Working with Fractions

Given: Tenzin drinks $\frac{1}{2}$ glass of milk every day.

In a week (7 days):
$$7 \times \frac{1}{2} = \frac{7}{2} = 3\frac{1}{2} \text{ glasses}$$

In January (31 days):
$$31 \times \frac{1}{2} = \frac{31}{2} = 15\frac{1}{2} \text{ glasses}$$

Answer: Tenzin drinks $3\frac{1}{2}$ glasses in a week and $15\frac{1}{2}$ glasses in January.

Given: The team makes 1 km in 8 days.

In one day:
$$\frac{1}{8} \text{ km}$$

In 5 days (one week):
$$5 \times \frac{1}{8} = \frac{5}{8} \text{ km}$$

Answer: In one day the team can make $\frac{1}{8}$ km; in a week (5 working days) they can make $\frac{5}{8}$ km of the water canal.

Given: 5 litres shared equally among 3 families.

Oil per family per week:
$$\frac{5}{3} = 1\frac{2}{3} \text{ litres}$$

Oil per family in 4 weeks:
$$4 \times \frac{5}{3} = \frac{20}{3} = 6\frac{2}{3} \text{ litres}$$

Answer: Each family gets $1\frac{2}{3}$ litres per week and $6\frac{2}{3}$ litres in 4 weeks.

Given: Moon sets $\frac{5}{6}$ hour later each day. Monday is the starting day.

Number of days from Monday to Thursday: 3 days.

Total extra hours by Thursday:
$$3 \times \frac{5}{6} = \frac{15}{6} = \frac{5}{2} = 2\frac{1}{2} \text{ hours}$$

Answer: The Moon will set $2\frac{1}{2}$ hours after 10 pm on Thursday, i.e., at 12:30 am.

Formula used: Multiply the whole number with the numerator; keep the denominator the same. Then convert the improper fraction to a mixed fraction.

(a) $7 \times \frac{3}{5} = \frac{7 \times 3}{5} = \frac{21}{5} = 4\frac{1}{5}$

(b) $4 \times \frac{1}{3} = \frac{4 \times 1}{3} = \frac{4}{3} = 1\frac{1}{3}$

(c) $\frac{9}{7} \times 6 = \frac{9 \times 6}{7} = \frac{54}{7} = 7\frac{5}{7}$

(d) $\frac{13}{11} \times 6 = \frac{13 \times 6}{11} = \frac{78}{11} = 7\frac{1}{11}$

Formula used: $\frac{1}{b} \times \frac{1}{d} = \frac{1}{b \times d}$

When a unit square is divided into $b$ rows and $d$ columns, it creates $b \times d$ equal parts, and one such part represents the product.

(a) $\frac{1}{3} \times \frac{1}{5} = \frac{1}{3 \times 5} = \frac{1}{15}$

(b) $\frac{1}{4} \times \frac{1}{3} = \frac{1}{4 \times 3} = \frac{1}{12}$

(c) $\frac{1}{5} \times \frac{1}{2} = \frac{1}{5 \times 2} = \frac{1}{10}$

(d) $\frac{1}{6} \times \frac{1}{5} = \frac{1}{6 \times 5} = \frac{1}{30}$

Now: $\frac{1}{12} \times \frac{1}{18} = \frac{1}{12 \times 18} = \frac{1}{216}$

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

The unit square is divided into rows equal to one denominator and columns equal to the other. The shaded region (numerator rows × numerator columns) gives the product.

(a) $\frac{2}{3} \times \frac{4}{5} = \frac{2 \times 4}{3 \times 5} = \frac{8}{15}$

(b) $\frac{1}{4} \times \frac{2}{3} = \frac{1 \times 2}{4 \times 3} = \frac{2}{12} = \frac{1}{6}$

(c) $\frac{3}{5} \times \frac{1}{2} = \frac{3 \times 1}{5 \times 2} = \frac{3}{10}$

(d) $\frac{4}{6} \times \frac{3}{5} = \frac{4 \times 3}{6 \times 5} = \frac{12}{30} = \frac{2}{5}$

Given: In 1 hour, $\frac{7}{10}$ of the tank is filled.

So in $t$ hours, the fraction filled $= t \times \frac{7}{10}$.

(a) $\frac{1}{3} \times \frac{7}{10} = \frac{7}{30}$ of the tank.

(b) $\frac{2}{3} \times \frac{7}{10} = \frac{14}{30} = \frac{7}{15}$ of the tank.

(c) $\frac{3}{4} \times \frac{7}{10} = \frac{21}{40}$ of the tank.

(d) $\frac{7}{10} \times \frac{7}{10} = \frac{49}{100}$ of the tank.

(e) For the tank to be full, we need the fraction filled $= 1$.
$$t \times \frac{7}{10} = 1 \implies t = 1 \div \frac{7}{10} = \frac{10}{7} = 1\frac{3}{7} \text{ hours}$$

Answer: The tap should run for $1\frac{3}{7}$ hours for the tank to be full.

Step 1: Land taken by government $= \frac{1}{6}$.

Land remaining with Somu $= 1 - \frac{1}{6} = \frac{5}{6}$.

Step 2: Somu gives half of $\frac{5}{6}$ to Krishna:
$$\text{Krishna's share} = \frac{1}{2} \times \frac{5}{6} = \frac{5}{12}$$

Step 3: Somu gives $\frac{1}{3}$ of $\frac{5}{6}$ to Bora:
$$\text{Bora's share} = \frac{1}{3} \times \frac{5}{6} = \frac{5}{18}$$

Step 4: Land Somu keeps:
$$\frac{5}{6} - \frac{5}{12} - \frac{5}{18}$$
LCM of 6, 12, 18 = 36.
$$= \frac{30}{36} - \frac{15}{36} - \frac{10}{36} = \frac{5}{36}$$

(a) Krishna got $\frac{5}{12}$ of the original land.

(b) Bora got $\frac{5}{18}$ of the original land.

(c) Somu kept $\frac{5}{36}$ of the original land for herself.

Given: Length $= 9\frac{3}{5}$ ft $= \frac{48}{5}$ ft; Breadth $= 3\frac{3}{4}$ ft $= \frac{15}{4}$ ft.

Formula: Area $=$ Length $\times$ Breadth.

$$\text{Area} = \frac{48}{5} \times \frac{15}{4}$$

Cancelling common factors: $\frac{48}{4} = 12$ and $\frac{15}{5} = 3$:

$$= 12 \times 3 = 36 \text{ sq ft}$$

Answer: The area of the rectangle is $36$ sq ft.

Given: 4 saplings in a row; distance between consecutive saplings $= \frac{3}{4}$ m.

Concept: With 4 saplings, there are $4 - 1 = 3$ gaps between them.

Distance from first to last sapling:
$$3 \times \frac{3}{4} = \frac{9}{4} = 2\frac{1}{4} \text{ m}$$

Answer: The distance between the first and last sapling is $2\frac{1}{4}$ m.

Calculating each quantity:

First quantity: $\frac{12}{15}$ of 500 g
$$= \frac{12}{15} \times 500 = \frac{12 \times 500}{15} = \frac{6000}{15} = 400 \text{ g}$$

Second quantity: $\frac{3}{20}$ of 4 kg $= \frac{3}{20}$ of 4000 g
$$= \frac{3}{20} \times 4000 = \frac{12000}{20} = 600 \text{ g}$$

Comparison: $600$ g > 400 g.

Answer: $\frac{3}{20}$ of 4 kg (= 600 g) is heavier than $\frac{12}{15}$ of 500 g (= 400 g).

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 $\frac{1}{n}$.