Filling and Lifting

Given: When the milk from all three glasses is poured into same-sized glasses, we can compare the levels directly.

From the picture, Ritu's same-sized glass shows the highest milk level.

Answer: Ritu drank the most milk.

Given: The three same-sized glasses show different levels of milk after pouring.

From the picture, Monu's same-sized glass shows the lowest milk level.

Answer: Monu drank the least milk.

Concept: We compare the levels of milk in the same-sized glasses after pouring.

From the picture, the order from most to least is: Ritu > Nita > Monu.

(a) Nita's glass holds more milk than Monu's glass.
(Reason: Nita's level is higher than Monu's level.)

(b) Monu's glass holds less milk than Nita's glass.
(Reason: Monu's level is lower than Nita's level.)

(c) Ritu's glass holds more milk than Nita's glass.
(Reason: Ritu's level is the highest of all three.)

Concept: The glass whose same-sized comparison glass has the highest level holds the most milk, and the one with the lowest level holds the least.

(a) ☑ Ritu's glass holds the most milk.

(b) ☑ Monu's glass holds the least milk.

Concept: Among the three vessels — small bowl, glass, and bottle — the bottle is the largest and the small bowl is the smallest. The glass is in between.

(a) The glass holds the least water.
(Note: Among glass and bottle, the glass holds less. If the small bowl is the reference unit, the glass holds more than the bowl but less than the bottle. Based on the activity context where the glass is being filled by bowls and the bottle by glasses, the glass holds less than the bottle.)

Standard expected answer based on the activity:
(a) The glass holds the least water. *(smallest vessel among glass and bottle)*
(b) The bottle holds the most water. *(largest vessel)*

This is an activity-based/open-ended question. Students should look around their homes and identify larger vessels.

Sample Answer: Vessels that can store more water than a bottle include:
- Bucket
- Drum / large container
- Matka (earthen pot)
- Large jug or pitcher
- Overhead water tank

*(Answers will vary depending on the bottle size and vessels available at home.)*

This is an activity-based question that depends on the sizes of the ladle and bowl shown in the picture.

Concept: A ladle is smaller than a bowl, so several ladles are needed to fill the bowl.

Sample Answer: Based on the picture, approximately 4 ladles fill the bowl.
*(Actual answer depends on the sizes shown in the figure. Students should pour water using the ladle into the bowl and count the number of times.)*

This is an activity-based question depending on the sizes shown in the picture.

Concept: A jug holds more water than a glass, so it can fill multiple glasses.

Sample Answer: Based on the picture, the jug can fill approximately 4 glasses.
*(Students should pour water from the jug into the glass and count how many full glasses are obtained.)*

Concept: To fill half a glass, we need a vessel that holds about half the capacity of the glass.

Answer: We can use a ladle to fill half of the glass, since a ladle holds less water than a full glass. Pour ladles one at a time until the glass is half full.

Answer: We would use the jug for distributing lemonade in glasses.

Reason: The jug holds more lemonade than a glass or a bowl, so it is easy to pour from the jug into each glass without having to refill frequently. It is also convenient to hold and pour.

Given: From Question 2, 1 jug fills 4 glasses.

Formula: Number of glasses = Number of jugs × Glasses per jug

$$\text{Number of glasses} = 3 \times 4 = 12$$

Answer: 3 jugs of lemonade can fill 12 glasses.

Given: From the activity, suppose 1 glass = 2 ladles (based on the picture).

Formula: Total ladles = Number of glasses × Ladles per glass

$$\text{Total ladles} = 4 \times 2 = 8$$

Answer: 8 ladles are needed to fill 4 glasses.
*(The exact number depends on the sizes shown in the figure. Students should use their measured value from the activity.)*

Answer: Yes, we can use a ladle, bowl, or glass to find out how much lemonade a jug can hold.

Method:
- Fill the ladle (or bowl or glass) with water/lemonade.
- Pour it into the jug.
- Count how many times you pour until the jug is full.
- That count tells you the capacity of the jug in terms of ladles (or bowls or glasses).

Conclusion: Utensils with less capacity can be used to measure a utensil of larger capacity by counting how many times the smaller vessel fills the larger one. The total quantity of liquid remains the same.

Given: Nita uses a 1 litre bottle to compare with the jug, glass, and bowl.

