Get the most accurate RBSE Solutions for Class 7 Mathematics Chapter 2 Fractions and Decimal Numbers here. Updated for the 2026-27 academic session, these solutions are based on the latest RBSE textbooks for Class 7 Mathematics. Our expert-created answers for Class 7 Mathematics are available for free download in PDF format.
Detailed Chapter 2 Fractions and Decimal Numbers RBSE Solutions for Class 7 Mathematics
For Class 7 students, solving RBSE textbook questions is the most effective way to build a strong conceptual foundation. Our Class 7 Mathematics solutions follow a detailed, step-by-step approach to ensure you understand the logic behind every answer. Practicing these Chapter 2 Fractions and Decimal Numbers solutions will improve your exam performance.
Class 7 Mathematics Chapter 2 Fractions and Decimal Numbers RBSE Solutions PDF
Rajasthan Board RBSE Class 7 Maths Chapter 2 Fractions And Decimal Numbers Ex 2.1
Question 1. Find 5 equivalent fractions of each fraction given below
(i) \( \frac{2}{8} \)
(ii) \( \frac{6}{7} \)
(iii) \( \frac{7}{4} \)
(iv) \( \frac{100}{45} \)
Answer:
(i) For \( \frac{2}{8} \):
Equivalent fraction: \( \frac{2}{8} \times \frac{2}{2} = \frac{4}{16} \)
Equivalent fraction: \( \frac{2}{8} \times \frac{3}{3} = \frac{6}{24} \)
Equivalent fraction: \( \frac{2}{8} \times \frac{4}{4} = \frac{8}{32} \)
Equivalent fraction: \( \frac{2}{8} \times \frac{5}{5} = \frac{10}{40} \)
Equivalent fraction: \( \frac{2}{8} \times \frac{6}{6} = \frac{12}{48} \)
So, five equivalent fractions for \( \frac{2}{8} \) are \( \frac{4}{16}, \frac{6}{24}, \frac{8}{32}, \frac{10}{40}, \frac{12}{48} \).
(ii) For \( \frac{6}{7} \):
Equivalent fraction: \( \frac{6}{7} \times \frac{2}{2} = \frac{12}{14} \)
Equivalent fraction: \( \frac{6}{7} \times \frac{3}{3} = \frac{18}{21} \)
Equivalent fraction: \( \frac{6}{7} \times \frac{4}{4} = \frac{24}{28} \)
Equivalent fraction: \( \frac{6}{7} \times \frac{5}{5} = \frac{30}{35} \)
Equivalent fraction: \( \frac{6}{7} \times \frac{6}{6} = \frac{36}{42} \)
So, five equivalent fractions for \( \frac{6}{7} \) are \( \frac{12}{14}, \frac{18}{21}, \frac{24}{28}, \frac{30}{35}, \frac{36}{42} \). It's helpful to remember that multiplying the numerator and denominator by the same non-zero number creates an equivalent fraction.
(iii) For \( \frac{7}{4} \):
Equivalent fraction: \( \frac{7}{4} \times \frac{2}{2} = \frac{14}{8} \)
Equivalent fraction: \( \frac{7}{4} \times \frac{3}{3} = \frac{21}{12} \)
Equivalent fraction: \( \frac{7}{4} \times \frac{4}{4} = \frac{28}{16} \)
Equivalent fraction: \( \frac{7}{4} \times \frac{5}{5} = \frac{35}{20} \)
Equivalent fraction: \( \frac{7}{4} \times \frac{6}{6} = \frac{42}{24} \)
So, five equivalent fractions for \( \frac{7}{4} \) are \( \frac{14}{8}, \frac{21}{12}, \frac{28}{16}, \frac{35}{20}, \frac{42}{24} \).
(iv) For \( \frac{100}{45} \):
First, simplify the fraction by dividing by 5: \( \frac{100 \div 5}{45 \div 5} = \frac{20}{9} \).
