NCERT Solutions for Class 6 Maths: Chapter 04 Operations on Fractions Set 13
Explore reliable textbook solutions for Chapter 04 Operations on Fractions Set 13 tailored for Class 6 learners. Utilizing these Maths answers ensures thorough preparation and strengthens foundational knowledge before final MSBSHSE evaluations.
Practice Class 6 Maths Solutions: Chapter 04 Operations on Fractions Set 13
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Question 1. Write the reciprocals of the following numbers:
1. 7
2. \( \frac{11}{3} \)
3. \( \frac{5}{13} \)
4. 2
5. \( \frac{6}{7} \)
Answer:
1. \( \frac{1}{7} \)
2. \( \frac{3}{11} \)
3. \( \frac{13}{5} \)
4. \( \frac{1}{2} \)
5. \( \frac{7}{6} \)
In simple words: The reciprocal of a number is 1 divided by that number, or for a fraction, it's flipping the numerator and denominator. When you multiply a number by its reciprocal, the result is 1.
🎯 Exam Tip: Remember to express mixed numbers as improper fractions before finding their reciprocals to avoid common errors.
Question 2. Carry out the following Divisions:
(i) \( \frac{2}{3} \div \frac{1}{4} \)
(ii) \( \frac{5}{9} \div \frac{3}{2} \)
(iii) \( \frac{3}{7} \div \frac{5}{11} \)
(iv) \( \frac{11}{12} \div \frac{4}{7} \)
Answer:
(i) \( \frac{2}{3} \div \frac{1}{4} = \frac{2}{3} \times \frac{4}{1} = \frac{2 \times 4}{3 \times 1} = \frac{8}{3} \)
(ii) \( \frac{5}{9} \div \frac{3}{2} = \frac{5}{9} \times \frac{2}{3} = \frac{5 \times 2}{9 \times 3} = \frac{10}{27} \)
(iii) \( \frac{3}{7} \div \frac{5}{11} = \frac{3}{7} \times \frac{11}{5} = \frac{3 \times 11}{7 \times 5} = \frac{33}{35} \)
(iv) \( \frac{11}{12} \div \frac{4}{7} = \frac{11}{12} \times \frac{7}{4} = \frac{11 \times 7}{12 \times 4} = \frac{77}{48} \)
In simple words: To divide a fraction by another fraction, you multiply the first fraction by the reciprocal of the second fraction. This changes the division problem into a multiplication problem, which is easier to solve.
🎯 Exam Tip: Always remember to invert the second fraction (the divisor) and then multiply. Simplify the resulting fraction if possible.
Question 3. There were 420 students participating in the Swachh Bharat Campaign. They cleaned \( \frac{42}{75} \) part of the town, Sevagram. What part of Sevagram did each student clean if the work was equally shared by all?
Answer:
Solution:
Total number of students = 420
Part of town cleaned by all the students = \( \frac{42}{75} \)
\( \therefore \) Part of town cleaned by one student
\( = \frac{42}{75} \div 420 \)
\( = \frac{42}{75} \times \frac{1}{420} \)
\( = \frac{42 \times 1}{75 \times 420} \)
\( = \frac{1 \times 1}{75 \times 10} \)
\( = \frac{1}{750} \)
\( \therefore \) Part of town cleaned by one student is \( \frac{1}{750} \)
In simple words: To find the share of work for each student, divide the total part of the town cleaned by the total number of students. This simplifies the fraction to show how much each student contributed.
🎯 Exam Tip: For word problems involving fractions and sharing, clearly identify the total quantity and the number of parts it's being divided into. Show all steps of division and simplification.
Maharashtra Board Class 6 Maths Chapter 4 Operations On Fractions Practice Set 13 Intext Questions And Activities
Question 1. Ramanujan's Magic square. (Textbook pg. no. 28)
| 22 | 12 | 18 | 87 |
| 88 | 17 | 9 | 25 |
| 10 | 24 | 89 | 16 |
| 19 | 86 | 23 | 11 |
• Add the four numbers in the rows, the columns and along the diagonals of this square.
• What is the sum?
• Is it the same every time?
• What is the peculiarity?
• Look at the numbers in the first row, 22 - 12 - 1887. Find out why this date is special.
Obtain and read a biography of the great Indian mathematician Srinivasa Ramanujan.
Answer:
Sum of the numbers in each row:
(i) 22 + 12 + 18 + 87 = 139
(ii) 88 + 17 + 9 + 25 = 139
(iii) 10 + 24 + 89 + 16 = 139
(iv) 19 + 86 + 23 + 11 = 139
Sum of the numbers along the diagonals:
(i) 22 + 17 + 89 + 11 = 139
(ii) 87 + 9 + 24 + 19 = 139
Sum of the numbers in each column:
(i) 22 + 88 + 10 + 19 = 139
(ii) 12 + 17 + 24 + 86 = 139
(iii) 18 + 9 + 89 + 23 = 139
(iv) 87 + 25 + 16 + 11 = 139
\( \therefore \) We observe that the sum of the numbers in each of the rows, the columns and along each diagonal remains the same every time. The numbers in the first row 22 - 12 - 1887 is the birth date of Srinivasa Ramanujan.
In simple words: Ramanujan's magic square is a special grid where the sum of numbers in every row, column, and both main diagonals is the same. The first row's numbers also encode a significant date, which is Srinivasa Ramanujan's birth date.
🎯 Exam Tip: When analyzing magic squares, systematically check all rows, columns, and diagonals to verify the constant sum. Pay attention to any special numbers or dates mentioned as they often reveal unique properties of the square.
Step-by-Step Textbook Answers: Class 6 Maths Chapter 04 Operations on Fractions Set 13
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