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Important Questions for CBSE Class 6 Maths Chapter 12 - Ratio and Proportion

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CBSE Class 6 Maths Important Questions Chapter 12 - Ratio and Proportion - Free PDF Download

Embark on a mathematical journey with Class 6 Maths Chapter 12 - Ratio and Proportion. Discover the fascinating world of mathematical relationships and proportions that govern various aspects of daily life. This chapter lays the foundation for understanding ratios, offering practical insights into real-world applications. Free PDF download of Important Questions with Solutions for CBSE Class 6 Maths Chapter 12 - Ratio and Proportion prepared by expert Mathematics teachers from the latest edition of CBSE(NCERT) books

Register online for Maths tuition on Vedantu.com to score more marks in your examination. You can also register Online for NCERT Class 6 Science tuition on Vedantu.com to score more marks in CBSE board examination. 


Vedantu is a platform that provides free CBSE Solutions (NCERT) and other study materials for students. Maths Students who are looking for the better solutions ,they can download Class 6 Maths NCERT Solutions to help you to revise complete syllabus and score more marks in your examinations. 


Download CBSE Class 6 Maths Important Questions 2024-25 PDF

Also, check CBSE Class 6 Maths Important Questions for other chapters:

CBSE Class 6 Maths Important Questions

Sl.No

Chapter No

Chapter Name

1

Chapter 1

Knowing Our Numbers

2

Chapter 2

Whole Numbers

3

Chapter 3

Playing with Numbers

4

Chapter 4

Basic Geometrical Ideas

5

Chapter 5

Understanding Elementary Shapes

6

Chapter 6

Integers

7

Chapter 7

Fractions

8

Chapter 8

Decimals

9

Chapter 9

Data Handling

10

Chapter 10

Mensuration

11

Chapter 11

Algebra

12

Chapter 12

Ratio and Proportion

13

Chapter 13

Symmetry

14

Chapter 14

Practical Geometry

Study Important Questions for Class 6 Mathematics Chapter 12 - Ratio and Proportion

1 Mark

1. In a ratio, the first term is called ____ and the second term is called ______.

Ans: Antecedent, Consequent

For example, in ratio $13:15$, 13 is Antecedent and 15 is Consequent.


2. Product of ______ = Product of extremes.
Ans: Means

In a proportion, the first and last terms are known as the extremes, while the second and third terms are known as the meAns:


3. If a, b, c, d are in proportion, then

  1. ac = bd

  2. ad = bc

  3. ab = cd

  4. None of these.

Ans: (b) ad = bc

If the ratio of the first two quantities equals the ratio of the last two quantities, the numbers a, b, c, and d are proportional.


4. A ratio has _____ units

Ans: No

Because a ratio is made up of similar quantities, the units cancel each other out, and thus there is no unit for a ratio.


  2 Mark

1. Convert 80:50 into simplification.                                             

Ans: Given ratio of 80:50

Expressing as fractions 

$\dfrac{80}{50}\ = \dfrac{8 \times 10}{5 \times 10} = \dfrac{8}{5}$


2. Find the ratio of 40 cm to 2.5 m.

Ans: let’s first convert 2.5 m into cm

2.5 m = $2.5 \times 100$ = 250 cm

Now, 

$40 cm:2.5 = 40 cm:250 cm $

$= 40:250$

$= \dfrac{40}{250}$

$= \dfrac{4}{25}$

The required ratio is $4:25$


3. The length and breadth of a field are $\mathbf{80\ \text{m}}$ and $\mathbf{30\ \text{m}}$. what is the ratio of the breadth and length of the park?       

Ans:  Given, 

Length of park = 80 m

Breadth of park = 30 m

Ratio of the breadth and length = $ \dfrac{Breadth}{Length} = \dfrac{30 m}{80 m}$

$=\dfrac{3}{8}$

Required ratio is $3:8$


4. If 30 oranges cost Rs. 120. What is the cost of 50 oranges? 

Ans: Here we will use the unitary method.

Cost of 30 oranges =  Rs.120

Now, cost of one orange = $\dfrac{120}{30} = Rs. 4 $    

Cost of 50 oranges = cost of one orange $\times $ cost of fifty oranges 

$= 4 \times  50$

$= Rs. 200$


3 Mark

1. Find the ratio of 45 minutes to an hour?

Ans: To find 45 min:1 hour

We know that one hour = 60 min

Therefore, 45 min:1 hour = 45 min:60 mnt

$= \dfrac{45}{60} $

Dividing numerator and denominator by 3

$\dfrac{45}{60} \div  \dfrac{3}{3} = \dfrac{15}{20}$

Dividing numerator and denominator by 5

$\dfrac{15}{20} \div  \dfrac{5}{5} = \dfrac{3}{4}$


2. Find the ratio 250 ml to 4 l.

Ans: To find 250 ml:4 l

First, let’s convert 4 l into ml

We know 

$1 l = 1000 ml$

$ \therefore 4 l = 1000 \times 4 ml$

$\Rightarrow 4 l = 4000 ml$

Now, 

$250 ml:4 l = 250 ml:4000 ml$

$= 250:4000$

$\Rightarrow \dfrac{250}{4000} \div \dfrac{10}{10} = \dfrac{25}{400} \div \dfrac{5}{5}$

$= \dfrac{5}{80} \div \dfrac{5}{5}$

$= \dfrac{1}{16}$

$= 1:16$

The required Ratio is 1:16.