Now, find equivalent fractions for \( \frac{20}{9} \):
Equivalent fraction: \( \frac{20}{9} \times \frac{2}{2} = \frac{40}{18} \)
Equivalent fraction: \( \frac{20}{9} \times \frac{3}{3} = \frac{60}{27} \)
Equivalent fraction: \( \frac{20}{9} \times \frac{4}{4} = \frac{80}{36} \)
Equivalent fraction: \( \frac{20}{9} \times \frac{5}{5} = \frac{100}{45} \) (This is the original fraction, so we need to continue)
Equivalent fraction: \( \frac{20}{9} \times \frac{6}{6} = \frac{120}{54} \)
So, five equivalent fractions for \( \frac{100}{45} \) are \( \frac{20}{9}, \frac{40}{18}, \frac{60}{27}, \frac{80}{36}, \frac{120}{54} \).
In simple words: To find equivalent fractions, multiply both the top number (numerator) and the bottom number (denominator) by the same whole number. You can pick any whole number except zero.
🎯 Exam Tip: Always make sure to multiply both the numerator and the denominator by the exact same number to keep the fraction's value the same.
Question 2. Use >, < and = for comparison of the following:
(i) \( \frac{3}{7} \text{ } \square \text{ } \frac{2}{5} \)
(ii) \( \frac{6}{8} \text{ } \square \text{ } \frac{12}{16} \)
(iii) \( \frac{11}{15} \text{ } \square \text{ } \frac{12}{17} \)
(iv) \( \frac{15}{40} \text{ } \square \text{ } \frac{3}{8} \)
Answer:
(i) Compare \( \frac{3}{7} \) and \( \frac{2}{5} \)
First, find the Least Common Multiple (LCM) of the denominators 7 and 5. The LCM of 7 and 5 is \( 7 \times 5 = 35 \).
Convert each fraction to an equivalent fraction with a denominator of 35.
For \( \frac{3}{7} \): Multiply the numerator and denominator by 5.
\( \frac{3}{7} = \frac{3 \times 5}{7 \times 5} = \frac{15}{35} \)
For \( \frac{2}{5} \): Multiply the numerator and denominator by 7.
\( \frac{2}{5} = \frac{2 \times 7}{5 \times 7} = \frac{14}{35} \)
Now, compare the fractions: \( \frac{15}{35} \) and \( \frac{14}{35} \).
Since \( 15 > 14 \), we have \( \frac{15}{35} > \frac{14}{35} \).
Therefore, \( \frac{3}{7} > \frac{2}{5} \).
(ii) Compare \( \frac{6}{8} \) and \( \frac{12}{16} \)
First, simplify both fractions if possible.
For \( \frac{6}{8} \): Divide the numerator and denominator by 2. \( \frac{6 \div 2}{8 \div 2} = \frac{3}{4} \)
For \( \frac{12}{16} \): Divide the numerator and denominator by 4. \( \frac{12 \div 4}{16 \div 4} = \frac{3}{4} \)
Since both fractions simplify to \( \frac{3}{4} \), they are equal.
Therefore, \( \frac{6}{8} = \frac{12}{16} \). Equal fractions have the same value, even if they look different.
(iii) Compare \( \frac{11}{15} \) and \( \frac{12}{17} \)
First, find the LCM of the denominators 15 and 17. The LCM of 15 and 17 is \( 15 \times 17 = 255 \).
Convert each fraction to an equivalent fraction with a denominator of 255.
For \( \frac{11}{15} \): Multiply the numerator and denominator by 17.
\( \frac{11}{15} = \frac{11 \times 17}{15 \times 17} = \frac{187}{255} \)
For \( \frac{12}{17} \): Multiply the numerator and denominator by 15.
\( \frac{12}{17} = \frac{12 \times 15}{17 \times 15} = \frac{180}{255} \)
Now, compare the fractions: \( \frac{187}{255} \) and \( \frac{180}{255} \).
