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RS Aggarwal Solutions Class 6 Chapter-10 Ratio, Proportion and Unitary Method (Ex 10C) Exercise 10.3

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RS Aggarwal Solutions Class 6 Chapter-10 Ratio, Proportion and Unitary Method (Ex 10C) Exercise 10.3 - Free PDF

Free PDF download of RS Aggarwal Solutions Class 6 Chapter-10 Ratio, Proportion and Unitary Method (Ex 10C) Exercise 10.3 solved by Expert Mathematics Teachers on Vedantu.com. All Exercise 10.3 Questions with Solutions for Class 6 Maths RS Aggarwal to help you to revise the complete Syllabus and Score More marks. Register for online coaching for IIT JEE (Mains & Advanced) and other engineering entrance exams.

You can also register Online for NCERT Class 6 Science tuition on Vedantu.com to score more marks in the CBSE board examination. Vedantu is a platform that provides free CBSE Solutions (NCERT) and other study materials for students.

RS Aggarwal Solutions Class 6 Chapter 10 - Free PDF

In our daily life, we often come across situations where we need to compare quantities in terms of their magnitudes/ measurements in the form of ratios and percentages. They come into use in our daily lives activities such as shopping, cooking, traveling, budget planning, etc.  RS Aggarwal Solution Class 6 Chapter 10 provides solutions for all questions from exercise  10.3 in a simplified manner so it is easier for students to understand the base concept and solve it quickly in the future. 

Vedantu provides RS Aggarwal Solutions to Class 6 Maths Chapter 10 Ratios, Proportions and unitary method Exercise 10.3 and all the other exercises for the students to learn and understand the chapter well. Given below is the link for  RS Aggarwal Solutions to Class 6 Maths Chapter 10 Ratios, Proportions and unitary method Exercise

Refer to, RS Aggarwal Class 6 Solutions Chapter-10 Ratio, Proportion and Unitary Method Vedantu for Solved exercises. 

Ratios and proportions are basically two parts of the same coin. The ratio of a and b is represented in the form of a:b, whereas a proportion states that two ratios are equal. Here, a and b are any two given integers. The ratio and proportion are the two important concepts, and it is important to understand the concept as it is useful in both mathematics as well as in science. 

The unitary method is where you know the value of one unit and then the value of more units.

For example, when you go to buy tomatoes, you ask the salesperson the value of 3 tomatoes and then he tells you that he is selling 10 tomatoes for 50 rs. 

Topics Covered Under Ratios, Proportions, and Unitary Method  

  1. Definition of Ratio and Proportion 

  2. Difference between Ratio and proportion 

  3. Applications of ratios and proportions 

  4. Definition of unitary method 

  5. Application of unitary method. 

Tips While Preparing Ratios, Proportion and Unitary Method Chapter for Exam

  1. Keep practicing daily  

  2. Do not tire yourself out and take breaks after each chapter to refresh your mind for the new chapter.  

  3. Your main aim is to focus on the basic concept and not to memorize it. 

  4. Solve previous year’s question papers and sample papers to boost your preparation. 

  5. Do not just practice once, do it at least twice or thrice. 

  6. Check out Vedantu’s solution for step by step procedure and also for question papers and sample papers.

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FAQs on RS Aggarwal Solutions Class 6 Chapter-10 Ratio, Proportion and Unitary Method (Ex 10C) Exercise 10.3

1. What is the main principle behind the unitary method used in RS Aggarwal Class 6, Chapter 10, Exercise 10C?

The core principle of the unitary method is to first determine the value of a single unit from a given quantity. Once the value of 'one unit' is known, you can then calculate the value of the required number of units by multiplying. This method is a practical application of the concept of proportion.

2. How do you correctly solve a word problem from Exercise 10C using the step-by-step unitary method?

To solve a word problem using the unitary method as per the RS Aggarwal solutions, follow these steps:

  • Step 1: Read the problem carefully to identify the value given for a specific number of units (e.g., the cost of 12 pens).
  • Step 2: Calculate the value of one single unit by dividing the total value by the number of units (e.g., find the cost of 1 pen).
  • Step 3: Multiply the value of the single unit by the number of units you need to find the value for (e.g., calculate the cost of 8 pens).

3. Why is finding the value of 'one unit' so crucial in the unitary method?

Finding the value of 'one unit' is the most critical step because it establishes a standard base for comparison. This single unit value acts as a constant multiplier. Once you have this base value, you can easily scale it up or down to find the value for any quantity, which makes this method efficient and less prone to errors for solving problems involving proportion.

4. How is the unitary method applied differently when calculating 'more' versus 'less' quantity?

The core process remains the same, but the final step changes. After finding the value of one unit:

  • To find the value of a larger quantity (more), you multiply the single unit value by the larger number.
  • To find the value of a smaller quantity (less), you still multiply the single unit value, but by the smaller number. The common first step is always to find the value of 'one' by division.

5. What is a common mistake students make while solving problems in RS Aggarwal Exercise 10C?

A very common mistake is confusing the operations of multiplication and division. Students sometimes directly multiply or divide the given numbers without first finding the value of the single unit. For example, if the cost of 15 oranges is given and the cost of 10 is asked, a student might incorrectly divide 15 by 10. The correct method is to always find the cost of 1 orange first and then multiply by 10.

6. How can the unitary method be used to solve problems involving time and distance?

The unitary method is very effective for time and distance problems. For instance, if a car travels 180 km in 3 hours, you can use the unitary method to find the distance it covers in 1 hour (its speed). To do this, you would calculate 180 km / 3 hours = 60 km per hour. This 'unit value' (speed) can then be used to determine the distance it would cover in 5 hours (60 km/hr * 5 hr = 300 km).