NCERT Solutions for Class 6 Maths Chapter 12 Ratio and Proportion

NCERT Solutions for Class 6 Maths Chapter 12 Ratio and Proportion

Edited By Ramraj Saini | Updated on Apr 05, 2024 11:05 AM IST

NCERT Solutions for Class 6 Maths Chapter 12 Ratio and Proportion are discussed in this article. These NCERT solutions are created by the expert team at careers360 considering the latest CBSE syllabus 2024-25. In our daily life, we compare things like he is taller than me by 10 cm, Ramesh got double of my marks in Maths. The first statement is a comparison which means that the difference in height of both is 10 cm. The second statement is showing the ratio of marks. You can say that the ratio of marks obtained by Ramesh and it is 2 : 1. In this article, you will get NCERT solutions for Class 6 Maths chapter 12 Ratio and Proportion.

It is advisable to also cover all the topics of NCERT Class 6 Maths Books to score well. If comparing quantities do not have the same unit, first convert them into the same unit and then find the ratio. Students must try to complete the NCERT Class 6 Maths Syllabus in at the earliest. You should try to solve all the NCERT questions by yourself so you won't these silly mistakes. You can take help from the NCERT Solutions for Class 6 .

NCERT Solutions for Class 6 Maths Chapter 12 Ratio and Proportion - Important Formulae

  1. The ratio of a number ‘a’ to another number ‘b’ where b ≠ 0 is written as a fraction a/b also represented as a : b. Here the first term is a and the second term is b.

  2. A ratio is said to be in the simplest form if its two terms have no common factor other than 1

  3. The ratio of two numbers is usually expressed in its simplest form.

  4. The ratio of two quantities is an abstract quantity, i.e., it has no units in itself.

  5. An equality of two ratios is called a proportion. If a : b = c : d , then we write a : b :: c : d.

  6. The numbers a, b, c, d are in proportion if the ratio of the first two is equal to the ratio of the last two, i.e., a : b = c : d.

  7. If four numbers a, b, c, d are in proportion, then a and d are known as extreme terms and b and c are called middle terms.

  8. Four numbers are in proportion if the product of extreme terms is equal to the product of middle terms, i.e., a : b :: c : d if and only if ad = bc.

  9. From the terms of a given proportion, we can make three more proportions.

  10. If a : b = b : c, then a, b, c are said to be in continued proportion.

  11. if a, b, and c are in continued proportion, i.e., a : b :: b : c, then b is called the mean proportional between a and c.

  12. Mean Proportion = b = √(ac)

  13. Value of a given number of articles = (Value of one article) x (Number of articles)

  14. Value of one article = (Value of given number of articles) / (Number of article).......... (Unitary Method)

For more, Download Ebook - NCERT Class 6 Maths: Chapterwise Important Formulas And Points

NCERT Solutions for Class 6 Maths Chapter 12 Ratio and Proportion - Important Points

Difference: The method of comparing quantities by taking the difference between them. It involves subtracting one quantity from another to determine the gap or distance between them.

Comparison by Ratio: A method of meaningful comparison between quantities by using division. It involves determining how many times one quantity is in relation to another quantity. The comparison is expressed in the form of a ratio, which represents the relative sizes or magnitudes of the quantities being compared.

Same Unit: In order to compare quantities by ratio, they must be expressed in the same unit. If the quantities are initially in different units, they need to be converted to a common unit before the ratio can be calculated.

Same Ratio in Different Situations: It is possible for the same ratio to occur in different contexts or situations. The ratio between two quantities remains constant regardless of the specific scenario in which it is being used.

Order of Ratios: The order in which quantities are taken to express their ratio is significant. The ratio 3:2 is distinct from the ratio 2:3, indicating that the position or arrangement of quantities in a ratio affects its meaning.

Ratio as a Fraction: A ratio can be treated as a fraction, with the numerator representing one quantity and the denominator representing another quantity. This allows for mathematical operations involving ratios, such as addition, subtraction, multiplication, and division.

Equivalent Ratios: Two ratios are considered equivalent if the fractions corresponding to them are equivalent. Ratios can be scaled up or down while maintaining their equivalence. For example, the ratio 3:2 is equivalent to 6:4 or 12:8.