3. Find the equivalent ratio of 75:100 

Ans:  

$=75:100$

$=\dfrac{75}{100} \div \dfrac{5}{5} = \dfrac{15}{20}$ 

$=\dfrac{15}{20} \div \dfrac{5}{5} $ 

$=\dfrac{3}{4}$

The equivalent ratios for 75:100 are 15:20 and 3:4


4. Are 20, 40, 60, 120 in proportion? 

Ans: First we will find 20:40 and $60:120$

$20:40=\dfrac{20}{40}=\dfrac{2}{4}=\dfrac{1}{2} $ 

$60:120=\dfrac{60}{120}=\dfrac{6}{12}=\dfrac{1}{2} $

The numbers in the simplest form are equal. 

Yes, 20, 40, 60, 120 are in proportion. 


5. A 15 men can reap a field in 25 days. In how many days can 20 men reap the same field?

Ans: $15$ men can reap fields in =  $25$days.. 

$1$ men can reap field in  = $(25 \times 15$ ) days

$20$ men can reap field in = $\dfrac{25\times 15}{20}$ days 

$=\dfrac{25\times 15}{20} $

$=\dfrac{5\times 3}{4} $

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

$=3\dfrac{3}{4} days $

Therefore, $20$ men can reap the same field in $=3\dfrac{3}{4}\text{days}$.


  4 Mark

1. Fill up the blanks $\mathbf{\dfrac{2}{3}=\dfrac{14}{}=\dfrac{6}{}}$ 

Ans: Let ,                                     

$\dfrac{2}{3}=\dfrac{14}{x} $

$2x=14\times 3 $

$x=\dfrac{14\times 3}{2}=7\times 3=21$

Let’s substitute 

$\dfrac{2}{3}=\dfrac{14}{}$

Similarly, 

$\dfrac{2}{3}=\dfrac{6}{} $

$2x=3\times 6 $

$x=\dfrac{3\times 6}{2}=3\times 3=9 $ 

$\dfrac{2}{3}=\dfrac{6}{} $

Hence, $\dfrac{2}{3}=\dfrac{14}{21}=\dfrac{6}{9}$


2. Two numbers are in the ratio 3:5 and their sum is 192. Find the numbers.

Ans: Let common ratio is $x$

Therefore, numbers are 3x and 5x

According to question, 

3x + 5x = 192

$\therefore 8x=192$

$x=\dfrac{192}{8}=24$

Required numbers are, 

$3x=3\times 24=72$ 

$5x=5\times 24=120 $


3. Compare the ratios \[\mathbf{\left( 1\text{ }:\text{ }2 \right)}\] and \[\mathbf{\left( 4\text{ }:\text{ }5 \right)}\]

Ans:  Given, $1:2= \dfrac{1}{2}$

$4:5= \dfrac{4}{5}$

LCM of $5$ and $2$ is $10$ 

Therefore, 

$\dfrac{1}{2} \times \dfrac{5}{5} = \dfrac{5}{10}$ 

$\dfrac{4}{5} \times \dfrac{2}{2} = \dfrac{8}{10}$

Here, $\dfrac{5}{10} < \dfrac{8}{10}$

$\dfrac{5}{10} = \dfrac{1}{2} \dfrac{8}{10} = \dfrac{4}{5}$

Therefore, $\dfrac{1}{2}<\dfrac{4}{5} $

$1:2<4:5$


4. If $\mathbf{x:63::36:81}$

Ans:  We can use formula 

Product of means = Product of extremes

$\Rightarrow 63 \times 36 = 81 \times x $

$\Rightarrow 81 \times x = 63 \times 36 $

$\Rightarrow x = \dfrac{63 \times 36}{81} $

$\Rightarrow \dfrac{7 \times 36}{9} $

$\Rightarrow 7 \times 4 $

$ \Rightarrow x=28 $


5 Mark

1. Divide \[\mathbf{\text{Rs}\text{. 2000/-}}\] between Asha and Kiran in the ratio 4:6.

Ans: Given amount = \[\text{Rs}\text{. 2000/-}\]

Given ratio = $4:6$

Let common ratio is $x$

Then, Asha’s share = $4x$

Similarly, Kiran’s share = $6x$

According to Question, 

$4x+6x=2000 $

$\Rightarrow 10x=2000 $

$\Rightarrow x = \dfrac{2000}{10} = 200$

Now, 

Asha’s share = $4\times 200 = \text{Rs}\text{.} 800$

Kiran’s share = $6\times 200 = \text{Rs}\text{.} 1200$


2. Divide $\mathbf{\text{Rs}\text{.} \text{500}}$ among $\mathbf{\text{A,B,C}}$ in the ratio \[\mathbf{1\text{ }:\text{ }2\text{ }:\text{ }3}\]

