Weight and Capacity

We check each item against its realistic weight:

1. Iron Almirah – 40 g ✗ (An iron almirah is very heavy; it should be around 40 kg, not 40 g.)
2. Bed – 60 kg ✓ (A bed typically weighs around 50–80 kg, so 60 kg is correct.)
3. Rice Bag – 5 kg ✓ (A standard rice bag of 5 kg is correct.)
4. Sofa – 30 g ✗ (A sofa is heavy furniture; it should be around 30 kg, not 30 g.)
5. Bucket – 1 kg 800 g ✓ (A filled or sturdy bucket can weigh around 1 kg 800 g.)
6. Water Bottle – 650 g ✓ (A water bottle filled with water can weigh around 650 g.)
7. Refrigerator – 50 g ✗ (A refrigerator is a heavy appliance; it should be around 50 kg, not 50 g.)

Summary: Items 2, 3, 5, and 6 are correctly recorded (✓). Items 1, 4, and 7 are incorrectly recorded (✗).

Given: Weight of sugar = 5 kg 50 g

Concept: $1 \text{ kg} = 1{,}000 \text{ g}$

Working:
$$5 \text{ kg } 50 \text{ g} = 5 \times 1{,}000 \text{ g} + 50 \text{ g}$$
$$= 5{,}000 \text{ g} + 50 \text{ g} = 5{,}050 \text{ g}$$

Answer: 5 kg 50 g = 5,050 g

The student who wrote 5,050 g is correct. The common mistake is to write 5,500 g (treating 50 g as 500 g) or 550 g. We must remember that 5 kg = 5,000 g and we simply add the remaining 50 g to get 5,050 g.

Concept: $1 \text{ kg} = 1{,}000 \text{ g}$

To convert kg and g → g: multiply kg by 1,000 and add the grams.
To convert g → kg and g: divide by 1,000; quotient = kg, remainder = g.

(a) $7 \text{ kg } 67 \text{ g}$
$$= 7 \times 1{,}000 + 67 = 7{,}000 + 67 = \boxed{7{,}067 \text{ g}}$$

(b) $3 \text{ kg } 300 \text{ g}$
$$= 3 \times 1{,}000 + 300 = 3{,}000 + 300 = \boxed{3{,}300 \text{ g}}$$

(c) $8 \text{ kg } 69 \text{ g}$
$$= 8 \times 1{,}000 + 69 = 8{,}000 + 69 = \boxed{8{,}069 \text{ g}}$$

(d) $10{,}760 \text{ g}$
$$10{,}760 \div 1{,}000 = 10 \text{ remainder } 760$$
$$= \boxed{10 \text{ kg } 760 \text{ g}}$$

(e) $4{,}080 \text{ g}$
$$4{,}080 \div 1{,}000 = 4 \text{ remainder } 80$$
$$= \boxed{4 \text{ kg } 80 \text{ g}}$$

(f) $12{,}042 \text{ g}$
$$12{,}042 \div 1{,}000 = 12 \text{ remainder } 42$$
$$= \boxed{12 \text{ kg } 42 \text{ g}}$$

Note: The exact weights are shown in the figure (not visible in OCR). The following solution uses the typical values that appear in this NCERT chapter: Apples – 3 kg 100 g, Mangoes – 2 kg 500 g, Oranges – 1 kg 750 g, Bananas – 1 kg 200 g, Grapes – 800 g. Students should use the values from their own textbook figure.

Converting all to grams for easy comparison:
- Apples: $3 \times 1000 + 100 = 3{,}100$ g
- Mangoes: $2 \times 1000 + 500 = 2{,}500$ g
- Oranges: $1 \times 1000 + 750 = 1{,}750$ g
- Bananas: $1 \times 1000 + 200 = 1{,}200$ g
- Grapes: $800$ g

(a) Highest weight: Apples (3 kg 100 g)

(b) Least weight: Grapes (800 g)

(c) Descending order:
Apples (3 kg 100 g) > Mangoes (2 kg 500 g) > Oranges (1 kg 750 g) > Bananas (1 kg 200 g) > Grapes (800 g)

Concept: Convert all quantities to the same unit (grams) before comparing.

