Access RBSE Solutions for Class 6 Chapter 12: Ratio and Proportion Embarking on the mathematical exploration of Class 6 within the Rajasthan Board of Secondary Education (RBSE), students encounter the intriguing concepts of ratios and proportions in Chapter 12. Aptly titled “Ratio and Proportion,” this chapter serves as a gateway to understanding the relationships between quantities and the principles of proportionality. To facilitate a thorough understanding of these essential mathematical concepts, we present comprehensive RBSE solutions tailored to Chapter 12. Let’s delve into the key ideas and explore how our solutions can make the study of ratios and proportions both accessible and engaging.
Table of Contents
1. The Significance of Mastering Ratios and Proportions:
Before delving into the solutions, it’s imperative to recognize the importance of ratios and proportions in the mathematical landscape. These concepts are foundational for various mathematical applications, from solving real-life problems to more advanced mathematical principles. A solid grasp of ratios and proportions is integral for building a robust mathematical foundation.
2. Overview of RBSE Class 6 Maths Chapter 12:
Chapter 12, “Ratio and Proportion,” is designed to introduce students to the fundamental concepts of ratios and proportions. The key topics covered include:
- Understanding ratios and their representation.
- Exploring proportions and their significance.
- Solving problems involving ratios and proportions.
- Applications of ratios and proportions in real-life scenarios.
3. RBSE Solutions: A Comprehensive Companion:
Our RBSE Class 6 Maths Chapter 12 solutions are crafted to be a comprehensive companion, ensuring that students not only solve problems but also understand the underlying principles of ratios and proportions. The solutions include:
- Clear explanations for each concept.
- Step-by-step solutions for problem-solving.
- Practical examples to illustrate the application of ratios and proportions.
- Exercises for practice to reinforce learning.
4. Facilitating Learning Through Our Solutions:
- Conceptual Clarity: Our solutions provide clear and concise explanations to help students grasp the fundamental concepts of ratios and proportions.
- Step-by-Step Approach: By offering a step-by-step approach to problem-solving, our solutions guide students through the intricacies of working with ratios and proportions.
- Real-Life Context: Real-life examples are seamlessly integrated into the solutions to demonstrate the relevance of ratios and proportions in everyday situations, making the learning experience both relatable and enjoyable.
RBSE Solutions for Class 6 MATHS Chapter 12 Exercise 12.1 Page No. 251
1. There are 20 girls and 15 boys in a class.
(a) What is the ratio of the number of girls to the number of boys?
(b) What is the ratio of the number of girls to the total number of students in the class?
Solutions:
Given
Number of girls = 20 girls
Number of boys = 15 boys
Total number of students = 20 + 15
= 35
(a) Ratio of the number of girls to the number of boys = 20 / 15 = 4 / 3
(b) Ratio of the number of girls to the total number of students = 20 / 35 = 4 / 7
2. Out of 30 students in a class, 6 like football, 12 like cricket and the remaining like tennis. Find the ratio of
(a) The number of students liking football to the number of students liking tennis.
(b) The number of students liking cricket to the total number of students.
Solutions:
Given
Number of students who like football = 6
Number of students who like cricket = 12
Number of students who like tennis = 30 – 6 – 12
= 12
(a) Ratio of the number of students liking football to the number of students liking tennis
= 6 / 12 = 1 / 2
(b) Ratio of the number of students liking cricket to the total number of students
= 12 / 30
= 2 / 5
3. See the figure and find the ratio of
(a) Number of triangles to the number of circles inside the rectangle.
(b) Number of squares to all the figures inside the rectangle.
(c) Number of circles to all the figures inside the rectangle.
Solutions:
Given in the figure
Number of triangles = 3
Number of circles = 2
Number of squares = 2
Total number of figures = 7
(a) Ratio of the number of triangles to the number of circles inside the rectangle
= 3 / 2
(b) Ratio of the number of squares to all the figures inside the rectangle
= 2 / 7
(c) Ratio of the number of circles to all the figures inside the rectangle
= 2 / 7
4. The distance travelled by Hamid and Akhtar in an hour is 9 km and 12 km, respectively. Find the ratio of the speed of Hamid to the speed of Akhtar.
