Maharashtra Board Class 6 Maths Chapter 5 Decimal Fractions Set 16 Solutions

Get the most accurate MSBSHSE Solutions for Class 6 Maths Chapter 5 Decimal Fractions Set 16 here. Updated for the 2026-27 academic session, these solutions are based on the latest MSBSHSE textbooks for Class 6 Maths. Our expert-created answers for Class 6 Maths are available for free download in PDF format.

Detailed Chapter 5 Decimal Fractions Set 16 MSBSHSE Solutions for Class 6 Maths

For Class 6 students, solving MSBSHSE textbook questions is the most effective way to build a strong conceptual foundation. Our Class 6 Maths solutions follow a detailed, step-by-step approach to ensure you understand the logic behind every answer. Practicing these Chapter 5 Decimal Fractions Set 16 solutions will improve your exam performance.

Class 6 Maths Chapter 5 Decimal Fractions Set 16 MSBSHSE Solutions PDF

Decimal Fractions Class 6 Maths Chapter 5 Practice Set 16 Solutions Maharashtra Board

Std 6 Maths Practice Set 16 Solutions Answers

 

Question 1. If, 317 x 45 = 14265, then 3.17 × 4.5 = ?
Answer:
Solution:
3.17 x 4.5
= 14.265
In simple words: The product of the numbers 3.17 and 4.5 is found by first multiplying the whole number parts (317 and 45) and then placing the decimal point based on the total number of decimal places in the original numbers.

🎯 Exam Tip: When multiplying decimals, count the total number of decimal places in the factors to accurately position the decimal point in the final product.

 

Question 2. If, 503 × 217 = 109151, then 5.03 × 2.17 = ?
Answer:
Solution:
5.03 x 2.17
= 10.9151
In simple words: To find the product of 5.03 and 2.17, multiply 503 by 217 and then adjust the decimal point according to the total number of decimal places in both factors.

🎯 Exam Tip: Understand that multiplying numbers with decimals involves the same multiplication process as whole numbers, with an additional step of correctly placing the decimal point based on the sum of decimal places in the original numbers.

 

Question 3.
(i) 2.7 x 1.4
(ii) 6.17 x 3.9
(iii) 0.57 x 2
(iv) 5.04 × 0.7
Answer:
Solution:
(i) 2.7 x 1.4
\( = \frac{27}{10} \times \frac{14}{10} \)
\( = \frac{27 \times 14}{10 \times 10} = \frac{378}{100} \)
= 3.75

(ii) 6.17 x 3.9
\( = \frac{617}{100} \times \frac{39}{10} \)
\( = \frac{617 \times 39}{100 \times 10} = \frac{24063}{1000} \)
= 24.063

(iii) 0.57 x 2
\( = \frac{57}{100} \times \frac{2}{1} \)
\( = \frac{57 \times 2}{100 \times 1} = \frac{114}{100} \)
= 1.14

(iv) 5.04 × 0.7
\( = \frac{504}{100} \times \frac{7}{10} \)
\( = \frac{504 \times 7}{100 \times 10} = \frac{3528}{1000} \)
= 3.528
In simple words: Each part of this question involves multiplying decimal numbers by converting them into fractions, performing the multiplication, and then converting the result back to a decimal.

🎯 Exam Tip: Remember that the number of zeros in the denominator of the fractional representation of a decimal determines the number of decimal places in the resulting decimal number.

 

Question 4. Virendra bought 18 bags of rice, each bag weighing 5.250 kg. How much rice did he buy altogether? If the rice costs Rs 42 per kg, how much did he pay for it?
Answer:
Solution:
Weight of one bag of rice = 5.250 kg
Number of bags of rice = 18
.. Total Weight = 18 × 5.250
\( = \frac{18}{1} \times \frac{525}{100} = \frac{18 \times 525}{1 \times 100} = \frac{9450}{100} = 94.5 \text{ kg} \)
Cost of 1 kg of rice = Rs 42
.. Cost of 94.5 kg of rice = 42 × 94.5
\( = \frac{42}{1} \times \frac{945}{10} = \frac{42 \times 945}{1 \times 10} = \frac{39690}{10} = 3969 \)
.. Total rice bought by Virendra is 94.5 kg, and the amount paid for it is Rs 3969.
In simple words: First, calculate the total weight of rice by multiplying the number of bags by the weight of each bag. Then, calculate the total cost by multiplying the total weight of rice by the cost per kilogram.

