Maharashtra State Board Class 11 (FYJC / HSC) Commerce Mathematics & Statistics Solutions Chapter 8 Linear Inequations Ex 8.1 - Digest Answers | Maharashtra Board Solutions

Q: Question 5: Rajiv obtained 70 and 75 marks in the first two unit tests. Find the minimum marks he should get in the third test to have an average of at least 60 marks.

Let x1, x2, x3 denote the marks in 1st, 2nd and 3rd unit test respectively. Then ≥ 60 ∴ ≥ 60 ∴ 145 + x3 ≥ 3(60) Subtracting 145 from both sides, we get x3 ≥ 180 – 145 ∴ x3 ≥ 35 Rajiv must obtain a minimum of 35 marks to maintain an average of at least 60 marks.

Q: Question 6: 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 the first four examinations are 87,

Let x1, x2, x3, x4, x5 denote the marks in five examinations. Then ≥ 90 ∴ ≥ 90 ∴ 368 + x5 ≥ 450 Subtracting 368 from both sides, we get ∴ x5 ≥ 82 Sunita must obtain a minimum of 82 marks in the 5th examination to get a grade of A.

Q: Question 7: Find all pairs of consecutive odd positive integers, both of which are smaller than 10 such that their sum is more than 11.

Let two consecutive positive integers be 2n – 1, 2n + 1 where n ≥ 1 ∈ Z, Given that 2n – 1 < 10 and 2n + 1 < 10 ∴ 2n < 11 and 2n < 9 ∴ 2n < 9 ∴ n < …..(i) Also, (2n – 1) + (2n + 1) > 11 ∴ 4n > 11 ∴ n > …….(ii) From (i) and (ii) Since, n is an integer, ∴ n = 3, 4 n = 3 give

Q: Question 9: 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 t

Let the shortest side be x. Then longest side length = 2x and third side length = x + 2 Perimeter = x + 2x + x + 2 = 4x +2 Given, perimeter > 166 ∴ 4x + 2 > 166 ∴ 4x > 164 ∴ x > 41 ∴ Minimum integer length of shortest side is 42 cm.

Balbharati Maharashtra State Board11th Commerce Maths Solution Book PdfChapter 8 Linear Inequations Ex 8.1 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 8 Linear Inequations Ex 8.1

Question 1 Maharashtra Board Solution

Write the inequations that represent the interval and state whether the interval is bounded or unbounded: (i) [-4, ]

Solution & Step-by-Step Answer:

[-4, ] Here, x takes values between -4 and including -4 and ∴ the required inequation is -4 ≤ x ≤ ∴ it is a bounded (closed) interval.

(ii) (0, 0.9]
Solution:
(0, 0.9]
Here, x takes values between 0 and 0.9, including 0.9 and excluding 0.
∴ the required inequation is 0 < x ≤ 0.9
∴ it is a bounded (semi-right closed) interval.

(iii) (-∞, ∞)
Solution:
(-∞, ∞)
Here, x takes values between -∞ and ∞
∴ the required inequation is -∞ < x < ∞
∴ it is an unbounded (open) interval.

(iv) [5, ∞)
Solution:
[5, ∞)
Here, x takes values between 5 and ∞ including 5.
∴ the required inequation is 5 ≤ x < ∞
∴ it is an unbounded (semi-left closed) interval.

(v) (-11, -2)
Solution:
(-11, -2)
Here, x takes values between -11 and -2
∴ the required inequation is -11 < x < -2
∴ it is a bounded (open) interval.

(vi) (-∞, 3)
Solution:
(-∞, 3)
Here, x takes values between -∞ and 3
∴ the required inequation is -∞ < x < 3
∴ it is an unbounded (open) interval.

