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# NCERT Solutions for Class 6 Maths Chapter 12 - Ratio and Proportion Exercise 12.1

Last updated date: 12th Aug 2024
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## NCERT Solutions for Maths Class 6 Chapter 12 Exercise 12.1 - FREE PDF Download

NCERT Solutions for Class 6 Maths Ratio and Proportion, introduces students to the concepts of comparing quantities through ratios and understanding their relationships through proportions. Exercise 12.1 focuses on the basics of ratios, how to express them, and the importance of simplifying them. These foundational concepts are crucial for solving more complex mathematical problems in higher grades. Students should concentrate on practising the given problems to master the techniques of finding and simplifying ratios, and solving proportions. Vedantu’s NCERT Solutions of class 6 maths exercise 12.1 solution offer step-by-step explanations, making it easier for students to grasp these concepts thoroughly. This exercise lays the groundwork for understanding real-world applications of ratios and proportions, helping students build a strong mathematical foundation.

Table of Content
1. NCERT Solutions for Maths Class 6 Chapter 12 Exercise 12.1 - FREE PDF Download
2. Glance on NCERT Solutions Maths Chapter 12 Exercise 12.1 Class 6 | Vedantu
3. Access NCERT Solutions for Maths Class 6 Chapter 12 - Ratio and Proportion
4. Class 6 Maths Chapter 12: Exercises Breakdown
5. CBSE Class 6 Maths Chapter 12 Other Study Materials
6. Chapter-Specific NCERT Solutions for Class 6 Maths
FAQs

## Glance on NCERT Solutions Maths Chapter 12 Exercise 12.1 Class 6 | Vedantu

• Class 6 Maths Exercise 12.1 Solution explains the concept of ratios and how to express them in different forms.

• A ratio is a way to compare two quantities by using division, showing how many times one value is contained within another.

• Simplifying ratios by dividing both terms by their greatest common factor is covered in the exercise.

• Two quantities can be compared only if they are in the same unit.

• Proportion is discussed as an equation that states two ratios are equal, and understanding this is crucial.

• Problems include finding missing values in proportional relationships, reinforcing the concept of proportion.

• The exercise involves converting ratios into percentages to understand their practical applications.

• There are 16 fully solved questions in Ratio and Proportion of Class 6 Maths Exercise 12.1 Solution.

## Access NCERT Solutions for Maths Class 6 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: It is given that there are 20 girls and 15 boys in a class.

The ratio of the number of girls to the number of boys is calculated by dividing

the number of girls by the number of boys in the class, which can be represented

as $\dfrac{{20}}{{15}}$.

Simplifying the ratio,

$\Rightarrow \text{Ratio of girls to that of boys} = \dfrac{{20}}{{15}}$.

Dividing the numerator and the denominator of the right-hand side by 5,

$\Rightarrow \text{Ratio of girls to that of boys} = \dfrac{4}{3}$

Therefore, the ratio of girls to that of the boys in a class is found to be

$4:3$.

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

Ans: The total number of students in the class is the sum of the number of girl and the number of boys in the class

$\Rightarrow {\text{Total Students in the class}} = {\text{Total number of girls }} + {\text{Total number of boys}}$

$\Rightarrow {\text{Total Students in the class}} = 20 + 15$

$\Rightarrow {\text{Total Students in the class}} = 35$

The ratio of girls to the total students is determined by dividing the number of girls by the total number of students.

${\text{Ratio of girls to the total students}} = \dfrac{{20}}{{35}}$

Divide the numerator and denominator of the right-hand side of the above equation by 5.

$\Rightarrow {\text{Ratio of girls to the total students}} = \dfrac{4}{7}$

Therefore, the ratio of girls to the total students is found to be $4:7$.

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: It is given that the total number of students in the class is 30.

The total number of students who like football is 6.

Also,  the total number of students who like cricket is 12.

Hence, the total number of students who like tennis will be the difference between the total number of students and the total students who like cricket and football, that is,

$\Rightarrow {\text{The number of students who like tennis}} = 30 - 6 - 12$

$\Rightarrow {\text{The number of students who like tennis}} = 12$

Thus, the ratio of students liking football to the number of students who like tennis is the number of students liking football divided by the number of students who like tennis, that is,

$\Rightarrow {\text{Ratio of students who like football to that of tennis}} = \dfrac{6}{{12}}$

Now, divide the numerator and denominator of the right-hand side by 3.