🎯 Exam Tip: Clearly state all given information and show each step of the calculation, ensuring correct decimal multiplication and unit consistency, especially when dealing with money and weight.

 

Question 5. Vedika has 23.5 metres of cloth. She used it to make 5 curtains of equal size. If each curtain required 4 metres 25 cm to make, how much cloth is left over?
Answer:
Solution:
We know, that 1 m = 100 cm
Cloth required to make 1 curtain = 4 m 25 cm
\( = 4 \text{ m} + \frac{25}{100} \text{ m} \)
= 4 m + 0.25 m
= 4.25 m
.. Cloth required to make 5 curtains = 5 × 4.25
\( = \frac{5}{1} \times \frac{425}{100} = \frac{5 \times 425}{1 \times 100} = \frac{2125}{100} \)
= 21.25 m
Cloth remaining with Vedika = Total cloth with Vedika – Cloth used
= 23.5 m - 21.25 m
= 2.25 m
.. The length of cloth remaining with Vedika is 2.25 m.
In simple words: First, convert the cloth required for one curtain from meters and centimeters to meters only. Then, calculate the total cloth needed for five curtains. Finally, subtract the used cloth from the total cloth Vedika had to find the remaining cloth.

🎯 Exam Tip: Remember to convert all measurements to a common unit (e.g., meters) before performing calculations. Pay close attention to decimal subtraction for accuracy in finding the remaining quantity.

MSBSHSE Solutions Class 6 Maths Chapter 5 Decimal Fractions Set 16

Students can now access the MSBSHSE Solutions for Chapter 5 Decimal Fractions Set 16 prepared by teachers on our website. These solutions cover all questions in exercise in your Class 6 Maths textbook. Each answer is updated based on the current academic session as per the latest MSBSHSE syllabus.

Detailed Explanations for Chapter 5 Decimal Fractions Set 16

Our expert teachers have provided step-by-step explanations for all the difficult questions in the Class 6 Maths 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 6 students who want to understand both theoretical and practical questions. By studying these MSBSHSE Questions and Answers your basic concepts will improve a lot.

Benefits of using Maths Class 6 Solved Papers

Using our Maths solutions regularly students will be able to improve their logical thinking and problem-solving speed. These Class 6 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 5 Decimal Fractions Set 16 to get a complete preparation experience.

FAQs

Where can I find the latest Maharashtra Board Class 6 Maths Chapter 5 Decimal Fractions Set 16 Solutions for the 2026-27 session?

The complete and updated Maharashtra Board Class 6 Maths Chapter 5 Decimal Fractions Set 16 Solutions is available for free on StudiesToday.com. These solutions for Class 6 Maths are as per latest MSBSHSE curriculum.

Are the Maths MSBSHSE solutions for Class 6 updated for the new 50% competency-based exam pattern?

Yes, our experts have revised the Maharashtra Board Class 6 Maths Chapter 5 Decimal Fractions Set 16 Solutions as per 2026 exam pattern. All textbook exercises have been solved and have added explanation about how the Maths concepts are applied in case-study and assertion-reasoning questions.

How do these Class 6 MSBSHSE solutions help in scoring 90% plus marks?

Toppers recommend using MSBSHSE language because MSBSHSE marking schemes are strictly based on textbook definitions. Our Maharashtra Board Class 6 Maths Chapter 5 Decimal Fractions Set 16 Solutions will help students to get full marks in the theory paper.

Do you offer Maharashtra Board Class 6 Maths Chapter 5 Decimal Fractions Set 16 Solutions in multiple languages like Hindi and English?

Yes, we provide bilingual support for Class 6 Maths. You can access Maharashtra Board Class 6 Maths Chapter 5 Decimal Fractions Set 16 Solutions in both English and Hindi medium.

Is it possible to download the Maths MSBSHSE solutions for Class 6 as a PDF?

Yes, you can download the entire Maharashtra Board Class 6 Maths Chapter 5 Decimal Fractions Set 16 Solutions in printable PDF format for offline study on any device.