Far and Near

We convert all measurements to the same unit (cm) before comparing.

(a) 456 cm ___ 5 m

Convert 5 m to cm: $5 \text{ m} = 5 \times 100 = 500 \text{ cm}$

Compare: $456 \text{ cm}$ vs $500 \text{ cm}$

456 < 500

\boxed{456 \text{ cm} < 5 \text{ m}}

---

(b) 55 cm + 200 cm ___ 200 cm + 54 cm

Left side: $55 + 200 = 255 \text{ cm}$

Right side: $200 + 54 = 254 \text{ cm}$

Compare: $255$ vs $254$

\boxed{55 \text{ cm} + 200 \text{ cm} > 200 \text{ cm} + 54 \text{ cm}}

---

(c) 6 m 5 cm ___ 6 m 50 cm

Convert both to cm:
- $6 \text{ m } 5 \text{ cm} = 600 + 5 = 605 \text{ cm}$
- $6 \text{ m } 50 \text{ cm} = 600 + 50 = 650 \text{ cm}$

Compare: $605$ vs $650$

\boxed{6 \text{ m } 5 \text{ cm} < 6 \text{ m } 50 \text{ cm}}

---

(d) 2 m 150 cm ___ 3 m 50 cm

Convert both to cm:
- $2 \text{ m } 150 \text{ cm} = 200 + 150 = 350 \text{ cm}$
- $3 \text{ m } 50 \text{ cm} = 300 + 50 = 350 \text{ cm}$

Compare: $350$ vs $350$

$$\boxed{2 \text{ m } 150 \text{ cm} = 3 \text{ m } 50 \text{ cm}}$$

---

(e) 238 cm ___ 138 cm + 1 m

Right side: $138 \text{ cm} + 1 \text{ m} = 138 + 100 = 238 \text{ cm}$

Compare: $238$ vs $238$

$$\boxed{238 \text{ cm} = 138 \text{ cm} + 1 \text{ m}}$$

Note: This question refers to a figure/table of statue heights that is not fully visible in the OCR. The standard data used in NCERT Class 5 for this question is:
- Statue of Unity (India): 182 m
- Spring Temple Buddha (China): 128 m
- Statue of Liberty (USA): 93 m
- Christ the Redeemer (Brazil): 38 m
- Motherland Calls (Russia): 85 m

The tallest statue in the world (as per the textbook context) is the Statue of Unity at 182 m.

---

(a) Difference between the tallest statue (Statue of Unity) and the Statue of Liberty:

$$182 \text{ m} - 93 \text{ m} = 89 \text{ m}$$

The difference is 89 m.

---

(b) Statues whose heights have the least difference:

Compare all pairs:
- Motherland Calls (85 m) and Statue of Liberty (93 m): $93 - 85 = 8$ m
- Spring Temple Buddha (128 m) and Statue of Unity (182 m): $182 - 128 = 54$ m

The least difference is between the Motherland Calls and the Statue of Liberty (difference = 8 m).

---

(c) Statues whose heights have the largest difference:

The largest difference is between the tallest and the shortest statues:
- Statue of Unity (182 m) and Christ the Redeemer (38 m): $182 - 38 = 144$ m

The largest difference is between the Statue of Unity and Christ the Redeemer (difference = 144 m).

---

(d) Height of which statue doubled equals the Statue of Unity (182 m)?

We need: $2 \times h = 182$

$$h = 182 \div 2 = 91 \text{ m}$$

The statue closest to 91 m is the Statue of Liberty (93 m). Based on the textbook data, the answer is the Statue of Liberty (approximately, depending on exact figures given in the figure).

Given: $1 \text{ km} = 1{,}000 \text{ m}$

Number of ropes needed $= 1{,}000 \div \text{length of each rope}$

| Length of rope | Calculation | Number of ropes needed |
|---|---|---|
| 1,000 m | $1{,}000 \div 1{,}000$ | 1 |
| 100 m | $1{,}000 \div 100$ | 10 |
| 10 m | $1{,}000 \div 10$ | 100 |
| 200 m | $1{,}000 \div 200$ | 5 |
| 500 m | $1{,}000 \div 500$ | 2 |
| 250 m | $1{,}000 \div 250$ | 4 |

Given: Total race distance $= 3 \text{ km} = 3{,}000 \text{ m}$

Water stations are placed every $500 \text{ m}$.

Number of water stations $= 3{,}000 \div 500 = 6$

However, a water station at the finish line (3,000 m) may not be needed as a separate station. Based on standard interpretation, stations are placed after the start and before or at the finish:

Positions from the starting point:
$$500 \text{ m},\ 1{,}000 \text{ m},\ 1{,}500 \text{ m},\ 2{,}000 \text{ m},\ 2{,}500 \text{ m},\ 3{,}000 \text{ m}$$

Number of water stations = 6

They will be placed at: 500 m, 1,000 m, 1,500 m, 2,000 m, 2,500 m, and 3,000 m from the starting point.