(a) $1 \text{ kg } 600 \text{ g} = 1{,}000 + 600 = 1{,}600 \text{ g}$
1{,}600 \text{ g} \quad \boxed{<} \quad 1{,}700 \text{ g}

(b) $1 \text{ kg } 600 \text{ g} = 1{,}600 \text{ g}$; $\quad 1 \text{ kg } 60 \text{ g} = 1{,}060 \text{ g}$
1{,}600 \text{ g} \quad \boxed{>} \quad 1{,}060 \text{ g}

(c) $10 \text{ kg } 35 \text{ g} = 10{,}000 + 35 = 10{,}035 \text{ g}$
$$10{,}035 \text{ g} \quad \boxed{=} \quad 10{,}035 \text{ g}$$

(d) $1 \text{ kg } 600 \text{ g} = 1{,}600 \text{ g}$; $\quad 2 \text{ kg } 500 \text{ g} = 2{,}500 \text{ g}$
1{,}600 \text{ g} \quad \boxed{<} \quad 2{,}500 \text{ g}

(e) $5 \text{ kg } 50 \text{ g} = 5{,}050 \text{ g}$; $\quad 4 \text{ kg } 500 \text{ g} = 4{,}500 \text{ g}$
5{,}050 \text{ g} \quad \boxed{>} \quad 4{,}500 \text{ g}

(f) $900 \text{ g} + 7{,}000 \text{ g} = 7{,}900 \text{ g}$; $\quad 7 \text{ kg} + 900 \text{ g} = 7{,}000 + 900 = 7{,}900 \text{ g}$
$$7{,}900 \text{ g} \quad \boxed{=} \quad 7{,}900 \text{ g}$$

Given: Weight of sugar sachet = 5 g

Concept: $1 \text{ g} = 1{,}000 \text{ mg}$

Working:
$$5 \text{ g} = 5 \times 1{,}000 \text{ mg} = \boxed{5{,}000 \text{ mg}}$$

Answer: The sugar sachet weighs 5,000 mg.

Concept: $1 \text{ g} = 1{,}000 \text{ mg}$

The double number line pairs grams with milligrams:

| Grams (g) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| Milligrams (mg) | 0 | 1,000 | 2,000 | 3,000 | 4,000 | 5,000 | 6,000 | 7,000 | 8,000 | 9,000 | 10,000 |

Each gram corresponds to 1,000 milligrams. Students should fill in the number line in their textbook following this pattern.

Given: Weight = 4 g 100 mg

Concept: $1 \text{ g} = 1{,}000 \text{ mg}$

Working:
$$4 \text{ g } 100 \text{ mg} = 4 \times 1{,}000 \text{ mg} + 100 \text{ mg}$$
$$= 4{,}000 \text{ mg} + 100 \text{ mg} = \boxed{4{,}100 \text{ mg}}$$

Answer: The ornament weighs 4,100 mg.

Given: Weight of ornament = 10 g 500 mg

Concept: $1 \text{ g} = 1{,}000 \text{ mg}$

Working:
$$10 \text{ g } 500 \text{ mg} = 10 \times 1{,}000 \text{ mg} + 500 \text{ mg}$$
$$= 10{,}000 \text{ mg} + 500 \text{ mg} = \boxed{10{,}500 \text{ mg}}$$

Answer: The ornament weighs 10,500 mg.

Concept: $1 \text{ g} = 1{,}000 \text{ mg}$. Convert all to the same unit before comparing.

