Posted by RBSE Solutions for Class 6 Maths Chapter 12: Ratio and Proportion offers comprehensive explanations and solutions to help students understand the concepts of ratio and proportion. This chapter introduces students to comparing quantities using ratios, understanding equivalent ratios, and solving problems based on proportions. These solutions make it easy for students to learn the concepts and apply them to real-life situations.
Key Concepts Covered in Chapter 12: Ratio and Proportion
- Introduction to Ratios
- Understanding Equivalent Ratios
- Definition and Use of Proportion
- Solving Ratio Problems
- Real-Life Applications of Ratios and Proportions
Mastering the concepts of ratio and proportion is essential for understanding measurements, comparisons, and relationships between quantities.
Detailed RBSE Class 6 Solutions for Chapter 12: Ratio and Proportion
Here are detailed, question-wise solutions for Chapter 12, designed to provide step-by-step explanations. Each solution is tailored to build a strong foundation in these concepts, enabling students to solve problems with confidence.
NCERT Solutions for Class 6 Chapter 12: Ratio and Proportion Exercise 12.1
Table of Contents
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
The total number of students = 20 + 15
= 35
(a) The ratio of the number of girls to the number of boys = 20 / 15 = 4 / 3
(b) The 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
The number of students who like football = 6
The number of students who like cricket = 12
The 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
= 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
The number of triangles = 3
The number of circles = 2
The number of squares = 2
The total number of figures = 7
(a) The ratio of the number of triangles to the number of circles inside the rectangle
= 3 / 2
(b) The ratio of the number of squares to all the figures inside the rectangle
= 2 / 7
(c) The ratio of the number of circles to all the figures inside the rectangle
= 2 / 7
4. Distances travelled by Hamid and Akhtar in an hour are 9 km and 12 km. 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 following 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 Seema = ₹ 50000
Money spent by Seema = ₹ 150000 – ₹ 50000 = ₹ 100000
(a) The ratio of the money earned to money saved = 150000 / 50000 = 15 / 5
= 3 / 1
(b) The ratio of the 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
The number of teachers in a school = 102
The number of students in a 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
The total number of students = 4320
The number of girls = 2300
The number of boys = 4320 – 2300
= 2020
(a) The 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) The ratio of the number of boys to the number of girls = 2020 / 2300
= (2 × 2 × 5 × 101) / (2 × 2 × 5 × 115)
= 101 / 115
(c) The 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 basketball, 800 opted cricket and the remaining opted table tennis. If a student can opt only one game, find the ratio of
(a) Number of students who opted basketball to the number of students who opted table tennis.
(b) Number of students who opted cricket to the number of students opting basketball.
(c) Number of students who opted basketball to the total number of students.
Solutions:
(a) The ratio of the number of students who opted basketball to the number of students who opted table tennis = 750 / 250 = 3 / 1
(b) The ratio of the number of students who opted cricket to the number of students opting basketball
= 800 / 750 = 16 / 15
(c) The ratio of the number of students who opted 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:
The cost of a dozen pens = ₹ 180
The cost of 1 pen = 180 / 12
= ₹ 15
The cost of 8 ball pens = ₹ 56
The 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.
| The breadth of the hall (in metres) | 10 | 40 | |
| The 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.
The number of pens Sheela has = 3 / 5 × 20
= 3 × 4
= 12
The number of pens Sangeeta has = 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 Bhoomika will get 4 / 9 of the total money.
The amount Shreya gets = 5 / 9 × 36
= 20
The amount Bhoomika gets = 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 the father after 10 years to the age of the son after 10 years.
(d) Age of the father to the age of the son when the 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) The son was 12 years old 2 years ago. So, the age of the father 2 years ago will be
= 42 – 2 = 40 years
Required ratio = 40 / 12 = (4 × 10) / (4 × 3) = 10 / 3
(c) After ten years age of the father = 42 + 10 = 52 years
After 10 years age of the son = 14 + 10 = 24 years
Required ratio = 52 / 24 = (4 × 13) / (4 × 6)
= 13 / 6
(d) 12 years ago, 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
NCERT Solutions for Class 6 Chapter 12: Ratio and Proportion Exercise 12.2
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
∴ These are in proportion.
(b) 33, 121, 9, 96
33 / 121 = 3 / 11
9 / 96 = 3 / 32
Hence, 33:121 ≠ 9: 96
∴ These are not in proportion.
(c) 24, 28, 36, 48
24 / 28 = 6 / 7
36 / 48 = 3 / 4
Hence, 24: 28 ≠ 36:48
∴ These are not in proportion.
(d) 32, 48, 70, 210
32 / 48 = 2 / 3
70 / 210 = 1 / 3
Hence, 32: 48 ≠ 70: 210
∴ These 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 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, it is true.
