CBSE Class 7 Mathematics Chapter 2: Fractions and Decimals NCERT Solutions

Solution:

To solve these problems, we perform addition or subtraction of fractions. When fractions have different denominators, we find a common denominator (Least Common Multiple - LCM) before performing the operation.

1. To find $2 - \frac{3}{5}$, we first express 2 as a fraction with denominator 5: $2 = \frac{2 \times 5}{5} = \frac{10}{5}$. Now, subtract: $\frac{10}{5} - \frac{3}{5} = \frac{10 - 3}{5} = \frac{7}{5}$. Converting this improper fraction to a mixed number gives $1\frac{2}{5}$.

2. To find $4 + \frac{7}{8}$, express 4 as a fraction with denominator 8: $4 = \frac{4 \times 8}{8} = \frac{32}{8}$. Now, add: $\frac{32}{8} + \frac{7}{8} = \frac{32 + 7}{8} = \frac{39}{8}$. Converting this improper fraction to a mixed number gives $4\frac{7}{8}$.

3. To find $\frac{3}{5} + \frac{2}{7}$, the LCM of 5 and 7 is 35. Convert the fractions: $\frac{3}{5} = \frac{3 \times 7}{5 \times 7} = \frac{21}{35}$ and $\frac{2}{7} = \frac{2 \times 5}{7 \times 5} = \frac{10}{35}$. Add the numerators: $\frac{21}{35} + \frac{10}{35} = \frac{21 + 10}{35} = \frac{31}{35}$.

4. To find $\frac{9}{11} - \frac{4}{15}$, the LCM of 11 and 15 is 165. Convert the fractions: $\frac{9}{11} = \frac{9 \times 15}{11 \times 15} = \frac{135}{165}$ and $\frac{4}{15} = \frac{4 \times 11}{15 \times 11} = \frac{44}{165}$. Subtract the numerators: $\frac{135}{165} - \frac{44}{165} = \frac{135 - 44}{165} = \frac{91}{165}$.

5. To find $\frac{7}{10} + \frac{2}{5} + \frac{3}{2}$, the LCM of 10, 5, and 2 is 10. Convert the fractions: $\frac{2}{5} = \frac{2 \times 2}{5 \times 2} = \frac{4}{10}$ and $\frac{3}{2} = \frac{3 \times 5}{2 \times 5} = \frac{15}{10}$. Add the numerators: $\frac{7}{10} + \frac{4}{10} + \frac{15}{10} = \frac{7 + 4 + 15}{10} = \frac{26}{10}$. Simplify the result: $\frac{26}{10} = \frac{13}{5}$. Converting to a mixed number gives $2\frac{3}{5}$.

6. To find $2\frac{2}{3} + 3\frac{1}{2}$, first convert the mixed fractions to improper fractions: $2\frac{2}{3} = \frac{(2 \times 3) + 2}{3} = \frac{8}{3}$ and $3\frac{1}{2} = \frac{(3 \times 2) + 1}{2} = \frac{7}{2}$. The LCM of 3 and 2 is 6. Convert the fractions: $\frac{8}{3} = \frac{8 \times 2}{3 \times 2} = \frac{16}{6}$ and $\frac{7}{2} = \frac{7 \times 3}{2 \times 3} = \frac{21}{6}$. Add the numerators: $\frac{16}{6} + \frac{21}{6} = \frac{16 + 21}{6} = \frac{37}{6}$. Converting to a mixed number gives $6\frac{1}{6}$.

7. To find $8\frac{1}{2} - 3\frac{5}{8}$, first convert the mixed fractions to improper fractions: $8\frac{1}{2} = \frac{(8 \times 2) + 1}{2} = \frac{17}{2}$ and $3\frac{5}{8} = \frac{(3 \times 8) + 5}{8} = \frac{29}{8}$. The LCM of 2 and 8 is 8. Convert the fractions: $\frac{17}{2} = \frac{17 \times 4}{2 \times 4} = \frac{68}{8}$. Subtract the numerators: $\frac{68}{8} - \frac{29}{8} = \frac{68 - 29}{8} = \frac{39}{8}$. Converting to a mixed number gives $4\frac{7}{8}$.