Lowest Form of a Ratio: A ratio can be expressed in its simplest or lowest form by reducing it to its smallest possible integer values. For instance, the ratio 50:15 can be simplified to 10:3 by dividing both terms by their greatest common divisor.

Proportion: Four quantities are said to be in proportion if the ratio of the first and second quantities is equal to the ratio of the third and fourth quantities. Proportions establish relationships and comparisons between multiple sets of quantities.

Order in Proportions: The order of terms in a proportion is crucial. The arrangement of quantities must remain consistent across the ratios to ensure a valid proportion. Altering the order of terms can result in a different comparison or relationship.

Unitary Method: The unitary method is a technique used to solve problems involving proportions and ratios. It involves finding the value of one unit and then calculating the value of the desired number of units based on that reference. It is often used in solving problems related to cost, price, or measurement.

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NCERT Solutions for Class 6 Maths Chapter 12 Ratio and Proportion

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NCERT Class 6 Maths Chapter 12 Ratio and Proportion Topic 12.2 Ratio

Question:1 In a class, there are 20 boys and 40 girls. What is the ratio of the number of boys to the number of girls?

Answer: number of boys = 20

number of girls = 40

\frac{number \ of\ boys}{number\ of\ girls}=\frac{20}{40}=\frac{2}{4}=\frac{1}{2}

So the required ratio is 1:2

Question:2 Ravi walks 6 km in an hour while Roshan walks 4 km in an hour What is the ratio of the distance covered by Ravi to the distance covered by Roshan?

Answer: The distance covered in one hour by Ravi = 6 Km

The distance covered in one hour by Roshan = 4 Km

\frac{ The \ distance \ covered \ by \ Ravi }{ The \ distance \ covered \ by\ Roshan}=\frac{6}{4}=\frac{3}{2}

So the required ratio is 3:2

Question:1 Saurabh takes 15 minutes to reach school from his house and Sachin takes one hour to reach school
from his house. Find the ratio of the time taken by Saurabh to the time taken by Sachin.

Answer: Time taken by Saurabh = 15 minutes

Time taken by Sachin= 1 hour = 60 minutes. To find ratios we have to convert the given quantities to the same units. Here we are expressing both the quantities in minutes

the ratio of the time taken by Saurabh to the time taken by Sachin= 15:60=1:4

Question:2 Cost of a toffee is 50 paise and cost of a chocolate is rupees 10. Find the ratio of the cost of a toffee to the cost of a chocolate.

Answer: Cost of toffee = 50 paise

cost of chocolate = 10 rupees

1rupee = 100 paise

Therefore, 10rupee = 1000 paise

So, the cost of chocolate = 1000 paise

the ratio of the cost of toffee to the cost of chocolate=50:1000=1:20

Question:3 In a school, there were 73 holidays in one year. What is the ratio of the number of holidays to the number of days in one year?

Answer: Number of holidays in a year = 73

Number of days in a year = 365

\frac{number \ of \ holidays }{number \ of \ days \ in \ one \ year}=\frac{73}{365}=\frac{73}{73\times5}=\frac{1}{5}

the ratio of the number of holidays to the number of days in one year= 1:5

Question:1 Find the ratio of number of notebooks to the number of books in your bag.

Answer: If there are 3 notebooks and 4 books in the bag then the ratio of the number of notebooks to the number of books =3:4

Question:2 Find the ratio of number of desks and chairs in your classroom.

Answer: If there are 8 desks and 32 chairs then the ratio of number of desks and chairs is 8:32=1:4

Question:3 Find the number of students above twelve years of age in your class. Then, find the ratio of the number of students with age above twelve years and the remaining students.

Answer: suppose there are 40 students in the class and 5 students are above 12 years, then there are 40-5=35 students below or equal to 12 years.

Then the ratio of the number of students with age above twelve years and the remaining students = 5:35=1:7

Question:4 Find the ratio of number of doors and the number of windows in your classroom.

Answer: If there are four windows and one door then the ratio of the number of doors and the number of windows =1:4

Question:5 Draw any rectangle and find the ratio of its length to its breadth.