Since \( 187 > 180 \), we have \( \frac{187}{255} > \frac{180}{255} \).
Therefore, \( \frac{11}{15} > \frac{12}{17} \).
(iv) Compare \( \frac{15}{40} \) and \( \frac{3}{8} \)
First, simplify the fraction \( \frac{15}{40} \). Divide the numerator and denominator by 5.
\( \frac{15 \div 5}{40 \div 5} = \frac{3}{8} \)
Since \( \frac{15}{40} \) simplifies to \( \frac{3}{8} \), they are equal.
Therefore, \( \frac{15}{40} = \frac{3}{8} \).
In simple words: To compare fractions, make their bottom numbers (denominators) the same using the LCM. Then, compare their top numbers (numerators). The fraction with the bigger numerator is larger. Sometimes, you can simplify the fractions first to make comparison easier.
🎯 Exam Tip: Always look to simplify fractions before finding the LCM, as this can make calculations much easier and reduce the chance of errors.
Question 3. Arrange the following in ascending order:
(i) \( \frac{1}{5}, \frac{3}{7}, \frac{7}{10} \)
(ii) \( \frac{2}{9}, \frac{2}{3}, \frac{8}{21} \)
Answer:
(i) Arrange \( \frac{1}{5}, \frac{3}{7}, \frac{7}{10} \) in ascending order.
First, find the LCM of the denominators 5, 7, and 10.
LCM of 5, 7, 10:
\( 5 = 5 \)
\( 7 = 7 \)
\( 10 = 2 \times 5 \)
LCM = \( 2 \times 5 \times 7 = 70 \).
Now, convert each fraction to an equivalent fraction with a denominator of 70.
For \( \frac{1}{5} \): Multiply numerator and denominator by 14 (since \( 70 \div 5 = 14 \)).
\( \frac{1}{5} = \frac{1 \times 14}{5 \times 14} = \frac{14}{70} \)
For \( \frac{3}{7} \): Multiply numerator and denominator by 10 (since \( 70 \div 7 = 10 \)).
\( \frac{3}{7} = \frac{3 \times 10}{7 \times 10} = \frac{30}{70} \)
For \( \frac{7}{10} \): Multiply numerator and denominator by 7 (since \( 70 \div 10 = 7 \)).
\( \frac{7}{10} = \frac{7 \times 7}{10 \times 7} = \frac{49}{70} \)
Now, arrange the equivalent fractions in ascending order (smallest to largest) by comparing their numerators:
\( \frac{14}{70}, \frac{30}{70}, \frac{49}{70} \)
So, the ascending order for the original fractions is: \( \frac{1}{5} < \frac{3}{7} < \frac{7}{10} \).
(ii) Arrange \( \frac{2}{9}, \frac{2}{3}, \frac{8}{21} \) in ascending order.
First, find the LCM of the denominators 9, 3, and 21.
LCM of 9, 3, 21:
\( 9 = 3 \times 3 \)
\( 3 = 3 \)
\( 21 = 3 \times 7 \)
LCM = \( 3 \times 3 \times 7 = 63 \).
Now, convert each fraction to an equivalent fraction with a denominator of 63.
For \( \frac{2}{9} \): Multiply numerator and denominator by 7 (since \( 63 \div 9 = 7 \)).
\( \frac{2}{9} = \frac{2 \times 7}{9 \times 7} = \frac{14}{63} \)
For \( \frac{2}{3} \): Multiply numerator and denominator by 21 (since \( 63 \div 3 = 21 \)).
\( \frac{2}{3} = \frac{2 \times 21}{3 \times 21} = \frac{42}{63} \)
For \( \frac{8}{21} \): Multiply numerator and denominator by 3 (since \( 63 \div 21 = 3 \)).