(b) 21: 6:: 35: 10
21 / 6 = 7 / 2
35 / 10 = 7 / 2
Hence, 21: 6 = 35: 10
Therefore, it is true.
(c) 12: 18 :: 28: 12
12 / 18 = 2 / 3
28 / 12 = 7 / 3
Hence, 12: 18 ≠ 28:12
Therefore, it is false.
(d) 8: 9:: 24: 27
We know that = 24 / 27 = (3 × 8) / (3 × 9)
= 8 / 9
Hence, 8: 9 = 24: 27
Therefore, it is true.
(e) 5.2: 3.9:: 3: 4
As 5.2 / 3.9 = 4/3
Hence, 5.2: 3.9 ≠ 3: 4
Therefore, it is false.
(f) 0.9: 0.36:: 10: 4
0.9 / 0.36 = 90 / 36
= 10 / 4
Hence, 0.9: 0.36 = 10: 4
Therefore, it is 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, it is true.
(b) 7.5 litres : 15 litres = 5 kg : 10 kg
7.5 / 15 = 1 / 2
5 / 10 = 1 / 2
Hence, it is true.
(c) 99 kg : 45 kg = ₹ 44 : ₹ 20
99 / 45 = 11 / 5
44 / 20 = 11 / 5
Hence, it is true.
(d) 32 m : 64 m = 6 sec : 12 sec
32 / 64 = 1 / 2
6 / 12 = 1 / 2
Hence, it is true.
(e) 45 km : 60 km = 12 hours : 15 hours
45 / 60 = 3 / 4
12 / 15 = 4 / 5
Hence, it is 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.
NCERT Solutions for Class 6 Maths Chapter 12: Ratio and Proportion Exercise 12.3
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 m cloth = 210 × 5 = 1050
∴ 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 in one day by her = 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:
Measure of rain in 3 days = 276 mm
Measure of rain in one day = 276 / 3
= 92 mm
So, 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 in ₹ 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 in ₹ 91.50 = 5 kg
Wheat purchased in ₹ 1 = 5 / 91.50 kg
So, wheat purchased in ₹ 183 = (5 / 91.50) × 183
= 10 kg
5. The temperature dropped 15 degree celsius 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, 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 has to pay for a whole year, if the rent per month remains 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 in₹ 180 = 4 dozens
= 4 × 12
= 48 bananas
Number of bananas bought in ₹ 1 = 48 / 180
So, number of bananas bought in ₹ 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
∴ Weight of 40 books is 5 kg
9. A truck requires 108 litres of diesel for covering 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
∴ 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 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.
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- Rbse Solutions for Class 6 Chapter 5: Understanding Elementary Shapes
- Rbse Solutions for Class 6 Chapter 6: Integers
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- RBSE Solutions for Class 6 Maths Chapter 10: Mensuration | Updated for 2024-25
- RBSE Solutions for Class 6 Maths Chapter 11: Algebra | Updated for 2024-25
- RBSE Solutions for Class 6 Maths Chapter 12: Ratio and Proportion | Updated for 2024-25
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- RBSE Solutions for Class 6 Maths Chapter 14: Practical Geometry | Updated for 2024-25
Benefits of RBSE Class 6 Chapter 12 Solutions
- Concept Clarity: Clear explanations aid in understanding and applying the concepts of ratio and proportion.
- Exam Preparation: Practicing these solutions enhances students’ skills, which is beneficial for exams.
- Real-Life Relevance: Ratio and proportion are practical concepts that students can apply in daily life, such as in recipes, maps, and financial calculations.
FAQs on RBSE Solutions for Class 6 Maths Chapter 12: Ratio and Proportion
Q1: What is a ratio, and why is it important?
A1: A ratio is a way to compare two quantities by showing how many times one value contains or is contained within the other. Ratios are important because they help in understanding relationships and making comparisons.
Q2: What does proportion mean in mathematics?
A2: Proportion is a statement that two ratios are equal. It is used to solve problems where two quantities vary in a consistent way.
Q3: Where can I find detailed RBSE solutions for Class 6 Chapter 12?
A3: You can access step-by-step solutions on rbsesolution.in, which provides detailed explanations for all questions in Chapter 12.
Q4: How are ratios useful in everyday life?
A4: Ratios are commonly used in various aspects of life, like cooking recipes, calculating speed, understanding map scales, and financial budgeting.
Q5: What are equivalent ratios?
A5: Equivalent ratios are ratios that express the same relationship between quantities. They are found by multiplying or dividing both terms of a ratio by the same number.
Conclusion
The RBSE Solutions for Class 6 Maths Chapter 12: Ratio and Proportion equip students with the necessary skills to solve real-life problems involving ratios and proportions. By practicing these solutions, students can gain confidence in handling mathematical relationships and comparisons, which are invaluable in both academic and practical settings.