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# NCERT Solutions for Class 6 Maths Chapter 12 Ratio and Proportion

Last updated date: 12th Aug 2024
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## NCERT Solutions for Maths Class 6 Chapter 12 Ratio and Proportion - FREE PDF Download

Embark on a mathematical exploration with Class 6 Maths Chapter 12 PDF Solutions  Ratio and Proportion. This chapter introduces students to the fundamental concepts of ratios and proportions, laying the groundwork for understanding relationships between quantities. Dive into a world where mathematical relationships unlock the secrets of proportionality, making math both engaging and practical.

Table of Content
1. NCERT Solutions for Maths Class 6 Chapter 12 Ratio and Proportion - FREE PDF Download
2. Glance on Class 6 Maths Chapter 12 - Ratio and Proportion
3. Access Exercise wise NCERT Solutions for Chapter 12 Maths Class 6
4. Exercises Under NCERT Solutions for Class 6 Maths Chapter 12 Ratio and Proportion
5. Access NCERT Solutions for Class 6 Maths Chapter 12 – Ratio and Proportion
5.1Exercise 12.1
5.2Exercise 12.2
5.3Exercise 12.3
6. Class 6 Maths Chapter 12: Exercises Breakdown
7. Other Study Material for CBSE Class 6 Maths Chapter 12
8. Chapter-Specific NCERT Solutions for Class 6 Maths
FAQs

NCERT Solutions for Class 6 Maths Chapter 12 Ratio and Proportion have been made available to you to download from the official Vedantu website. You now have the provision of downloading the entire chapter of Ratio and Proportion Class 6 at your convenience and preparing the Ratio and Proportion Questions for Class 6 for your examination. The NCERT Solution Class 6 Math Chapter 12 are being prepared by experienced teachers who know the right way to present the children’s topics.

## Glance on Class 6 Maths Chapter 12 - Ratio and Proportion

• Ratio and Proportion is mainly showing the difference between two numbers.

• Ratio: If the total number of boys in a class is 40 and the total number of girls in a class is 50. Then the ratio between the number of boys and girls in a class is 40 / 50, which is 4:5.

• The number of students in a class is equal to the sum of the number of boys in a class and the number of girls in a class.

• The ratio between the number of students in a class and the number of girls in a class is 90 / 50 which is 9: 5

• Proportion is if the two sets of numbers are in the same ratio, then the numbers are said to be the proportion.

• The unitary Method is, we need to find the answer for one product and can calculate it for many.

• Explore practical applications, such as using ratios and proportions in recipes, maps, and scale drawings, enhancing problem-solving skills.

• There are three exercises (31 fully solved questions) in ratio and proportion questions in class 6 math chapter 12.

## Access Exercise wise NCERT Solutions for Chapter 12 Maths Class 6

 Current Syllabus Exercises of Class 6 Maths Chapter 12 NCERT Solutions of Class 10 Maths Ratio and Proportion Exercise 12.1 NCERT Solutions of Class 10 Maths Ratio and Proportion Exercise 12.2 NCERT Solutions of Class 10 Maths Ratio and Proportion Exercise 12.3

## Exercises Under NCERT Solutions for Class 6 Maths Chapter 12 Ratio and Proportion

Chapter 12 of Class 6 Maths focuses on understanding and applying the concepts of ratio and proportion. The chapter includes three key exercises: Exercise 12.1, Exercise 12.2, and Exercise 12.3.

• Exercise 12.1 introduces students to the basic concept of ratios, teaching them to compare two quantities and express them in ratio form, as well as simplify these ratios.

• Exercise 12.2 covers equivalent ratios, showing students how to find equivalent ratios by multiplying or dividing both terms by the same number, and includes problems on finding missing terms and verifying equivalent ratios.

• Exercise 12.3 deals with proportions, helping students identify and solve problems involving proportional relationships, and apply these concepts to real-life scenarios such as scaling recipes or maps. These exercises are essential for building a solid foundation in ratios and proportions, which are crucial for advanced mathematical concepts and practical applications.

## Access NCERT Solutions for Class 6 Maths Chapter 12 – Ratio and Proportion

### Exercise 12.1

1. There are 20 girls and 15 boys in a class.

(a) What is the ratio of number of girls to the number of boys?

Ans: Given: Number of girls is 20 and number of boys is 15.

Thus, the ratio of girls to that of boys we have,

$\dfrac{{Number\;of\;girls}}{{Number\;of\;boys}} = \dfrac{{20}}{{15}} = \dfrac{4}{3} = 4:3$

(b) What is the ratio of girls to the total number of students in the class?

Ans: Given: Number of girls is 20

The total number of students in the class $= {\text{ }}20 + 15{\text{ }} = 35$

$\dfrac{{Number\;of\;girls}}{{Total\;Number\;of\;students}} = \dfrac{{20}}{{35}}$

$= \dfrac{{20 \div 5}}{{35 \div 5}}$

$= \dfrac{4}{7} = 4:7$

2. Out of 30 students in a class, 6 like football, 12 like cricket and remaining like tennis. Find the ratio of:

(a) Number of students liking football to number of students liking tennis.

Ans: Total number of students = 30

Number of students like football = 6

Number of students like cricket = 12

Thus, the number of students like tennis = 30 – 6 – 12 = 12

Therefore,

Ratio of number of students liking football to number of students liking tennis $= \dfrac{6}{{12}} = \dfrac{1}{2} = 1:2$

(b) Number of students liking cricket to total number of students.

