Maharashtra Board Class 6 Maths Chapter 11 Ratio Proportion Set 28 Solutions

Get the most accurate MSBSHSE Solutions for Class 6 Maths Chapter 11 Ratio Proportion Set 28 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 11 Ratio Proportion Set 28 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 11 Ratio Proportion Set 28 solutions will improve your exam performance.

Class 6 Maths Chapter 11 Ratio Proportion Set 28 MSBSHSE Solutions PDF

Ratio-Proportion Class 6 Maths Chapter 11 Practice Set 28 Solutions Maharashtra Board

Question 1. In each example below, find the ratio of the first number to the second:
(i) 24, 56
(ii) 63,49
(iii) 52, 65
(iv) 84, 60
(v) 35, 65
(vi) 121, 99
Answer:
i. 24, 56
\( \frac{24}{56} = \frac{24 \div 8}{56 \div 8} = \frac{3}{7} \)
= 3:7
ii. 63,49
\( \frac{63}{49} = \frac{63 \div 7}{49 \div 7} = \frac{9}{7} \)
= 9:7
iii. 52, 65
\( \frac{52}{65} = \frac{52 \div 13}{65 \div 13} = \frac{4}{5} \)
= 4:5
iv. 84, 60
\( \frac{84}{60} = \frac{84 \div 12}{60 \div 12} = \frac{7}{5} \)
= 7:5
v. 35, 65
\( \frac{35}{65} = \frac{35 \div 5}{65 \div 5} = \frac{7}{13} \)
= 7:13
vi. 121, 99
\( \frac{121}{99} = \frac{121 \div 11}{99 \div 11} = \frac{11}{9} \)
= 11:9
In simple words: To find the ratio, divide both numbers by their greatest common divisor (GCD) to simplify the fraction to its lowest terms. Then, express this simplified fraction as a ratio.

🎯 Exam Tip: Always simplify ratios to their lowest possible terms to ensure full marks. Make sure to find the GCD correctly.

Question 2. Find the ratio of the first quantity to the second.
(i) 25 beads, 40 beads
(ii) Rs 40, Rs 120
(iii) 15 minutes, 1 hour
(iv) 30 litres, 24 litres
(v) 99 kg, 44000 grams
(vi) 1 litre, 250 ml
(vii) 60 paise, 1 rupee
(viii) 750 grams, \( \frac{1}{2} \) kg
(ix) 125 cm, 1 metre
Answer:
i. Required Ratio = \( \frac{25}{40} = \frac{25 \div 5}{40 \div 5} = \frac{5}{8} \)
ii. Required Ratio = \( \frac{40}{120} = \frac{40 \div 40}{120 \div 40} = \frac{1}{3} \)
iii. 1 hour = 60 minutes
Required Ratio = \( \frac{15}{60} = \frac{15 \div 15}{60 \div 15} = \frac{1}{4} \)
iv. Required Ratio = \( \frac{30}{24} = \frac{30 \div 6}{24 \div 6} = \frac{5}{4} \)
v. 99 kg = 99 x 1000 grams = 99000 grams
Required Ratio = \( \frac{99000}{44000} = \frac{99000 \div 1000}{44000 \div 1000} = \frac{99}{44} \)
= \( \frac{99 \div 11}{44 \div 11} = \frac{9}{4} \)
vi. 1 litre, 250 ml
1 litre = 1000 ml
Required Ratio = \( \frac{1000}{250} = \frac{1000 \div 10}{250 \div 10} = \frac{100}{25} \)
= \( \frac{100 \div 25}{25 \div 25} = \frac{4}{1} \)
vii. 60 paise, 1 rupee
1 rupee = 100 paise
Required Ratio = \( \frac{60}{100} = \frac{60 \div 20}{100 \div 20} = \frac{3}{5} \)
viii. 750 grams, \( \frac{1}{2} \) kg
\( \frac{1}{2} \) kg = \( \frac{1}{2} \) x 1000 grams = 500 grams
Required Ratio = \( \frac{750}{500} = \frac{750 \div 10}{500 \div 10} = \frac{75}{50} \)
= \( \frac{75 \div 25}{50 \div 25} = \frac{3}{2} \)
ix. 125 cm, 1 metre
1 metre = 100 cm
Required Ratio = \( \frac{125}{100} = \frac{125 \div 25}{100 \div 25} = \frac{5}{4} \)
In simple words: To find the ratio of different quantities, first convert them to the same units. Then, express the first quantity as a fraction of the second and simplify it to its lowest terms.