Solutions:
We know that the speed of a certain object is the distance travelled by that object in an hour
Distance travelled by Hamid in one hour = 9 km
Distance travelled by Akhtar in one hour = 12 km
Speed of Hamid = 9 km/hr
Speed of Akhtar = 12 km/hr
The ratio of the speed of Hamid to the speed of Akhtar = 9 / 12 = 3 / 4
5. Fill in the blanks:
15 / 18 = ☐ / 6 = 10 / ☐ = ☐ / 30 [Are these equivalent ratios?]
Solutions:
15 / 18 = (5 × 3) / (6 × 3)
= 5 / 6
5 / 6 = (5 × 2) / (6 × 2)
= 10 / 12
5 / 6 = (5 × 5) / (6 × 5)
= 25 / 30
Hence, 5, 12 and 25 are the numbers which come in the blanks, respectively.
Yes, all are equivalent ratios.
6. Find the ratio of the following:
(a) 81 to 108
(b) 98 to 63
(c) 33 km to 121 km
(d) 30 minutes to 45 minutes
Solutions:
(a) 81 / 108 = (3 × 3 × 3 × 3) / (2 × 2 × 3 × 3 × 3)
= 3 / 4
(b) 98 / 63 = (14 × 7) / (9 × 7)
= 14 / 9
(c) 33 / 121 = (3 × 11) / (11 × 11)
= 3 / 11
(d) 30 / 45 = (2 × 3 × 5) / (3 × 3 × 5)
= 2 / 3
7. Find the ratio of the following:
(a) 30 minutes to 1.5 hours
(b) 40 cm to 1.5 m
(c) 55 paise to ₹ 1
(d) 500 ml to 2 litres
Solutions:
(a) 30 minutes to 1.5 hours
30 min = 30 / 60
= 0.5 hours
Required ratio = (0.5 × 1) / (0.5 × 3)
= 1 / 3
(b) 40 cm to 1.5 m
1.5 m = 150 cm
Required ratio = 40 / 150
= 4 / 15
(c) 55 paise to ₹ 1
₹ 1 = 100 paise
Required ratio = 55 / 100 = (11 × 5) / (20 × 5)
= 11 / 20
(d) 500 ml to 2 litres
1 litre = 1000 ml
2 litre = 2000 ml
Required ratio = 500 / 2000 = 5 / 20 = 5 / (5 × 4)
= 1 / 4
8. In a year, Seema earns ₹ 1,50,000 and saves ₹ 50,000. Find the ratio of
(a) Money that Seema earns to the money she saves
(b) Money that she saves to the money she spends.
Solutions:
Money earned by Seema = ₹ 150000
Money saved by her = ₹ 50000
Money spent by her = ₹ 150000 – ₹ 50000 = ₹ 100000
(a) Ratio of money earned to money saved = 150000 / 50000 = 15 / 5
= 3 / 1
(b) Ratio of money saved to money spent = 50000 / 100000 = 5 / 10
= 1 / 2
9. There are 102 teachers in a school of 3300 students. Find the ratio of the number of teachers to the number of students.
Solutions:
Given
Number of teachers in the school = 102
Number of students in the school = 3300
The ratio of the number of teachers to the number of students = 102 / 3300
= (2 × 3 × 17) / (2 × 3 × 550)
= 17 / 550
10. In a college, out of 4320 students, 2300 are girls. Find the ratio of
(a) Number of girls to the total number of students.
(b) Number of boys to the number of girls.
(c) Number of boys to the total number of students.