$\Rightarrow {\text{Ratio of students who like football to that of tennis}} = \dfrac{2}{4}$

Therefore, the ratio of students who like football to that of tennis is $1:2$.

(b). The number of students liking cricket to a total number of students.

Ans: It is known that the total number of students in the class is 30.

Also, the number of students who like cricket is 12.

Thus, the ratio of the number of students liking cricket to the total number of students is obtained by dividing the number of students who like cricket by the total number of students.

$\Rightarrow {\text{Ratio of students who like cricket to the total number of students}} = \dfrac{{12}}{{30}}$

Now, divide the numerator and denominator of the right-hand side by 6.

$\Rightarrow {\text{Ratio of students who like cricket to the total number of students}} = \dfrac{2}{5}$

Therefore, the ratio of the number of students liking cricket to the total number of students is $2:5$.

3. See the figure and find the ratio of:

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

Ans: It is observed from the figure that, there are three triangles, two squares and two circles respectively.

Thus, the ratio of triangles to that of the number of circles inside the rectangle is determined by dividing the number of triangles by the number of circles found inside the rectangle.

The number of triangles is 3 and the number of circles inside the rectangle is 2.

$\Rightarrow {\text{Ratio of traingles to that of circles}} = \dfrac{3}{2}$

Therefore, the ratio of the triangles to the number of circles present inside the rectangle is found to be $3:2$.

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

Ans: It is observed from the figure that, there are three triangles, two squares and two circles respectively.

The total number of figures inside the rectangle is 7.

And, there are 2 squares inside the rectangle.

Thus, the ratio of the number of squares to that of the total number of figures inside the rectangle is determined by dividing the number of squares by the total number of figures found inside the rectangle.

$\Rightarrow {\text{Ratio of number of squares to all the figures inside the rectangle}} = \dfrac{2}{7}$

Therefore, the ratio of the number of squares to the total number of figures presents inside the rectangle is found to be $2:7$.

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

Ans: It is observed from the figure that, there are three triangles, two squares and two circles respectively.

The total number of figures inside the rectangle is 7.

And, there are 2 circles inside the rectangle.

Thus, the ratio of the number of circles to that of the total number of figures inside the rectangle is determined by dividing the number of circles by the total number of figures found inside the rectangle.

$\Rightarrow {\text{Ratio of number of circles to all the figures inside the rectangle}} = \dfrac{2}{7}$

Therefore, the ratio of the number of circles to the total number of figures presents inside the rectangle is found to be $2:7$.

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 Akhtar.

Ans: It is given that Hamid and Akhtar travel the distance of 9 km and 12 km in an hour.

It is known that the speed is calculated as the distance divided by time, that is,

${\text{Speed}} = \dfrac{{{\text{Distance}}}}{{{\text{Time}}}}$

First, calculate the respective speeds of Hamid and Akhtar with the given information.

The speed of Hamid is calculated as follows,

$\Rightarrow {\text{Speed of Hamid}} = \dfrac{{9{\text{ km}}}}{{1{\text{ h}}}}$

$\Rightarrow {\text{Speed of Hamid}} = 9{\text{ km/h}}$

And, the speed of Akhtar is calculated as follows,

$\Rightarrow {\text{Speed of Akhtar}} = \dfrac{{12{\text{ km}}}}{{1{\text{ h}}}}$

$\Rightarrow {\text{Speed of Akhtar}} = 12{\text{ km/h}}$

Now, the ratio of the speed of Hamid to that of the speed of Akhtar is calculated by dividing the speed of Hamid by the speed of Akhtar.

$\Rightarrow {\text{Ratio of the speed of Hamid to that of the speed of Akhtar}} = \dfrac{9}{{12}}$

On dividing the numerator and denominator of the right-hand side of the above equation by 3,

$\Rightarrow {\text{Ratio of the speed of Hamid to that of the speed of Akhtar}} = \dfrac{3}{4}$

Therefore, the ratio of the speed of Hamid to that of the speed of Akhtar is found to be $3:4$.