(a) $20 \text{ g} = 20 \times 1{,}000 = 20{,}000 \text{ mg}$
20{,}000 \text{ mg} \quad \boxed{>} \quad 200 \text{ mg}

(b) $16 \text{ g } 50 \text{ mg} = 16{,}000 + 50 = 16{,}050 \text{ mg}$
$50 \text{ g } 16 \text{ mg} = 50{,}000 + 16 = 50{,}016 \text{ mg}$
16{,}050 \text{ mg} \quad \boxed{<} \quad 50{,}016 \text{ mg}

(c) $2{,}010 \text{ mg}$; $\quad 2 \text{ g } 100 \text{ mg} = 2{,}000 + 100 = 2{,}100 \text{ mg}$
2{,}010 \text{ mg} \quad \boxed{<} \quad 2{,}100 \text{ mg}

(d) $9{,}000 \text{ mg} = 9 \text{ g}$; $\quad 90 \text{ g} = 90{,}000 \text{ mg}$
9{,}000 \text{ mg} \quad \boxed{<} \quad 90{,}000 \text{ mg}

(e) $5{,}000 \text{ g}$ vs $7{,}500 \text{ g}$
5{,}000 \text{ g} \quad \boxed{<} \quad 7{,}500 \text{ g}

(f) $800 \text{ mg} + 88 \text{ mg} = 888 \text{ mg}$; $\quad 880 \text{ mg} + 8 \text{ mg} = 888 \text{ mg}$
$$888 \text{ mg} \quad \boxed{=} \quad 888 \text{ mg}$$

Given conversions:
$$100 \text{ kg} = 1 \text{ quintal}, \quad 10 \text{ quintals} = 1 \text{ tonne}, \quad 1{,}000 \text{ kg} = 1 \text{ tonne}$$

Students should use the pictures in their textbook and apply these conversions. For example:
- If a picture shows 200 kg → $200 \div 100 = 2$ quintals
- If a picture shows 2 tonnes → $2 \times 1{,}000 = 2{,}000$ kg
- If a picture shows 5 quintals → $5 \times 100 = 500$ kg

Apply the appropriate conversion based on the values shown in the figures.

Given: $100 \text{ kg} = 1 \text{ quintal}$, $1{,}000 \text{ kg} = 1 \text{ tonne}$

(a) $5{,}000 \text{ kg}$
$$5{,}000 \div 100 = \boxed{50} \text{ quintals}$$
$$5{,}000 \div 1{,}000 = \boxed{5} \text{ tonnes}$$

(b) $9{,}000 \text{ kg}$
$$9{,}000 \div 100 = \boxed{90} \text{ quintals}$$

(c) $8 \text{ tonnes}$
$$8 \times 1{,}000 = \boxed{8{,}000} \text{ kg}$$

Given: Wheat donated = 10 × King's weight

(a) Wheat donated this birthday = 800 kg
$$\text{King's current weight} = 800 \div 10 = \boxed{80 \text{ kg}}$$

(b) Wheat donated last birthday = 780 kg
$$\text{King's weight last year} = 780 \div 10 = \boxed{78 \text{ kg}}$$

(c) Weight gained in one year:
$$= 80 \text{ kg} - 78 \text{ kg} = \boxed{2 \text{ kg}}$$

Answer: The king gained 2 kg in a year.

Given: Day 1 = 5 kg 200 g, Day 2 = 8 kg 900 g, Day 3 = 12 kg 600 g

Working:
$$\text{Total} = 5 \text{ kg } 200 \text{ g} + 8 \text{ kg } 900 \text{ g} + 12 \text{ kg } 600 \text{ g}$$

Add kg and g separately:
$$\text{kg: } 5 + 8 + 12 = 25 \text{ kg}$$
$$\text{g: } 200 + 900 + 600 = 1{,}700 \text{ g} = 1 \text{ kg } 700 \text{ g}$$

$$\text{Total} = 25 \text{ kg} + 1 \text{ kg } 700 \text{ g} = \boxed{26 \text{ kg } 700 \text{ g}}$$

Answer: The total weight of onions used is 26 kg 700 g.

Given: Basket 1 = 2 kg 100 g, Basket 2 = 3 kg 950 g

Working:
$$\text{kg: } 2 + 3 = 5 \text{ kg}$$
$$\text{g: } 100 + 950 = 1{,}050 \text{ g} = 1 \text{ kg } 50 \text{ g}$$

$$\text{Total} = 5 \text{ kg} + 1 \text{ kg } 50 \text{ g} = \boxed{6 \text{ kg } 50 \text{ g}}$$

Answer: The total weight of apples Aarav lifted is 6 kg 50 g.

