# Class 11 RD Sharma Solutions – Chapter 15 Linear Inequations – Exercise 15.4

• Last Updated : 03 Mar, 2021

### Question 1: Find all pairs of consecutive odd positive integers, both of which are smaller than 10, such that their sum is more than 11.

Solution:

Let x and x + 2 be the two consecutive odd positive integers.

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Given that the integers are smaller than 10 and their sum is more than 11.

Therefore,

x + 2 < 10 and x + (x + 2) > 11

x < 10 – 2 and 2x + 2 > 11

x < 8 and 2x > 11 – 2

x < 8 and 2x > 9

x < 8 and x > 9/2

9/2 < x < 8

Therefore, the two consecutive odd positive integers are x = 5, 7

The pairs of consecutive odd integers are

Let x = 5, then (x + 2) = (5 + 2) = 7.

Let x = 7, then (x + 2) = (7 + 2) = 9.

Therefore, the required pairs of odd integers are (5, 7) and (7, 9).

### Question 2: Find all pairs of consecutive odd natural numbers, both of which are larger than 10, such that their sum is less than 40.

Solution:

Let x and x + 2 be the two consecutive odd positive integers.

Given that both the odd natural numbers are greater than 10 and their sum is less than 40.

Therefore,

x > 10 and x + x + 2 <40

x > 10 and 2x < 38

x > 10 and x < 38/2

x > 10 and x < 19

10 < x < 19

Therefore, the two consecutive odd positive integers are x = 11, 13, 15, 17

The pairs of consecutive odd integers are

Let x = 11, then (x + 2) = (11 + 2) = 13

Let x = 13, then (x + 2) = (13 + 2) = 15

Let x = 15, then (x + 2) = (15 + 2) = 17

Let x = 17, then (x + 2) = (17 + 2) = 19.

Therefore, the required pairs of odd natural numbers are (11, 13), (13, 15), (15, 17), and (17, 19)

### Question 3: Find all pairs of consecutive even positive integers, both of which are larger than 5, such that their sum is less than 23.

Solution:

Let x and x + 2 be the two consecutive even positive integers.

Given that both the even integers are greater than 5 and their sum is less than 23.

Therefore,

x > 5 and x + x + 2 < 23

x > 5 and 2x < 21

x > 5 and x < 21/2

5 < x < 21/2

5 < x < 10.5

Therefore, the consecutive even positive integers are x = 6, 8, 10

The pairs of consecutive even integers are

Let x = 6, then (x + 2) = (6 + 2) = 8

Let x = 8, then (x + 2) = (8 + 2) = 10

Let x = 10, then (x + 2) = (10 + 2) = 12.

Therefore, the required pairs of even positive integer are (6, 8), (8, 10), and (10, 12)

### Question 4. The marks scored by Rohit in two tests were 65 and 70. Find the minimum marks he should score in the third test to have an average of at least 65 marks.

Solution:

Given: marks scored by Rohit in two tests are 65 and 70.

Let marks scored by Rohit in the third test be x.

The average marks in the three papers ≥ 65

Average = (marks in 1st two papers + marks in the third test)/3

(65 + 70 + x)/3 ≥ 65

(135 + x)/3 ≥ 65

(135 + x) ≥ 65 × 3

(135 + x) ≥ 195

x ≥ 195 – 135

x ≥ 60

Since, the minimum marks to get an average of 65 marks is 60.

Therefore, the minimum marks required in the third test is 60.

### Question 5: A solution is to be kept between 86° and 95°F. What is the range of temperature in degree Celsius, if the Celsius (C)/Fahrenheit (F) conversion formula is given by F = 9/5C + 32.

Solution:

Let us consider F1 = 86° F and F2 = 95°

Given, F = 9/5C + 32

F1 = 9/5 C1 + 32

F1 – 32 = 9/5 C1

C1 = 5/9 (F1 – 32)

C1 = 5/9 (86 – 32)

C1 = 5/9 (54)

C1 = 5 × 6

C1 = 30° C

Now,

F2 = 9/5 C2 + 32

F2 – 32 = 9/5 C2

C2 = 5/9 (F2 – 32)

C2 = 5/9 (95 – 32)

C2 = 5/9 (63)

C2 = 5 × 7

C2 = 35° C

Therefore, the range of temperature of the solution is 30°C and 35°C.

### Question 6: A solution is to be kept between 30°C and 35°C. What is the range of temperature in degrees Fahrenheit?

Solution:

Let us consider C1 = 30°C and C2 = 35°C

We know, F = 9/5C + 32

F1 = 9/5 C1 + 32

= 9/5 × 30 + 32

= 9 × 6 + 32

= 54 + 32

= 86°F

Now,

F2 = 9/5 C2 + 32

= 9/5 × 35 + 32

= 9 × 7 + 32

= 63 + 32

= 95°F

Therefore, the range of temperature of the solution is 86°F and 95°F.