5. Fill in the following blanks:

$\dfrac{{15}}{{18}} = \dfrac{{\boxed{}}}{6} = \dfrac{{10}}{{\boxed{}}} = \dfrac{{\boxed{}}}{{30}}$ Are these equivalent ratios?

Ans: The ratio given is $\dfrac{{15}}{{18}}$.

Simplify the given ratio by dividing the numerator and denominator of the above expression by 3.

$\dfrac{{15}}{{18}} = \dfrac{5}{6}$

Now, multiply the simplified ratio by 2 in both numerator and denominator.

$\Rightarrow \dfrac{{5 \times 2}}{{6 \times 2}} = \dfrac{{10}}{{12}}$

Further, multiply the simplified equation $\dfrac{5}{6}$ by 5 on both the numerator and the denominator.

$\Rightarrow \dfrac{{5 \times 5}}{{6 \times 5}} = \dfrac{{25}}{{30}}$

Therefore, the blanks can be filled as follows,

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

Yes, the given ratios are found to be equivalent ratios.

6. Find the ratio of the following:

(a). The ratio of 81 to 108

Ans: The ratio of 81 to 108 is obtained by dividing 81 by 108 and is represented as $\dfrac{{81}}{{108}}$.

Both the numerator and denominator of the above expression are divisible by 27.

On dividing the obtained ratio is,

$\Rightarrow \dfrac{{81}}{{108}} = \dfrac{3}{4}$

The ratio of 81 to 108 is $3:4$.

(b). Ratio of 98 to 63

Ans: The ratio of 98 to 63 is obtained by dividing 98 by 63 and is represented as $\dfrac{{98}}{{63}}$.

Both the numerator and denominator of the above expression are divisible by 7.

On dividing by 7 the obtained ratio is,

$\Rightarrow \dfrac{{98}}{{63}} = \dfrac{{14}}{9}$

The ratio of 98 to 63 is $14:9$.

(c). Ratio of 33 km to 121 km

Ans: The ratio of 33 km to 121 km is obtained by dividing 33 by 121 and is represented as, $\dfrac{{33{\text{ km}}}}{{121{\text{ km}}}}$.

Both the numerator and denominator of the above expression are divisible by 11.

On dividing by 11 the obtained ratio is,

$\Rightarrow \dfrac{{33}}{{121}} = \dfrac{3}{{11}}$

The ratio of 33 km to 121 km is $3:11$.

(d). Ratio of 30 minutes to 45 minutes

Ans: The ratio of 30 minutes to 45 minutes is obtained by dividing 30 by 45 and is represented as $\dfrac{{30}}{{45}}$.

Both the numerator and denominator of the above expression are divisible by 15.

On dividing by 15the obtained ratio is,

$\Rightarrow \dfrac{{30}}{{45}} = \dfrac{2}{3}$

Therefore, the ratio of 30minutes to 45 minutes is found to be$2:3$.

7. Find the ratio of the following:

(a). Ratio of 30 minutes to 1.5 hour

Ans: It is known that 1.5 hour equals 90 minutes.

The ratio of 30 minutes to 1.5 hour is obtained by dividing 30 by 90 and is represented as, $\frac{{30{\text{minutes}}}}{{{\text{90 minutes}}}}$.

Both the numerator and denominator of the above expression are divisible by 30.

On dividing by 30 the obtained ratio is, $\Rightarrow \frac{{30}}{{90}} = \frac{1}{3}$

Therefore, the ratio of 30 minutes to 1 hour is formed to be 1:3.

(b). Ratio of 40 cm to $1.5$m.

Ans: It is known that 1 m is 100 cm.

Hence,

$1.5{\text{ m}} = 1.5 \times 100$

$= 150{\text{ cm}}$

The ratio of 40 cm to $1.5$ m is obtained by dividing 40 by 150 and is represented as $\dfrac{{40}}{{150}}$.

Both the numerator and denominator of the above expression are divisible by 10.

