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

NCERT solutions for class 6 maths chapter 12 Ratio and Proportion: 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 CBSE NCERT solutions for class 6 maths chapter 12 ratio and proportion. The second important topic of this chapter is proportion. If the ratio of the first two quantities is equal to the next two quantities then the 4 quantities are said to be in proportion. There are 15 questions explained in exercise 12.2 and exercise 12.3 with solutions for NCERT for class 6 maths chapter 12 Ratio and Proportion which will give you more clarity of the concept. For the comparison using ratios, the comparing quantities must have the same unit otherwise it will not give the right ratio. If comparing quantities do not have the same unit, first convert them into the same unit and then find the ratio. For example, if you want to find a ratio of 500 grams and 1 kilogram. So 1kg=1000g and the now the ratio is 500: 10000 or 1:2. Here order is very important if you have asked about the ratio of 1Kg and 500g, it would be 2:1. Since 1:2 not equal to 2:1 be careful about the order of the ratio. You should try to solve all the NCERT questions by yourself so you won't these silly mistakes. You can take help from the CBSE NCERT solutions for class 6 maths chapter 12 ratio and proportion. You can get NCERT solutions from class 6 to 12 for science and maths by clicking on the above link.

The main topics of NCERT Solutions for Class 6 Maths Chapter 12 are given below

12.1 Introduction

12.2 Ratio

12.3 Proportion

12.4 Unitary Method

## Solutions of NCERT for class 6 maths chapter 12 ratio and proportion topic 12.2 ratio

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

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

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

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

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

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

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

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

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

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

NCERT solutions for class 6 maths chapter 12 ratio and proportion-Exercise: 12.1

Given,

Number of boys = 15

Number of girls = 20

So,

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

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

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:

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

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:

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

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:

(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.

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:

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

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:

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

Equating all the fraction, we get

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

Yes, They are equivalent ratios.

Question:

(a) $\inline 81 \; to \; 108$

(b) $\inline 98 \; to \; 63$

(c) $\inline 33\; km$ to $\inline 121\; km$

(d) $\inline 30$ minutes to $\inline 45$ minutes

(a)Ratio  of  $\inline 81 \; to \; 108$

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

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

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

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

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

(d) The ratio of  $\inline 30$ minutes to $\inline 45$ minutes

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

Question:7  Find the ratio of the following:

(a) $\inline 30$ minutes to$\inline 1.5$ hours

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

(c) $\inline 55$ paise to $\inline Rs.1$

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

(a) $\inline 30$ minutes to$\inline 1.5$ hours

As we know,

$1 \:hour = 60\:minutes$

So,

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

Hence the ratio of  $\inline 30$ minutes to$\inline 1.5$ hours:

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

(b) $\inline 40\; cm$ to $1.5m$

As we know,

$1 \:m= 100\:cm$

So,

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

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

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

(c) $\inline 55$ paise to $\inline Rs.1$

As we know,

$1 \:rupee= 100\:paise$

Hence the ratio of  $\inline 55$ paise to $\inline Rs.1$

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

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

As we know,

$1 \:litre= 1000\:mL$

So

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

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

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

Money that Seema earns = $\inline Rs.1,50,000$

the money that Seema saves.= $\inline Rs.50,000$

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.

Money that Seema earns = $\inline Rs.1,50,000$

the money that Seema saves.= $\inline Rs.50,000$

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.

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

(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.

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:

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

(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.

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:

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

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:

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

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

Given

Total number of pens = 20

The required ratio between Sheela and Sangeeta = 3 :

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.

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.

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

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

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

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

Given, Present age of father = $\inline 42$years and that of his son =  $\inline 14$ 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 solutions for class 6 maths chapter 12 ratio and proportion topic 12.3 proportion

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

Solutions of NCERT for class 6 maths chapter 12 ratio and proportion-Exercise: 12.2

(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)  $33,44,75,100$

(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.

(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)  $0.9:0.36::10:4$

(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:

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

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

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

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

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

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

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

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

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

Hence The statement is True.

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

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

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

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

Hence The statement is True.

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

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

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

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

Hence The statement is True.

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

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

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

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

Hence The statement is True.

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

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

(b) $\inline 39$ litres : $\inline 65$ litres and $\inline 6$ bottles :$\inline 10$ bottles

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

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

(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) $\inline 39$ litres: litres and $\inline 6$ bottles :$\inline 10$ 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) $\inline 2\; kg:80\; kg\; and\; 25\; g:625\; g$

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

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

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

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

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

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

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.

## Solutions for class 6 maths chapter 12 ratio and proportions topic 12.4 unitary method

Question:2 Read the table and fill in the boxes.

 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

## CBSE NCERT solutions for class 6 maths chapter 12 ratio and proportion-Exercise: 12.3

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.

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.

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.

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

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.

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

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.

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.

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.

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.

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.

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.

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.

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.

NCERT Solutions for Class 6 Mathematics - Chapter-wise

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

## 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.

• It will help you understand the concept as well as solving some real-life problems.

Happy learning!!!