Answer: Suppose a rectangle has a length of 10 cm and a breadth of 7 cm then the ratio of its length to its breadth=10:7

NCERT Class 6 Maths Chapter 12 Ratio and Proportion Exercise 12.1

Question: 1(a) There are 20 girls and 15 boys in a class. What is the ratio of number of girls to the number of boys?

Answer: Given,

Number of boys = 15

Number of girls = 20

So,

The ratio of the number of girls to the number of boys:

=\frac{20}{15}=\frac{4}{3}=4:3

Question: 1(b) There are 20 girls and 15 boys in a class. What is the ratio of number of girls to the total number of students in the class?

Answer: Given,

Number of boys = 15

Number of girls = 20

Total number of students = 15 + 20 = 35.

the ratio of the number of girls to the total number of students in the class:

=\frac{20}{20+15}=\frac{20}{35}=\frac{4}{7}=4:7

Question: 2(a) Out of 30 students in a class, 6 like football, 12 like cricket and remaining like tennis. Find the ratio of Number of students liking football to number of students liking tennis.

Answer: Given,

Total Number of a student = 30

Number of students who like football = 6

Number of students who like cricket = 12

Number of remaining student wh play tennis = 30 - 6 - 12

= 12

Now,

The ratio of Number of students liking football to the number of students liking tennis:

=\frac{6}{12}=\frac{1}{2}=1:2

Question: 2 (b) Out of 30 students in a class, 6 like football, 12 like cricket and remaining like tennis. Find the ratio of Number of students liking cricket to total number of students.

Answer: Given,

Total Number of a student = 30

Number of students who like football = 6

Number of students who like cricket = 12

Number of remaining student wh play tennis = 30 - 6 - 12

= 12

Now, The ratio of Number of students liking cricket to the total number of students:

=\frac{12}{30}=\frac{6}{15}=\frac{2}{5}=2:5

Question: 3 See the figure and find the ratio of

1643779781572

(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.

Answer: From the figure, we can see that inside the rectangle,

Number of triangles = 3

Number of squares = 2

Number of circles = 2

So,

(a) The number of triangles to the number of circles inside the rectangle:

\frac{3}{2}=3:2

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

\frac{2}{7}=2:7

(c) The number of circles to all the figures inside the rectangle:

\frac{2}{7}=2:7

Question: 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.

Answer: As we know,

speed=\frac{distance}{time}

So,

Speed of Hamid :

speed=\frac{distance}{time}=\frac{9km}{1hour}=9km/h

Speed of Akhtar :

speed=\frac{distance}{time}=\frac{12km}{1hour}=12km/h

Hence, the ratio of the speed of Hamid to the speed of Akhtar:

\frac{9}{12}=\frac{3}{4}=3:4 .

Question: 5 Fill in the following blanks:

\frac{15}{18}=\frac{\square }{6}=\frac{10}{\square}=\frac{\square }{30} [Are these equivalent ratios?]

Answer: Equating all the fraction, we get

\frac{15}{18}=\frac{5 }{6}=\frac{10}{12}=\frac{25 }{30}

Yes, They are equivalent ratios.

Question: 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

Answer: (a)Ratio of 81 \; to \; 108

= \frac{81}{108}=\frac{27}{36}=\frac{3}{4}=3:4

(b) Ratio of 98 \; to \; 63

= \frac{98}{63}=\frac{14}{9}=14:9

(c) Ratio of 33\; km to 121\; km

= \frac{33}{121}=\frac{3}{11}=3:11

(d) The ratio of 30 minutes to 45 minutes

= \frac{30}{45}=\frac{6}{9}=\frac{2}{3}=2:3

Question: 7 Find the ratio of the following:

(a) 30 minutes to 1.5 hours

(b) 40\; cm to 1.5 \; cm

(c) 55 paise to Rs.1

(d) 500\; mL to 2\; litres

Answer: (a) 30 minutes to 1.5 hours

As we know,

1 \:hour = 60\:minutes

So,

1.5 \:hour =1.5\times 60=90\:minutes

Hence the ratio of 30 minutes to 1.5 hours:

\frac{30}{90}=\frac{3}{9}=\frac{1}{3}=1:3

(b) 40\; cm to 1.5m

As we know,

1 \:m= 100\:cm

So,

1.5 \:m=1.5\times 100=150\:cm

Hence the ratio of 40\; cm to 1.5 \:m=1.5\times 100=150\:m

\frac{40}{150}=\frac{4}{15}=4:15.