On dividing by 10the obtained ratio is,

$\Rightarrow \dfrac{{40}}{{150}} = \dfrac{4}{{15}}$

Therefore, the ratio of 40 cm to $1.5$ m is found to be $4:15$.

(c). Ratio of 55 paise to ₹.1.

Ans: It is known that Re.1  is 100 paise.

The ratio of 55 paise to Re.1 is obtained by dividing 55 by 100 and is represented as $\dfrac{{55}}{{100}}$.

Both the numerator and denominator of the above expression are divisible by 5.

On dividing by 5the obtained ratio is,

$\Rightarrow \dfrac{{55}}{{100}} = \dfrac{{11}}{{20}}$

Therefore, the ratio of 55 paise to Re.1is found to be$11:20$.

(d). Ratio of 500 ml to 2 litres.

Ans: It is known that 1 litre is equal to 1000 ml.

Thus, 2 litres will be equal to,

$2{\text{ liter}} = 2 \times 1000{\text{ ml}}$

$= 2000{\text{ ml}}$

The ratio of 500 ml to 2 litres is obtained by dividing 500 ml by 2000 ml and is represented as $\dfrac{{500}}{{2000}}$.

Both the numerator and denominator of the above expression are divisible by 500.

On dividing by 500 the obtained ratio is,

$\Rightarrow \dfrac{{500}}{{2000}} = \dfrac{1}{4}$

Therefore, the ratio of 500 ml to 2 litres is found to be $1:4$.

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: The total earning of Seema is 1,50,000 and her saving is 50,000.

The ratio of money that Seema earns to the money saved is determined by dividing the money earned by the money saved.

$\Rightarrow {\text{Ratio of money earned to the total money saved}} = \dfrac{{1,50,000}}{{50,000}}$

$\Rightarrow {\text{Ratio of money earned to the total money saved}} = \dfrac{3}{1}$

Therefore, the ratio of money that Seema earns to the money saved is found to be $3:1$.

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

Ans: The total earning of Seema is 1,50,000 and her saving is 50,000.

Thus, the total money spent will be the difference between her earnings and her savings.

${\text{Total money spent}} = 1,50,000 - 50,000$

${\text{Total money spent}} = 1,00,000$

The ratio of money that Seema saves to the money she spends is determined by dividing the money saved by the money spent by her.

$\Rightarrow {\text{Ratio of money saved to the total money spent}} = \dfrac{{50,000}}{{1,00,000}}$

$\Rightarrow {\text{Ratio of money saved to the total money spent}} = \dfrac{1}{2}$

Therefore, the ratio of money that Seema saves to the money spends is found to be $1:2$.

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: The total number of teachers in a school is 102.

The total number of students that are there in the school is 3300.

The ratio of the numbers of teachers to the numbers of the students can be found out by dividing the number of teachers by the total number of students.

Now,

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

$= \dfrac{{17}}{{550}}$

$= 17:550$

Therefore, the ratio of the number of teachers to the number of students is found out to be $17:550$.

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: The total numbers of students that are there in the college $= 4320$

The total numbers of girls among all the students $= 2300$

The ratio of the number of girls to the total number of students in the college $= \dfrac{{2300}}{{4320}}$

Dividing the numerator and denominator of the fraction by 20, we get

$\Rightarrow$ The ratio of number of girls to the total number of students in the college $= \dfrac{{115}}{{216}} = 115:216$

Therefore, the ratio of the number of girl students to the total number of students in the college is $115:216$.

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

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

$\Rightarrow$ Total number of boys $= 4320 - 2300 = 2020$

The ratio of number of boy students to the number of girl students $= \dfrac{{2020}}{{2300}}$

Dividing the numerator and denominator of the fraction by 20, we get

$\Rightarrow$ The ratio of number of boy students to the number of students $= \dfrac{{101}}{{115}} = 101:115$

Therefore, the ratio of number of boy students to the number of girl students in the college is $101:115$.

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

Ans: The total number of boys among all students in the college $= 2020$

The total number of students that are there in the college $= 4320$

The ratio of the number of boy students to the number of students $= \dfrac{{2020}}{{4320}}$

Dividing the numerator and denominator of the fraction by 20, we get

$\Rightarrow$ The ratio of number of boy student to the number of students $= \dfrac{{101}}{{216}} = 101:216$

Therefore, the ratio of the number of boys to the number of students in the college is $101:216$.