Ans: Total number of students = 30

Number of students like football = 6

Number of students like cricket = 12

Ratio of number of students liking cricket to total number of students $=\dfrac{12}{30}=\dfrac{2}{5}=2:5$

3. See the figure and find the ratio of:

(a). Number of triangles to the number of circles inside rectangle.

Ans: In the given figure,

The total number of triangles =3

The total number of circles =2

Therefore,

Required ratio of the number of triangles to the number of circles is

$\dfrac{{Number\;of\;Triangle}}{{Number\;of\;circle}} = \dfrac{3}{2} = 3:2$

(b). Number of squares to all the figures inside the rectangle.

Ans: In the given figure,

The total number of square = 2

The total number of figures =7

Therefore,

Required ratio of the number of square to the total number of figures is

$\dfrac{{Number\;of\;square}}{{Total\;Number\;of\;figures}} = \dfrac{2}{7} = 2:7$

(c). Number of circles to all the figures inside the rectangle.

Ans: In the given figure,

The total number of circles = 2

The total number of figures =7

Therefore,

Required ratio of the number of circles to the total number of figures is

$\dfrac{{Number\;of\;circles}}{{Total\;Number\;of\;figures}} = \dfrac{2}{7} = 2:7$

4. Distances travelled by Hamid and Akhtar in an hour are 9 km and 12 km. Find the ratio of speed of Hamid to the speed of Akhtar.

Ans: Given;

$Speed\;of\;Hamid = \dfrac{{Distance}}{{Time}} = \dfrac{9}{1} = 9km/h$

And distances travelled by Hamid and Akhtar in an hour = 12Km

$\therefore Speed\;of\;Akhtar = \dfrac{{Distance}}{{Time}} = \dfrac{{12}}{1} = 12km/h$

Therefore, Ratio of speed of Hamid to that of speed of Akhtar

$\therefore \dfrac{{Speed\;of\;Hamid}}{{Speed\;of\;Akhtar}} = \dfrac{9}{{12}} = \dfrac{{9 \div 3}}{{12 \div 3}} = \dfrac{3}{4} = 3:4$

5. Fill in the following blanks:

$\dfrac{{15}}{{18}} = \dfrac{{\boxed{}}}{6} = \dfrac{{10}}{{\boxed{}}} = \dfrac{{\boxed{}}}{{30}}$

Ans: In order to get the missing numbers,

Let consider the fact that

$18 \div 3 = 6$

$\therefore 15 \div 3 = 5$

$\dfrac{{15}}{{18}} = \dfrac{{\boxed5}}{6} = \dfrac{{10}}{{\boxed{}}} = \dfrac{{\boxed{}}}{{30}}$

Similarly,

$5 \times 2 = 10$

$\therefore 6 \times 2 = 12$

$\dfrac{{15}}{{18}} = \dfrac{{\boxed5}}{6} = \dfrac{{10}}{{\boxed{12}}} = \dfrac{{\boxed{}}}{{30}}$

Similarly, for third missing number compare

$\dfrac{5}{6} = \dfrac{{\boxed{}}}{{30}}$

Thus,

$6 \times 5 = 30$

$\therefore 5 \times 5 = 25$

Hence, after Ansving all equation we get,

$\dfrac{{15}}{{18}} = \dfrac{{\boxed5}}{6} = \dfrac{{10}}{{\boxed{12}}} = \dfrac{{\boxed{25}}}{{30}}$

Yes, these are equivalent ratios.

6. Find the ratio of the following:

a) 81 to 108

Ans:  Required ratio $= \dfrac{{81}}{{108}}$

Take LCM of 81 and 108

$81 = 3 \times 3 \times 3 \times 3$

$108 = 3 \times 3 \times 3 \times 2 \times 2$

$HCF\;of\;81\;and\;108 = 27$

$Ratio = \dfrac{{81 \div 27}}{{108 \div 27}} = \dfrac{3}{4} = 3:4$

b) 98 to 63

Ans:  Required ratio $= \dfrac{{98}}{{63}}$

Take LCM of 98 and 63

$98 = 2 \times 7 \times 7$

$63 = 3 \times 3 \times 7$

$HCF\;of\;98\;and\;63 = 7$

$Ratio = \dfrac{{98 \div 7}}{{63 \div 7}} = \dfrac{{14}}{9} = 14:9$

c) 33 km to 121 km

Ans:  Required ratio $= \dfrac{{33}}{{121}}$

Take LCM of 11 and 121

$33 = 3 \times 11$

$121 = 11 \times 11$

$HCF\;of\;33\;and\;121 = 11$

$Ratio = \dfrac{{33 \div 11}}{{121 \div 11}} = \dfrac{3}{{11}} = 3:11$

d) 30 minutes to 45 minutes

Ans:  Required ratio $= \dfrac{{30}}{{45}}$

Take LCM of 30 and 45

$30 = 2 \times 3 \times 5$

$45 = 3 \times 3 \times 5$

$HCF\;of\;30\;and\;45 = 15$

$Ratio = \dfrac{{30 \div 15}}{{45 \div 15}} = \dfrac{2}{3} = 2:3$

7. Find the ratio of the following:

(a) 30 minutes to 1.5 hour

Ans: 30 minutes to 1.5 hour

$1.5hours=1.5\times 60=90minutes~~~~$

[1hour= 60 minutes]

