Working with Fractions — NCERT Solutions

Formula used (Brahmagupta's division rule): $\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c} = \frac{a \times d}{b \times c}$

1. $3 \div \frac{7}{9} = 3 \times \frac{9}{7} = \frac{27}{7} = 3\frac{6}{7}$

2. $\frac{14}{4} \div 2 = \frac{14}{4} \times \frac{1}{2} = \frac{14}{8} = \frac{7}{4} = 1\frac{3}{4}$

3. $\frac{2}{3} \div \frac{2}{3} = \frac{2}{3} \times \frac{3}{2} = \frac{6}{6} = 1$

4. $\frac{14}{6} \div \frac{7}{3} = \frac{14}{6} \times \frac{3}{7} = \frac{14 \times 3}{6 \times 7} = \frac{42}{42} = 1$

5. $\frac{4}{3} \div \frac{3}{4} = \frac{4}{3} \times \frac{4}{3} = \frac{16}{9} = 1\frac{7}{9}$

6. $\frac{7}{4} \div \frac{1}{7} = \frac{7}{4} \times \frac{7}{1} = \frac{49}{4} = 12\frac{1}{4}$

7. $\frac{8}{2} \div \frac{4}{15} = \frac{8}{2} \times \frac{15}{4} = \frac{8 \times 15}{2 \times 4} = \frac{120}{8} = 15$

8. $\frac{1}{5} \div \frac{1}{9} = \frac{1}{5} \times \frac{9}{1} = \frac{9}{5} = 1\frac{4}{5}$

9. $\frac{1}{6} \div \frac{11}{12} = \frac{1}{6} \times \frac{12}{11} = \frac{12}{66} = \frac{2}{11}$

10. $3\frac{2}{3} \div 1\frac{3}{8}$: Convert to improper fractions: $3\frac{2}{3} = \frac{11}{3}$; $1\frac{3}{8} = \frac{11}{8}$.
$$\frac{11}{3} \div \frac{11}{8} = \frac{11}{3} \times \frac{8}{11} = \frac{8}{3} = 2\frac{2}{3}$$

(a) We need to find how many $\frac{1}{4}$ m pieces are in 8 m, which means dividing 8 by $\frac{1}{4}$.

Correct option: (iii) $8 \div \frac{1}{4}$

$$8 \div \frac{1}{4} = 8 \times 4 = 32 \text{ bags}$$

(b) We need to find the length of ribbon per badge when $\frac{1}{2}$ m is divided among 8 badges.

Correct option: (iv) $\frac{1}{2} \div 8$

$$\frac{1}{2} \div 8 = \frac{1}{2} \times \frac{1}{8} = \frac{1}{16} \text{ m per badge}$$

(c) We need to find how many $\frac{1}{6}$ kg portions are in 5 kg, which means dividing 5 by $\frac{1}{6}$.

Correct option: (iii) $5 \div \frac{1}{6}$

$$5 \div \frac{1}{6} = 5 \times 6 = 30 \text{ loaves}$$

Given: $\frac{1}{4}$ kg flour makes 12 rotis.

Flour per roti:
$$\frac{1}{4} \div 12 = \frac{1}{4} \times \frac{1}{12} = \frac{1}{48} \text{ kg}$$

Flour for 6 rotis:
$$6 \times \frac{1}{48} = \frac{6}{48} = \frac{1}{8} \text{ kg}$$

Alternatively: 6 rotis is half of 12 rotis, so flour needed $= \frac{1}{2} \times \frac{1}{4} = \frac{1}{8}$ kg.

Answer: $\frac{1}{8}$ kg of flour is used to make 6 rotis.

Evaluating each term:

$$1 \div \frac{1}{6} = 1 \times 6 = 6$$

$$1 \div \frac{1}{10} = 1 \times 10 = 10$$

$$1 \div \frac{1}{13} = 1 \times 13 = 13$$

$$1 \div \frac{1}{9} = 1 \times 9 = 9$$

$$1 \div \frac{1}{2} = 1 \times 2 = 2$$

Sum:
$$6 + 10 + 13 + 9 + 2 = 40$$

Answer: The friend should say the sum is $40$.

Given: Total pages = 400.

Pages read yesterday:
$$\frac{1}{5} \times 400 = 80 \text{ pages}$$

Pages read today:
$$\frac{3}{10} \times 400 = 120 \text{ pages}$$

Total pages read:
$$80 + 120 = 200 \text{ pages}$$

Pages remaining:
$$400 - 200 = 200 \text{ pages}$$

Answer: Mira needs to read $200$ more pages to finish the novel.

Given: Mileage = 16 km per litre; Petrol = $2\frac{3}{4} = \frac{11}{4}$ litres.