11. Out of 1800 students in a school, 750 opted for basketball, 800 opted cricket and remaining opted table tennis. If a student can opt only one game, find the ratio of:

(a). A number of students who opted basketball to the number of students who opted table tennis.

Ans: It is given that the total number of students in the school is 1800, of which 750 opted for basketball, 800 opted for cricket and the remaining opted for table tennis as their game.

The number of students who opted for basketball is 750 and the number of students opting for cricket is 800.

Thus, the number of students who opted for table tennis as their game is the difference between the total number of students in the school and the total number of students opting for basketball and cricket.

$\Rightarrow {\text{The number of students who opted table tennis}} = 1800 - \left( {750 + 800} \right)$

$\Rightarrow {\text{The number of students who opted table tennis}} = 250$

Hence, the ratio of the number of students who opted for basketball to the number of students who opted for table tennis is the number of students who opted for basketball divided by the number of students who opted for table tennis, that is, $\dfrac{{750}}{{250}}$.

$\Rightarrow {\text{Ratio of the students who opted for basketball to that of table tennis}} = \dfrac{{750}}{{250}}$ Now, divide the numerator and the denominator of the right-hand side of the above equation by 250.

$\Rightarrow {\text{Ratio of the students who opted basketball to that of table tennis}} = \dfrac{3}{1}$ Therefore, the ratio of a number of students who opted for basketball to the number of students who opted for table tennis is found to be $3:1$.

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

Ans: It is given that the total number of students in the school is 1800, of which 750 opted for basketball, 800 opted for cricket and the remaining opted for table tennis as their game.

The number of students who opted for basketball is 750 and the number of students opting for cricket is 800.

So, the ratio of the number of students who opted for cricket to the number of students who opted for basketball is the number of students who opted for cricket divided by the number of students who opted for basketball, that is represented as $\dfrac{{800}}{{750}}$.

$\Rightarrow {\text{Ratio of the number of students who opted cricket to those who opted basketball}} = \dfrac{{800}}{{750}}$

Now, divide the numerator and the denominator of the right-hand side of the above equation by 50.

$\Rightarrow {\text{Ratio of the number of students who opted cricket to those who opted basketball}} = \dfrac{{16}}{{15}}$ Therefore, the ratio of the number of students who opted cricket to the number of students opting basketball as their game is obtained as $16:15$.

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

Ans: It is given that the total number of students in the school is 1800, of which 750 opted for basketball, 800 opted for cricket and the remaining opted for table tennis as their game.

The number of students who opted for basketball is 750 and the total number of students is 1800.

So, the ratio of the number of students who opted for basketball to the total number of students is the number of students who opted for basketball divided by the total number of students, that is $\dfrac{{750}}{{1800}}$.

$\Rightarrow {\text{Ratio of the students who opted basketball to the total students}} = \dfrac{{750}}{{1800}}$

Now, divide the numerator and the denominator of the right-hand side of the above equation by 150.

$\Rightarrow {\text{Ratio of the students who opted basketball to the total students}} = \dfrac{5}{{12}}$

Therefore, the ratio of a number of students who opted basketball to the total number of students is found to be $5:12$.

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 ballpen.

Ans: In a dozen pens there are 12 pens.

The cost of dozen pens $= ₹180$

$\therefore$ The cost of 1 pen $= \dfrac{{₹180}}{{12}} = ₹15$

The cost of 8 ball pens $= ₹56$

Now, the cost of 1 ball pen $= \dfrac{{₹56}}{8}$

Dividing the terms, we get

The cost of 1 ball pen $= ₹7$

Now, in order to find the ratio of the cost of 1 pen to the cost of 1 ballpen we have to divide the cost of one pen by the cost of one ball pen.

So, the ratio of the cost of one pen to the cost of a ballpen $= \dfrac{{₹15}}{{₹7}} = 15:7$

Therefore, the ratio of the cost of a pen to the cost of a ballpen is found out to be $15:7$.