Now, ratio of 30 minutes to 1.5 hour $= 30minutes{\text{ }}:{\text{ }}1.5hour$

$=30minutes\text{ }:\text{ 90}minutes$

$=\dfrac{30}{1.5}=\dfrac{3}{9}=\dfrac{3\div 3}{9\div 3}=\dfrac{1}{3}=1:3$

(b) 40 cm to 1.5 m

Ans: $1.5m = 1.5 \times 100cm = 150cm\;\;\;\;\left( {1m = 100cm} \right)$

Now, ratio of 40 cm to 1.5 m

$=40cm:1.5m$

$=40cm:150cm$

$=\dfrac{4}{15}=\dfrac{4}{15}=4:15$

(c)55 paise to Re. 1

Ans: As we know that,

1 Re. 1 = 100 paise

Now, ratio of 55 paise to Re. $1 = 55paise:{\text{ }}100{\text{ }}paise$

$= \dfrac{{55}}{{100}} = \dfrac{{55 \div 5}}{{100 \div 5}} = \dfrac{{11}}{{20}} = 11:20$

(d) 500 ml to 2 liters

Ans: As we know that,

$1litre = 1000ml$

$2liters{\text{ }} = 2 \times 1000ml = 2000 ml$

Now, ratio of 500 ml to 2 liters = 500 ml: 2 liters

$2liters\text{ }=2\times 1000ml=2000ml$

$=\dfrac{5ml}{20ml}=\dfrac{5}{20}=\dfrac{1}{4}=1:4$

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.

Ans: Total earning =  1,50,000 and Saving =  50,000

$Money\;spent = 1,50,000 - 50,000 = 1,00,000$

Therefore, the ratio of money earned to money saved

Ratio of money earns to money saves$=\dfrac{150000}{50000}=\dfrac{3}{1}=3:1$

(b) Money that she saves to the money she spends.

Ans: Money spend by seema = Earned money – Saved money

$\text{Required ratio} = \dfrac{{Money\;saved\;by\;seema}}{{Money\;spend\;by\;seema}}$

Ratio of money saved to money spend$=\dfrac{5000}{100000}=\dfrac{1}{2}=1:2$

9. There are 102 teachers in a school of 3300 students. Find the ratio of the number of teachers to the number of students.

Ans:   Given; Number of teachers =102

Number of students =3300

$\text{Required ratio} = \dfrac{{Number{\text{ }}of{\text{ }}teachers}}{{Number{\text{ }}of{\text{ Students}}}}\;$

$= \dfrac{{102}}{{3300}}$

Take LCM of 102 and 3300

$102 = 2 \times 3 \times 17$

$3300 = 2 \times 3 \times 2 \times 5 \times 5 \times 11$

HCF$\;of\;102\;and\;3300 = 2 \times 3 = 6$

Ratio = $\dfrac{{102 \div 6}}{{3300 \div 6}} = \dfrac{{17}}{{550}} = 17:550$

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.

Ans: Given number of girls = 2300

And total number of students = 4320

$\text{Required ratio}=\dfrac{Number\text{ }of\text{ girls}}{Number\text{ }of\text{ Students}}\$

$=\dfrac{230}{432}=\dfrac{115}{216}$

(b) Number of boys to the number of girls.

Ans: Number of boys = Total number of students - Number of girls

$= 4320 - 2300 = 2020$

$\text{Required ratio}=\dfrac{Number\text{ }of\text{ boys}}{Number\text{ }of\text{ girls}}\$

$=\dfrac{202}{230}=\dfrac{101}{115}$

(c) Number of boys to the total number of students.

Ans: Number of boys = Total number of students - Number of girls

$= 4320 - 2300 = 2020$

$\text{Required ratio}=\dfrac{Number\text{ }of\text{ boys}}{Number\text{ }of\text{ Students}}\$

$=\dfrac{202}{432}=\dfrac{101}{216}$

11. Out of 1800 students in a school, 750 opted basketballs, 800 opted cricket and 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.

Ans: Given; Total number of students = 1800

Number of students who opted basketball = 750

Number of students who opted cricket =800

Therefore, number of students opted tennis $= 1800--\left( {750{\text{ }} + 800} \right){\text{ }} = 250$

$\text{Required ratio}=\dfrac{Number\ of\ students\text{ }who\text{ }opted\text{ }basketball}{Number\ of\ students\text{ }who\text{ }opted\text{ table-tennis}}\$

$=\dfrac{75}{25}=\dfrac{3}{1}=3:1$

(b) Number of students who opted cricket to the number of students opting basketball.

Ans:  $\text{Required ratio}=\dfrac{Number\ of\ students\text{ }who\text{ }opted\ cricket\text{ }}{Number\ of\ students\text{ }who\text{ }opted\text{ }basketball}\$

$=\dfrac{80}{75}=\dfrac{16}{15}=16:15$

(c) Number of students who opted basketball to the total number of students.

Ans:  $\text{Required ratio}=\dfrac{Number\ of\ students\text{ }who\text{ }opted\ basketball\text{ }}{Total\text{ }number\ of\ students}$

$=\dfrac{75}{180}=\dfrac{5}{12}=5:12$

12. Cost of a dozen pens is 180 and cost of 8 ball pens is 56. Find the ratio of the cost of a pen to  the cost of a ball pen.