\( \frac{8}{21} = \frac{8 \times 3}{21 \times 3} = \frac{24}{63} \)
Now, arrange the equivalent fractions in ascending order (smallest to largest) by comparing their numerators:
\( \frac{14}{63}, \frac{24}{63}, \frac{42}{63} \)
So, the ascending order for the original fractions is: \( \frac{2}{9} < \frac{8}{21} < \frac{2}{3} \).
In simple words: To put fractions in order, first make all their bottom numbers (denominators) the same by finding their LCM. Then, just arrange them by their top numbers (numerators) from smallest to largest for ascending order, or largest to smallest for descending order.
🎯 Exam Tip: Always clearly state the LCM and show the conversion of each fraction to its equivalent form before arranging them. This shows your working steps clearly.
Question 4. Solve:
(i) \( 2 + \frac{3}{5} \)
(ii) \( 4 + \frac{7}{8} \)
(iii) \( \frac{3}{5} + \frac{2}{7} \)
(iv) \( 8\frac{1}{2} - 3\frac{5}{8} \)
(v) \( 2\frac{2}{3} + 3\frac{1}{2} \)
(vi) \( \frac{7}{10} + \frac{2}{5} + \frac{3}{2} \)
Answer:
(i) Solve \( 2 + \frac{3}{5} \)
Write 2 as a fraction: \( 2 = \frac{2}{1} \).
Now add the fractions: \( \frac{2}{1} + \frac{3}{5} \)
Find the LCM of 1 and 5, which is 5.
Convert \( \frac{2}{1} \) to an equivalent fraction with a denominator of 5:
\( \frac{2}{1} = \frac{2 \times 5}{1 \times 5} = \frac{10}{5} \)
Add the fractions: \( \frac{10}{5} + \frac{3}{5} = \frac{10+3}{5} = \frac{13}{5} \)
The answer is \( \frac{13}{5} \). This can also be written as a mixed number: \( 2\frac{3}{5} \).
(ii) Solve \( 4 + \frac{7}{8} \)
Write 4 as a fraction: \( 4 = \frac{4}{1} \).
Now add the fractions: \( \frac{4}{1} + \frac{7}{8} \)
Find the LCM of 1 and 8, which is 8.
Convert \( \frac{4}{1} \) to an equivalent fraction with a denominator of 8:
\( \frac{4}{1} = \frac{4 \times 8}{1 \times 8} = \frac{32}{8} \)
Add the fractions: \( \frac{32}{8} + \frac{7}{8} = \frac{32+7}{8} = \frac{39}{8} \)
The answer is \( \frac{39}{8} \). This can also be written as a mixed number: \( 4\frac{7}{8} \).
(iii) Solve \( \frac{3}{5} + \frac{2}{7} \)
Find the LCM of the denominators 5 and 7. The LCM is \( 5 \times 7 = 35 \).
Convert each fraction to an equivalent fraction with a denominator of 35.
For \( \frac{3}{5} \): \( \frac{3}{5} = \frac{3 \times 7}{5 \times 7} = \frac{21}{35} \)
For \( \frac{2}{7} \): \( \frac{2}{7} = \frac{2 \times 5}{7 \times 5} = \frac{10}{35} \)
Add the fractions: \( \frac{21}{35} + \frac{10}{35} = \frac{21+10}{35} = \frac{31}{35} \)
The answer is \( \frac{31}{35} \).
(iv) Solve \( 8\frac{1}{2} - 3\frac{5}{8} \)
First, convert the mixed numbers to improper fractions.
\( 8\frac{1}{2} = \frac{(8 \times 2) + 1}{2} = \frac{16 + 1}{2} = \frac{17}{2} \)
\( 3\frac{5}{8} = \frac{(3 \times 8) + 5}{8} = \frac{24 + 5}{8} = \frac{29}{8} \)
Now subtract: \( \frac{17}{2} - \frac{29}{8} \)
Find the LCM of the denominators 2 and 8. The LCM is 8.