From the picture:
- The glass is smaller than the 1 litre bottle.
- The bowl is smaller than the 1 litre bottle.
- The jug holds exactly the same amount as the 1 litre bottle.

(a) The glass holds ☑ less than 1 litre.

(b) The bowl holds ☑ less than 1 litre.

(c) The jug holds ☑ exactly 1 litre.

Activity-based question. Students should take their 1 litre bottle and pour water into various vessels at home.

Method: Pour 1 litre of water from the bottle into the vessel. If the vessel is exactly full (neither overflowing nor partially empty), it holds exactly 1 litre.

Sample Answer: Vessels that are commonly exactly 1 litre:
- A standard 1-litre milk packet or bottle
- Some medium-sized jugs
- Certain cooking oil bottles

*(Answers will vary based on vessels available at home.)*

Activity-based question. Students should use the 1 litre bottle to check.

More than 1 litre (sample answers):
- Bucket
- Large jug
- Cooking pot
- Water bottle (large)

Less than 1 litre (sample answers):
- Drinking glass
- Small bowl
- Cup / mug
- Ladle

*(Answers will vary based on vessels available at home.)*

Concept:
- 1 litre = 2 half litres = 4 quarter litres
- From the picture, the mug is filled exactly halfway when 1 litre is poured, and the glass is filled exactly when a quarter litre is poured.

(a) The mug holds ☑ half litre of water.

(b) The glass holds ☑ quarter litre of water.

Answer: A bucket holds ☑ more than 1 litre of water.

Reason: A standard bucket holds about 10–15 litres of water, which is much more than 1 litre.

Answer: A mug holds ☑ less than half a litre of water.

Reason: A standard mug holds about 200–300 ml, which is less than half a litre (500 ml).
*(Students should verify this by pouring water from a half-litre measure into their mug.)*

Answer: A glass holds ☑ more than a quarter litre of water.

Reason: A standard drinking glass holds about 200–250 ml, and a quarter litre = 250 ml. Most glasses hold approximately equal to or slightly less than a quarter litre.

*(Note: Answers may vary. Students should verify by pouring a quarter litre of water into their glass at home.)*

Answer: Chintu is holding 3 textbooks in one hand and a pencil box in the other hand.

The hand holding the 3 textbooks is lower because the textbooks are heavier than the pencil box.

Concept: Heavier objects pull the hand downward due to their greater weight. This is the same principle used in a pan-balance — the heavier side goes down.

Concept: When you hold two objects, the heavier one makes your hand go lower. Compare by feel.

Sample Observations:

| Things to Compare | Observation |
|---|---|
| Lunch box and pencil box | Lunch box is heavier than pencil box |
| School bag and lunch bag | School bag is heavier than lunch bag |
| Apple and watermelon | Watermelon is heavier than apple |
| Balloon filled with air and a basketball | Basketball is heavier than the balloon |

*(Actual observations may vary slightly based on the specific items, but the general answers above are expected.)*

Concept: A simple pan-balance is used. Place the object on one side and add coins or erasers on the other side until both sides are balanced. The number of coins/erasers needed equals the weight of the object in those units.

Method:
1. Place the object (e.g., pencil) on one pan of the balance.
2. Add coins/erasers one by one to the other pan.
3. Count the coins/erasers when the balance is level.

Sample Table (answers will vary):

| Objects | How many coins or erasers balance the object? |
|---|---|
| Pencil | 3 coins |
| Ping Pong Ball | 5 coins |

*(Students should fill in actual values from their activity.)*

Concept: A matchbox filled with sand acts as a standard unit of weight. Place the object on one pan and add matchboxes on the other pan until balanced.

Method:
1. Fill a matchbox with sand — this is your unit weight.
2. Place the object on one pan.
3. Add matchboxes one by one to the other pan until balanced.
4. Count the matchboxes.

Sample Table (answers will vary):

| Objects | Guess (matchboxes) | Actual (matchboxes) |
|---|---|---|
| Pencil box | 8 | 10 |
| A spoon | 1 | 2 |
| 4 marbles | 2 | 3 |

Order from lightest to heaviest (sample):
Spoon → 4 Marbles → Pencil box

*(Students should record their own measured values and arrange accordingly.)*

Answer: A bag is heavier or lighter depending on:
- The number of books and notebooks it contains.
- The size and weight of each book.
- Other items like water bottles, lunch boxes, pencil boxes, etc.

