RBSE Solutions Class 5 Maths Chapter 3 Multiplication and Division Exercise 3.2

Get the most accurate RBSE Solutions for Class 5 Mathematics Chapter 3 Multiplication and Division here. Updated for the 2026-27 academic session, these solutions are based on the latest RBSE textbooks for Class 5 Mathematics. Our expert-created answers for Class 5 Mathematics are available for free download in PDF format.

Detailed Chapter 3 Multiplication and Division RBSE Solutions for Class 5 Mathematics

For Class 5 students, solving RBSE textbook questions is the most effective way to build a strong conceptual foundation. Our Class 5 Mathematics solutions follow a detailed, step-by-step approach to ensure you understand the logic behind every answer. Practicing these Chapter 3 Multiplication and Division solutions will improve your exam performance.

Class 5 Mathematics Chapter 3 Multiplication and Division RBSE Solutions PDF

 

Question 1. Solve this
(i) 255 ÷ 15
(ii) 312 ÷ 12
(iii) 640 ÷ 16
(iv) 702 ÷ 13
(v) 357 ÷ 21
(vi) 770 ÷ 28
(vii) 952 ÷ 34
(viii) 847 ÷ 18
(ix) 656 ÷ 23
(x) 945 ÷ 35
Answer:
(i) For \( 255 \div 15 \): Quotient = 17, Remainder = 0.
(ii) For \( 312 \div 12 \): Quotient = 26, Remainder = 0.
(iii) For \( 640 \div 16 \): Quotient = 40, Remainder = 0.
(iv) For \( 702 \div 13 \): Quotient = 54, Remainder = 0.
(v) For \( 357 \div 21 \): Quotient = 17, Remainder = 0.
(vi) For \( 770 \div 28 \): Quotient = 27, Remainder = 14.
(vii) For \( 952 \div 34 \): Quotient = 28, Remainder = 0.
(viii) For \( 847 \div 18 \): Quotient = 47, Remainder = 1.
(ix) For \( 656 \div 23 \): Quotient = 28, Remainder = 12.
(x) For \( 945 \div 35 \): Quotient = 27, Remainder = 0. When dividing numbers, the quotient is the result of the division and the remainder is what is left over.
In simple words: For each problem, divide the first number by the second number. The main answer is the quotient, and any amount left behind is the remainder.

🎯 Exam Tip: Always double-check your division by multiplying the quotient by the divisor and adding the remainder; the result should equal the original dividend.

 

Question 3. Chhaya bought 20 copies for Rs. 360. What is the price of a single copy?
Answer: To find the price of a single copy, we need to divide the total cost by the number of copies. So, we calculate Rs. \( 360 \div 20 \). The cost of one copy is Rs. 18. This helps us to find the unit price of any item.
In simple words: Divide the total money Chhaya spent (Rs. 360) by the number of copies she bought (20). The answer, Rs. 18, is the price for one copy.

🎯 Exam Tip: When you need to find the price of one item from a total cost for many, always divide the total cost by the number of items.

 

Question 4. Dheeraj has 864 bananas. How many dozen bananas does he have? (1 dozen = 12 objects)
Answer: We know that one dozen is equal to 12 objects. To find the number of dozens Dheeraj has, we need to divide his total bananas by 12. So, \( 864 \div 12 = 72 \). Therefore, Dheeraj has 72 dozens of bananas. This type of division helps to group individual items into common units.
In simple words: One dozen means 12. To find how many dozens Dheeraj has, divide his total 864 bananas by 12. He has 72 dozens of bananas.

🎯 Exam Tip: Remember standard units like 'dozen' (12 items) for quick calculations. This prevents confusion in quantity problems.

 

Question 5. 702 people can sit in 27 buses. How many people can sit in one bus.
Answer: To find how many people can sit in one bus, we divide the total number of people by the total number of buses. The calculation is \( 702 \div 27 = 26 \). Thus, 26 people can sit comfortably in each bus. This helps determine the capacity of each vehicle.
In simple words: If 702 people use 27 buses, divide 702 by 27 to find out how many people can sit in just one bus. The answer is 26 people per bus.

🎯 Exam Tip: For problems involving equal distribution, division is the key operation. Make sure to identify what represents the total and what represents the number of groups.

