcm respectively.
Since the three lengths are to be cut from a single piece of board of length 91 cm,
X cm + (x+3) cm + 2xcm ≤91 cm
⇒ 4x + 3 ≤ 91
⇒ 4x ≤ 91- 3
⇒ 4x ≤ 88
⇒ 4x/4 ≤ 88/4
⇒ x ≤ 22 …(1)
Also, the third piece is at least 5 cm longer than the second piece.
∴ 2x ≥ (x + 3) + 5
⇒ 2x ≥ x + 8
⇒ x ≥8… (2)
From (1) and (2), we obtain
8 ≤ x ≤ 22
Thus, the possible length of the shortest board is greater than or equal to 8 cm but less than or equal to 22 cm.
| Miscellaneous Exercise on Chapter 5 |
1. 2 ≤ 3x – 4 ≤ 5.
Ans: The inequality given is,
2 ≤ 3x – 4 ≤ 5
⇒ 2 ≤ 3x – 4 ≤ 5
⇒ 2 + 4 ≤ 3x – 4 + 4 ≤ 5 + 4
⇒6 ≤ 3x ≤ 9
⇒6/3 ≤ 3x /3 ≤ 9/3
⇒ 2 ≤ x ≤ 3
Hence, all real numbers x greater than or equal to 2, but less than or equal to 3 are solutions of given equality.
x ∈ [2, 3]
2. 6 ≤ – 3(2x – 4) < 12.
Ans: The inequality given is.
6 ≤ – 3(2x – 4) < 12
⇒ 6 ≤ – 3(2x – 4) < 12
Dividing the inequality by 3, we get.
⇒ 2 ≤ – (2x – 4) < 4
Multiplying the inequality by -1,
⇒ – 2 ≥ 2x – 4 > – 4 [multiplying the inequality with -1 changes the inequality sign.]
⇒ – 2 + 4 ≥ 2x – 4 + 4 > – 4 + 4
⇒ 2 ≥ 2x > 0
Dividing the inequality by 2,
⇒ 0 < x ≤ 1
Hence, all real numbers x greater than 0, but less than or equal to 1 are solutions of given equality.
x ∈ (0, 1)
Ans:
⇒ 2 ≥ x ≥ – 4
Thus, the solution set for the given inequalities [- 4, 2].
Ans:
⇒- 75 <3( x- 2)≤0
⇒ – 25 < x – 2≤0
⇒ – 25 + 2 < x ≤2
⇒ – 23 < x ≤2
Thus, the solution set for the given inequalities ( – 23,2].
Ans:
Ans:
Thus, the solution set for the given inequalities [1.11/3]
Solve the inequalities in Exercises 7 to 10 and represent the solution graphically on number line.
7. 5x + 1 > – 24 5x – 1 < 24.
Ans: 5x + 1 > – 24
⇒ 5x > – 25
⇒ x>-5… (1)
5x – 1 < 24
⇒ 5x < 25
⇒ x<5 … (2)
From (1) and (2), it can be concluded that the solution set for the given system of inequalities is (-5, 5). The solution of the given system of inequalities can be represented on number line as
8. 2(x – 1) < x + 5, 3(x + 2) > 2 – x.
Ans: 2(x – 1) < x + 5
⇒ 2x – 2 < x + 5
⇒ 2x – x < 5 + 2
⇒ x<7… (1)
3(x + 2) > 2 – x
⇒ 3x + 6 > 2 – x
⇒ 3x + x > 2 – 6
⇒ 4x > – 4
⇒ x>-1… (2)
From (1) and (2), it can be concluded that the solution set for the given system of inequalities is (-1, 7). The solution of the given system of inequalities can be represented on number line as
9. 3x – 7 > 2(x – 6) , 6 – x > 11 – 2x.
Ans: 3x – 7 > 2(x – 6)
⇒ 3x – 7 > 2x – 12
⇒3x – 2x > – 12 +7
⇒ x > -5 … (1)
6 – x > 11 – 2x
⇒ -x + 2x > 11 – 6
⇒ x > 5 … (2)
From (1) and (2), it can be concluded that the solution set for the given system of inequalities is (5,∞) The solution of the given system of inequalities can be represented on number line as
10. 5(2x – 7) – 3 (2x + 3) ≤ 0 , 2x + 19 ≤ 6x + 47.
Ans: 5(2x – 7) – 3(2x + 3) ≤ 0
⇒ 10x – 35 – 6x – 9 ≤ 0
⇒ 4x – 44 ≤ 0
⇒ 4x ≤ 44
⇒ x ≤ 11 …(1)
2x + 19 ≤ 6x + 47
⇒ 19 – 47 ≤ 6x – 2x
⇒ – 28 ≤ 4x
⇒ – 7 ≤ x …(2)
From (1) and (2), it can be concluded that the solution set for the given system of inequalities is [-7, 11]. The solution of the given system of inequalities can be represented on number line as
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
Ans: Since the solution is to be kept between 68°F and 77°F,
68 < F < 77
Thus, the required range of temperature in degree Celsius is between 20°C and 25°C.
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?
Ans: Let x litres of 2% boric acid solution is required to be added.
Then, total mixture = (x+ 640) litres
This resulting mixture is to be more than 4% but less than 6% boric acid.
∴ 2%x+ 8% of 640 > 4% of (x+ 640)
And, 2%x + 8% of 640 < 6% of (x + 640)
2%x + 8% of 640 > 4% of (x + 640)
⇒ 2x + 5120 > 4x + 2560
⇒ 5120 – 2560 > 4x – 2x
⇒ 5120 – 2560 > 2x
⇒2560> 2x
⇒ 1280 > x
2% X + 8% of 640 < 6% of (x + 640)
⇒ 2x + 5120 < 6x + 3840
⇒ 5120 – 3840< 6x – 2x
⇒ 1280 < 4x
⇒ 320 < x
∴ 320 < x < 1280
Thus, the number of litres of 2% of boric acid solution that is to be added will have to be more than 320 litres but less than 1280 litres.
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?
Ans: Let x litres of water is required to be added.
Then, total mixture = (x+ 1125) litres
It is evident that the amount of acid contained in the resulting mixture is 45% of 1125 litres.
This resulting mixture will contain more than 25% but less than 30% acid content.
∴ 30% of (1125 + x) > 45% of 1125
And, 25% of (1125 + x) < 45% of 1125
30% of (1125 + x) > 45% of 1125
∴ 562.5 < x < 900
Thus, the required number of litres of water that is to be added will have to be more than 562.5 but less than 900.
14. IQ of a person is given by the formula 
where MA is mental age and CA is chronological age. If 80 ≤ IQ ≤ 140 for a group of 12 years old children, find the range of their mental age.
Ans: It is given that for a group of 12 years old children, 80 ≤ 1Q ≤140…(i)
For a group of 12 years old children, CA = 12 years
Putting this value of IQ in (i), we obtain