Convert \( \frac{17}{2} \) to an equivalent fraction with a denominator of 8:
\( \frac{17}{2} = \frac{17 \times 4}{2 \times 4} = \frac{68}{8} \)
Subtract the fractions: \( \frac{68}{8} - \frac{29}{8} = \frac{68-29}{8} = \frac{39}{8} \)
The answer is \( \frac{39}{8} \). This can also be written as a mixed number: \( 4\frac{7}{8} \).
(v) Solve \( 2\frac{2}{3} + 3\frac{1}{2} \)
First, convert the mixed numbers to improper fractions.
\( 2\frac{2}{3} = \frac{(2 \times 3) + 2}{3} = \frac{6 + 2}{3} = \frac{8}{3} \)
\( 3\frac{1}{2} = \frac{(3 \times 2) + 1}{2} = \frac{6 + 1}{2} = \frac{7}{2} \)
Now add: \( \frac{8}{3} + \frac{7}{2} \)
Find the LCM of the denominators 3 and 2. The LCM is \( 3 \times 2 = 6 \).
Convert each fraction to an equivalent fraction with a denominator of 6.
For \( \frac{8}{3} \): \( \frac{8}{3} = \frac{8 \times 2}{3 \times 2} = \frac{16}{6} \)
For \( \frac{7}{2} \): \( \frac{7}{2} = \frac{7 \times 3}{2 \times 3} = \frac{21}{6} \)
Add the fractions: \( \frac{16}{6} + \frac{21}{6} = \frac{16+21}{6} = \frac{37}{6} \)
The answer is \( \frac{37}{6} \). This can also be written as a mixed number: \( 6\frac{1}{6} \). Remember to simplify fractions to their simplest form if possible.
(vi) Solve \( \frac{7}{10} + \frac{2}{5} + \frac{3}{2} \)
Find the LCM of the denominators 10, 5, and 2. The LCM is 10.
Convert each fraction to an equivalent fraction with a denominator of 10.
For \( \frac{7}{10} \): This fraction already has a denominator of 10.
For \( \frac{2}{5} \): Multiply numerator and denominator by 2.
\( \frac{2}{5} = \frac{2 \times 2}{5 \times 2} = \frac{4}{10} \)
For \( \frac{3}{2} \): Multiply numerator and denominator by 5.
\( \frac{3}{2} = \frac{3 \times 5}{2 \times 5} = \frac{15}{10} \)
Add the fractions: \( \frac{7}{10} + \frac{4}{10} + \frac{15}{10} = \frac{7+4+15}{10} = \frac{26}{10} \)
Simplify the answer by dividing both numerator and denominator by 2.
\( \frac{26}{10} = \frac{26 \div 2}{10 \div 2} = \frac{13}{5} \)
The answer is \( \frac{13}{5} \). This can also be written as a mixed number: \( 2\frac{3}{5} \).
In simple words: To add or subtract fractions, you must first make sure they have the same bottom number (denominator). Find the smallest common multiple (LCM) for all denominators and change each fraction to have that LCM as its new bottom number. Then, you can add or subtract the top numbers (numerators) and keep the bottom number the same. Always convert mixed numbers to improper fractions before calculations.
🎯 Exam Tip: When dealing with mixed numbers, always convert them to improper fractions first before performing addition or subtraction. Also, remember to simplify your final answer to its lowest terms.
Question 5. A rectangular photo whose length \( 2\frac{3}{4} \) inch and breadth \( \frac{7}{6} \) inch. Find its perimeter.
Answer:
Length of the rectangular photo \( (l) = 2\frac{3}{4} \) inch.
Convert the mixed number to an improper fraction: \( 2\frac{3}{4} = \frac{(2 \times 4) + 3}{4} = \frac{8+3}{4} = \frac{11}{4} \) inch.
Breadth of the rectangular photo \( (b) = \frac{7}{6} \) inch.
The formula for the perimeter of a rectangle is \( P = 2(l + b) \).