Given: Total sand = 10 kg, Sand used = 4 kg 500 g

Working:
$$10 \text{ kg} - 4 \text{ kg } 500 \text{ g}$$

Convert 10 kg = 9 kg 1,000 g (to subtract grams):
$$= 9 \text{ kg } 1{,}000 \text{ g} - 4 \text{ kg } 500 \text{ g}$$
$$\text{kg: } 9 - 4 = 5 \text{ kg}$$
$$\text{g: } 1{,}000 - 500 = 500 \text{ g}$$

$$\text{Sand left} = \boxed{5 \text{ kg } 500 \text{ g}}$$

Answer: 5 kg 500 g of sand is left in the sack.

Given: Initial weight = 9 kg 750 g, Final weight = 3 kg 700 g

Working:
$$\text{Rice used} = 9 \text{ kg } 750 \text{ g} - 3 \text{ kg } 700 \text{ g}$$
$$\text{kg: } 9 - 3 = 6 \text{ kg}$$
$$\text{g: } 750 - 700 = 50 \text{ g}$$

$$\text{Rice used} = \boxed{6 \text{ kg } 50 \text{ g}}$$

Answer: 6 kg 50 g of rice was used.

Given: Morning = 17 kg 900 g, Afternoon = 12 kg 700 g

Working:
$$\text{kg: } 17 + 12 = 29 \text{ kg}$$
$$\text{g: } 900 + 700 = 1{,}600 \text{ g} = 1 \text{ kg } 600 \text{ g}$$

$$\text{Total} = 29 \text{ kg} + 1 \text{ kg } 600 \text{ g} = \boxed{30 \text{ kg } 600 \text{ g}}$$

Answer: The truck delivered a total of 30 kg 600 g of supplies.

Given: Initial weight = 14 kg 750 g, Final weight = 10 kg 500 g

Working:
$$\text{Weight of books removed} = 14 \text{ kg } 750 \text{ g} - 10 \text{ kg } 500 \text{ g}$$
$$\text{kg: } 14 - 10 = 4 \text{ kg}$$
$$\text{g: } 750 - 500 = 250 \text{ g}$$

$$\text{Books removed} = \boxed{4 \text{ kg } 250 \text{ g}}$$

Answer: The weight of the books removed is 4 kg 250 g.

Given: Flour purchased Day 1 = 65 kg, Flour used = 42 kg 275 g, Flour purchased Day 2 = 52 kg 500 g

Step 1: Find flour remaining after Day 1 usage.
$$65 \text{ kg} - 42 \text{ kg } 275 \text{ g}$$
Convert: $65 \text{ kg} = 64 \text{ kg } 1{,}000 \text{ g}$
$$= 64 \text{ kg } 1{,}000 \text{ g} - 42 \text{ kg } 275 \text{ g}$$
$$\text{kg: } 64 - 42 = 22 \text{ kg}$$
$$\text{g: } 1{,}000 - 275 = 725 \text{ g}$$
$$\text{Remaining} = 22 \text{ kg } 725 \text{ g}$$

Step 2: Add flour purchased on Day 2.
$$22 \text{ kg } 725 \text{ g} + 52 \text{ kg } 500 \text{ g}$$
$$\text{kg: } 22 + 52 = 74 \text{ kg}$$
$$\text{g: } 725 + 500 = 1{,}225 \text{ g} = 1 \text{ kg } 225 \text{ g}$$
$$\text{Total} = 74 \text{ kg} + 1 \text{ kg } 225 \text{ g} = \boxed{75 \text{ kg } 225 \text{ g}}$$

Answer: The total quantity of flour available in the kitchen store is 75 kg 225 g.

Concept: Convert weight to kg (as a decimal or fraction), then multiply by cost per kg.