13. Consider the statement: Ratio of breadth and length of a ball 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: The ratio of breadth to length is given as $2:5$.

So, in the fraction form, the ratio of breadth to length is given as $\dfrac{2}{5}$.

To find the possible breadths and lengths of the hall, we have to find the equivalent ratios by multiplying $\dfrac{2}{5}$ to the given length and breadth.

So, the equivalent ratio of breadth to the length of the hall $= \dfrac{{10}}{{25}} = \dfrac{2}{5}$

Now, let the breadth in the first blank be $x$

So, the equivalent ratios of breadth to the length of the hall will be,

$\dfrac{x}{{50}} = \dfrac{2}{5}$

Now, cross multiplying the terms, we get

$\Rightarrow 5x = 2\left( {50} \right)$

$\Rightarrow 5x = 100$

Dividing both sides by 5, we get

$\Rightarrow x = \dfrac{{100}}{5}$

$\Rightarrow x = 20$

So, if the length of the hall is 50 m, then the breadth of the hall is found out to be 20 m.

Let the length in the second blank be $y$.

So, the equivalent ratios of breadth to the length of the hall will be,

$\dfrac{{40}}{y} = \dfrac{2}{5}$

Now, cross multiplying the terms, we get

$\Rightarrow 40\left( 5 \right) = 2y$

$\Rightarrow 200 = 2y$

Dividing both sides by 2, we get

$\Rightarrow y = \dfrac{{200}}{2}$

$\Rightarrow y = 100$

So, if the breadth of the hall is 40 m, then the breadth of the hall is found out to be 100 m.

Therefore, the table can be completed as follows,

 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 ${\text{3}}:{\text{2}}$.

Ans: It is given that the ratio between Sheela and Sangeeta is $3:2$.

So, the total parts in which the pens are divided $= 3 + 2 = 5$

Now, the total parts of pens with Sheela $= \dfrac{3}{5}$

Also, the total parts of pens with Sangeeta $= \dfrac{2}{5}$

So, the total number of pens that Sheela has $= \dfrac{3}{5} \times 20$

Multiplying the terms, we get

$\Rightarrow$ The total number of pens that Sheela has $= \dfrac{{60}}{5} = 12$

Now, the total number of pens that Sangeeta has $= \dfrac{2}{5} \times 20$

Multiplying the terms, we get

$\Rightarrow$ The total number of pens that Sangeeta has $= \dfrac{{40}}{5} = 8$

Therefore, among 20 pens, Sheela will have 12 pens and Sangeeta will have 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: It is given that Shreya’s age is 15 years and Bhoomika’s age is 12 years.

So, the ratio of Shreya’s age to Bhoomika’s age $= \dfrac{{15}}{{12}}$

Simplifying the ratio, we get

$\Rightarrow$ The ratio of Shreya’s age to Bhoomika’s age $= \dfrac{5}{4} = 5:4$

Now, the $36$ is divided between Shreya and Bhoomika in the ratio $5:4$.

So, the total parts in which $36$ can be divided $= 5 + 4 = 9$

Now, the total share of Shreya out of $36$$= \dfrac{5}{9} \times 36 Multiplying the terms, we get \Rightarrow The total share of Shreya out of 36$$ = \dfrac{{180}}{9} = 20$

Also, the total share of Bhoomika out of $36$$= \dfrac{4}{9} \times 36 Multiplying the terms, we get \Rightarrow The total share of Bhoomika out of 36$$ = \dfrac{{144}}{9} = 16$

Therefore, the total share of Shreya is $20$ and the share of Bhoomika is $16$.

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: It is given that the present age of the father is 42 years and his son’s age is 14 years.

The ratio of a father’s present age to that of a son is found by dividing the present age of the father by the son’s age, and is given by, $\dfrac{{42}}{{14}}$.

Thus,

$\Rightarrow {\text{Ratio of father's present age to that of the son}} = \dfrac{{42}}{{14}}$

Now, divide the right-hand side of the above equation by 14 on both the numerator and the denominator.

$\Rightarrow {\text{Ratio of father's present age to that of the son}} = \dfrac{3}{1}$

Therefore, the ratio of the father’s present age to that of the son is found to be $3:1$.