Question 2 Maharashtra Board Solution

Solve the following inequations (i) 3x – 36 > 0

Solution & Step-by-Step Answer:

3x – 36 > 0 Adding 36 both sides, we get 3x – 36 + 36 > 0 + 36 ∴ 3x > 36 Dividing both sides by 3, we get ∴ x > 12 ∴ x takes all real values more than 12 ∴ Solution set = (12, ∞)

(ii) 7x – 25 ≤ -4
Solution:
7x – 25 ≤ -4
Adding 25 on both sides, we get
7x – 25 + 25 ≤ -4 + 25
∴ 7x ≤ 21
Dividing both sides by 7, we get
x ≤ 3
∴ x takes all real values less or equal to 3.
∴ Solution Set = (-∞, 3]

(iii) 0 < < 3
Solution:
0 < < 3
0 < x – 5 < 12
Adding 5 on both sides, we get
5 < x < 17
x takes all real values between 5 and 17.
∴ Solution set = (5, 17)

(iv) |7x – 4| < 10
Solution:
|7x – 4| < 10
-10 < 7x – 4 < 10 …….[|x| < k is same as -k < x < k]
Adding 4 on both sides, we get
-6 < 7x < 14
Dividing both sides by 7, we get

∴ < x < 2
∴ x takes all real values between and 2.
∴ Solution set = (, 2)

Question 3 Maharashtra Board Solution

Sketch the graph which represents the solution set for the following inequations: (i) x > 5

Solution & Step-by-Step Answer:

x > 5 Here, x takes all real values that are greater than 5. ∴ Solution set represents the unbounded (open) interval (5, ∞) ∴ the required graph of the solution set is as follows:

(ii) x ≥ 5
Solution:
x ≥ 5
Here, x takes all real values that are greater than or equal to 5
∴ Solution set represents the unbounded (semi-left closed) interval [5, ∞)
∴ the required graph of the solution set is as follows:

(iii) x < 3
Solution:
x < 3
Here, x takes all real values that are less than 3.
∴ Solution set represents the unbounded (open) interval (-∞, 3)
∴ the required graph of the solution set is as follows:

(iv) x ≤ 3
Solution:
x ≤ 3
Here, x takes all real values less than and including 3
∴ Solution set represents the unbounded (semi-right closed) interval (-∞, 3]
∴ the required graph of the solution set is as follows:

(v) -4 < x < 3
Solution:
-4 < x < 3
Here, x takes all real values between -4 and 3.
∴ Solution set represents the bounded (open) interval (-4, 3)
∴ the required graph of the solution set is as follows:

(vi) -2 ≤ x < 2.5
Solution:
-2 ≤ x < 2.5
Here, x takes all values between -2 and 2.5 including -2.
∴ Solution set represents the bounded (semi-left closed) interval [-2, 2.5)
∴ the required graph of the solution set is as follows.

(vii) -3 ≤ x ≤ 1
Solution:
-3 ≤ x ≤ 1
Here, x takes all real values between -3 and 1 including -3 and 1
∴ Solution set represents the bounded (closed) interval [-3, 1]
∴ the required graph of the solution set is as follows:

(viii) |x| < 4
Solution:
|x| < 4 ⇒ -4 < x < 4
Here, x takes all real values between -4 and 4.
∴ Solution set represents bounded (open) interval (-4, 4)
∴ the required graph of the solution set is as follows:

(ix) |x| ≥ 3.5
Solution:
|x| ≥ 3.5 ⇒ x ≥ 3.5 or x ≤ -3.5
Here, x takes values greater than or equal to 3.5 or it takes values less than or equal to -3.5
∴ Solution set represents the unbounded (semi-left closed) interval [3.5, ∞) or the unbounded (semi-right closed) interval (-∞, -3.5]
∴ x ∈ (-∞, -3.5] ∪ [3.5, ∞)
∴ the required graph of the solution set is as follows:

Question 4 Maharashtra Board Solution

Solve the inequations: (i) 5x + 7 > 4 – 2x

Solution & Step-by-Step Answer:

5x + 7 > 4 – 2x Adding 2x on both sides, we get 7x + 7 > 4 Subtracting 7 from both sides, we get 7x > -3 Dividing by 7 on both sides, we get ∴ x > i.e., x takes all real values greater than ∴ the solution set is (, ∞)