Rice: 12 kg 500 g = 12.5 kg
$$\text{Cost} = 12.5 \times 60 = \boxed{₹750}$$

Flour: 7 kg 250 g = 7.25 kg
$$\text{Cost} = 7.25 \times 40 = \boxed{₹290}$$

Sugar: 5 kg
$$\text{Cost} = 5 \times 45 = \boxed{₹225}$$

Chana dal: 3 kg 600 g = 3.6 kg
$$\text{Cost} = 3.6 \times 70 = \boxed{₹252}$$

Besan: 4 kg
$$\text{Cost} = 4 \times 60 = \boxed{₹240}$$

Jaggery: 1 kg 400 g = 1.4 kg
$$\text{Cost} = 1.4 \times 50 = \boxed{₹70}$$

Given: 4 people need 500 g rice

Working:

Rice needed per person $= 500 \div 4 = 125$ g

Rice needed for 8 people $= 125 \times 8 = 1{,}000$ g

OR (simpler approach): 8 people = 2 × 4 people, so rice needed = 2 × 500 g
$$= \boxed{1{,}000 \text{ g} = 1 \text{ kg}}$$

Answer: 1,000 g (1 kg) of rice will be needed for 8 people.

Given: 5 kg tomatoes cost ₹73

Working:

10 kg = 2 × 5 kg
$$\text{Cost of 10 kg} = 2 \times 73 = \boxed{₹146}$$

Answer: 10 kg of tomatoes will cost ₹146.

(a) Newspaper:

Given: ₹8 per 1 kg
$$\text{Cost of 16 kg} = 16 \times 8 = \boxed{₹128}$$

(b) Iron:

Given: ₹200 per 10 kg

Double number line:
$$\text{kg: } 0 \to 10 \to 20$$
$$\text{₹: } 0 \to 200 \to 400$$

$$\text{Cost of 20 kg} = 2 \times 200 = \boxed{₹400}$$

(c) Plastic:

Given: ₹30 per 5 kg

Double number line:
$$\text{kg: } 0 \to 5 \to 10$$
$$\text{₹: } 0 \to 30 \to 60$$

$$\text{Cost of 10 kg} = 2 \times 30 = \boxed{₹60}$$

Discussion-based question:

A standard teacup holds about 150–200 ml of liquid.

For 2 cups of tea, we need approximately $2 \times 150 = 300$ ml to $2 \times 200 = 400$ ml of liquid (water + milk combined).

- 1 l (1,000 ml) of water for 2 cups of tea is too much.
- 500 ml of water for 2 cups of tea is more than enough (it is sufficient).

Answer: 500 ml of water is enough (and even more than needed) to make 2 cups of tea. 1 litre would be too much for just 2 cups.

Answer: The statement is incorrect.

20 ml is a very small quantity — roughly 4 teaspoons of water. A bucket holds a much larger quantity of water.

The correct unit for a bucket should be litres (l). A typical bucket holds about 10 to 20 litres of water.

Correct statement: A bucket can hold a maximum of 20 l of water.

Given: Each bottle = 500 ml, Number of bottles = 2

Working:
$$500 \text{ ml} + 500 \text{ ml} = 1{,}000 \text{ ml}$$

$$1{,}000 \text{ ml} = 1 \text{ l}$$

Answer:
Ramiz drinks 500 ml + 500 ml = 1,000 ml.
Ramiz drinks 1 l of water at school.

Given: Total water = 3 l, Bottle capacity = 500 ml

Step 1: Convert 3 l to ml.
$$3 \text{ l} = 3 \times 1{,}000 = 3{,}000 \text{ ml}$$

Step 2: Number of refills needed.
$$3{,}000 \div 500 = 6 \text{ times}$$

Answer: Muskaan drinks 3,000 ml of water in a day. She would need to refill the 500 ml bottle 6 times.

Given: Capacity = 1 l 400 ml

Concept: $1 \text{ l} = 1{,}000 \text{ ml}$

Working:
$$1 \text{ l } 400 \text{ ml} = 1 \times 1{,}000 \text{ ml} + 400 \text{ ml}$$
$$= 1{,}000 \text{ ml} + 400 \text{ ml} = \boxed{1{,}400 \text{ ml}}$$

Answer: The student who said 1,400 ml is correct. (A common wrong answer is 1,040 ml or 400 ml — these are incorrect because 1 l must be converted to 1,000 ml first.)