Distance:
$$16 \times \frac{11}{4} = \frac{176}{4} = 44 \text{ km}$$

Answer: The car will go $44$ km using $2\frac{3}{4}$ litres of petrol.

Given: Train time $= 5\frac{1}{6} = \frac{31}{6}$ hours; Plane time $= \frac{1}{2}$ hour.

Time saved by plane:
$$\frac{31}{6} - \frac{1}{2} = \frac{31}{6} - \frac{3}{6} = \frac{28}{6} = \frac{14}{3} = 4\frac{2}{3} \text{ hours}$$

Answer: The plane saves $4\frac{2}{3}$ hours.

Given: Mariam and cousins ate $\frac{4}{5}$ of the cake.

Remaining cake:
$$1 - \frac{4}{5} = \frac{1}{5}$$

Each friend's share (shared equally among 3 friends):
$$\frac{1}{5} \div 3 = \frac{1}{5} \times \frac{1}{3} = \frac{1}{15}$$

Answer: Each of Mariam's three friends got $\frac{1}{15}$ of the cake.

Analysis of the fractions:

$\frac{565}{465} > 1$ (since numerator > denominator)

$\frac{707}{676} > 1$ (since numerator > denominator)

Key principle: When both fractions are greater than 1, their product is greater than each individual fraction.

• Since $\frac{707}{676} > 1$: Product $= \frac{565}{465} \times \frac{707}{676} > \frac{565}{465}$ ✓
• Since $\frac{565}{465} > 1$: Product $= \frac{565}{465} \times \frac{707}{676} > \frac{707}{676}$ ✓
• Both fractions are greater than 1, so their product is also greater than 1. ✓

Correct options: (a), (c), and (e)

The product is greater than $\frac{565}{465}$, greater than $\frac{707}{676}$, and greater than 1.

Note: The figure is not available in the OCR text. However, based on the context of the chapter (fraction multiplication using unit squares), a typical such figure shows a square divided into equal parts with some parts shaded.

General approach:

• Count the total number of equal parts the square is divided into (= total parts).
• Count the number of shaded parts (= shaded parts).
• Fraction shaded $= \dfrac{\text{number of shaded parts}}{\text{total number of parts}}$.

For example, if the square is divided into 16 equal parts and 6 are shaded:
$$\text{Fraction shaded} = \frac{6}{16} = \frac{3}{8}$$

(Students should apply this method to the actual figure in their textbook.)

Note: The exact figure (Fig. 8.7) is not available in the OCR text. Based on the typical version of this problem in the textbook, the ants split equally at each branching point.

General approach: At each split point, the group divides into 2 equal parts. If there are $n$ split points along the path to a food source, the fraction reaching that source is $\frac{1}{2^n}$.

For example, if the ants split at 3 points before reaching the mango tree:
$$\text{Fraction at mango tree} = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}$$

(Students should trace the path in Fig. 8.7 in their textbook and multiply $\frac{1}{2}$ for each equal split along the path to each food source.)

Step 1: $1 - \frac{1}{2} = \frac{1}{2}$

Step 2: $\left(1 - \frac{1}{2}\right) \times \left(1 - \frac{1}{3}\right) = \frac{1}{2} \times \frac{2}{3} = \frac{2}{6} = \frac{1}{3}$

Step 3: $\left(1 - \frac{1}{2}\right) \times \left(1 - \frac{1}{3}\right) \times \left(1 - \frac{1}{4}\right) \times \left(1 - \frac{1}{5}\right)$
$$= \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \frac{4}{5} = \frac{1 \times 2 \times 3 \times 4}{2 \times 3 \times 4 \times 5} = \frac{1}{5}$$

Step 4: $\left(1 - \frac{1}{2}\right) \times \left(1 - \frac{1}{3}\right) \times \cdots \times \left(1 - \frac{1}{10}\right)$
$$= \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \frac{4}{5} \times \frac{5}{6} \times \frac{6}{7} \times \frac{7}{8} \times \frac{8}{9} \times \frac{9}{10}$$

This is a telescoping product — every numerator cancels with the previous denominator:
$$= \frac{1}{10}$$

General Statement:
$$\left(1 - \frac{1}{2}\right) \times \left(1 - \frac{1}{3}\right) \times \left(1 - \frac{1}{4}\right) \times \cdots \times \left(1 - \frac{1}{n}\right) = \frac{1}{n}$$

Explanation: Each factor $\left(1 - \frac{1}{k}\right) = \frac{k-1}{k}$. The product becomes:
$$\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \cdots \times \frac{n-1}{n}$$
In this telescoping product, all intermediate terms cancel, leaving $\frac{1}{n}$.