Solutions:
Given
Total number of students = 4320
Number of girls = 2300
Number of boys = 4320 – 2300
= 2020
(a) Ratio of the number of girls to the total number of students = 2300 / 4320
= (2 × 2 × 5 × 115) / (2 × 2 × 5 × 216)
= 115 / 216
(b) Ratio of the number of boys to the number of girls = 2020 / 2300
= (2 × 2 × 5 × 101) / (2 × 2 × 5 × 115)
= 101 / 115
(c) Ratio of the number of boys to the total number of students = 2020 / 4320
= (2 × 2 × 5 × 101) / (2 × 2 × 5 × 216)
= 101 / 216
11. Out of 1800 students in a school, 750 opted for basketball, 800 opted for cricket, and the remaining opted for table tennis. If a student can opt for only one game, find the ratio of
(a) The number of students who opted for basketball to the number of students who opted for table tennis.
(b) The number of students who opted for cricket to the number of students opting for basketball.
(c) The number of students who opted for basketball to the total number of students.
Solutions:
Given,
Number of students in the school = 1800
The number of students who opted for basketball = 750
The number of students who opted for cricket = 800
The number of students who opted for table tennis = 1800 – (750 + 800) = 1800 – 1550 = 250
(a) Ratio of the number of students who opted for basketball to the number of students who opted for table tennis = 750 / 250 = 3 / 1
(b) Ratio of the number of students who opted for cricket to the number of students who opted for basketball
= 800 / 750 = 16 / 15
(c) Ratio of the number of students who opted for basketball to the total number of students
= 750 / 1800 = 25 / 60 = 5 / 12
12. Cost of a dozen pens is ₹ 180, and the cost of 8 ball pens is ₹ 56. Find the ratio of the cost of a pen to the cost of a ball pen.
Solutions:
Cost of a dozen pens = ₹ 180
Cost of 1 pen = 180 / 12
= ₹ 15
Cost of 8 ball pens = ₹ 56
Cost of 1 ball pen = 56 / 8
= ₹ 7
Hence, the required ratio is 15 / 7.
13. Consider the statement: The ratio of breadth and length of a hall is 2:5. Complete the following table that shows some possible breadths and lengths of the hall.
| Breadth of the hall (in metres) | 10 | 40 | |
| Length of the hall (in metres) | 25 | 50 |
Solutions:
(i) Length = 50 m
Breadth / 50 = 2 / 5
By cross multiplication
5× breadth = 50 × 2
Breadth = (50 × 2) / 5
= 100 / 5
= 20 m
(ii) Breadth = 40 m
40 / Length = 2 / 5
By cross multiplication
2 × Length = 40 × 5
Length = (40 × 5) / 2
Length = 200 / 2
Length = 100 m
14. Divide 20 pens between Sheela and Sangeeta in a ratio of 3:2.
Solutions:
Terms of 3:2 = 3 and 2
The sum of these terms = 3 + 2
= 5
Now, Sheela will get 3 / 5 of the total pens, and Sangeeta will get 2 / 5 of the total pens
Number of pens with Sheela = 3 / 5 × 20
= 3 × 4
= 12
Number of pens with Sangeeta = 2 / 5 × 20
= 2 × 4
= 8
15. Mother wants to divide ₹ 36 between her daughters Shreya and Bhoomika in the ratio of their ages. If the age of Shreya is 15 years and the age of Bhoomika is 12 years, find how much Shreya and Bhoomika will get.
Solutions:
Ratio of ages = 15 / 12
= 5 / 4
Hence, the mother wants to divide ₹ 36 in the ratio of 5:4
Terms of 5:4 are 5 and 4
The sum of these terms = 5 + 4
= 9
Here, Shreya will get 5 / 9 of the total money, and Sangeeta will get 4 / 9 of the total money
The amount Shreya get = 5 / 9 × 36
= 20
The amount Sangeeta get = 4 / 9 × 36
= 16
Therefore, Shreya will get ₹ 20, and Sangeeta will get ₹ 16.