 

Question 6. Onkar has 400 rupees. If the cost of cloth is Rs. 30 per meter then how many meters of cloth can he buy and how much money will be left?
Answer: To find the amount of cloth Onkar can buy, we divide his total money by the cost per meter. We perform the division \( 400 \div 30 \). The quotient is 13, which means Onkar can buy 13 meters of cloth. The remainder is 10, so Rs. 10 will be left over with him. This calculation helps manage budgets effectively by showing both how much can be bought and what remains.
In simple words: Divide Onkar's Rs. 400 by the Rs. 30 cost per meter of cloth. He can buy 13 meters of cloth, and Rs. 10 will remain.

🎯 Exam Tip: In word problems asking for "how much is left," remember to identify and state the remainder from your division calculation clearly.

 

Question 7. How many garlands can be made from 648 flowers, if each garland is made of 24 flowers?
Answer: To determine how many garlands can be made, we divide the total number of flowers by the number of flowers required for each garland. The calculation is \( 648 \div 24 = 27 \). Therefore, 27 garlands can be created from 648 flowers. This shows how division helps with practical projects involving grouping items.
In simple words: To find out how many garlands you can make, divide the total number of flowers (648) by the number of flowers needed for one garland (24). You will get 27 garlands.

🎯 Exam Tip: For problems involving making equal-sized groups from a larger total, always use division to find the number of groups.

 

Question 8. How many years are there in 936 months? Find out.
Answer: We know that one year consists of 12 months. To convert 936 months into years, we divide the total number of months by 12. So, we calculate \( 936 \div 12 = 78 \). This means that 936 months is equivalent to 78 years. Such conversions are often used in time-related problems.
In simple words: There are 12 months in one year. So, to find how many years are in 936 months, divide 936 by 12. The answer is 78 years.

🎯 Exam Tip: Always recall the basic conversion factors for time units (like months to years) before attempting to solve such problems. Be precise with your division.

Free study material for Mathematics

RBSE Solutions Class 5 Mathematics Chapter 3 Multiplication and Division

Students can now access the RBSE Solutions for Chapter 3 Multiplication and Division prepared by teachers on our website. These solutions cover all questions in exercise in your Class 5 Mathematics textbook. Each answer is updated based on the current academic session as per the latest RBSE syllabus.

Detailed Explanations for Chapter 3 Multiplication and Division

Our expert teachers have provided step-by-step explanations for all the difficult questions in the Class 5 Mathematics chapter. Along with the final answers, we have also explained the concept behind it to help you build stronger understanding of each topic. This will be really helpful for Class 5 students who want to understand both theoretical and practical questions. By studying these RBSE Questions and Answers your basic concepts will improve a lot.

Benefits of using Mathematics Class 5 Solved Papers

Using our Mathematics solutions regularly students will be able to improve their logical thinking and problem-solving speed. These Class 5 solutions are a guide for self-study and homework assistance. Along with the chapter-wise solutions, you should also refer to our Revision Notes and Sample Papers for Chapter 3 Multiplication and Division to get a complete preparation experience.

FAQs

Where can I find the latest RBSE Solutions Class 5 Maths Chapter 3 Multiplication and Division Exercise 3.2 for the 2026-27 session?

The complete and updated RBSE Solutions Class 5 Maths Chapter 3 Multiplication and Division Exercise 3.2 is available for free on StudiesToday.com. These solutions for Class 5 Mathematics are as per latest RBSE curriculum.

Are the Mathematics RBSE solutions for Class 5 updated for the new 50% competency-based exam pattern?

Yes, our experts have revised the RBSE Solutions Class 5 Maths Chapter 3 Multiplication and Division Exercise 3.2 as per 2026 exam pattern. All textbook exercises have been solved and have added explanation about how the Mathematics concepts are applied in case-study and assertion-reasoning questions.

How do these Class 5 RBSE solutions help in scoring 90% plus marks?

Toppers recommend using RBSE language because RBSE marking schemes are strictly based on textbook definitions. Our RBSE Solutions Class 5 Maths Chapter 3 Multiplication and Division Exercise 3.2 will help students to get full marks in the theory paper.

Do you offer RBSE Solutions Class 5 Maths Chapter 3 Multiplication and Division Exercise 3.2 in multiple languages like Hindi and English?

Yes, we provide bilingual support for Class 5 Mathematics. You can access RBSE Solutions Class 5 Maths Chapter 3 Multiplication and Division Exercise 3.2 in both English and Hindi medium.

Is it possible to download the Mathematics RBSE solutions for Class 5 as a PDF?

Yes, you can download the entire RBSE Solutions Class 5 Maths Chapter 3 Multiplication and Division Exercise 3.2 in printable PDF format for offline study on any device.