Solution:

To arrange fractions in descending order, we need to compare them. This is easiest when they have a common denominator. We find the Least Common Multiple (LCM) of the denominators and convert each fraction to an equivalent fraction with that LCM.

(i) For the fractions $\frac{2}{9}, \frac{2}{3}, \frac{8}{21}$:

The denominators are 9, 3, and 21. The LCM of 9, 3, and 21 is 63.

Convert each fraction to an equivalent fraction with a denominator of 63:

• $\frac{2}{9} = \frac{2 \times 7}{9 \times 7} = \frac{14}{63}$
• $\frac{2}{3} = \frac{2 \times 21}{3 \times 21} = \frac{42}{63}$
• $\frac{8}{21} = \frac{8 \times 3}{21 \times 3} = \frac{24}{63}$

Now, arrange these equivalent fractions in descending order based on their numerators: $\frac{42}{63} > \frac{24}{63} > \frac{14}{63}$.

Therefore, the original fractions in descending order are: $\frac{2}{3} > \frac{8}{21} > \frac{2}{9}$.

(ii) For the fractions $\frac{1}{5}, \frac{3}{7}, \frac{7}{10}$:

The denominators are 5, 7, and 10. The LCM of 5, 7, and 10 is 70.

Convert each fraction to an equivalent fraction with a denominator of 70:

• $\frac{1}{5} = \frac{1 \times 14}{5 \times 14} = \frac{14}{70}$
• $\frac{3}{7} = \frac{3 \times 10}{7 \times 10} = \frac{30}{70}$
• $\frac{7}{10} = \frac{7 \times 7}{10 \times 7} = \frac{49}{70}$

Now, arrange these equivalent fractions in descending order based on their numerators: $\frac{49}{70} > \frac{30}{70} > \frac{14}{70}$.

Therefore, the original fractions in descending order are: $\frac{7}{10} > \frac{3}{7} > \frac{1}{5}$.

Solution:

To determine if this is a magic square, we need to calculate the sum of numbers for each row, each column, and both diagonals. If all these sums are equal to the sum of the first row ($\frac{15}{11}$), then it is a magic square.

Sum of Rows:

• Row 1: $\frac{4}{11} + \frac{9}{11} + \frac{2}{11} = \frac{4+9+2}{11} = \frac{15}{11}$ (Given)
• Row 2: $\frac{3}{11} + \frac{5}{11} + \frac{1}{11} = \frac{3+5+1}{11} = \frac{9}{11}$
• Row 3: $\frac{8}{11} + \frac{1}{11} + \frac{6}{11} = \frac{8+1+6}{11} = \frac{15}{11}$

Since the sum of Row 2 ($\frac{9}{11}$) is not equal to the sum of Row 1 ($\frac{15}{11}$), this square is not a magic square. However, let's complete the calculations for all sums to be thorough.

Sum of Columns:

• Column 1: $\frac{4}{11} + \frac{3}{11} + \frac{8}{11} = \frac{4+3+8}{11} = \frac{15}{11}$
• Column 2: $\frac{9}{11} + \frac{5}{11} + \frac{1}{11} = \frac{9+5+1}{11} = \frac{15}{11}$
• Column 3: $\frac{2}{11} + \frac{1}{11} + \frac{6}{11} = \frac{2+1+6}{11} = \frac{9}{11}$

Column 3 sum ($\frac{9}{11}$) is also not equal to $\frac{15}{11}$.

Sum of Diagonals:

• Diagonal 1 (top-left to bottom-right): $\frac{4}{11} + \frac{5}{11} + \frac{6}{11} = \frac{4+5+6}{11} = \frac{15}{11}$
• Diagonal 2 (top-right to bottom-left): $\frac{2}{11} + \frac{5}{11} + \frac{8}{11} = \frac{2+5+8}{11} = \frac{15}{11}$

Since the sums of Row 2 and Column 3 are $\frac{9}{11}$, which is different from the sum of Row 1, Row 3, Column 1, Column 2, and both diagonals (all $\frac{15}{11}$), this is not a magic square.