Ans: Given, cost of 12 pens = Rs 180

The cost of 1 pen = $\dfrac{{Total\;cost\;of\;pen}}{{Number\;of\;pen}} = \dfrac{{180}}{{12}} = 15$

Also, given cost of 8 pens = Rs 56

The cost of 1 ball pen = $\dfrac{{Total\;cost\;of\;1\;ball\;pen}}{{Number\;of\;pen}} = \dfrac{{56}}{8} = 7$

Now, $\text{Required ratio} = \dfrac{{cost\;of\;1\;pen}}{{cost\;of\;1\;ball\;pen}} = \dfrac{{15}}{7} = 15:7$

13. Consider the statement: 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 meters) 10 40 Length of the hall (in meters) 25 50

Ans: Given, Ratio of breadth and length of a hall = 2:5

Also, given breath of the hall = 10

And length of the hall = 25

Then, Ratio of breadth to length $= \dfrac{{10}}{{25}} = \dfrac{2}{5} = 2:5$

Other equivalent ratios are therefore for first missing number we have,

$\dfrac{{10}}{{25}} = \dfrac{{\boxed{}}}{{50}}$

$\dfrac{{10 \times 2}}{{25 \times 2}} = \dfrac{{\boxed{20}}}{{50}}$

Similarly, for second missing number we have,

$\dfrac{{10}}{{25}} = \dfrac{{20 \times 2}}{{50 \times 2}} = \dfrac{{40}}{{\boxed{100}}}$

Thus, the complete table is

 Breadth of the hall (in meters) 10 20 40 Length of the hall (in meters) 25 50 100

14. Divide 20 pens between Sheela and Sangeeta in the ratio 3: 2.

Ans:  Given ratio = 3:2

Sum of the ratio  $= 3 + {\text{ }}2 = 5$

Therefore, part of Sheela $= \dfrac{3}{5}$

And part of Sangeeta $= \dfrac{2}{5}$

Total number of pens = 20

Number of pens for Sheela $= \dfrac{3}{5} \times 20 = 12$

Number of pens for Sangeeta $= \dfrac{2}{5} \times 20 = 8$

Hence, out of 20 pens Sheela gets 12 pens and Sangeeta gets 8 pens.

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 age of Bhoomika is 12 years, find how much Shreya and Bhoomika will get.

Ans:  Given: Shreya’s age = 15 years

And Bhoomika age = 12 years,

Ratio of their age $= \dfrac{{15}}{{12}} = \dfrac{5}{4} = 5:4$

Now, mother wants to divide Rs 36 between her daughters in the ratio of their ages.

Then sum of the parts of ratio $= 5 + 4 = 9$

$\therefore Shreya's\;share = \dfrac{5}{9} \times 36 = Rs\;20$

$\therefore Bhoomika\;share = \dfrac{4}{9} \times 36 = Rs\;16$

Hence, Shreya gets Rs 20 and Bhoomika gets Rs 16.

16. Present age of father is 42 years and that of his son is 14 years. Find the ratio of:

(a) Present age of father to the present age of son.

Ans: Given: Present age of father = 42 years

And Present age of son = 14 years

$\text{Required ratio} = \dfrac{{Present{\text{ }}age{\text{ }}of{\text{ }}father}}{{Present{\text{ }}age{\text{ }}of{\text{ son}}}}$

$= \dfrac{{42}}{{14}} = \dfrac{{42 \div 14}}{{14 \div 14}} = \dfrac{3}{1} = 3:1$

(b) Age of the father to the age of the son, when son was 12 years old.

Ans:  When son was 12 years old

Then, father’s age $= 42 - {\text{ }}2 = 40$

$\therefore \text{Required ratio} = \dfrac{{{\text{2}}\;{\text{years}}\;{\text{back }}father\;age}}{{{\text{2}}\;{\text{years}}\;{\text{back son age}}}}$

$= \dfrac{{40}}{{12}} = \dfrac{{40 \div 4}}{{12 \div 4}} = \dfrac{{10}}{3} = 10:3$

(c) Age of father after 10 years to the age of son after 10 years.

Ans: Age of father after 10 years $= 42 + 10 = 52\;years$

And the age of son after 10 years $= 14 + 10 = 24\;years$

$\therefore \text{Required ratio}= \dfrac{{After\;10\;{\text{years}}\;father\;age}}{{After\;10\;{\text{years}}\;{\text{son age}}}}$

$= \dfrac{{52}}{{24}} = \dfrac{{52 \div 4}}{{24 \div 4}} = \dfrac{{13}}{6} = 13:6$

(d) Age of father to the age of son when father was 30 years old.

Ans: When father was 30 years old then the age of son is 2 years.

$42 - 12 = 30\;years$

$\therefore 14 - 12 = 2\;years$

$\therefore \text{Required ratio} = \dfrac{{{\text{12}}\;{\text{years}}\;{\text{back }}father\;age}}{{{\text{12}}\;{\text{years}}\;{\text{back son age}}}}$

$= \dfrac{{30}}{2} = \dfrac{{30 \div 2}}{{2 \div 2}} = \dfrac{{15}}{1} = 15:1$

### Exercise 12.2

1. Determine the following are in proportion:

(a) 15, 45, 40, 120

Ans: We have, 15, 45, 40, 120

$\therefore Ratio\ of\ 15\ to\ 45=\dfrac{15}{45}=\dfrac{15\div 15}{45\div15}=\dfrac{1}{3}=1:3$

$And\ Ratio\ of\ 40\ to\ 120=\dfrac{4}{12}=\dfrac{1}{3}=1:3$

Since, 15:45 = 40: 120

Therefore, 15, 45, 40 and 120 are in proportion.