(ii) 3x + 1 ≥ 6x – 4
Solution:
3x + 1 ≥ 6x – 4
Subtracting 3x from both sides, we get
1 ≥ 3x – 4
Adding 4 on both sides, we get
5 ≥ 3x
Dividing by 3 on both sides, we get
≥ x
i.e., x ≤
i.e., x takes all real values less than or equal to .
∴ the solution set is (-∞, ]

(iii) 4 – 2x < 3(3 – x)
Solution:
4 – 2x < 3(3 – x)
∴ 4 – 2x < 9 – 3x
Adding 3x on both sides, we get
4 + x < 9
Subtracting 4 from both sides, we get
x < 5
i.e., x takes all real values less than 5
∴ the solution set is (-∞, 5)

(iv) x – 6 ≤ x – 7
Solution:
x – 6 ≤ x – 7
Multiplying by 4 on both sides, we get
3x – 24 ≤ 4x – 28
Subtracting 3x from both sides, we get
-24 ≤ x – 28
Adding 28 on both the sides, we get
∴ 4 ≤ x i.e., x ≥ 4
i.e., x takes all real values greater or equal to 4.
∴ the solution set is [4, ∞)

(v) -8 ≤ -(3x – 5) < 13
Solution:
-8 < -(3x – 5) < 13 Multiplying by -1 throughout (so inequality sign changes) 8 ≥ 3x – 5 > -13
i.e., -13 < 3x – 5 ≤ 8
Adding 5 on both the sides, we get
-8 < 3x ≤ 13
Dividing, by 3 on both sides, we get
∴ < x ≤
i.e., x takes all real values between and including .
∴ the solution set is

(vi) -1 < 3 – ≤ 1
Solution:
-1 < 3 – ≤ 1
Subtracting 3 from both sides, we get
-4 < – < -2 Multiplying by -1 throughout (so inequality sign changes) ∴ 4 > > 2
i.e., 2 < < 4
Multiplying by 5 on both sides, we get
10 < x < 20
i.e., x takes all real values between 10 and 20.
∴ the solution set is (10, 20)

(vii) 2|4 – 5x| ≥ 9
Solution:
2|4 – 5x | ≥ 9
∴ |4 – 5x| ≥
∴ 4 – 5x ≥ or 4 – 5x ≤ – ……[|x| ≥ a implies x ≤ -a or x ≥ a]
Subtracting 4 from both sides, we get
-5x ≥ or -5x ≤
Divide by -5 (so inequality sign changes)
∴ x ≤ or x ≥
∴ x takes all real values less than or equal to
or it takes all real values greater or equal to .
∴ the solution set is (-∞, ] or [, ∞)

(viii) |2x + 7| ≤ 25
Solution:
|2x + 7| < 25
∴ -25 ≤ 2x + 7 ≤ 25 …..[|x| ≤ a implies -a ≤ x ≤ a]
Subtracting 7 from both sides, we get
-32 ≤ 2x ≤ 18
Dividing by 2 on both sides, we get
-16 ≤ x ≤ 9
∴ x can take all real values between -16 and 9 including -16 and 9.
∴ the solution set is [-16, 9]

(ix) 2|x + 3| > 1
Solution:
2|x + 3| > 1
Dividing by 2 on both sides, we get
|x + 3| >
∴ x + 3 < – or x + 3 > …..[|x| > a implies x < -a or x > a]
Subtracting 3 from both sides, we get
x < – 3 – or x > -3 +
∴ x < or x >
∴ x can take all real values less or it can take values greater than .
∴ Solution set is (-∞, ) ∪ (, ∞)

(x) < 0
Solution:
< 0
Since < 0, when a > 0 and b < 0 or a < 0 and b > 0
∴ either x + 5 > 0 and x – 3 < 0
or x + 5 < 0 and x – 3 > 0
Case I:
x + 5 > 0 and x – 3 < 0 ∴ x > -5 and x < 3
∴ -5 < x < 3
∴ solution set = (-5, 3)
Case II:
x + 5 < 0 and x – 3 > 0
∴ x < -5 and x > 3
which is not possible
∴ solution set = Φ
∴ solution set of the given inequation is (-5, 3)

