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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

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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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Delve into the world of ratios and proportions with Class 6 Maths Chapter 12 Ratios and Proportions. This exploration offers valuable insights into understanding and applying mathematical relationships. As you navigate through this chapter, you'll unravel the secrets of proportionality and enhance your foundational mathematical knowledge with Vedantu:


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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.

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

1. What types of important questions are usually asked from CBSE Class 6 Maths Chapter 12, Ratio and Proportion?

For the Class 6 exams, you can expect a mix of questions from this chapter. These typically include:

  • 1-mark questions: Finding the ratio in its simplest form or identifying if two ratios are in proportion.
  • 2-mark questions: Solving problems involving equivalent ratios or applying the unitary method for simple cases.
  • 3-mark questions: Solving multi-step word problems that require you to form a ratio and then solve for an unknown quantity.

Focusing on these types will help you cover the most frequently asked questions for the 2025-26 session.

2. How do these important questions relate to the official CBSE syllabus for this chapter?

These important questions are designed to cover all the core concepts mentioned in the CBSE Class 6 Maths syllabus. They test your understanding of comparing quantities to form a ratio, understanding proportion as an equality of two ratios, and applying the unitary method to solve practical problems.

3. Why are word problems based on the unitary method considered so important in this chapter?

Word problems using the unitary method are crucial because they test your ability to apply concepts to real-life situations. They check if you can first determine the value of a single unit (like the cost of one pen) and then calculate the value for the required number of units (like the cost of 10 pens). These questions often carry higher marks as they assess both calculation skills and logical reasoning.

4. What is a common mistake to avoid in Ratio and Proportion questions to not lose marks?

A very common mistake is writing the terms of a ratio in the wrong order. For example, if a question asks for the ratio of 'girls to boys', you must write the number of girls first. Reversing the order will result in an incorrect answer. Another frequent error is forgetting to convert quantities to the same unit (e.g., metres and centimetres) before calculating their ratio.

5. In an exam, how can I quickly tell if a question is about 'ratio' or 'proportion'?

It's simple! A question about a ratio will usually ask you to compare just two quantities. For example, 'Find the ratio of 20 pencils to 10 erasers'. A question about proportion will involve four quantities and ask you to check if two ratios are equal. It might give you three terms and ask you to find the fourth to make the statement true, like 'Are 2, 4, 6, and 12 in proportion?'.

6. Are questions on 'unit rate' important for the Class 6 exam?

Yes, understanding 'unit rate' is important as it often appears in short-answer questions. A unit rate is a practical application of ratio where you find the value for 'one' unit, such as cost per item or distance per hour. Mastering this helps in solving unitary method problems more efficiently and accurately in your exam.

7. What is the best way to use these important questions for exam preparation?

For effective preparation, first try to solve each important question on your own without looking at the solution. Once you have an answer, compare your method with the provided steps. This helps you understand the correct way to present your answer and identify any conceptual gaps in your understanding before the actual exam.