(b) 33, 121, 9, 96

Ans: We have, 33, 121, 9, 96

$\therefore Ratio\;of\;33\;to\;121 = \dfrac{{33}}{{121}} = \dfrac{{33 \div 11}}{{121 \div 11}} = \dfrac{3}{{11}} = 3:11$

And $\;Ratio\;of\;9\;to\;96 = \dfrac{9}{{96}} = \dfrac{{9 \div 3}}{{96 \div 3}} = \dfrac{3}{{32}} = 3:32$

Since, $33:121 \ne 9:96$

Therefore, 33, 121, 9, 96 are not in proportion.

(c) 24, 28, 36, 48

Ans: We have, 24, 28, 36, 48

Ratio $\;of\;24\;to\;28 = \dfrac{{24}}{{28}} = \dfrac{{24 \div 4}}{{28 \div 4}} = \dfrac{6}{7} = 6:7$

And $\;Ratio\;of\;36\;to\;48 = \dfrac{{36}}{{48}} = \dfrac{{36 \div 6}}{{48 \div 6}} = \dfrac{6}{7} = 6:8$

Since, $24:28 \ne 36:48$

Therefore, 24, 28, 36 and 48 are not in proportion.

(d) 32, 48, 70, 210

Ans: We have, 32, 48, 70, 210

$\therefore Ratio\ of\ 32\ to\ 48=\dfrac{32}{48}=\dfrac{32\div 16}{48\div 16}=\dfrac{2}{3}=2:3$

$And\ Ratio\ of\ 70\ to\ 210=\dfrac{7}{21}=\dfrac{7\div 7}{21\div 7}=\dfrac{1}{3}=1:3$

$Since,\ 32:49\ne 70:210$

Therefore, 32, 48, 70 and 210are not in proportion.

(e) 4, 6, 8, 12

Ans: We have, 4, 6, 8, 12

$\therefore Ratio\;of\;4\;to\;6 = \dfrac{4}{6} = \dfrac{{4 \div 2}}{{6 \div 3}} = \dfrac{2}{3} = 2:3$

And $\;Ratio\;of\;8\;to\;12 = \dfrac{8}{{12}} = \dfrac{{8 \div 4}}{{12 \div 4}} = \dfrac{2}{3} = 2:3$

Since, $\;4:6 = 8:12$

Therefore, 4, 6, 8 and 12 are in proportion.

(f) 33, 44, 75, 100

Ans: We have, 33, 44, 75, 100

Ratio $\;of\;33\;to\;44 = \dfrac{{33}}{{44}} = \dfrac{{33 \div 11}}{{44 \div 11}} = \dfrac{3}{4} = 3:4$

And $\;Ratio\;of\;75\;to\;100 = \dfrac{{75}}{{100}} = \dfrac{{75 \div 25}}{{100 \div 25}} = \dfrac{3}{4} = 3:4$

Since,$\;33:44 = 75:100$

Therefore, 33, 44, 75 and 100 are in proportion.

2. Write True (T) or False (F) against each of the following statement:

(a) 16: 24: 20: 30

Ans:  $\dfrac{{16}}{{24}} = \dfrac{{20}}{{30}}$

$\dfrac{{16 \div 8}}{{24 \div 8}} = \dfrac{{20 \div 10}}{{30 \div 10}}$

$\dfrac{2}{3} = \dfrac{2}{3}$

Hence, it is True.

(b) 21: 6: 35: 10

Ans: $\dfrac{{21}}{6} = \dfrac{{35}}{{10}}$

$\dfrac{{21 \div 3}}{{6 \div 3}} = \dfrac{{35 \div 5}}{{10 \div 5}}$

$\dfrac{7}{2} = \dfrac{7}{2}$

Hence, it is True.

(c) 12: 18: 28: 12

Ans:  $\dfrac{{12}}{{18}} = \dfrac{{28}}{{12}}$

$\dfrac{{12 \div 6}}{{18 \div 6}} = \dfrac{{28 \div 4}}{{12 \div 4}}$

$\dfrac{2}{3} \ne \dfrac{7}{3}$

Hence, it is false.

(d) 8: 9: 24: 27

Ans: $\dfrac{8}{9} = \dfrac{{24}}{{27}}$

$\dfrac{8}{9} = \dfrac{{24 \div 3}}{{27 \div 3}}$

$\dfrac{8}{9} = \dfrac{8}{9}$

Hence, it is True.

(e) 5.2: 3.9: 3: 4

Ans: $\dfrac{{5.2}}{{3.9}} = \dfrac{3}{4}$

$\dfrac{{5.2 \div 1.3}}{{3.9 \div 1.3}} = \dfrac{3}{4}$

$\dfrac{4}{3} \ne \dfrac{3}{4}$

Hence, it is false.

(f) 0.9: 0.36: 10: 4

Ans: $\dfrac{{0.9}}{{0.36}} = \dfrac{{10}}{4}$

$\dfrac{{0.9 \div 0.18}}{{0.36 \div 0.18}} = \dfrac{{10 \div 2}}{{4 \div 2}}$

$\dfrac{5}{2} = \dfrac{5}{2}$

Hence, it is True.

3. Are the following statements true?

(a) 40 persons: 200 persons = Rs15: Rs 75

Ans: We have,

40 persons : 200 persons =$\dfrac{4}{20}=\dfrac{4\div 4}{20\div 4}=\dfrac{1}{5}$

Rs15:Rs 75= $\dfrac{{15}}{{75}} = \dfrac{{15 \div 15}}{{75 \div 15}} = \dfrac{1}{5}$

Hence, the statement is true.