(xi) > 0
Solution:
> 0
Since > 0,
when a > 0 and b > 0 or a < 0 and b < 0 b
∴ either x – 2 > 0 and x + 5 > 0
or x – 2 < 0 and x + 5 < 0 Case I: x – 2 > 0 and x + 5 > 0
∴ x > 2 and x > -5
∴ x > 2
∴ solution set = (2, ∞)
Case II:
x – 2 < 0 and x + 5 < 0
∴ x < 2 and x < -5
∴ x < -5
∴ solution set = (-∞, -5)
∴ the solution set of the given inequation is (-∞, -5) ∪ (2, ∞)

Question 5 Maharashtra Board Solution

Rajiv obtained 70 and 75 marks in the 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 & Step-by-Step Answer:

Let x1, x2, x3 denote the marks in 1st, 2nd and 3rd unit test respectively. Then ≥ 60 ∴ ≥ 60 ∴ 145 + x3 ≥ 3(60) Subtracting 145 from both sides, we get x3 ≥ 180 – 145 ∴ x3 ≥ 35 Rajiv must obtain a minimum of 35 marks to maintain an average of at least 60 marks.

Question 6 Maharashtra Board Solution

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 the first four examinations are 87, 92, 94, and 95, find the minimum marks that Sunita must obtain in the fifth examination to get a grade ‘A’ in the course.

Solution & Step-by-Step Answer:

Let x1, x2, x3, x4, x5 denote the marks in five examinations. Then ≥ 90 ∴ ≥ 90 ∴ 368 + x5 ≥ 450 Subtracting 368 from both sides, we get ∴ x5 ≥ 82 Sunita must obtain a minimum of 82 marks in the 5th examination to get a grade of A.

Question 7 Maharashtra Board Solution

Find all pairs of consecutive odd positive integers, both of which are smaller than 10 such that their sum is more than 11.

Solution & Step-by-Step Answer:

Let two consecutive positive integers be 2n – 1, 2n + 1 where n ≥ 1 ∈ Z, Given that 2n – 1 < 10 and 2n + 1 < 10 ∴ 2n < 11 and 2n < 9 ∴ 2n < 9 ∴ n < …..(i) Also, (2n – 1) + (2n + 1) > 11 ∴ 4n > 11 ∴ n > …….(ii) From (i) and (ii) Since, n is an integer, ∴ n = 3, 4 n = 3 gives 2n – 1 = 5, 2n + 1 = 7 and n = 4 gives 2n – 1 = 7, 2n + 1 = 9 ∴ The pairs of positive consecutive integers are (5, 7) and (7, 9).

Question 8 Maharashtra Board Solution

Find all pairs of consecutive even positive integers, both of which are larger than 5 such that their sum is less than 23.

Solution & Step-by-Step Answer:

Let 2n, 2n + 2 be two positive consecutive integers where n ≥ 1 ∈ Z. Given that 2n > 5 and 2n + 2 > 5 ∴ n > and 2n > 3 ∴ n > and n > ∴ n > ……(i) Also (2n) + (2n + 2) < 23 ∴ 4n + 2 < 23 ∴ 4n < 21 ∴ n < ……(ii) From (i) and (ii) and n is an integer. ∴ n = 3, 4, 5 n = 3 gives 2n = 6, 2n + 2 = 8 n = 4 gives 2n = 8, 2n + 2 = 10 n = 5 gives 2n = 10, 2n + 2 = 12 ∴ The pairs of positive even consecutive integers are (6, 8) (8, 10), (10, 12)

Question 9 Maharashtra Board Solution

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 integer length of the shortest side.

Solution & Step-by-Step Answer:

Let the shortest side be x. Then longest side length = 2x and third side length = x + 2 Perimeter = x + 2x + x + 2 = 4x +2 Given, perimeter > 166 ∴ 4x + 2 > 166 ∴ 4x > 164 ∴ x > 41 ∴ Minimum integer length of shortest side is 42 cm.

Chapter 8 Linear Inequations Ex 8.1 Solutions