### Question 7: To receive grade ‘A’ in a course, one must obtain an average of 90 marks or more in five papers each of 100 marks. If Shikha scored 87, 95, 92, and 94 marks in the first four papers each the minimum marks that she must score in the last paper to get grade ‘A’ in the course.

Solution:

Given: marks scored by Shikha in the first four papers are 87, 95, 92, and 94.

Let marks scored by Shikha in the fifth test be x.

The average marks in the five papers ≥ 90

Average = (marks in 1st four papers + marks in the fifth test)/5

(87 + 95 + 92 + 94 + x)/5 ≥ 90

182 + 186 + x ≥ 90 × 5

368 + x ≥ 450

x ≥ 450 – 368

x ≥ 82

Since, the minimum marks to get an average of 0 marks is 82.

Therefore, the minimum marks required in the fifth paper is 82.

### Question 8: A company manufactures cassettes and its cost and revenue functions for a week are C = 300 + (3/2)x and R = 2x respectively, where x is the number of cassettes produced and solid in a weel. How many cassettes must be sold for the company to realize a profit?

Solution:

Given: Cost = 300 + (3/2)x and Revenue = 2x

As profit = Revenue – Cost

Therefore, to earn a profit, the revenue should be greater than the cost.

Revenue > Cost

2x > 300 + (3/2)x

2x – (3/2)x > 300

(4x – 3x) > 600

x > 600

Therefore, the manufacture must sell more than 600 cassettes to gain profit.

### Question 9: The longest side of a triangle is three times the shortest side and the third side is 2 cm shorter than the longest side if the perimeter of the triangles at least 61cm. Find the minimum length of the shortest side.

Solution:

Let the length of the shortest side be x.

Given, the longest side of a triangle is three times the shortest side = 3x

and the third side is 2 cm less than the longest side = 3x – 2

Also given that the perimeter of the triangle ≥ 61

Therefore,

x + 3x – 2 + 3x ≥ 61

7x ≥ 61 + 2

7x ≥ 63

x ≥ 63/7

x ≥ 9

Therefore, the minimum length of the shortest side is 9cm.

### Question 10: How many liters of water will have to be added to 1125 liters of the 45% solution of acid so that the resulting mixture will contain more than 25% but less than 30% acid content?

Solution:

Let the quantity of water to be added in liters be x

Therefore,

25% of (1125 + x) < 45% of 1125

25/100(1125 + x) < 45/100 1125

1125 + x < (45 × 1125)/25

1125 + x < 45 × 45

1125 + x < 2025

x < 2025 – 1125

x < 900   ……. (1)

Also given that 45% of 1125 < 30% of (1125 + x)

Therefore,

45/100 × 1125 < 30/100 (1125 + x)

45/30 × 1125 < 1125 + x

3/2 × 1125 < 1125 + x

1687.5 < 1125 + x

1687.5 – 1125 < x

562.5 < x   ……. (2)

Now, by using equation 1 and 2, we get

562.5 < x < 900

Therefore, the quantity of water to be added will be between 562.5 liters and 900 liters.

### Question 11: A Solution of 8% boric acid is 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 there are 640 liters of the 8% solution, how many liters of 2% solution will have to be added?

Solution:

Let the 2% solution be added to 640 liters of the 8% solution be x.

Therefore, the total quantity of mixture = (640 + x)

The total acid in (640 + x) liters of mixture is

(2/100)x + (8/100)640

Given that the resulting mixture should be more than 4% and less than 6%.

4/100(640 + x) < (2x/100 + (8 ×640)/100 < 6/100(640 + x)

4(640 + x) < (2x + 8640) < 6(640 + x)

2560 + 4x < 2x + 8640 and 2x + 8640 < 3840 + 6x

2560 – 8640 < 2x – 4x and 2x – 6x < 3840 – 8640

x < 1280 and x > 320

Therefore, more than 320 liters but less than 1280 liters of 2% to be added.

### Question 12: The water acidity in a pool is considered normal when the average pH reading of three daily measurements is between 7.2 and 7.8. If the first two pH readings are 7.48 and 7.85, find the range of the pH value for the third reading that will result in the acidity level being normal.

Solution:

Let x be the pH value of third reading.

Therefore,

7.2 < (7.48 + 7.85 + x)/3 < 7.8

21.6 < 7.48 + 7.85 + x < 23.4

21.6 < 15.33 + x < 23.4

21.6 – 15.33 < x and x < 23.4 – 15.33

6.27 < x and x < 8.07

Therefore, the range of pH values for the third reading lies between 6.27 and 8.07.

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