(c) 55 paise to Rs.1

As we know,

1 \:rupee= 100\:paise

Hence the ratio of 55 paise to Rs.1

\frac{55}{100}=\frac{11}{20}=11:20

(d) 500\; mL to 2\; litres

As we know,

1 \:litre= 1000\:mL

So

2 \:litre= 2\times1000=2000\:mL

Hence the ratio of 500\; mL to 2\; litres :

\frac{500}{2000}=\frac{1}{4}=1:4

Question: 8(a) In a year, Seema earns Rs.1,50,000 and saves Rs.50,000 . Find the ratio of Money that Seema earns to the money she saves.

Answer: Money that Seema earns = Rs.1,50,000

the money that Seema saves.= Rs.50,000

So, The ratio of Money that Seema earns to the money she saves:

=\frac{150000}{50000}=\frac{3}{1}=3:1 .

Hence the required ratio is 3:1.

Question: 8(b) In a year, Seema earns Rs. 1,50,000 and saves Rs. 50,000 . Find the ratio of Money that she saves to the money she spends.

Answer: Money that Seema earns = Rs.1,50,000

the money that Seema saves.= Rs.50,000

The amount of money Seema spends = Rs\: 150,000-Rs\:50,000=100,000.

So, The ratio of Money that she saves to the money she spends.

=\frac{50000}{100000}=\frac{1}{2}=1:2 .

Hence the required ratio is 1:2.

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

Answer: Given

Number of Teacher = 102

Number of students = 3300

So, the ratio of the number of teachers to the number of students:

\frac{102}{3300}=\frac{17}{550}=17:550.

Hence the required ratio is 17 : 550

Question: 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.

Answer: Given

Total number of students = 4320

Number of girls = 2300

The number of boys = 4320 - 2300

= 2020

So,

the ratio of

(a) Number of girls to the total number of students:

=\frac{2300}{4320}=\frac{230}{432}=\frac{115}{216}=115:216

(b) The number of boys to the number of girls:

=\frac{2020}{2300}=\frac{101}{115}=101:115

(c) The number of boys to the total number of students:

=\frac{2020}{4320}=\frac{101}{216}=101:216

Question: 11 Out of 1800 students in a school, 750 opted basketball, 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.

(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.

Answer: Total number of students = 1800

Number of students who opted for basketball = 750

Number of students who opted for cricket= 800

Number of students who opted for Table Tennis = 1800 - 750 - 800

= 250

Now,

The ratio of

(a) The number of students who opted basketball to the number of students who opted table tennis:

\frac{750}{250}=\frac{3}{1}=3:1

(b) The number of students who opted cricket to the number of students opting basketball:

\frac{800}{750}=\frac{16}{15}=16:15

(c) The number of students who opted basketball to the total number of students:

\frac{750}{1800}=\frac{5}{12}=5:12

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

Answer: Cost of 12 ( a dozen ) pens = Rs 180

Cost of 1 pen = 180 / 12 = Rs 15

Cost of 8 ball pens = Rs 56

Cost of 1 ball pen = 56 / 8 = Rs 7

So,

the ratio of the cost of a pen to the cost of a ball pen:

=\frac{15}{7}=15:7 .

Question: 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.

1643779858442

Answer: Given Breadth and Length is in proportion 2:5,

Maintaining that proportion, we get.

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


Question: 14 Divide 20 pens between Sheela and Sangeeta in the ratio of 3:2,

Answer: Given

Total number of pens = 20

The required ratio between Sheela and Sangeeta = 3 : 2

On adding the numbers in ratio we get 3 + 2 = 5.

So

Sheela will have 3/5 of the total pen :

=\frac{3}{5}\times20=3\times4=12

and Sangeeta will have 2/5 of the total pen:

=\frac{2}{5}\times20=2\times4=8

Hence Sheela will get 12 pens and Sangeeta will get 8 pens.