🎯 Exam Tip: Unit conversion is crucial for ratio problems involving different measurement units. Always convert to the smaller unit to avoid decimals and simplify calculations.

Question 3. Reema has 24 notebooks and 18 books. Find the ratio of notebooks to books.
Answer:
Ratio of notebooks to books
= \( \frac{\text{Total number of notebooks}}{\text{Total number of books}} = \frac{24}{18} = \frac{24 \div 6}{18 \div 6} = \frac{4}{3} \)
\( \therefore \) The ratio of notebooks to books with Reema is \( \frac{4}{3} \).
In simple words: The ratio of notebooks to books is found by dividing the number of notebooks by the number of books and simplifying the resulting fraction.

🎯 Exam Tip: Clearly identify which quantity is the numerator and which is the denominator based on the order asked in the question (e.g., "notebooks to books").

Question 4. 30 cricket players and 20 kho-kho players are training on a field. What is the ratio of cricket players to the total number of players?
Answer:
Total number of players = Cricket players + Kho-kho players
= 30 + 20 = 50
Ratio of cricket players to the total number of players
= \( \frac{\text{Number of cricket players}}{\text{Total number of players}} = \frac{30}{50} = \frac{30 \div 10}{50 \div 10} = \frac{3}{5} \)
\( \therefore \) The ratio of cricket players to the total number of players is \( \frac{3}{5} \).
In simple words: To find the ratio of a specific group to the total, first calculate the total number of items, then divide the number of items in the specific group by the total and simplify.

🎯 Exam Tip: When dealing with ratios involving a 'total', always ensure the total is calculated correctly by summing up all relevant categories. Simplify the final fraction.

Question 5. Snehal has a red ribbon that is 80 cm long and a blue ribbon 2.20 m long. What is the ratio of the length of the red ribbon to that of the blue ribbon?
Answer:
1 metre = 100 cm
Length of the red ribbon = 80 cm
Length of the blue ribbon = 2.20 m = 2.20 x 100 cm
= \( \frac{220}{100} \times \frac{100}{1} = \frac{220 \times 100}{100 \times 1} \) = 220 cm
\( \therefore \) Ratio of length of the red ribbon to that of the blue ribbon
= \( \frac{\text{Length of red ribbon}}{\text{Length of blue ribbon}} = \frac{80 \text{ cm}}{220 \text{ cm}} = \frac{80 \div 20}{220 \div 20} = \frac{4}{11} \)
\( \therefore \) The ratio of the length of the red ribbon to that of the blue ribbon is \( \frac{4}{11} \).
In simple words: To compare lengths given in different units, convert them to a common unit (e.g., centimetres), then form the ratio and simplify the fraction.

🎯 Exam Tip: Pay close attention to unit consistency. Mismatched units are a common source of error. Always convert all quantities to the same unit before forming the ratio.