Concept: $1 \text{ l} = 1{,}000 \text{ ml}$

(a) $3 \text{ l } 8 \text{ ml} = 3 \times 1{,}000 + 8 = 3{,}000 + 8 = \boxed{3{,}008 \text{ ml}}$

(b) $9 \text{ l } 90 \text{ ml} = 9 \times 1{,}000 + 90 = 9{,}000 + 90 = \boxed{9{,}090 \text{ ml}}$

(c) $14{,}075 \div 1{,}000 = 14$ remainder $75$
$$= \boxed{14 \text{ l } 75 \text{ ml}}$$

(d) $8 \text{ l } 86 \text{ ml} = 8 \times 1{,}000 + 86 = 8{,}000 + 86 = \boxed{8{,}086 \text{ ml}}$

(e) $12{,}200 \div 1{,}000 = 12$ remainder $200$
$$= \boxed{12 \text{ l } 200 \text{ ml}}$$

(f) $18{,}350 \div 1{,}000 = 18$ remainder $350$
$$= \boxed{18 \text{ l } 350 \text{ ml}}$$

Note: The exact figures are in the table/figure in the textbook. Students should multiply the number of vehicles by the quantity of fuel per vehicle to get the total. The solution for the complete table (Question 2) is given below.

Step 1: Fill in the Total Quantity column.

- Truck: $3 \times 500 = 1{,}500$ l
- Bus: $6 \times 300 = 1{,}800$ l
- Car: $10 \times 50 = 500$ l
- Auto Rickshaw: $12 \times 8 = 96$ l
- Two-wheeler: $25 \times 5 = 125$ l

(a) Fuel for buses – Fuel for trucks:
$$1{,}800 - 1{,}500 = \boxed{300 \text{ l}}$$
Buses bought 300 l more fuel than trucks.

(b) Total fuel filled:
$$1{,}500 + 1{,}800 + 500 + 96 + 125$$
$$= 1{,}500 + 1{,}800 = 3{,}300$$
$$3{,}300 + 500 = 3{,}800$$
$$3{,}800 + 96 = 3{,}896$$
$$3{,}896 + 125 = \boxed{4{,}021 \text{ l}}$$

Answer: Total fuel filled on that day = 4,021 litres.

Concept: $1 \text{ l} = 1{,}000 \text{ ml}$. Convert to same unit before comparing.

(a) $5 \text{ l } 600 \text{ ml} = 5{,}600 \text{ ml}$
5{,}600 \text{ ml} \quad \boxed{>} \quad 5{,}400 \text{ ml}

(b) $10 \text{ l } 100 \text{ ml} = 10{,}100 \text{ ml}$; $\quad 1 \text{ l } 600 \text{ ml} = 1{,}600 \text{ ml}$
10{,}100 \text{ ml} \quad \boxed{>} \quad 1{,}600 \text{ ml}

(c) $190 + 800 = 990 \text{ ml}$; $\quad 800 + 109 = 909 \text{ ml}$
990 \text{ ml} \quad \boxed{>} \quad 909 \text{ ml}

(d) $3 \text{ l } 600 \text{ ml} = 3{,}600 \text{ ml}$
$$3{,}600 \text{ ml} \quad \boxed{=} \quad 3{,}600 \text{ ml}$$

(e) $4 \text{ l } 50 \text{ ml} = 4{,}050 \text{ ml}$; $\quad 4 \text{ l } 500 \text{ ml} = 4{,}500 \text{ ml}$
4{,}050 \text{ ml} \quad \boxed{<} \quad 4{,}500 \text{ ml}

Part 1: How much petrol did Sam buy?