Question: 15 Mother wants to divide Rs.36 between her daughters Shreya and Bhoomika in the ratio of their ages. If age of Shreya is 15 years and age of Bhoomika is 12 years, find how much Shreya and Bhoomika will get.

Answer: Given,

Total money = Rs 36

Bhoomikas age = 12 years

Shreya's age = 15 years.

Now According to the question,

we are dividing 36 in ratio 15 : 12.

So, the sum of number in ratio = 15 + 12 = 27

Hence

amount of money Shreya gets:

=\frac{15}{27}\times36

=\frac{15}{3}\times4

=5\times4

=20 Rs .

Amount of money Sangeeta gets :

=\frac{12}{27}\times36=\frac{12}{3}\times4=4\times4=16.

Hence Shreya and Sangeeta get 20 Rs and 16 Rs respectively.

Question:1 6 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.

(b) Age of the father to the age of 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.

Answer: Given, Present age of father = 42 years and that of his son = 14 years.

The ratio of

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

=\frac{42}{14}=\frac{3}{1}=3:1

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

=\frac{42-2}{14-2}=\frac{40}{12}=\frac{10}{3}=10:3

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

=\frac{42+10}{14+10}=\frac{52}{24}=\frac{13}{6}=13:6.

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

=\frac{42-12}{14-12}=\frac{30}{2}=\frac{15}{1}=15:1

NCERT Class 6 Maths Chapter 12 Ratio and Proportion Topic 12.3 Proportion

Question: Check whether the given ratios are equal, i.e. they are in proportion. If yes, then write them in the proper form.
1. 1 : 5 and 3 : 15
2. 2 : 9 and 18 : 81
3. 15 : 45 and 5 : 25
4. 4 : 12 and 9 : 27
5. ` 10 to ` 15 and 4 to 6

Answer: 1. 1 : 5 and 3 : 15

\frac{3}{15}=\frac{3}{3\times5}=\frac{1}{5}

So the ratios are in proportion
2. 2 : 9 and 18 : 81

\frac{2}{9}=\frac{2\times9}{9\times9}=\frac{18}{81}

So So the ratios are in proportion
3. 15 : 45 and 5 : 25

\\\frac{15}{45}=\frac{1}{3}\\\frac{5}{25}=\frac{1}{5}

The given ratios are not equal, so they are not in proportion
4. 4 : 12 and 9 : 27

\\\frac{4}{12}=\frac{1}{3}\\\frac{9}{27}=\frac{1}{3}

The given ratios are equal, so they are in proportion
5. ` 10 to ` 15 and 4 to 6

\\\frac{10}{15}=\frac{2}{3}\\\frac{4}{6}=\frac{2}{3}

The given ratios are equal, so they are in proportion

NCERT Class 6 Maths Chapter 12 Ratio and Proportion Exercise 12.2

Question: 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

Answer: (a) 15,45,40,120

\frac{15}{45}=\frac{1}{3}........(1)

\frac{40}{120}=\frac{1}{3}........(2)

Since (1) and (2) are equal, Yes they are in proportion.

(b) 33,121,9,96

\frac{33}{121}=\frac{3}{11}........(1)

\frac{9}{96}=\frac{3}{32}........(2)

Since (1) and (2) are not equal, No they are not in proportion.

(c) 24,28,36,48

\frac{24}{28}=\frac{6}{7}........(1)

\frac{36}{48}=\frac{3}{4}........(2)

Since (1) and (2) are not equal, No they are not in proportion.

(d) 32,48,70,210

\frac{32}{48}=\frac{2}{3}........(1)

\frac{70}{210}=\frac{1}{3}........(2)

Since (1) and (2) are not equal, No they are not in proportion.

(e) 4,6,8,12

\frac{4}{6}=\frac{2}{3}........(1)

\frac{8}{12}=\frac{2}{3}........(2)

Since (1) and (2) are equal, Yes they are in proportion.

