Solutions: Section 7.8

Page 179: Addition

Q1. Brahmagupta’s Method:

• a. $\frac{2}{7} + \frac{5}{7} + \frac{6}{7} = \frac{13}{7}$
• b. $\frac{3}{4} + \frac{1}{3} = \frac{9}{12} + \frac{4}{12} = \frac{13}{12}$
• c. $\frac{2}{3} + \frac{5}{6} = \frac{4}{6} + \frac{5}{6} = \frac{9}{6} = \frac{3}{2}$

Q2. Paint Mixture: $\frac{2}{3} + \frac{3}{4} = \frac{8}{12} + \frac{9}{12} = \frac{17}{12}$ litres ($1 \frac{5}{12}$ litres).

Q3. Lace: Geeta $\frac{2}{5}$, Shamim $\frac{3}{4}$. Total: $\frac{2}{5} + \frac{3}{4} = \frac{8}{20} + \frac{15}{20} = \frac{23}{20}$ meters. Since $\frac{23}{20} > 1$, yes, it is sufficient to cover 1 meter.

Page 181: Subtraction (Simple)

Q1. $\frac{5}{8} - \frac{3}{8} = \frac{2}{8} = \frac{1}{4}$ Q2. $\frac{7}{9} - \frac{5}{9} = \frac{2}{9}$ Q3. $\frac{10}{27} - \frac{1}{27} = \frac{9}{27} = \frac{1}{3}$

Page 182: Subtraction (Brahmagupta)

Q1.

• a. $\frac{8}{15} - \frac{3}{15} = \frac{5}{15} = \frac{1}{3}$
• b. $\frac{2}{5} - \frac{4}{15} = \frac{6}{15} - \frac{4}{15} = \frac{2}{15}$
• d. $\frac{2}{3} - \frac{1}{2} = \frac{4}{6} - \frac{3}{6} = \frac{1}{6}$

Q2. Subtract as indicated:

• a. $\frac{13}{4}$ from $\frac{10}{3}$: $\frac{10}{3} - \frac{13}{4} = \frac{40}{12} - \frac{39}{12} = \frac{1}{12}$
• b. $\frac{18}{5}$ from $\frac{23}{3}$: $\frac{23}{3} - \frac{18}{5} = \frac{115}{15} - \frac{54}{15} = \frac{61}{15}$

Q3. Problems:

• a. Jaya’s walk: Total $\frac{7}{10}$. Auto $\frac{1}{2}$. Walk = $\frac{7}{10} - \frac{1}{2} = \frac{7}{10} - \frac{5}{10} = \frac{2}{10} = \frac{1}{5}$ km.
• b. Jeevika vs Namit: Jeevika: $\frac{10}{3} = \frac{40}{12}$. Namit: $\frac{13}{4} = \frac{39}{12}$. $\frac{39}{12} < \frac{40}{12}$, so Namit takes less time. Difference: $\frac{1}{12}$ minute.