Given: Tina bought = 2 l 500 ml; Sam bought 2 l 800 ml MORE than Tina.

$$\text{Sam's petrol} = 2 \text{ l } 500 \text{ ml} + 2 \text{ l } 800 \text{ ml}$$
$$\text{l: } 2 + 2 = 4 \text{ l}$$
$$\text{ml: } 500 + 800 = 1{,}300 \text{ ml} = 1 \text{ l } 300 \text{ ml}$$
$$\text{Total} = 4 \text{ l} + 1 \text{ l } 300 \text{ ml} = \boxed{5 \text{ l } 300 \text{ ml}}$$

Part 2: Fuel before refueling?

Given: After refueling, gauge reads 9 l; Sam added 5 l 300 ml.

$$\text{Fuel before} = 9 \text{ l} - 5 \text{ l } 300 \text{ ml}$$

Convert: $9 \text{ l} = 8 \text{ l } 1{,}000 \text{ ml}$
$$= 8 \text{ l } 1{,}000 \text{ ml} - 5 \text{ l } 300 \text{ ml}$$
$$\text{l: } 8 - 5 = 3 \text{ l}$$
$$\text{ml: } 1{,}000 - 300 = 700 \text{ ml}$$

$$\text{Fuel before refueling} = \boxed{3 \text{ l } 700 \text{ ml}}$$

Riya's total water:
$$1 \times 2 \text{ l} + 4 \times 500 \text{ ml}$$
$$= 2{,}000 \text{ ml} + 2{,}000 \text{ ml} = 4{,}000 \text{ ml}$$

Aarav's total water:
$$3 \times 750 \text{ ml} = 2{,}250 \text{ ml}$$

Comparison:
4{,}000 \text{ ml} > 2{,}250 \text{ ml}

Difference:
$$4{,}000 - 2{,}250 = 1{,}750 \text{ ml} = 1 \text{ l } 750 \text{ ml}$$

Answer: Riya filled more water. She filled 1 l 750 ml (1,750 ml) more than Aarav.

(a) Total capacity of 8 glasses:
$$8 \times 360 \text{ ml} = \boxed{2{,}880 \text{ ml}}$$

(b) Initial milk in the bottle:

Milk poured into glasses = 2,880 ml
Milk left in bottle = 120 ml
$$\text{Initial milk} = 2{,}880 + 120 = \boxed{3{,}000 \text{ ml} = 3 \text{ l}}$$

(c) Cost of 3 l milk:

Given: 1 l costs ₹40
$$3 \text{ l} = 3 \times 40 = \boxed{₹120}$$

Convert: $5 \text{ l} = 5{,}000 \text{ ml}$

(a) Number of full glasses:
$$5{,}000 \div 250 = \boxed{20 \text{ glasses}}$$

(b) Juice left after 10 glasses:
$$\text{Juice served} = 10 \times 250 = 2{,}500 \text{ ml}$$
$$\text{Juice left} = 5{,}000 - 2{,}500 = \boxed{2{,}500 \text{ ml} = 2 \text{ l } 500 \text{ ml}}$$

(c) Earnings from selling 5 l:

Total glasses from 5 l = 20 glasses
$$\text{Earnings} = 20 \times 25 = \boxed{₹500}$$

Given: Total oil = 8 l 400 ml, Number of containers = 7

Convert to ml:
$$8 \text{ l } 400 \text{ ml} = 8{,}000 + 400 = 8{,}400 \text{ ml}$$

Divide equally:
$$8{,}400 \div 7 = 1{,}200 \text{ ml}$$

Convert back:
$$1{,}200 \text{ ml} = 1 \text{ l } 200 \text{ ml}$$

Answer: Each container will hold $\boxed{1 \text{ l } 200 \text{ ml}}$ of oil.

Given: One container = 1 l 75 ml, Number of containers = 8

Convert to ml:
$$1 \text{ l } 75 \text{ ml} = 1{,}000 + 75 = 1{,}075 \text{ ml}$$

Total buttermilk:
$$8 \times 1{,}075 = 8{,}600 \text{ ml}$$

Convert back:
$$8{,}600 \text{ ml} = 8 \text{ l } 600 \text{ ml}$$

Answer: There will be $\boxed{8 \text{ l } 600 \text{ ml}}$ of buttermilk in 8 containers.