(f) 33,44,75,100

\frac{33}{44}=\frac{3}{4}........(1)

\frac{75}{100}=\frac{3}{4}........(2)

Since (1) and (2) are equal, Yes they are in proportion.

Question: 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

Answer: (a) 16:24::20:30

\frac{16}{24}=\frac{2}{3}.......(1)

\frac{20}{30}=\frac{2}{3}.......(2)

As we can see (1) and (2) are equal So, They are in proportion.

16:24::20:30

Hence the statement is True.

(b) 21:6::35:10

\frac{21}{6}=\frac{7}{2}.......(1)

\frac{35}{10}=\frac{7}{2}.......(2)

As we can see (1) and (2) are equal So, They are in proportion.i.e

. 21:6::35:10 .

Hence the statement is True.

(c) 12:18::28:12

\frac{12}{18}=\frac{2}{3}.......(1)

\frac{28}{12}=\frac{7}{3}.......(2)

As we can see (1) and (2) are not equal So, They are not in proportion.

Hence the statement is False.

(d) 8:9::24:27

\frac{8}{9}=\frac{8}{9}.......(1)

\frac{24}{27}=\frac{8}{9}.......(2)

As we can see (1) and (2) are equal So, They are in proportion.i.e.

8:9::24:27

Hence the statement is True.

(e) 5.2:3.9::3:4

\frac{5.2}{3.9}=\frac{4}{3}.......(1)

\frac{3}{4}=\frac{3}{4}.......(2)

As we can see (1) and (2) are not equal So, They are not in proportion.

Hence the statement is False.

(f) 0.9:0.36::10:4

\frac{0.9}{0.36}=\frac{10}{4}.......(1)

\frac{10}{4}=\frac{10}{4}.......(2)

As we can see (1) and (2) are equal So, They are in proportion.i.e.

0.9:0.36::10:4

Hence the statement is True.

Question: 3 Are the following statements true?

(a) 40 persons : 200 persons = Rs.15:Rs.75

(b) 7.5 litres : 15 litres = 5\; kg:10\; kg

(c) 99\; kg:45\; kg = Rs.44\; :Rs.\; 20

(d) 32\; m:64\; m=6\; sec:12\; sec

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

Answer: (a) 40 persons : 200 persons = Rs.15:Rs.75

\frac{40}{200}=\frac{1}{5}.........(1)

\frac{15}{75}=\frac{1}{5}.........(2)

As we can see (1) is equal to (2), They are in proportion.

Hence The statement is True.

(b) 7.5 litres : 15 litres = 5\; kg:10\; kg

\frac{7.5}{15}=\frac{1}{2}.........(1)

\frac{5}{10}=\frac{1}{2}.........(2)

As we can see (1) is equal to (2), They are in proportion.

Hence The statement is True.

(c) 99\; kg:45\; kg = Rs.44\; :Rs.\; 20

\frac{99}{45}=\frac{11}{5}.........(1)

\frac{44}{20}=\frac{11}{5}.........(2)

As we can see (1) is equal to (2), They are in proportion.

Hence The statement is True.

(d) 32\; m:64\; m=6\; sec:12\; sec

\frac{32}{64}=\frac{1}{2}.........(1)

\frac{6}{12}=\frac{1}{2}.........(2)

As we can see (1) is equal to (2), They are in proportion.

Hence The statement is True.

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

\frac{45}{60}=\frac{3}{4}.........(1)

\frac{12}{15}=\frac{4}{5}.........(2)

As we can see (1) is not equal to (2), They are not in proportion.

Hence the statement is False.

Question: 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\; Rs.40:Rs.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\; Rs.4:Rs.50

Answer: (a) 25\; cm:1\; m\; and\; Rs.40:Rs.160

\frac{25}{100}=\frac{1}{4}...........(1)

\frac{40}{160}=\frac{1}{4}...........(2)

As we can see (1) and (2) are equal, they are in proportion.

Middle Terms: 1 m and Rs 40

Extreme Terms: 25 cm and Rs 160.

(b) 39 litres: litres and 6 bottles : 10 bottles

\frac{39}{65}=\frac{3}{5}...........(1)

\frac{6}{10}=\frac{3}{5}...........(2)

As we can see (1) and (2) are equal, they are in proportion.