16. Present age of the father is 42 years, and that of his son is 14 years. Find the ratio of
(a) Present age of the father to the present age of the son
(b) Age of the father to the age of the son, when the son was 12 years old
(c) Age of father after 10 years to the age of son after 10 years
(d) Age of father to the age of son when father was 30 years old
Solutions:
(a) Present age of father = 42 years
Present age of son = 14 years
Required ratio 42 / 14
= 3 / 1
(b) 2 years ago, the son was 12 years old. So, 2 years ago, the age of the father was
= 42 – 2 = 40 years
Required ratio = 40 / 12 = (4 × 10) / (4 × 3) = 10 / 3
(c) After ten years, the age of the father = 42 + 10 = 52 years
After 10 years, the age of the son = 14 + 10 = 24 years
Required ratio = 52 / 24 = (4 × 13) / (4 × 6)
= 13 / 6
(d) 12 years ago, the age of the father was 30
At that time, the age of the son = 14 – 12
= 2 years
Required ratio = 30 / 2 = (2 × 15) / 2
= 15 / 1
RBSE Solutions for Class 6 MATHS Chapter 12 Exercise 12.2 Page No. 255
1. Determine if the following are in proportion.
(a) 15, 45, 40, 120
(b) 33, 121, 9, 96
(c) 24, 28, 36, 48
(d) 32, 48, 70, 210
(e) 4, 6, 8, 12
(f) 33, 44, 75, 100
Solutions:
(a) 15, 45, 40, 120
15 / 45 = 1 / 3
40 / 120 = 1 / 3
Hence, 15:45 = 40:120
∴ They are in proportion.
(b) 33, 121, 9, 96
33 / 121 = 3 / 11
9 / 96 = 3 / 32
Hence 33:121 ≠ 9: 96
∴ They are not in proportion.
(c) 24, 28, 36, 48
24 / 28 = 6 / 7
36 / 48 = 3 / 4
Hence, 24:28 ≠ 36:48
∴ They are not in proportion.
(d) 32, 48, 70, 210
32 / 48 = 2 / 3
70 / 210 = 1 / 3
Hence, 32:48 ≠ 70:210
∴ They are not in proportion.
(e) 4, 6, 8, 12
4 / 6 = 2 / 3
8 / 12 = 2 / 3
Hence 4: 6 = 8: 12
∴ These are in a proportion
(f) 33, 44, 75, 100
33/ 44 = 3/ 4
75 / 100 = 3 / 4
Hence, 33:44 = 75:100
∴ These are in proportion.
2. Write True (T) or False ( F ) against each of the following statements :
(a) 16:24 :: 20:30
(b) 21:6 :: 35:10
(c) 12:18 :: 28:12
(d) 8:9 :: 24:27
(e) 5.2:3.9 :: 3:4
(f) 0.9:0.36 :: 10:4
Solutions:
(a) 16:24 :: 20:30
16 / 24 = 2 / 3
20 / 30 = 2 / 3
Hence, 16:24 = 20:30
Therefore, True.
(b) 21:6 :: 35:10
21 / 6 = 7 / 2
35 / 10 = 7 / 2
Hence, 21:6 = 35:10
Therefore, True.
(c) 12:18 :: 28:12
12 / 18 = 2 / 3
28 / 12 = 7 / 3
Hence, 12:18 ≠ 28:12
Therefore, False,
(d) 8:9:: 24:27
We know that = 24 / 27 = (3 × 8) / (3 × 9)
= 8 / 9
Hence, 8:9 = 24:27
Therefore, True.