(b)  7.5 liters: 15 liters = 5 kg= 10 kg

Ans: We have,

$7.5liters:15liters = \dfrac{{7.5}}{{15}} = \dfrac{{75}}{{150}} = \dfrac{{75 \div 75}}{{150 \div 75}} = \dfrac{1}{2}$

$5{\text{ }}kg:10kg = \dfrac{5}{{10}} = \dfrac{{5 \div 5}}{{10 \div 5}} = \dfrac{1}{2}$

Hence, the statement is true.

(c)  99 kg: 45 kg = Rs 44: Rs 20

Ans: We have,

${\text{99 }}kg:45kg = \dfrac{{99}}{{45}} = \dfrac{{99 \div 9}}{{45 \div 9}} = \dfrac{{11}}{5}$

$Rs44:Rs{\text{ }}20 = \dfrac{{44}}{{20}} = \dfrac{{44 \div 4}}{{20 \div 4}} = \dfrac{{11}}{5}$

Hence, the statement is true.

(d) 32 m: 64 m = 6 sec.: 12 sec.

Ans: We have,

$32m:{\text{ }}64m{\text{ }} = \dfrac{{32}}{{64}} = \dfrac{{32 \div 32}}{{64 \div 32}} = \dfrac{1}{2}$

$6sec.{\text{ }}:12sec. = \dfrac{6}{{12}} = \dfrac{1}{2}$

Hence, the statement is true.

(e) 45 km: 60 km = 12 hours: 15 hours

Ans: We have,

$45km:60km = \dfrac{{45}}{{60}} = \dfrac{{45 \div 15}}{{60 \div 15}} = \dfrac{3}{4}$

$12{\text{ }}hours{\text{ }}:{\text{ }}15hours = \dfrac{{12}}{{15}} = \dfrac{{12 \div 3}}{{15 \div 3}} = \dfrac{4}{5}$

Hence, it is false statement.

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 Rs40:Rs160

Ans: ${\text{25 cm : 1 m = 25 cm : (1 }} \times {\text{ 100) cm = 25 cm : 100 cm = }}\dfrac{{25}}{{100}} = \dfrac{1}{4} = 1:4$

$Rs40:Rs160=\dfrac{4}{16}=\dfrac{4\div 4}{16\div 4}=\dfrac{1}{4}=1:4$

Since the ratios are equal, therefore these are in proportion.

Middle terms are 1 m and Rs 40 and Extreme terms are 25 cm and Rs 160.

(b) 39 liters: 65 liters and 6 bottles: 10 bottles

Ans: ${\text{39 liters : 65 liters}} = \dfrac{{39}}{{65}} = \dfrac{{39 \div 13}}{{65 \div 13}} = \dfrac{3}{5} = 3:5$

${\text{6 bottles : 10 bottles = }}\dfrac{6}{{10}} = \dfrac{{6 \div 2}}{{10 \div 2}} = \dfrac{3}{5} = 3:5$

Since the ratios are equal, therefore these are in proportion.

Middle terms are 65 liters and 6 bottles and extreme terms are 39 liters and 10 bottles.

(c) 2 kg: 80 kg and 25 g: 625 g

Ans: ${\text{2 kg : 80 kg = }}\dfrac{2}{{80}} = \dfrac{{2 \div 2}}{{80 \div 2}} = \dfrac{1}{{40}} = 1:40$

${\text{25 g : 625 g = }}\dfrac{{25}}{{625}} = \dfrac{{25 \div 25}}{{625 \div 25}} = \dfrac{1}{{25}} = 1:25$

Since the ratios are not equal, therefore these are not in proportion

(d) 200 ml: 2.5 ml and RS4: Rs 50

Ans: $200ml:2.5ml\text{= 200 ml : (25 }\times \text{ 1000) liters = 200 ml : 2500 ml =}\dfrac{2}{25}=\dfrac{2}{25}=2:25$

$RS4:Rs50=\dfrac{4}{50}=\dfrac{4\div 2}{50\div 2}=\dfrac{2}{25}=2:25$

Since the ratios are equal, therefore these are in proportion.

Middle terms are 2.5 liters and Rs 4 and Extreme terms are 200 ml and Rs 50.

### Exercise 12.3

1. If the cost of 7 m of cloth is Rs 1470, find the cost of 5 m of cloth.

Ans:   Given: the cost of 7 m of cloth = Rs 1470

Cost of 1 m of cloth $= \dfrac{{1470}}{7} = Rs 210$

Therefore, cost of 5 m of cloth $= 210 \times 5 = Rs 1050$

Hence, the cost of 5 m of cloth is Rs 1050.

2. Ekta earns Rs 3000 in 10 days. How much will she earn in 30 days?

Ans:  Earning of 1 day $= \dfrac{{3000}}{{10}} = Rs 300$

Earning of 30 days $= Rs300 \times 30 = Rs 9000$

Thus, the earning of 30 days is Rs 9,000.

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.

Ans:   Given; rainfall in 3 days = 276 mm

Rain in 1 day $= \dfrac{{276}}{3} = 92mm$

Rain in 7 days $= 92 \times 7 = 644mm$

Thus, the rain in 7 days is 644 mm.

4. Cost of 5 kg of wheat is Rs 91.50.

(a) What will be the cost of 8 kg of wheat?