Given: Total race distance $= 3 \text{ km} = 3{,}000 \text{ m}$

Children stand at every $300 \text{ m}$.

Number of positions $= 3{,}000 \div 300 = 10$

Positions from the starting point:
$$300 \text{ m},\ 600 \text{ m},\ 900 \text{ m},\ 1{,}200 \text{ m},\ 1{,}500 \text{ m},\ 1{,}800 \text{ m},\ 2{,}100 \text{ m},\ 2{,}400 \text{ m},\ 2{,}700 \text{ m},\ 3{,}000 \text{ m}$$

Number of children needed = 10

They will stand at: 300 m, 600 m, 900 m, 1,200 m, 1,500 m, 1,800 m, 2,100 m, 2,400 m, 2,700 m, and 3,000 m from the starting point.

Given: Total race distance $= 3 \text{ km} = 3{,}000 \text{ m}$

Flags are placed every $50 \text{ m}$.

Total number of flag positions (not counting the start) $= 3{,}000 \div 50 = 60$

Flags are placed alternately (Red, Blue, Red, Blue, …) starting from the first position at 50 m.

Total flags = 60

Since flags alternate:
- Red flags = $60 \div 2 = 30$
- Blue flags = $60 \div 2 = 30$

30 red flags and 30 blue flags are needed till the finish line.

Note: The specific sub-questions for this table are not fully visible in the OCR. Below are the key facts that can be derived from the table, which are typically asked:

Total distance of the journey (Dibrugarh to Kanniyakumari):
$$= 4{,}187 \text{ km}$$

Distance between consecutive stations (for reference):
- Dibrugarh to Dimapur: $306 \text{ km}$
- Dimapur to Guwahati: $556 - 306 = 250 \text{ km}$
- Guwahati to Jalpaiguri Road: $983 - 556 = 427 \text{ km}$
- Jalpaiguri Road to Bhubaneswar: $2{,}007 - 983 = 1{,}024 \text{ km}$
- Bhubaneswar to Visakhapatnam: $2{,}450 - 2{,}007 = 443 \text{ km}$
- Visakhapatnam to Vijayawada JN: $2{,}800 - 2{,}450 = 350 \text{ km}$
- Vijayawada JN to Coimbatore JN: $3{,}675 - 2{,}800 = 875 \text{ km}$
- Coimbatore JN to Kanniyakumari: $4{,}187 - 3{,}675 = 512 \text{ km}$

Students should use the table to answer specific questions about distances between stations as directed by the teacher.

Concept used: $10 \text{ mm} = 1 \text{ cm}$

Convert each length to mm for ease of measurement:

1. $5 \text{ cm } 5 \text{ mm} = 50 + 5 = 55 \text{ mm}$
2. $3 \text{ cm } 6 \text{ mm} = 30 + 6 = 36 \text{ mm}$
3. $8 \text{ cm } 3 \text{ mm} = 80 + 3 = 83 \text{ mm}$
4. $36 \text{ mm} = 3 \text{ cm } 6 \text{ mm}$
5. $67 \text{ mm} = 6 \text{ cm } 7 \text{ mm}$

Steps to draw each line:
- Place the scale on the notebook.
- Mark the starting point at 0.
- Mark the ending point at the required length.
- Join the two points with a sharp pencil to draw the line.

How to draw 36 mm and 67 mm lines:

For 36 mm: Since most scales show both cm and mm markings, place the scale and mark from 0 to the 36 mm mark (which is the same as 3 cm 6 mm mark). Draw the line.

For 67 mm: Mark from 0 to the 67 mm mark (which is the same as 6 cm 7 mm). Draw the line.

*Alternatively*, if the scale only shows cm, convert: $36 \text{ mm} = 3.6 \text{ cm}$ and $67 \text{ mm} = 6.7 \text{ cm}$, and mark accordingly between the cm markings.

Relationships used:
$$10 \text{ mm} = 1 \text{ cm}, \quad 100 \text{ cm} = 1 \text{ m}, \quad 1{,}000 \text{ m} = 1 \text{ km}$$

mm ↔ cm number line:
For every 1 cm, there are 10 mm.
- 1 cm = 10 mm
- 2 cm = 20 mm
- 5 cm = 50 mm
- 10 cm = 100 mm

cm ↔ m number line:
For every 1 m, there are 100 cm.
- 1 m = 100 cm
- 2 m = 200 cm
- 5 m = 500 cm
- 10 m = 1,000 cm

m ↔ km number line:
For every 1 km, there are 1,000 m.
- 1 km = 1,000 m
- 2 km = 2,000 m
- 5 km = 5,000 m