(e) 5.2:3.9 :: 3: 4
As 5.2 / 3.9 = 4 / 3
Hence, 5.2: 3.9 ≠ 3: 4
Therefore, False,
(f) 0.9:0.36 :: 10:4
0.9 / 0.36 = 90 / 36
= 10 / 4
Hence, 0.9: 0.36 = 10: 4
Therefore, True,
3. Are the following statements true?
(a) 40 persons: 200 persons = ₹ 15 : ₹ 75
(b) 7.5 litres: 15 litres = 5 kg : 10 kg
(c) 99 kg: 45 kg = ₹ 44 : ₹ 20
(d) 32 m: 64 m = 6 sec : 12 sec
(e) 45 km : 60 km = 12 hours : 15 hours
Solutions:
(a) 40 persons : 200 persons = ₹ 15 : ₹ 75
40 / 200 = 1 / 5
15 / 75 = 1 / 5
Hence, True.
(b) 7.5 litres : 15 litres = 5 kg : 10 kg
7.5 / 15 = 1 / 2
5 / 10 = 1 / 2
Hence, True.
(c) 99 kg : 45 kg = ₹ 44 : ₹ 20
99 / 45 = 11 / 5
44 / 20 = 11 / 5
Hence, True.
(d) 32 m : 64 m = 6 sec : 12 sec
32 / 64 = 1 / 2
6 / 12 = 1 / 2
Hence, True.
(e) 45 km : 60 km = 12 hours : 15 hours
45 / 60 = 3 / 4
12 / 15 = 4 / 5
Hence, False.
4. Determine if the following ratios form a proportion. Also, write the middle terms and extreme terms where the ratios form a proportion.
(a) 25 cm : 1 m and ₹ 40 : ₹ 160
(b) 39 litres : 65 litres and 6 bottles : 10 bottles
(c) 2 kg : 80 kg and 25 g : 625 g
(d) 200 mL : 2.5 litre and ₹ 4 : ₹ 50
Solutions:
(a) 25 cm : 1 m and ₹ 40 : ₹ 160
25 cm = 25 / 100 m
= 0.25 m
0.25 / 1 = 1 / 4
40 / 160 = 1 / 4
Yes, these are in proportion.
Middle terms are 1 m, ₹ 40, and extreme terms are 25 cm, ₹ 160.
(b) 39 litres : 65 litres and 6 bottles : 10 bottles
39 / 65 = 3 /5
6 / 10 = 3 / 5
Yes, these are in proportion.
Middle terms are 65 litres, 6 bottles, and extreme terms are 39 litres, 10 bottles.
(c) 2 kg : 80 kg and 25 g : 625 g
2 / 80 = 1 / 40
25 / 625 = 1 / 25
No, these are not in proportion.
(d) 200 mL : 2.5 litre and ₹ 4 : ₹ 50
1 litre = 1000 ml
2.5 litre = 2500 ml
200 / 2500 = 2 / 25
4 / 50 = 2 / 25
Yes, these are in proportion.
Middle terms are 2.5 litres, ₹ 4 and extreme terms are 200 ml, ₹ 50.
RBSE Solutions for Class 6 MATHS Chapter 12 Exercise 12.3 Page No. 259
1. If the cost of 7 m of cloth is ₹ 1470, find the cost of 5 m of cloth.
Solutions:
Given
Cost of 7 m cloth = ₹ 1470
Cost of 1 m cloth = 1470 / 7
= ₹ 210
So, cost of 5 cloth = 210 × 5 = 1050
∴ The cost of 5 m cloth is ₹ 1050.
2. Ekta earns ₹ 3000 in 10 days. How much will she earn in 30 days?
Solutions:
Money earned by Ekta in 10 days = ₹ 3000
Money earned by her in one day = 3000 / 10
= ₹ 300
So, money earned by her in 30 days = 300 × 30
= ₹ 9000.
3. If it has rained 276 mm in the last 3 days, how many cm of rain will fall in one full week (7 days)? Assume that the rain continues to fall at the same rate.
Solutions:
The measure of rain in 3 days = 276 mm
The measure of rain in one day = 276 / 3
= 92 mm
So, the measure of rain in one week, i.e. 7 days = 92 × 7
= 644 mm
= 644 / 10
= 64.4 cm
4. Cost of 5 kg of wheat is ₹ 91.50.
(a) What will be the cost of 8 kg of wheat?
