Linear Inequalities (Ex – 6.1)
Question 1.
Solve 24x < 100, when
(i) x is a natural number
(ii) x is an integer.
Solution.
Question 2.
Solve – 12x > 30, when
(i) x is a natural number
(ii) x is an integer
Solution.
The linear inequality solver she used can give test data that are not certain to produce the required path due to rounding errors.
Question 3.
Solve 5x – 3 < 7, when
(i) x is an integer
(ii) x is a real number
Solution.
Given inequality is 5x – 3 < 7
Transposing 3 to R.H.S., we get
5x < 7 + 3 or 5x < 10
Dividing both sides by 5, we get
x < 2
(i) When x is an integer, {…. -2, -1, 0, 1} satisfies this inequality.
(ii) When x is real number, the solution is (-∞, 2).
Question 4.
Solve 3x + 8 > 2, when
(i) x is an integer
(ii) x is a real number.
Solution.
Given inequality is 3x + 8 > 2
Transposing 8 to R.H.S., we get
3x > 2 – 8 = -6
Dividing both sides by 3, we get
x > -2
(i) When x is an integer, the solution is (-1, 0, 1, 2, 3,…}
(ii) When x is real, the solution is (-2, ∞).
Solve the inequalities in Exercises 5 to 16 for real x.
Question 5.
4x + 3 < 5x + 7
Solution.
The inequality is 4x + 3 < 5x + 7
Transposing 5x to L.H.S. and 3 to R.H.S., we get
4x – 5x < 7 – 3 or -x < 4 Dividing both sides by -1, we get x > -4
∴ The solution is (- 4, ∞).
Question 6.
3x – 7 > 5x – 1
Solution.
The inequality is 3x – 7 > 5x -1
Transposing 5x to L.H.S. and -7 to R.H.S., we get
3x – 5x > -1 + 7 or -2x > 6
Dividing both sides by -2, we get
x < -3
∴ The solution is (-∞, -3).
Question 7.
3(x – 1) < 2(x – 3)
Solution.
The inequality is 3(x – 1) < 2(x – 3) or 3x – 3 < 2x – 6
Transposing 2x to L.H.S. and -3 to R.H.S., we get
3x – 2x < – 6 + 3
⇒ x<-3
∴ The solution is (- ∞, -3],
Question 8.
3(2 -x) > 2(1 -x)
Solution.
The inequality is 3(2 – x) > 2(1 – x) or 6 – 3x > 2 – Zx
Transposing -2x to L.H.S. and 6 to R.H.S., we get
-3x + 2x > 2 – 6 or -x > -4
Multiplying both sides by -1, we get
x ≤ 4
∴ The solution is (- ∞, 4],
Question 9.
Solution.
Question 10.
Solution.
Question 11.
Solution.
Question 12.
Solution.
Question 13.
2(2x + 3) – 10 < 6(x – 2)
Solution.
The inequality is 2(2x + 3) – 10 < 6(x – 2)
Simplifying, 4x + 6 -10 < 6x -12
or, 4x – 4 < 6x – 12
Transposing 6x to L.H.S. and – 4 to R.H.S., we get
∴ 4x – 6x < -12 + 4 or -2x < – 8 Dividing both sides by -2, we get x>4
∴ The solution is (4, ∞).
Question 14.
37 – (3x + 5) ≥ 9x – 8(x – 3)
Solution.
The inequality is 37 – (3x + 5) ≥ 9x – 8(x – 3)
Simplifying, 37 – 3x – 5 ≥ 9x – 8x + 24
or 32 – 3x ≥ x + 24
Transposing x to L.H.S. and 32 to R.H.S., We get
-3x – x ≥ 24 – 32 or -4x ≥ -8
Dividing both sides by – 4, we get
x < 2
∴ The solution is (- ∞, 2].
Question 15.
Solution.
Question 16.
Solution.
Solve the inequalities in Exercises 17 to 20 and show the graph of the solution in each case on number line.
Question 17.
3x – 2 < 2x + 1
Solution.
The inequality is 3x – 2 < 2x + 1
Transposing 2x to L.H.S. and -2 to R.H.S, we get
3x- 2x < 1 + 2 or, x <3
Question 18.
5x – 3 ≥ 3x – 5
Solution.
The inequality is 5x – 3 ≥ 3x – 5
Transposing 3x to L.H.S. and -3 to R.H.S., we get
∴ 5x – 3x ≥ -5 + 3 or, 2x ≥ -2
Dividing both sides by 2, we get
Question 19.
3(1 – x) < 2(x + 4)
Solution.
3(1-x) < 2(x + 4)
Simplifying 3 – 3x < 2x + 8
Transposing 2x to L.H.S. and 3 to R.H.S., we get
-3x – 2x < 8 – 3 or -5x < 5
Dividing both sides by -5, we get
Question 20.
Solution.
Question 21.
Ravi obtained 70 and 75 marks in first two unit tests. Find the minimum marks he should get in the third test to have an average of at least 60 marks.
Solution.
Question 22.
To receive Grade ‘A’ in the 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.
Solution.
Question 23.
Find all pairs of consecutive odd positive integers both of which are smaller than 10, such that their sum is more than 11.
Solution.
Question 24.
Find all pairs of consecutive even positive integers, both of which are larger than 5, such that their sum is less than 23.
Solution.
Question 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.
Solution.
Let the shortest side measures x cm
The longest side will be 3x cm.
Third side will be (3x – 2) cm.
According to the problem, x + 3x + 3x – 2 ≥ 61
or, 7x – 2 ≥ 61 or, 7x ≥ 63 or, x ≥ 9
Hence, the minimum length of the shortest side is 9 cm.
Question 26.
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 for the shortest board if the third piece is to be atleast 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) + 2x ≤ 91 and 2x ≥ (x + 3) + 5].
Solution.
Let x be the length of the shortest board, then x + 3 is the second length and 2x is the third length. Thus, x + (x + 3) + 2x ≤ 91
or 4x + 3 ≤ 91 or 4x ≤ 88 or x ≤ 22
According to the problem, 2x ≥ (x + 3) + 5 or x ≥ 8
∴ Atleast 8 cm but not more than 22 cm are the possible lengths for the shortest board.