*(Students should fill in the blanks on the number lines in their textbook using the above relationships.)*

Concept: $10 \text{ mm} = 1 \text{ cm}$, $100 \text{ cm} = 1 \text{ m}$

---

(a) $4 \text{ cm } 5 \text{ mm} = \underline{\hspace{1cm}} \text{ mm}$

$$4 \text{ cm } 5 \text{ mm} = (4 \times 10) + 5 = 40 + 5 = \boxed{45 \text{ mm}}$$

---

(b) $89 \text{ mm} = \underline{\hspace{1cm}} \text{ cm } \underline{\hspace{1cm}} \text{ mm}$

$$89 \text{ mm} = 80 \text{ mm} + 9 \text{ mm} = 8 \text{ cm} + 9 \text{ mm} = \boxed{8 \text{ cm } 9 \text{ mm}}$$

---

(c) $234 \text{ cm} = \underline{\hspace{1cm}} \text{ mm}$

$$234 \text{ cm} = 234 \times 10 = \boxed{2{,}340 \text{ mm}}$$

*(Note: The "= 8 cm 9 mm" in the original appears to be a hint for part (b), not part (c).)*

---

(d) $514 \text{ mm} = \underline{\hspace{1cm}} \text{ cm } \underline{\hspace{1cm}} \text{ mm}$

$$514 \div 10 = 51 \text{ remainder } 4$$

$$514 \text{ mm} = \boxed{51 \text{ cm } 4 \text{ mm}}$$

---

(e) $6 \text{ m } 34 \text{ cm} = \underline{\hspace{1cm}} \text{ cm}$

$$6 \text{ m } 34 \text{ cm} = (6 \times 100) + 34 = 600 + 34 = \boxed{634 \text{ cm}}$$

---

(f) $20 \text{ m } 12 \text{ cm} = \underline{\hspace{1cm}} \text{ cm}$

$$20 \text{ m } 12 \text{ cm} = (20 \times 100) + 12 = 2{,}000 + 12 = \boxed{2{,}012 \text{ cm}}$$

---

(g) $397 \text{ m} = \underline{\hspace{1cm}} \text{ cm}$

$$397 \text{ m} = 397 \times 100 = \boxed{39{,}700 \text{ cm}}$$

Given:
- Length of first ribbon roll $= 3 \text{ m } 75 \text{ cm}$
- Length of second ribbon roll $= 2 \text{ m } 25 \text{ cm}$

To find: Total length of ribbon

Solution:

Add the two lengths:

$$\begin{array}{r}$$
3 \text{ m } 75 \text{ cm} \\
+ \quad 2 \text{ m } 25 \text{ cm} \\
\hline
5 \text{ m } 100 \text{ cm}
\end{array}

Since $100 \text{ cm} = 1 \text{ m}$:

$$5 \text{ m } 100 \text{ cm} = 5 \text{ m} + 1 \text{ m} = 6 \text{ m}$$

Rani has $\boxed{6 \text{ m}}$ of ribbon in total.

Given:
- Distance from Bhopal to Sanchi $= 48 \text{ km } 700 \text{ m}$
- Distance from Bhopal to waterfall $= 17 \text{ km } 900 \text{ m}$

To find: Distance from waterfall to Sanchi

Solution:

Distance from waterfall to Sanchi $=$ Distance (Bhopal to Sanchi) $-$ Distance (Bhopal to waterfall)

$$\begin{array}{r}$$
48 \text{ km } 700 \text{ m} \\
- \quad 17 \text{ km } 900 \text{ m} \\
\hline
\end{array}

Since 700 \text{ m} < 900 \text{ m}, we borrow 1 km = 1,000 m from the km column:

$$48 \text{ km } 700 \text{ m} = 47 \text{ km } 1{,}700 \text{ m}$$

$$\begin{array}{r}$$
47 \text{ km } 1{,}700 \text{ m} \\
- \quad 17 \text{ km } \quad 900 \text{ m} \\
\hline
30 \text{ km } \quad 800 \text{ m}
\end{array}

Sanchi is $\boxed{30 \text{ km } 800 \text{ m}}$ from the waterfall.

Given:
- First section $= 2 \text{ km } 300 \text{ m}$
- Second section $= 2 \text{ km } 650 \text{ m}$

To find: Total distance covered

Solution:

$$\begin{array}{r}$$
2 \text{ km } 300 \text{ m} \\
+ \quad 2 \text{ km } 650 \text{ m} \\
\hline
4 \text{ km } 950 \text{ m}
\end{array}

The total distance covered by the Gulmarg Gondola cable car is $\boxed{4 \text{ km } 950 \text{ m}}$.

Concept: Convert to the same unit before comparing.