Substitute the values of length and breadth into the formula:
\( P = 2 \left( \frac{11}{4} + \frac{7}{6} \right) \)
First, add the fractions inside the bracket. Find the LCM of the denominators 4 and 6.
Multiples of 4: 4, 8, **12**, 16...
Multiples of 6: 6, **12**, 18...
The LCM of 4 and 6 is 12.
Convert the fractions to equivalent fractions with a denominator of 12:
\( \frac{11}{4} = \frac{11 \times 3}{4 \times 3} = \frac{33}{12} \)
\( \frac{7}{6} = \frac{7 \times 2}{6 \times 2} = \frac{14}{12} \)
Now add the equivalent fractions:
\( \frac{33}{12} + \frac{14}{12} = \frac{33+14}{12} = \frac{47}{12} \)
Now substitute this sum back into the perimeter formula:
\( P = 2 \times \frac{47}{12} \)
Multiply:
\( P = \frac{2 \times 47}{12} = \frac{94}{12} \)
Simplify the fraction by dividing both numerator and denominator by 2:
\( P = \frac{94 \div 2}{12 \div 2} = \frac{47}{6} \)
Convert the improper fraction to a mixed number:
\( \frac{47}{6} = 7 \frac{5}{6} \) inch.
The perimeter of the rectangular photo is \( 7\frac{5}{6} \) inches. Knowing common geometric formulas like perimeter is useful for various real-world applications.
In simple words: To find the perimeter of a rectangle, first turn any mixed numbers into simple fractions. Then, add the length and breadth together, making sure they have the same bottom number. Finally, multiply that sum by 2 to get the total perimeter.
🎯 Exam Tip: Always convert mixed numbers to improper fractions before performing calculations involving addition, subtraction, multiplication, or division. Remember to simplify your final answer and express it in a suitable form (improper or mixed number, as appropriate).
Question 6. Sheela took \( 3\frac{3}{5} \) hours in whitewashing a shop and Necia completed whitewashing of similar shop in \( 3\frac{5}{7} \) hours. Who took more time and how much?
Answer:
Time taken by Sheela = \( 3\frac{3}{5} \) hours.
Convert to an improper fraction: \( \frac{(3 \times 5) + 3}{5} = \frac{15+3}{5} = \frac{18}{5} \) hours.
Time taken by Necia = \( 3\frac{5}{7} \) hours.
Convert to an improper fraction: \( \frac{(3 \times 7) + 5}{7} = \frac{21+5}{7} = \frac{26}{7} \) hours.
To compare the times, we need to find the LCM of the denominators 5 and 7.
The LCM of 5 and 7 is \( 5 \times 7 = 35 \).
Convert Sheela's time to an equivalent fraction with a denominator of 35:
\( \frac{18}{5} = \frac{18 \times 7}{5 \times 7} = \frac{126}{35} \) hours.
Convert Necia's time to an equivalent fraction with a denominator of 35:
\( \frac{26}{7} = \frac{26 \times 5}{7 \times 5} = \frac{130}{35} \) hours.
Now, compare the times: \( \frac{126}{35} \) and \( \frac{130}{35} \).
Since \( 130 > 126 \), Necia took more time.
To find out how much more time, subtract Sheela's time from Necia's time:
Difference in time = \( \frac{130}{35} - \frac{126}{35} = \frac{130-126}{35} = \frac{4}{35} \) hours.
So, Necia took \( \frac{4}{35} \) hours more than Sheela. This problem shows how comparing fractions helps in daily life situations involving time.
In simple words: First, change the mixed numbers for both Sheela's and Necia's time into simple fractions. Then, make sure both fractions have the same bottom number (denominator) by finding their LCM. Compare the top numbers to see who took longer. To find out how much longer, subtract the smaller time from the larger time.