Ans:  Given amount = $\text{Rs}\text{.} \text{500}$

Given ratio = \[1\text{ }:\text{ }2\text{ }:\text{ }3\]

Let common ratio = $x$

A’s Share = $x$

B’s Share = $2x$

C’s Share = $3x$

According to question 

$x+2x+3x=500 $

$ \Rightarrow  6x = 500 $

$ \Rightarrow  x = \dfrac{500}{6} $

$ \Rightarrow  x = 83.33 $

Now, 

A’s Share = $\text{Rs}\text{.} 83.33$

B’s  share = $\text{Rs}\text{.} (2 \times  83.33) = \text{Rs}. 166.66$ 

C’s Share = $\text{Rs}\text{.} 3 \times  83.33 =\text{Rs}\text{.} 250 $


3. If $\mathbf{4:x::x:36}$. Find the value of $\mathbf{x}$.

Ans:  We will use formula 

Product of means = Product of extremes

$x \times x=4\times 36 $

${{x}^{2}}=144 $

$ x=\sqrt{144} $

$ x=12$ 


4. If 10, 15, x are in proportion. Find the value of x.

Ans: Given, 10, 15, x are in proportion.

\[\Rightarrow \text{10:15::15:x}\]

We will use formula 

Product of means = Product of extremes

$ 15\times 15=10 \times x $

$10x=15\times  15 $

$ \Rightarrow x=\dfrac{15 \times 15}{10} $

$  =\dfrac{15\times 3}{2} $

$ =\dfrac{45}{2} $

$ \Rightarrow x=22.5$ 


5. Find the ratio of the price of coffee powder to that of milk powder when coffee powder cost  24/- per 100\,g and milk powder cost 180/- per kg.

Ans:  Given, cost of $100 g$ coffee powder =$\text{Rs}\text{.} 24 $

Cost of $1\,g$ coffee powder =$\text{Rs}\text{.} \dfrac{24}{100} $

Cost of $1000 g$ coffee powder =$ \text{Rs}\text{.} \left( \dfrac{24}{100} \times 1000 \right)= \text{Rs}\text{.} 240 $

Cost of $1 \text{kg}$of coffee powder = $\text{Rs}\text{.} 240$
Cost of $1\ \text{kg}$ of milk powder = $\text{Rs}\text{.} 180 $

Cost of $1 \text{kg}$ of coffee:Cost of $1 \text{kg}$ of milk powder

$=\text{Rs}\text{.} 240 :\text{Rs}.180$

$=\dfrac{240}{180}$

$=\dfrac{24}{18}=\dfrac{8}{6}$

$=\dfrac{4}{3}=4:3$


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Important Related Links for CBSE Class 6 Maths 

CBSE Class 6 Maths Study Materials

CBSE Class 6 Maths NCERT Solutions

NCERT Class 6 Maths Book PDF

CBSE Class 6 Maths Formulas

Revision Notes for CBSE Class 6 Maths

CBSE Sample Papers for Class 6 Maths



Conclusion

Reviewing all the crucial questions for Class 6 Maths Chapter 12, Ratios and Proportions, provides students with a solid grasp of the chapter's topics. The extra and important questions for Class 6 Maths Chapter 12 Ratios and Proportions engage in a concept-focused discussion, encompassing all chapter themes. This question-and-answer method proves time-saving during exam prep, offering an efficient way to revise the chapter and enhance understanding. Practising these important questions streamlines preparation and boosts confidence for the upcoming exams.

FAQs on Important Questions for CBSE Class 6 Maths Chapter 12 - Ratio and Proportion

1. How do you learn ratios and proportions for a Class 6?

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2. What is the importance of studying ratio and proportion?

Students' grasp of numerous areas in mathematics and science depends on their ability to use ratios and proportions. They play a crucial role in the development of slope, constant rate of change, and other concepts and abilities that are essential to understanding and mastering algebra. Ratios and proportions are useful in everyday life as well as in your scholastic development.

3. What is a unit rate?

An item's unit rate is its price for one of it. This is expressed as a ratio with a one as the denominator. For instance, if you covered 70 yards in 10 seconds, you covered 7 yards on average every second. 70 yards in 10 seconds and 7 yards in 1 second are both ratios, although only the latter is a unit rate.

4. How do you solve ratio and proportion questions?

In order to solve the questions on ratio and proportion, you must have your fundamentals right. You must have the basic knowledge of what ratio and proportion really is and what is the difference between them. To attain perfection in exams and in your concept, you can learn from Vedantu.com that gives you spectacular knowledge about the topics. It gives you important questions on whatever topic you need that helps you excel in your exams.

5. What is ratio formula?

Ratios contrast two figures by ordinarily dividing them. A/B would be your formula if you were comparing one data point (A) to another data point (B). This indicates that you are multiplying information A by information B. For instance, your ratio will be 5/10 if A is 5 and B is 10.