---

(a) 11 mm and 1 cm

Convert: $1 \text{ cm} = 10 \text{ mm}$

Compare: $11 \text{ mm}$ vs $10 \text{ mm}$

Bigger: $\mathbf{11 \text{ mm}}$ (circle this)

Difference: $11 - 10 = \boxed{1 \text{ mm}}$

---

(b) 26 mm and 2 cm

Convert: $2 \text{ cm} = 20 \text{ mm}$

Compare: $26 \text{ mm}$ vs $20 \text{ mm}$

Bigger: $\mathbf{26 \text{ mm}}$ (circle this)

Difference: $26 - 20 = \boxed{6 \text{ mm}}$

---

(c) 20 cm and 201 mm

Convert: $20 \text{ cm} = 200 \text{ mm}$

Compare: $200 \text{ mm}$ vs $201 \text{ mm}$

Bigger: $\mathbf{201 \text{ mm}}$ (circle this)

Difference: $201 - 200 = \boxed{1 \text{ mm}}$

---

(d) 1,020 mm and 1 m

Convert: $1 \text{ m} = 100 \text{ cm} = 1{,}000 \text{ mm}$

Compare: $1{,}020 \text{ mm}$ vs $1{,}000 \text{ mm}$

Bigger: $\mathbf{1{,}020 \text{ mm}}$ (circle this)

Difference: $1{,}020 - 1{,}000 = \boxed{20 \text{ mm}}$

---

(e) 2 m and 245 cm

Convert: $2 \text{ m} = 200 \text{ cm}$

Compare: $200 \text{ cm}$ vs $245 \text{ cm}$

Bigger: $\mathbf{245 \text{ cm}}$ (circle this)

Difference: $245 - 200 = \boxed{45 \text{ cm}}$

---

(f) 5,678 m and 6 km

Convert: $6 \text{ km} = 6{,}000 \text{ m}$

Compare: $5{,}678 \text{ m}$ vs $6{,}000 \text{ m}$

Bigger: $\mathbf{6 \text{ km}}$ (circle this)

Difference: $6{,}000 - 5{,}678 = \boxed{322 \text{ m}}$

---

(g) 6 km 1,480 m and 7 km 479 m

Convert both to metres:
- $6 \text{ km } 1{,}480 \text{ m} = 6{,}000 + 1{,}480 = 7{,}480 \text{ m}$
- $7 \text{ km } 479 \text{ m} = 7{,}000 + 479 = 7{,}479 \text{ m}$

Compare: $7{,}480 \text{ m}$ vs $7{,}479 \text{ m}$

Bigger: $\mathbf{6 \text{ km } 1{,}480 \text{ m}}$ (circle this)

Difference: $7{,}480 - 7{,}479 = \boxed{1 \text{ m}}$

Given: Cloth needed for 1 shirt $= 1 \text{ m } 80 \text{ cm}$

Number of shirts $= 20$

To find: Total cloth needed for 20 shirts

Solution:

$$20 \times 1 \text{ m } 80 \text{ cm} = (20 \times 1 \text{ m}) + (20 \times 80 \text{ cm})$$

$$= 20 \text{ m} + 1{,}600 \text{ cm}$$

Convert $1{,}600 \text{ cm}$ to metres: $1{,}600 \div 100 = 16 \text{ m}$

$$= 20 \text{ m} + 16 \text{ m} = \boxed{36 \text{ m}}$$

36 m of cloth will be needed to make shirts for 20 children.

Given: Cost of 5 m cloth $= ₹100$

To find: Cost of 1 m cloth

Solution:

$$\text{Cost of 1 m} = ₹100 \div 5 = \boxed{₹20}$$

₹20 is needed to buy 1 m of cloth.

Part 1: Length of thread needed

Given: 1 m thread is needed for 50 cm of sari border.

For 5 m sari border:

$5 \text{ m} = 500 \text{ cm}$

Number of 50 cm sections in 500 cm $= 500 \div 50 = 10$

Thread needed $= 10 \times 1 \text{ m} = \boxed{10 \text{ m}}$

---

Part 2: Cost of thread

Given: Cost of 1 m thread $= ₹50$

Thread needed $= 10 \text{ m}$

$$\text{Total cost} = 10 \times ₹50 = \boxed{₹500}$$

Anita needs 10 m of thread, which will cost ₹500.

Given: Total length of road $= 12 \text{ km } 600 \text{ m}$

Number of days $= 6$

To find: Length of road laid each day

Solution:

Convert total length to metres:
$$12 \text{ km } 600 \text{ m} = 12{,}000 + 600 = 12{,}600 \text{ m}$$

Length laid each day:
$$12{,}600 \div 6 = 2{,}100 \text{ m}$$

Convert back: $2{,}100 \text{ m} = 2 \text{ km } 100 \text{ m}$

The workers lay $\boxed{2 \text{ km } 100 \text{ m}}$ of road each day.