Ans: Given: Cost of 5 kg of wheat=Rs 91.50

Cost of 1 kg of wheat $= \dfrac{{Rs91.50}}{5} = Rs 18.3$

Cost of 8 kg of wheat $= Rs 18.3 \times 8 = Rs 146.4$

Hence, the cost of 8 kg of wheat is Rs 146.4.

(b) What quantity of wheat can be purchased in Rs 183?

Ans:   Given, in Rs 91.50, quantity of wheat that can be purchased =5 Kg

Therefore, In Rs 1, quantity of wheat that can be purchased $= \dfrac{5}{{91.50}} = \dfrac{{5 \times 10}}{{915}} = \dfrac{{50}}{{915}}Kg$

Therefore, In Rs 183, quantity of wheat that can be purchased $= \dfrac{{50}}{{915}}Kg \times 183 = 10Kg$

Hence, 10 kg wheat can be purchased in Rs 61.

5. The temperature dropped 15 degrees 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?

Ans:  Degree of temperature dropped in last 30 days = 15 degrees

$\therefore$Degree of temperature dropped in 1 days $= \dfrac{{15}}{{30}} = \dfrac{{15 \div 15}}{{30 \div 15}} = \dfrac{1}{2}\deg ree$

$\therefore$Degree of temperature dropped in last 10 days $= \dfrac{1}{2} \times 10 = 5\deg ree$

Thus, 5-degree Celsius temperature dropped in 10 days.

6. Shains pays Rs 15000 as rent for 3 months. How much does she has to pay for a whole year, if the rent per month remains same?

Ans:  Rent paid for 3 months = Rs 15000

$\therefore$Rent paid for 1 months $= \dfrac{{Rs 15000}}{3} = Rs 5000$

$\therefore$Rent paid for 12 months $= Rs 5000 \times 12 = Rs 60,000$

Thus, the total rent of one year is Rs 60,000.

7. Cost of 4 dozen bananas is  180. How many bananas can be purchased for Rs 90?

Ans:  Cost of 4 dozen bananas = Rs 180

We know that 1 dozen = 12 units

4 dozen $= 4 \times 12 = 48$

$\therefore$Cost of 48 bananas = Rs 180

$\therefore$ Given, in Rs 180, number of bananas that can be purchased = 48

$\therefore$In Rs 1, number of bananas that can be purchased $= \dfrac{{48}}{{180}} = \dfrac{4}{15}$

$\therefore$From Rs 90. number of bananas can be purchased $= \dfrac{4}{15} \times 90 = \dfrac{{360}}{15} = 24$

Thus, 24 bananas can be purchased for Rs 90.

8. The weight of 72 books is 9 kg what is the weight of 40 such books?

Ans:  The weight of 72 books = 9 kg

$\therefore$ The weight of 1 book$= \dfrac{9}{{72}}Kg$

$\therefore$The weight of 40 books$= \dfrac{9}{{72}}Kg \times 40 = \dfrac{{360}}{{72}} = \dfrac{{360 \div 72}}{{72 \div 72}} = 5Kg$

Thus, the weight of 40 books is 5 kg.

9. A truck requires 108 liters 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?

Ans:  For covering 594 km, a truck will be required diesel = 108 liters

$\therefore$For covering 1 km, a truck will be required diesel $= \dfrac{{108}}{{594}}L = \dfrac{{108 \div 54}}{{594 \div 54}} = \dfrac{2}{{11}}L$

$\therefore$ For covering 1650 km, a truck will be required diesel$= \dfrac{2}{{11}} \times 1650L = 2 \times 150 = 300L$

Thus, 300 L diesel will be required by the truck to cover a distance of 1650 km.

10. Raju purchases 10 pens for Rs150 and Manish buys 7 pens for Rs84. Can you say who got the pen cheaper?

Ans:  Raju purchase 10 pens for = Rs 150

$\therefore$Cost of 1 pen $= \dfrac{{150}}{{10}} = \operatorname{Rs} 15$

For Manish, cost of 7 pens = Rs 84

$\therefore$ Cost of 1 pen $= \dfrac{{84}}{7} = Rs12$

Here,    $Rs12 < \operatorname{Rs} 15$

Thus, Manish got the pens cheaper.

11. Anish made 42 runs in 6 overs and Anup made 63 runs in 7 overs. Who made more runs per over?

Ans:  Anish made in 6 overs = 42 runs

$\therefore$Anish made in 1 overs $= \dfrac{{42}}{6} = 7$

Anup made in 7 overs = 63

$\therefore$Anup made in 1 overs$= \dfrac{{63}}{7} = 9$

Here, $9 > 7$

Thus, Anup made more runs per over.

## Class 6 Maths Chapter 12: Exercises Breakdown

 Exercise Number of Questions Exercise 12.1 16 Questions & Solutions Exercise 12.2 4 Questions & Solutions Exercise 12.3 11 Questions & Solutions

## Conclusion

In conclusion, Class 6 Maths Ch 12 Ratio and Proportion is an essential part of your mathematical journey. This chapter helps build a strong foundation by teaching how to compare quantities using ratios and how to determine equality through proportions. It is important to focus on understanding the key concepts and practising the methods to simplify ratios and solve proportion problems effectively.

From the previous year’s question papers, typically, around 2-3, questions were asked from Class 6 Maths Ch 12, emphasizing its significance. These questions often involve practical applications of ratios and proportions, so ensure you are comfortable with both the theory and practical aspects.

