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.