A bag with more books or heavier items will be heavier than a bag with fewer or lighter items.

Activity-based answer. Students should count the books in their bag and in their friend's bag.

Expected finding: The bag with more books (or heavier books) will generally be heavier. If one bag has 6 books and another has 4 books, the bag with 6 books is likely heavier.

Answer: Yes, we can make the two bags approximately the same weight by moving a book from the heavier bag to the lighter bag.

Method:
1. Identify the heavier bag and the lighter bag.
2. Move one book from the heavier bag to the lighter bag.
3. Check again using the pan-balance or by feel.
4. Repeat if needed until both bags feel approximately equal in weight.

Conclusion: By transferring books, we can balance the weight of the two bags.

Activity-based question. Students should look for objects at home with '1 kg' written on them.

Sample Answer: Objects that are commonly 1 kilogram:
- 1 kg salt packet
- 1 kg sugar packet
- 1 kg rice packet
- 1 kg flour (atta) packet
- 1 kg dal packet

*(Answers will vary based on what is available at home.)*

Concept: Use the feel of a 1 kg salt packet as a reference. Compare other objects to it.

General guidelines:
- Objects that feel lighter than the 1 kg packet → less than 1 kg
- Objects that feel heavier than the 1 kg packet → more than 1 kg

Sample answers (based on common objects shown in such activities):
- A pencil → less than 1 kg
- A school bag (full) → more than 1 kg
- An apple → less than 1 kg
- A brick → more than 1 kg
- A notebook → less than 1 kg

*(Students should tick based on their own comparison with the 1 kg packet.)*

Given: From the picture, two daal packets together balance a 1 kg weight on the pan-balance.

Concept: If 2 packets = 1 kg, then each packet = $\frac{1}{2}$ kg.

$$2 \times \text{weight of each packet} = 1 \text{ kg}$$
$$\text{Weight of each packet} = \frac{1}{2} \text{ kg}$$

Answer: Each daal packet weighs ☑ half kilogram.

Given: From the picture, four tea packets together balance a 1 kg weight on the pan-balance.

Concept: If 4 packets = 1 kg, then each packet = $\frac{1}{4}$ kg.

$$4 \times \text{weight of each packet} = 1 \text{ kg}$$
$$\text{Weight of each packet} = \frac{1}{4} \text{ kg}$$

Answer: Each tea packet weighs ☑ quarter kilogram.

Given: 3 balls — one is heavier, two are equal in weight. Only a pan-balance is available (no weights).

Answer: You will have to weigh only 1 time.

Explanation: Place one ball on each pan of the balance (any two balls).
- If one side goes down → that ball is the heavy ball. Done in 1 weighing.
- If both sides are equal → the third ball (not on the balance) is the heavy ball. Done in 1 weighing.

In both cases, the heavy ball is identified in just one weighing.

Method (using the balance only once):

Step 1: Label the three balls as Ball A, Ball B, and Ball C.

Step 2: Place Ball A on the left pan and Ball B on the right pan.

Step 3: Observe the balance:
- If left side (Ball A) goes down → Ball A is the heavy ball.
- If right side (Ball B) goes down → Ball B is the heavy ball.
- If both sides are equal → Ball C (the one not placed on the balance) is the heavy ball.

Conclusion: In just one weighing, we can identify the heavy ball.

Given: Three balls — Red, Orange, Green — all different weights. Pan-balance available, no standard weights.

Strategy: We need at least 2 weighings to find the order of all three.

Step 1 — First Weighing:
Place the Red ball on the left pan and the Orange ball on the right pan.
- Suppose Red goes down → Red is heavier than Orange.

Step 2 — Second Weighing:
Place the Red ball on the left pan and the Green ball on the right pan.
- Case A: If Red goes down → Red is heavier than Green too.
- Now compare Orange and Green: place Orange vs Green.
- Whichever goes down is heavier.
- This gives the full order.
- Case B: If Green goes down → Green is heavier than Red.
- We already know Red > Orange, and Green > Red.
- So order is: Green (heaviest) > Red > Orange (lightest).

General Method:
1. Weigh any two balls (e.g., Red vs Orange).
2. Weigh the heavier one against the third ball (Green).
3. From the results, determine the complete order: heaviest → middle → lightest.

Conclusion: Using 2 to 3 weighings, we can find which ball is the heaviest and which is the lightest by comparing them on the pan-balance two at a time.