Middle Terms: 65 litres and 6 bottles

Extreme Terms: 39 litres and 10 bottles.

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

\frac{2}{80}=\frac{1}{40}...........(1)

\frac{25}{626}=\frac{1}{25}...........(2)

As we can see (1) and (2) are not equal, they are not in proportion.

(d) 200\; mL:2.5\; litre\; and\; Rs.4:Rs.50

\frac{200}{2500}=\frac{2}{25}...........(1)

\frac{4}{50}=\frac{2}{50}...........(2)

As we can see (1) and (2) are equal, they are in proportion.

Middle Terms: 2.5 litres and Rs 4

Extreme Terms:200 mL and Rs 50.

NCERT Class 6 Maths Chapter 12 Ratio and Proportion Topic 12.4 Unitary Method

Question:2 Read the table and fill in the boxes

25882

Answer:

Time Distance travelled by Karan Distance travelled by Kriti
2 hours 8 6
1 hour 4 3
4 hours 16 12

Distance travelled in 1 hour will be half of the distance travelled in 2 hours. Distance travelled in 4 hours will be double of the distance travelled in 2 hours

NCERT Class 6 Maths Chapter 12 Ratio and Proportion Exercise 12.3

Question: 1 If the cost of 7\; m of cloth is Rs.1470, find the cost of 5\; m of cloth.

Answer: Given,

Cost of 7 m cloth = Rs 1470

So

Cost of 1 m cloth :

=\frac{1470}{7}=Rs\:210

So,

Cost of 5 m cloth :

=Rs\:210\times5=Rs\:1050

Hence the cost of 5 m cloth is Rs 1050.

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

Answer: Given

Amount of money earned in 10 days:

=Rs \:3000

So,

Amount of money earned in 1 day :

=\frac{3000}{10}=Rs\:300

So,

Amount of Money earned in 30 days :

30\times\:300=Rs\:9000

Hence Ekta will earn 9000 Rs in 30 days.

Question: 3 If it has rained 276\; mm in the last 3\; days, 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.

Answer: Given

The measure of rain in 3 days :

=276\; mm

So,

The measure of rain in 1 day :

=\frac{276}{3}=92\; mm

And Hence,

The measure of rain in 7 days :

=7\times92=644\; mm

Therefore, 644 mm rain will fall in a week.

Question: 4(a) Cost of 5 kg of wheat is Rs. 91.50.

What will be the cost of 8\; kg of wheat?

Answer: Given,

The cost of 5 kg of wheat:

=Rs \:91.50

So,

The cost of 1 kg of wheat :

=\frac{91.50}{5}=Rs\:18.30

And Hence,

The cost of 8 kg of wheat :

=8\times\:18.30=Rs\:146.40

Therefore, the cost of 8 kg of wheat is Rs 146.40.

Question: 4(b) Cost of 5\; kg of wheat is Rs.91.50.

What quantity of wheat can be purchased in Rs.183?

Answer: Given,

The cost of 5 kg of wheat:

=Rs \:91.50

So,

The cost of 1 kg of wheat :

=\frac{91.50}{5}=Rs\:18.30

So, The amount of wheat which can be bought in Rs 183:

=\frac{183}{18.3}=10kg

Hence 10 kg of wheat can be bought in Rs 183.

Question: 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?

Answer: Temperature drop in 30 days :

=15^o

So, Temperature drop in 1 day :

=\frac{15}{30}=\frac{1}{2}=0.5^o

And Hence, The Temperature drop in 10 days :

=10\times0.5^o=5^o

Hence if the temperature rate remains the same, there will be a drop of 5 degrees in the next 10 days.

Question: 6 Shaina 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?

Answer: Given

Rent of 3 months :

=Rs \:15000

So,

Rent of 1 month :

=\frac{15000}{3}=Rs \:5000

And Hence Using unity principle

Rent of 1 year (12 months ) :

=12\times\:5000=Rs\:60000

Therefore, The total rent for one year is Rs 60000.

