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RD Sharma Class 7 Maths Solutions Chapter 10 - Unitary Method

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RD Sharma Solutions for Class 7 Maths Chapter 10 - Unitary Method - Free PDF Download

Free PDF download of RD Sharma Solutions for Class 7 Maths Chapter 10 - Unitary Method solved by Expert Mathematics Teachers on Vedantu.com. All Chapter 10 - Unitary Method Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register for online coaching for IIT JEE (Mains & Advanced), NEET, Engineering and Medical entrance exams.

Class 7 RD Sharma Textbook Solutions Chapter 10 - Unitary Method

In this Chapter 10 - Unitary Method, several exercise questions with solutions for RD Sharma Class 7 Maths are given to help the students and understand the concepts better. 


We have provided step by step solutions for all exercise questions given in the pdf of Class 7 RD Sharma Chapter 10 - Unitary Method. All the Exercise questions with solutions in Chapter 10 - Unitary Method are given below:

Exercise 10.1


At Vedantu, students can also get Class 7 Maths Revision Notes, Formula and Important Questions and also students can refer the complete Syllabus for Class 7 Maths, Sample Paper and Previous Year Question Paper to prepare for their exams to score more marks.

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FAQs on RD Sharma Class 7 Maths Solutions Chapter 10 - Unitary Method

1. How is the unitary method used to solve problems in RD Sharma Class 7 Chapter 10?

The unitary method is a fundamental technique for solving problems involving ratio and proportion. The process in RD Sharma solutions involves two main steps: First, you find the value of a single unit by division. Second, you find the value of the required number of units by multiplication. For example, if 5 pens cost ₹50, the cost of one pen is ₹50 ÷ 5 = ₹10. The cost of 8 pens would then be ₹10 × 8 = ₹80.

2. What are the two main types of variation problems solved using the unitary method in this chapter?

The solutions in this chapter primarily deal with two types of variation, which are crucial for setting up the problem correctly:

  • Direct Variation: When one quantity increases, the other quantity also increases. For example, more money buys more items.
  • Inverse Variation: When one quantity increases, the other quantity decreases. For example, more workers take less time to complete a job.

Identifying the correct variation is the first step to solving these problems accurately.

3. What is a common mistake students make when deciding between direct and inverse variation in unitary method problems?

A frequent error is misinterpreting the relationship between the quantities. Students often default to multiplication for the final step, even in cases of inverse variation. For instance, in a problem where 10 men finish a job in 20 days, a student might incorrectly calculate that 20 men would take more time. The key is to logically assess the situation: more men will do the work faster, indicating an inverse relationship. Therefore, one must first multiply to find the work-effort for one man (10 × 20) and then divide by the new number of men.

4. How do you solve a problem where more people result in a shorter duration, like a classic hostel food problem?

This is a typical inverse variation problem. To solve it using the unitary method, follow these steps. For example, if food is sufficient for 50 students for 45 days:

  • Step 1: Find how long the food would last for 1 student. Since it's an inverse relation, for one student, the food will last longer. So, you multiply: 45 days × 50 students.
  • Step 2: Calculate for the required number of students (e.g., 75). Now, divide the result from Step 1 by the new number of students: (45 × 50) ÷ 75 = 30 days.

Thus, for 75 students, the food would last for 30 days.

5. How does mastering the unitary method in Class 7 help with more advanced topics like Percentage, and Profit & Loss?

Mastering the unitary method builds a strong foundation for many advanced mathematical concepts. It teaches the core skill of finding a 'per unit' value, which is essential for:

  • Percentages: To find a percentage of a number, we often first find the value of 1%.
  • Profit and Loss: To calculate profit or loss, you must first determine the cost price or selling price of a single item.
  • Speed, Distance, and Time: Calculating speed requires finding the distance covered per unit of time (e.g., km per hour).

This method strengthens the logical reasoning needed for complex ratio and proportion problems in higher classes.

6. How can the unitary method be used to solve complex 'work and time' problems involving different types of workers, like men and women?

For problems with different worker types (e.g., 12 men or 18 women can do a work in 14 days), the solution involves standardising the unit of work. First, you establish an equivalence: 12 men = 18 women, which simplifies to 1 man = 1.5 women. Next, convert all workers into a single equivalent unit. For instance, if you need to find the time for 8 men and 16 women, you would convert the men to women (8 men = 8 × 1.5 = 12 women) and add them, resulting in a total of 28 women. Finally, apply the standard inverse variation rule for work and time.

7. Why are RD Sharma solutions for the Unitary Method chapter beneficial for building a strong foundation for competitive exams?

RD Sharma solutions are highly beneficial because they offer a wider variety and complexity of problems than standard textbooks. They include many Higher Order Thinking Skills (HOTS) questions that challenge students to apply the unitary method in diverse, non-obvious scenarios. Practising these questions helps build the speed, accuracy, and problem-solving aptitude required for future competitive exams, where questions on 'Time & Work' or 'Pipes & Cisterns' are very common.

8. Can the unitary method be applied to problems involving speed, distance, and time?

Yes, absolutely. The unitary method is perfect for solving problems related to speed, distance, and time. For instance, if a train covers 240 km in 4 hours, you can find its speed per unit time (1 hour) using the unitary method: 240 km ÷ 4 hours = 60 km/hr. Once you have this unit value (speed), you can easily calculate the distance it would cover in 7 hours (60 × 7 = 420 km) or the time it would take to travel 300 km (300 ÷ 60 = 5 hours).