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

Ans: It is given that the present age of the father is 42 years and his son’s age is 14 years. Thus, when the son was 12 years, that is, 2 years ago, the father’s age would be,

$\left( {42 - 2} \right) = 40$ years.

The ratio of the ages of father to that of a son, in this case, would be the ratio of father’s age two years ago divided by the age of the son two years ago, that is, $\dfrac{{40}}{{12}}$

${\text{Ratio of the age of father to that of son two years ago}} = \dfrac{{40}}{{12}}$

Now, divide the numerator and the denominator of the right-hand side of the above equation by 4.

$\Rightarrow \dfrac{{40}}{{12}} = \dfrac{{10}}{3}$

Therefore, the ratio of age of the father to the age of the son, when the son was 12 years old is found to be $10:3$.

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

Ans: It is given that the present age of the father is 42 years and his son’s age is 14 years. After 10 years, the age of father and son will be 10 more than their present ages.

$\Rightarrow {\text{Father's Age after 10 years}} = 42 + 10$

$\Rightarrow {\text{Father's Age after 10 years}} = 52$

Also,

$\Rightarrow {\text{Son's Age after 10 years}} = 14 + 10$

$\Rightarrow {\text{Son's Age after 10 years}} = 24$

Now, the ratio of their ages after 10 years will be the father’s age after 10 years divided by the son’s age after 10 years, that is, $\dfrac{{52}}{{24}}$.

$\Rightarrow {\text{Ratio of their ages after 10 years}} = \dfrac{{52}}{{24}}$

Divide the numerator and denominator of the right-hand side of the above equation by 4.

$\Rightarrow {\text{Ratio of their ages after 10 years}} = \dfrac{{13}}{6}$

Therefore, the ratio of the age of the father after 10 years to the age of the son after 10years is found to be $13:6$.

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

Ans: It is given that the present age of the father is 42 years and his son’s age is 14 years. When the father was 30 years old, that is, 12 years ago, his son’s age will be $\left( {14 - 12} \right)$ years.

$\Rightarrow {\text{Son's age when his father was 30 years old}} = 14 - 12$

$\Rightarrow {\text{Son's age when his father was 30 years old}} = 2$

Thus, the son’s age was 2 years when his father was 30 years old.

The ratio of their ages is the age of father 12 years ago divided by the age of son 12 years ago, that is, $\dfrac{{30}}{2}$.

$\Rightarrow {\text{Ratio of their ages 12 years ago}} = \dfrac{{30}}{2}$

Now, divide the numerator and the denominator of the right-hand side of the above equation by 15.

$\Rightarrow {\text{Ratio of their ages 12 years ago}} = \dfrac{{15}}{1}$

Therefore, the ratio of the age of the father to the age of the son when the father was 30 years old is found to be $15:1$.

## Conclusion

NCERT Solutions for Maths Exercise 12.1 Class 6 Chapter 12 - Ratio and Proportion provide a solid foundation for understanding the basics of ratios and proportions. This exercise focuses on comparing quantities, simplifying ratios, and solving proportion problems. By practising class 6 maths ch 12 ex 12.1 solutions, students can build their problem-solving skills and prepare effectively for exams. Vedantu’s expertly crafted solutions ensure that students grasp these fundamental concepts with ease, setting a strong base for more advanced topics in mathematics. Consistent practice and thorough understanding are key to mastering this chapter and excelling in future mathematical challenges.

## Class 6 Maths Chapter 12: Exercises Breakdown

 Exercise Number of Questions Exercise 12.2 4 Questions and Solutions Exercise 12.3 11 Questions and Solutions

## CBSE Class 6 Maths Chapter 12 Other Study Materials

 S.No. Important Links for Chapter 12 Ratio and Proportion 1 Class 6 Ratio and Proportion Important Questions 2 Class 6 Ratio and Proportion Revision Notes 3 Class 6 Ratio and Proportion Important Formulas 4 Class 6 Ratio and Proportion NCERT Exemplar Solution

## 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 Exercise 12.1

1. How many questions are present in NCERT Solutions for Class 6 Maths Ch 12 Ex 12.1?

Here is a break-up of the questions included in the NCERT solutions for class 6 maths ch 12 ex 12.1, Ratio and Proportion.