(b) What quantity of wheat can be purchased for ₹ 183?
Solutions:
(a) Cost of 5 kg wheat = ₹ 91.50.
Cost of 1 kg wheat = 91.50 / 5
= ₹ 18.3
So, the cost of 8 kg wheat = 18.3 × 8
= ₹ 146.40
(b) Wheat purchased for ₹ 91.50 = 5 kg
Wheat purchased for ₹ 1 = 5 / 91.50 kg
So, wheat purchased for ₹ 183 = (5 / 91.50) × 183
= 10 kg
5. The temperature dropped 150 C in the last 30 days. If the rate of temperature drop remains the same, how many degrees will the temperature drop in the next ten days?
Solutions:
Temperature drop in 30 days = 150 C
Temperature drop in 1 day = 15 / 30
= (1 / 2)0 C
So, the temperature drop in next 10 days = (1 / 2) × 10
= 50 C
∴ The temperature drop in the next 10 days will be 50 C
6. Shaina pays ₹ 15000 as rent for 3 months. How much does she have to pay for a whole year if the rent per month remains the same?
Solutions:
Rent paid by Shaina in 3 months = ₹ 15000
Rent for 1 month = 15000 / 3
= ₹ 5000
So, rent for 12 months, i.e. 1 year = 5000 × 12
= ₹ 60,000
∴ Rent paid by Shaina in 1 year is ₹ 60,000
7. Cost of 4 dozen bananas is ₹ 180. How many bananas can be purchased for ₹ 90?
Solutions:
Number of bananas bought for ₹ 180 = 4 dozens
= 4 × 12
= 48 bananas
Number of bananas bought for ₹ 1 = 48 / 180
So, number of bananas bought for ₹ 90 = (48 / 180) × 90
= 24 bananas
∴ 24 bananas can be purchased for ₹ 90
8. The weight of 72 books is 9 kg. What is the weight of 40 such books?
Solutions:
Weight of 72 books = 9 kg
Weight of 1 book = 9 / 72
= 1 / 8 kg
So, weight of 40 books = (1 / 8) × 40
= 5 kg
∴ The weight of 40 books is 5 kg.
9. A truck requires 108 litres of diesel to cover a distance of 594 km. How much diesel will be required by the truck to cover a distance of 1650 km?
Solutions:
Diesel required for 594 km = 108 litres
Diesel required for 1 km = 108 / 594
= 2 / 11 litre
So, diesel required for 1650 km = (2 / 11) × 1650
= 300 litres
∴ The diesel required by the truck to cover a distance of 1650 km is 300 litres.
10. Raju purchases 10 pens for ₹ 150, and Manish buys 7 pens for ₹ 84. Can you say who got the pens cheaper?
Solutions:
Pens purchased by Raju for ₹ 150 = 10 pens
Cost of 1 pen = 150 / 10
= ₹ 15
Pens purchased by Manish for ₹ 84 = 7 pens
Cost of 1 pen = 84 / 7
= ₹ 12
∴ Pens purchased by Manish are cheaper than those of Raju.
11. Anish made 42 runs in 6 overs, and Anup made 63 runs in 7 overs. Who made more runs per over?
Solutions:
Runs made by Anish in 6 overs = 42
Runs made by Anish in 1 over = 42 / 6
= 7
Runs made by Anup in 7 overs = 63
Runs made by Anup in 1 over = 63 / 7
= 9
∴ Anup scored more runs than Anish.
As students navigate the world of ratios and proportions in RBSE Class 6 Maths Chapter 12, our SEO-friendly article underscores the significance of mastering these foundational concepts. Our comprehensive solutions not only guide students through problem-solving but also provide conceptual clarity, a step-by-step approach, and real-life applications to make the learning experience enriching. Navigating ratios and proportions is a pivotal aspect of mathematical education, and our solutions are crafted to make this journey both accessible and academically rewarding for RBSE Class 6 students.
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