🎯 Exam Tip: When comparing or subtracting mixed numbers, it's safest to convert them to improper fractions first to avoid common errors with borrowing or comparing whole parts separately. Clearly state the LCM and show your steps.
Question 7. Distributing the birthday cake among Reena, Teena and Meena; \( \frac{2}{5} \) part and Teena given \( \frac{1}{3} \) part and remaining part was given to Meena. Find the Meena's share.
Answer:
Part of cake Reena received = \( \frac{2}{5} \)
Part of cake Teena received = \( \frac{1}{3} \)
Let the total cake be represented as 1 whole.
First, find the total share received by Reena and Teena together.
Total share of Reena and Teena = \( \frac{2}{5} + \frac{1}{3} \)
Find the LCM of the denominators 5 and 3. The LCM is \( 5 \times 3 = 15 \).
Convert each fraction to an equivalent fraction with a denominator of 15:
For Reena's share: \( \frac{2}{5} = \frac{2 \times 3}{5 \times 3} = \frac{6}{15} \)
For Teena's share: \( \frac{1}{3} = \frac{1 \times 5}{3 \times 5} = \frac{5}{15} \)
Add their shares: \( \frac{6}{15} + \frac{5}{15} = \frac{6+5}{15} = \frac{11}{15} \)
This means Reena and Teena together received \( \frac{11}{15} \) of the cake.
Meena received the remaining part of the cake. To find Meena's share, subtract the combined share of Reena and Teena from the total cake (1 whole).
Meena's share = \( 1 - \frac{11}{15} \)
Write 1 as a fraction with a denominator of 15: \( 1 = \frac{15}{15} \).
Meena's share = \( \frac{15}{15} - \frac{11}{15} = \frac{15-11}{15} = \frac{4}{15} \)
So, Meena's share of the cake is \( \frac{4}{15} \). This problem illustrates how fractions are used in sharing and division scenarios.
In simple words: To find Meena's share, first add the parts of the cake that Reena and Teena received. Make sure their fractions have the same bottom number before adding. Then, subtract this total from the whole cake (which is 1) to find what is left for Meena.
🎯 Exam Tip: When dealing with parts of a whole, remember that the "whole" is always represented by 1. Always find a common denominator (LCM) before adding or subtracting fractions.
Free study material for Mathematics
RBSE Solutions Class 7 Mathematics Chapter 2 Fractions and Decimal Numbers
Students can now access the RBSE Solutions for Chapter 2 Fractions and Decimal Numbers prepared by teachers on our website. These solutions cover all questions in exercise in your Class 7 Mathematics textbook. Each answer is updated based on the current academic session as per the latest RBSE syllabus.
Detailed Explanations for Chapter 2 Fractions and Decimal Numbers
Our expert teachers have provided step-by-step explanations for all the difficult questions in the Class 7 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 7 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 7 Solved Papers
Using our Mathematics 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 2 Fractions and Decimal Numbers to get a complete preparation experience.
FAQs
The complete and updated RBSE Solutions Class 7 Maths Chapter 2 Fractions and Decimal Numbers Exercise 2.1 is available for free on StudiesToday.com. These solutions for Class 7 Mathematics are as per latest RBSE curriculum.
Yes, our experts have revised the RBSE Solutions Class 7 Maths Chapter 2 Fractions and Decimal Numbers Exercise 2.1 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.
Toppers recommend using RBSE language because RBSE marking schemes are strictly based on textbook definitions. Our RBSE Solutions Class 7 Maths Chapter 2 Fractions and Decimal Numbers Exercise 2.1 will help students to get full marks in the theory paper.
Yes, we provide bilingual support for Class 7 Mathematics. You can access RBSE Solutions Class 7 Maths Chapter 2 Fractions and Decimal Numbers Exercise 2.1 in both English and Hindi medium.
Yes, you can download the entire RBSE Solutions Class 7 Maths Chapter 2 Fractions and Decimal Numbers Exercise 2.1 in printable PDF format for offline study on any device.