• Exercise 12.1 has 16 questions

• Exercise 12.2 has 11 questions

• Exercise 12.3 has 4 questions

2. How many questions have long solutions (more than 180 words) in NCERT Solutions for Class 6 Maths Chapter 12 ratio and proportion?

There are about six questions in NCERT Solutions for Class 6 Maths Chapter 12 Ratio and Proportion whose solutions exceed 180 words. These long answer questions are very important for the exams and therefore need to be clearly understood. Vedantu's NCERT solutions PDF provides you with the perfect guidance and support to get a clear idea of all the answers and the correct method of solving the problems.

3. What are the topics covered in NCERT Solutions for Class 6 Maths Chapter 12?

Class 6 Maths Chapter 12 talks in-depth about ratios, proportions, as well as the unitary method. We get equivalent ratios through multiplication or division of the denominator and numerator by a constant value. When the value of two ratios is equal, they are called proportionate. Here, the symbol ‘::’ is used to denote the same. The unitary method refers to the process in which the value of a single unit is calculated first to find out the value of the required number of units.

4. What are the real-time applications of the unitary method?

The following are a few real-time applications of the unitary method:

• To calculate the speed for a particular distance, when the value of speed and distance are in different units.

• To calculate the number of persons required for the completion of work.

• To calculate the area of a square of any given length when the ratio of the area with its side is known.

• To calculate the price of a given quantity of objects, when the price and quantity of objects are known by different units.

5. What are the most important definitions that come in Class 6 Maths Chapter 12?

The most important definitions in Class 6 Maths Chapter 12 are as follows:

• Ratio- The comparison of two values with the help of division is called a ratio.

• Proportion- when two ratios have equal values, they are known to be Proportional.

• Unitary Method- the calculation of the value of one unit to find out the total number of units is known as the Unitary method.

6. What is a ratio in class 6 math exercise 12.1?

A ratio is a way to compare two quantities by dividing one by the other. It shows how many times one value contains another. For example, the ratio of 2 to 3 is written as 2:3 or 2/3​. Ratios are fundamental in understanding relationships between numbers.In maths class 6 chapter 12 exercise 12.1 are used in various mathematical calculations and real-life scenarios.

7. How do you express a Ratio in Class 6th Maths Chapter 12 Exercise 12.1?

A ratio can be expressed in the form a​/b or as a:b. Both notations indicate the same relationship between two quantities. For example, a ratio of 4 to 5 can be written as 4:5 or 4/5​. Class 6th Maths Chapter 12 Exercise 12.1 helps in comparing quantities directly and understanding their proportionate relationships. Expressing ratios correctly is crucial for solving ratio problems accurately.

8. What is a proportion in 12.1 class 6 maths?

In 12.1 class 6 maths proportion is an equation that states two ratios are equal. It can be written as $\frac{a}{b}= \frac{c}{d}$​ or a:b = c:d. Proportions are used to find unknown values when the relationship between quantities is consistent. They are essential in solving problems involving scales and maps. Understanding proportions helps in maintaining consistent relationships between different sets of quantities.

9. How do you simplify a ratio in maths class 6 chapter 12 exercise 12.1?

To simplify a ratio, divide both terms by their greatest common divisor (GCD). For example, to simplify the ratio 8:12, find the GCD of 8 and 12, which is 4, and then divide both terms by 4, resulting in the simplified ratio 2:3. Simplifying ratios makes them easier to work with and compare. This process helps in reducing ratios to their simplest form for clarity and ease of use.

10. What is the importance of ratios in daily life in class 6 ex 12.1?

In class 6 ex 12.1 Ratios are used in various real-life situations, such as cooking recipes, where ingredients must be mixed in specific proportions. They are also important in comparing prices, understanding map scales, and creating models. Ratios help in making informed decisions by comparing different quantities effectively. They are fundamental in fields like finance, engineering, and science, making them essential for everyday calculations.

11. What should students focus on class 6 maths ex 12.1?

Students should focus on practicing problems, understanding the concepts of ratios and proportions, and applying them to real-life situations.