Question 6. Shubham's age today is 12 years and his father's is 42 years. Shubham's mother is younger than his father by 6 years. Find the following ratios.
(i) Ratio of Shubham's age today to his mother's age today.
(ii) Ratio of Shubham's mother's age today to his father's age today.
(iii) The ratio of Shubham's age to his mother's age when Shubham was 10 years old.
Answer:
Shubham's age today = 12 years
Shubham's father's age = 42 years
Shubham's mother age = Shubham's father's age - 6 years
= 42 years - 6 years = 36 years
i. Ratio of Shubham's age today to his mother's age today
= \( \frac{\text{Shubham's age today}}{\text{Shubham's Mother's age today}} = \frac{12}{36} = \frac{12 \div 12}{36 \div 12} = \frac{1}{3} \)
\( \therefore \) The ratio of Shubham's age today to his mother's age today is \( \frac{1}{3} \).
ii. Ratio of Shubham's mother age today to his father's age today
= \( \frac{\text{Shubham's mother's age today}}{\text{Shubham's father's age today}} = \frac{36}{42} = \frac{36 \div 6}{42 \div 6} = \frac{6}{7} \)
\( \therefore \) The ratio of Shubham's mother's age today to his father's age today is \( \frac{6}{7} \).
iii. Shubham's age today is 12 years and his mothers age is 36 years.
Hence when Shubham's age was 10 years, his mother's age was 34 years (i.e. 36 - 2 years).
Ratio of Shubham's age to his mother's age when Shubham was 10 years old
= \( \frac{\text{Shubham's age}}{\text{Shubham's mother's age}} = \frac{10}{34} = \frac{10 \div 2}{34 \div 2} = \frac{5}{17} \)
\( \therefore \) The ratio of Shubham's age to his mother's age when Shubham was 10 years old is \( \frac{5}{17} \).
In simple words: This problem involves calculating ages based on given information and then finding various ratios between them, ensuring to use the correct ages for each specific ratio calculation.

🎯 Exam Tip: Read multi-part ratio problems carefully. Break down the problem into smaller steps (e.g., calculate all ages first) before forming the ratios. Ensure you are using the correct age for the specified time (present or past).

Maharashtra Board Class 6 Maths Chapter 11 Ratio-Proportion Practice Set 28 Intext Questions And Activities

Question 1. In the figure, colour some squares with any colour you like and leave some blank. (Textbook pg. no. 57)

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक 4x4 ग्रिड को दर्शाता है जिसमें कुल 16 वर्ग हैं। छात्रों को इन वर्गों में से कुछ को रंगना है और कुछ को खाली छोड़ना है ताकि वे विभिन्न अनुपातों की गणना कर सकें।
(i) Count all the boxes and write the number.
(ii) Count the colored ones and write the number.
(iii) Count the blank ones and write the number.
(iv) Find the ratio of the colored boxes to the blank ones.
(v) Find the ratio of the colored boxes to the total boxes.
(vi) Find the ratio of the blank boxes to the total boxes.
Answer:
i. The number of all boxes is 16.
ii. The number of colored boxes is 7.
iii. The number of blank boxes is 9.
iv. Ratio of the colored boxes to the blank ones
= \( \frac{\text{Number of coloured boxes}}{\text{Number of blank boxes}} = \frac{7}{9} \)
v. Ratio of the colored boxes to the total boxes
= \( \frac{\text{Number of coloured boxes}}{\text{Number of total boxes}} = \frac{7}{16} \)
vi. Ratio of the blank boxes to the total boxes
= \( \frac{\text{Number of blank boxes}}{\text{Number of total boxes}} = \frac{9}{16} \)
In simple words: This activity demonstrates ratios using a grid. Students count total, colored, and blank boxes, then form ratios for colored to blank, colored to total, and blank to total, simplifying each ratio.

🎯 Exam Tip: When working with visual aids like grids, accurately count each category (total, colored, blank) before setting up the ratios. Ensure the ratios are expressed in simplest form.

MSBSHSE Solutions Class 6 Maths Chapter 11 Ratio Proportion Set 28

Students can now access the MSBSHSE Solutions for Chapter 11 Ratio Proportion Set 28 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 11 Ratio Proportion Set 28

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.

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FAQs

Where can I find the latest Maharashtra Board Class 6 Maths Chapter 11 Ratio Proportion Set 28 Solutions for the 2026-27 session?

The complete and updated Maharashtra Board Class 6 Maths Chapter 11 Ratio Proportion Set 28 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 11 Ratio Proportion Set 28 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.

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