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

Answer: Given,

Number of bananas we can buy in Rs 180 = 4 dozen = 12 x 4 = 48

The number of bananas we can buy in Rs 1 :

=\frac{48}{180}=\frac{4}{15}

So, the number of bananas we can buy in Rs 90:

=90\times\frac{4}{15}=24

Hence we can buy 24 bananas in Rs 90.

Question: 8 The weight of 72 books is 9\; kg. What is the weight of 40 such books?

Answer: Given,

The weight of 72 books = 9 kg

So, The weight of 1 book :

=\frac{9}{72}=\frac{1}{8}kg

And hence,

The weight of 40 such books:

=40\times\frac{1}{8}=5kg

Hence, the weight of 40 books will be 5 kg.

Question: 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?

Answer: Given

Diesel requires for covering 594 km = 108 litres

So,

Diesel requires for covering 1 km :

=\frac{108}{594}=\frac{2}{11}L

And hence,

Diesel requires for covering 1650 km :

=1650\times\frac{2}{11}=300L

Hence The truck will require 300 litres of diesel to cover the distance of 1650 km.

Question: 10 Raju purchases 10 pens for Rs.150 and Manish buys 7 pens for Rs.84. Can you say who got the pens cheaper?

Answer: Cost of Raju's 10 pens = Rs 150

Cost of Raju's 1 pen :

=\frac{150}{10}=Rs \:15

And

Cost of Manish's 7 pens = Rs 84

Cost of Manish's 1 pen;

=\frac{84}{7}=Rs \:12

As we can see the cost of Raju's 1 pen is 15 and the cost of Manish's 1 pen is 12, Manish got the pen at a cheaper rate.

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

Answer: Anish's Case:

runs in 6 overs = 42

So, runs in 1 over:

=\frac{42}{6}=7

Anup's Case:

Run in 7 overs = 63

So, Runs in 1 over :

=\frac{63}{7}=9

As we can see Anup made more runs per over.

Ratio and Proportion Class 6 Maths Chapter 12-Topics

  • Ratio
  • Proportion
  • Unitary Method

Also Check

NCERT Books and NCERT Syllabus

NCERT Solutions for Class 6 Mathematics Chapter Wise

Chapters No. Chapters Name
Chapter - 1 Knowing Our Numbers
Chapter - 2 Whole Numbers
Chapter - 3 Playing with Numbers
Chapter - 4 Basic Geometrical Ideas
Chapter - 5 Understanding Elementary Shapes
Chapter - 6 Integers
Chapter - 7 Fractions
Chapter - 8 Decimals
Chapter - 9 Data Handling
Chapter -10 Mensuration
Chapter -11 Algebra
Chapter -12 Ratio and Proportion
Chapter -13 Symmetry
Chapter -14 Practical Geometry

Key Features of NCERT Solutions for Class 6 Maths Chapter 12:

Expertly Crafted Solutions: The NCERT Solutions for class 6 chapter 12 maths are skillfully developed to offer clear and accurate explanations for all the problems and concepts discussed in the chapter.

Conceptual Clarity: The NCERT Maths book Class 6 Chapter 12 solutions aim to enhance students' conceptual understanding by presenting complex topics in a simplified and easy-to-understand manner.

Comprehensive Coverage: The Maths Class 6 Chapter 12 solutions cover all the important topics and subtopics, ensuring that students grasp the complete essence of the chapter.

Exam-Focused Approach: The NCERT Class 6 Maths Chapter 12 solutions are designed with a focus on exam preparation, equipping students with essential tools and techniques to tackle questions effectively and score well in their exams.

NCERT Solutions for Class 6 - Subject Wise

Benefits of NCERT Solutions for Class 6 Maths Chapter 12 Ratio and Proportion-

  • You will get NCERT Solutions for this chapter very easily.
  • You will get some new way to solve some specific problems.
  • All the questions in CBSE NCERT solutions for Class 6 Maths chapter 12 ratio and proportion are explained in a step-by-step manner so it will be very easy for you to understand the concept. Students can practice ratio and proportion questions for class 6 from PDF provided at careers360 website.
  • It is going to help you with your homework as all the practice questions given below every topic are also covered in this article.
  • It will help you understand the concept as well as solving some real-life problems.

Happy learning!

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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