## Chapter-Specific NCERT Solutions for Class 6 Maths

Given below are the chapter-wise NCERT Solutions for Class 6 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

## FAQs on NCERT Solutions for Class 6 Maths Chapter 12 Ratio and Proportion

1. What is the Ratio?

The ratio is the comparison between two entities of the same unit. For example, the height of two buildings, the length of two clothes, and the area of two playgrounds. CBSE Class 6 Maths Chapter 12 solutions deal with problems on ratio. NCERT Solutions for the chapter can be obtained on the Vedantu website.

2. What is the Proportion?

When two ratios are compared, the comparison is called a proportion. Therefore, there are four separate entities involved in proportion. All four entities have to have the same units. Class 6 Maths Ratio and Proportion solution deal with several problems related to the concept of proportion.

The best reference guide for the students to score high marks in the exams are the NCERT Solutions provided by Vedantu. We provide detailed explanations of the solutions covering each and every concept in the chapter. Students can access the solutions when they face difficulty in answering exercise-wise problems. They help to clarify the students' doubts instantly. These solutions are available both online and offline providing chapter wise and exercise wise solutions helping the students irrespective of their intelligence quotient.

4. Do I need to make notes while referring to NCERT Solutions of Maths for Chapter 12 of Class 6 ?

It is highly recommended for students to make notes while referring to the NCERT Solutions of Chapter 12 in Maths of Class 6. Making notes in your own words is a best practice as it will help to quickly recall the concepts studied earlier. For some problems, writing down the methods in an easier way will help the students to remember for a long time.  Therefore, this will help the students to complete the revision at twice the pace and memorise twice the amount of information compared to studying without notes.

5. What are the best ways to learn concepts in Class 6 Maths of Chapter 12?

The best way for the students to learn the concepts in NCERT solutions maths is by practicing constantly and with due diligence. It is very important for the students to build a strong knowledge and have a clear understanding of all the topics in the chapter before attempting exercise problems. Students must continuously revise the theories, formulas, and problems. Having a strong conceptual foundation will help the students to build confidence in themselves to solve any type of questions that occur in exams.

6. How CBSE Students can utilize Maths NCERT Solutions effectively for Chapter 12 of Class 6 Maths?

There are some effective methods for the students to follow. The first step that the students have to do is to understand the theory lying behind any concept or problem. Next is to go through the formulas and memorise them.  Students have to understand the derivation as well as the ideology behind the formulas too. Then, they have to apply these formulas and concepts to the exercise problems. At last, students should use the NCERT solutions of Chapter 12 Of Class 6 Maths to check if their answers and methods are correct or not.

7. Are NCERT Solutions of Chapter 12 in Maths of Class 6 sufficient for CBSE Exams?

Yes. NCERT Solutions for Chapter 12 of Class 6 Maths are enough for the preparation for the CBSE exams as the majority of the questions for the exam are asked from the Maths NCERT books. Maths NCERT books follow the CBSE curriculum strictly. The Vedantu NCERT solutions for Maths are designed in such a way that it will help the students to prepare well for the exam. We provide various study tips and interesting facts to make learning easier for the students.  The solutions are free of cost and also available on Vedantu Mobile app.

8. What is a ratio in class 6 maths chapter 12 pdf solution?

In class 6 maths chapter 12 pdf solutions, a ratio is a comparison between two quantities showing the relative sizes of each. For example, if there are 4 apples and 2 oranges, the ratio of apples to oranges is 4:2, which simplifies to 2:1. Ratios help in understanding how one quantity relates to another in terms of magnitude, allowing for easy comparisons and proportional reasoning.

9. How is a ratio expressed in class 6 chapter 12 maths?

In class 6 chapter 12 maths a ratio is expressed using the colon symbol ':'. For instance, the ratio of 5 to 10 is written as 5:10, which simplifies to 1:2. This means for every 1 unit of the first quantity, there are 2 units of the second quantity. Ratios are a fundamental way to express relationships in mathematics and are used in various applications like scaling recipes or models.

10. What is a proportion in ratio and proportion questions class 6?

In ratio and proportion questions class 6, a proportion is an equation that states that two ratios are equal. For example, the ratio 3:6 is proportional to 1:2 because 3/6 equals 1/2. Proportions are used to solve problems where you need to find an unknown quantity, maintaining the same ratio. This concept is crucial in ensuring consistency and accuracy in calculations involving equivalent relationships.

11. Why is understanding ratios and proportions important?

Understanding ratios and proportions is important because they are widely used in everyday life and advanced mathematics. For instance, they are essential in cooking for scaling recipes, in map reading for understanding distances, and in science for calculating concentrations. Mastery of these concepts helps students solve practical problems effectively and lays the groundwork for more complex mathematical concepts.

12. What should students focus on in chapter 12 class 6 maths?

In chapter 12 class 6 maths Students should focus on understanding the fundamental concepts of ratios and proportions, learning to simplify ratios, and solving proportion problems. Key areas include practising problems on finding equivalent ratios, setting up and solving proportions, and applying these concepts to real-world scenarios. Paying attention to examples and practising exercises will help solidify these skills.

13. How many questions were asked from this chapter in previous exams for class 6 math chapter 12?

Typically, around 2–3 questions from class 6 math chapter 12 have been asked in previous exams. These questions often test students' understanding of basic concepts and their ability to apply ratios and proportions in various scenarios. Reviewing past exam questions and practising similar problems can help students prepare effectively and perform well in exams.