Welcome! Access the best MSBSHSE Solutions for Class 7 Maths Chapter 9 Set 37 Direct Proportion and Inverse Proportion right here. Designed for the 2026-27 term, these answers follow the newest MSBSHSE curriculum guidelines for Class 7 Maths. Get our expert-verified study solutions for Class 7 Maths as free downloadable PDFs.
MSBSHSE Textbook Solutions for Class 7 Maths Chapter 9 Set 37 Direct Proportion and Inverse Proportion
Want to build a strong core? For Class 7 students, working through MSBSHSE textbook questions is essential. Our Class 7 Maths solutions use clear, step-by-step methods so you grasp the reasoning behind each answer. Practicing these Chapter 9 Set 37 Direct Proportion and Inverse Proportion solutions will help you score better marks in exams.
Get Chapter 9 Set 37 Direct Proportion and Inverse Proportion MSBSHSE Solution PDF for Class 7 Maths
Question 1. If 7 kg onions cost Rs 140, how much must we pay for 12 kg onions?
Answer:
Let the cost of 12 kg onions be Rs x.
The quantity of onions and their cost are in direct proportion.
\[ \frac{7}{140} = \frac{12}{x} \]
\( \therefore 7x = 12 \times 140 \) ....(Multiplying both sides by 140x)
\( \therefore x = \frac{12 \times 140}{7} \)
\( = 240 \) We must pay Rs 240 for 12 kg onions.
In simple words: The problem is about direct proportion. As the quantity of onions increases, their cost also increases proportionally. Set up a ratio of quantities to costs to find the unknown cost for 12 kg of onions.
🎯 Exam Tip: Ensure that the units are consistent on both sides of the proportion. Clearly state the direct proportional relationship to earn full marks.
Question 2. If Rs 600 buy 15 bunches of feed, how many will Rs 1280 buy?
Answer:
Let the bunches of feed bought for Rs 1280 be x.
The quantity of feed bought and their cost are in direct proportion.
\[ \frac{600}{15} = \frac{1280}{x} \]
\( \therefore 600x = 1280 \times 15 \) ....(Multiplying both sides by 15x)
\( \therefore x = \frac{1280 \times 15}{600} = 32 \) 32 bunches of feed can be bought for Rs 1280.
In simple words: This is a direct proportion problem. More money means more bunches of feed. Set up a ratio comparing the cost to the number of bunches to solve for the unknown quantity.
🎯 Exam Tip: Write down the proportional relationship clearly before setting up the equation. Double-check your calculations, especially during multiplication and division steps.
Question 3. For 9 cows, 13 kg 500 g of food supplement are required every day. In the same proportion, how much will be needed for 12 cows?
Answer:
Let the food supplement required for 12 cows be x kg.
The quantity of food supplement required and the number of cows are in direct proportion.
\[ \frac{13kg\,500gram}{9} = \frac{x\,kg}{12} \] (13kg 500 gram = 13.5 kg)
\[ \therefore \frac{13.5}{9} = \frac{x}{12} \]
\( \therefore 13.5 \times 12 = 9x \) ....(Multiplying both sides by 9 x 12)
\( \therefore x = \frac{13.5 \times 12}{9} \)
\( \therefore x = 18 \) The food supplement required for 12 cows is 18 kg.
In simple words: This problem involves direct proportion, as more cows will require more food supplement. Convert grams to kilograms for consistency, then set up and solve the proportional equation.
🎯 Exam Tip: Always convert all quantities to a single, consistent unit (e.g., kg) before performing calculations. This minimizes errors in proportional reasoning.
Question 4. The cost of 12 quintals of soyabean is Rs 36,000. How much will 8 quintals cost?
Answer:
Let the cost of 8 quintals of soyabean be Rs x.
The quantity of soyabeans and their cost are in direct proportion.
\[ \therefore \frac{12}{36000} = \frac{8}{x} \]
\( \therefore 12x = 8 \times 36000 \) ....(Multiplying both sides by 36000x)
\( \therefore x = \frac{8 \times 36000}{12} = 24000 \) The cost of 8 quintals of soyabean is Rs 24000.
In simple words: This problem demonstrates direct proportion; fewer quintals of soyabean will cost less. Set up a ratio of quantities to costs to determine the cost for 8 quintals.
🎯 Exam Tip: Clearly define your variable (x) at the beginning of the solution. Show all steps of cross-multiplication and division for clarity and to avoid calculation mistakes.
Question 5. Two mobiles cost Rs 16,000. How much money will be required to buy 13 such mobiles ?
Answer:
Let the cost of 13 mobiles be Rs x.
The quantity of mobiles and their cost are in direct proportion.
\[ \therefore \frac{2}{16000} = \frac{13}{x} \]
\( \therefore 2x = 13 \times 16000 \) ....(Multiplying both sides by 16000x)
\( \therefore x = \frac{13 \times 16000}{2} = 104000 \) Rs 104000 will be required to buy 13 mobiles.
In simple words: This is a direct proportion scenario where more mobiles mean a higher total cost. Set up the ratio of the number of mobiles to their cost to find the price for 13 mobiles.
🎯 Exam Tip: Always write the final answer in a complete sentence, including the units (e.g., Rs) to make it clear and comprehensive.
Free MSBSHSE Textbook Explanations: Class 7 Maths
Accessing Official MSBSHSE Solutions for Chapter 9 Set 37 Direct Proportion and Inverse Proportion
Explore expert-verified MSBSHSE Solutions for Chapter 9 Set 37 Direct Proportion and Inverse Proportion right here on our portal. Designed by experienced educators, these answers address every exercise question found within your Class 7 Maths textbook, fully updated to match the latest academic standards and active MSBSHSE syllabus guidelines.
Deep-Dive Explanations & Answer Guides
Our faculty has formulated detailed, step-by-step explanations for challenging problems across the Class 7 Maths chapter. Beyond giving direct outcomes, we break down the underlying theories to foster genuine topic comprehension. Ideal for Class 7 learners tackling both theoretical and numerical questions, reviewing these MSBSHSE Questions and Answers significantly strengthens fundamental concepts.
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Using our Maths solutions regularly students will be able to improve their logical thinking and problem-solving speed. These Class 7 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 9 Set 37 Direct Proportion and Inverse Proportion to get a complete preparation experience.
FAQs
The complete and updated Maharashtra Board Class 7 Chapter 9 Set 37 Direct Proportion and Inverse Proportion Solutions is available for free on StudiesToday.com. These solutions for Class 7 Maths are as per latest MSBSHSE curriculum.
Yes, our experts have revised the Maharashtra Board Class 7 Chapter 9 Set 37 Direct Proportion and Inverse Proportion 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.
Toppers recommend using MSBSHSE language because MSBSHSE marking schemes are strictly based on textbook definitions. Our Maharashtra Board Class 7 Chapter 9 Set 37 Direct Proportion and Inverse Proportion Solutions will help students to get full marks in the theory paper.
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