Sol: Distance walked in 1 day = 3 km
Number of days = 15
Total distance = 3 × 15 = 45 km
The man will walk 45 km in 15 days.
Class 5 Exam > Class 5 Notes > Mathematics (Maths Mela) Class 5 - New NCERT > Worksheet Solutions: Far and Near
Worksheet Solutions: Far and Near | Mathematics (Maths Mela) Class 5 - New NCERT PDF Download
Q1: 250 cm = ? m
(a) 2.5 m
(b) 25 m
(c) 0.25 m
(d) 2500 m
Sol: (a) 2.5 m
1 meter = 100 centimeters
So, 250 ÷ 100 = 2.5 meters
Q2: A rope is 600 cm long. If cut into 10 equal pieces, the length of each piece is:
(a) 60 m
(b) 60 cm
(c) 6 m
(d) 600 m
Sol: (b) 60 cm
Total length = 600 cm
Number of pieces = 10
Each piece = 600 ÷ 10 = 60 cm
Q3: 5 km = ? m
(a) 500
(b) 5000
(c) 50
(d) 50,000
Sol: (b) 5000
1 kilometer = 1000 meters
So, 5 × 1000 = 5000 meters
Q4: A train is 250 m long. What is its length in kilometers?
(a) 0.025 km
(b) 2.5 km
(c) 0.25 km
(d) 25 km
Sol: (c) 0.25 km
1 kilometer = 1000 meters
So, 250 ÷ 1000 = 0.25 kilometers
Q5: A ribbon is 36 m long. If each piece is 4 m long, how many pieces can be cut?
(a) 8
(b) 9
(c) 10
(d) 12
Sol: (b) 9
Total length = 36 meters
Each piece = 4 meters
Number of pieces = 36 ÷ 4 = 9
Q6: Convert the following into centimeters.
(i) 66 mm
(ii) 32 dm
(iii) 4 cm 8 mm
(iv) 63 km
(v) 16 m
(vi) 1.5 dm
Sol: We use:
1 cm = 10 mm
1 dm = 10 cm
1 m = 100 cm
1 km = 1000 m = 100,000 cm
(i) 66 mm
Since 10 mm = 1 cm
66 mm = 66 ÷ 10 = 6.6 cm
(ii) 32 dm
Since 1 dm = 10 cm
32 dm = 32 × 10 = 320 cm
(iii) 4 cm 8 mm
8 mm = 0.8 cm
So, 4 cm + 0.8 cm = 4.8 cm
(iv) 63 km
1 km = 100,000 cm
63 km = 63 × 100,000 = 6,300,000 cm
(v) 16 m
1 m = 100 cm
16 m = 16 × 100 = 1600 cm
(vi) 1.5 dm
1 dm = 10 cm
1.5 dm = 1.5 × 10 = 15 cm
Q7: Convert the following into kilometers.
(i) 14 m
(ii) 59 hm
(iii) 136 dam
(iv) 207 dam
(v) 5600 cm
(vi) 10000 cm
Sol: We use:
1 km = 1000 m
1 hm (hectometer) = 100 m
1 dam (decameter) = 10 m
1 m = 100 cm
(i) 14 m
1 km = 1000 m
14 m = 14 ÷ 1000 = 0.014 km
(ii) 59 hm
1 hm = 100 m
59 hm = 59 × 100 = 5900 m
5900 ÷ 1000 = 5.9 km
(iii) 136 dam
1 dam = 10 m
136 dam = 136 × 10 = 1360 m
1360 ÷ 1000 = 1.36 km
(iv) 207 dam
207 dam = 207 × 10 = 2070 m
2070 ÷ 1000 = 2.07 km
(v) 5600 cm
1 m = 100 cm → 5600 cm = 5600 ÷ 100 = 56 m
Now, 56 ÷ 1000 = 0.056 km
(vi) 10000 cm
10000 ÷ 100 = 100 m
100 ÷ 1000 = 0.1 km
Q8: Compare the following using >, < or = sign:
(i) 702 cm .......... 503 cm
(ii) 2 m .......... 9 m
(iii) 800 cm .......... 80 m
(iv) 702 cm .......... 5 m
(v) 8 km .......... 7000 m
(vi) 625 cm .......... 9 m
(vii) 10 m .......... 4 cm
(viii) 1000 m .......... 1000 km
(ix) 100 cm .......... 100 m
(x) 1 km .......... 1000 m
Sol: (i) 702 cm > 503 cm
So, 702 cm > 503 cm
(ii) 2 m … 9 m
Clearly, 2 m < 9 m
(iii) 800 cm … 80 m
Convert: 800 cm = 8 m.
So, 8 m < 80 m
→ 800 cm < 80 m
(iv) 702 cm … 5 m
Convert: 702 cm = 7 m 2 cm = 7.02 m.
So, 7.02 m > 5 m
→ 702 cm > 5 m
(v) 8 km … 7000 m
Convert: 8 km = 8000 m.
So, 8000 m > 7000 m
→ 8 km > 7000 m
(vi) 625 cm … 9 m
Convert: 625 cm = 6 m 25 cm = 6.25 m.
So, 6.25 m < 9 m
→ 625 cm < 9 m
(vii) 10 m … 4 cm
Convert: 10 m = 1000 cm.
So, 1000 cm > 4 cm
→ 10 m > 4 cm
(viii) 1000 m … 1000 km
Convert: 1000 m = 1 km.
So, 1 km < 1000 km
→ 1000 m < 1000 km
(ix) 100 cm … 100 m
Convert: 100 cm = 1 m.
So, 1 m < 100 m
→ 100 cm < 100 m
(x) 1 km … 1000 m
Convert: 1 km = 1000 m.
So, 1 km = 1000 m
Q9: A road between two villages is 8 km 250 m long. Out of this, workers have repaired 5 km 750 m. How much road is still left to repair?
Sol: Total length of road = 8 km 250 m
Repaired length = 5 km 750 m
First, convert both into metres:
8 km 250 m = 8000 + 250 = 8250 m
5 km 750 m = 5000 + 750 = 5750 m
Length left to repair = 8250 – 5750 = 2500 m
Now convert back into km and m:
2500 m = 2 km 500 m
2 km 500 m of road is still left to repair.
Q10: A cloth merchant bought a roll of cloth 45 m long. He sold 12 m 40 cm to one customer and 18 m 60 cm to another. How much cloth is left with him?
Sol: Total length of cloth = 45 m
Sold cloth:
To first customer = 12 m 40 cm = 12.40 m
To second customer = 18 m 60 cm = 18.60 m
Total cloth sold = 12.40 m + 18.60 m = 31.00 m
Cloth left = 45.00 m – 31.00 m = 14.00 m
14 m cloth is left with the merchant.
Q11: The distance from a school to the railway station is 4 km 300 m. The bus travels this distance in the morning and returns in the afternoon. What is the total distance covered by the bus in a day?
Sol: Distance from school to railway station = 4 km 300 m
Since the bus goes in the morning and comes back in the afternoon, the distance is covered two times.
Convert into metres first:
4 km 300 m = 4000 + 300 = 4300 m
Total distance = 2 × 4300 = 8600 m
Now convert back into km and m:
8600 m = 8 km 600 m
The bus covers 8 km 600 m in a day.
Q12: A ribbon is 96 m long. It is cut into 12 equal pieces. What is the length of each piece?
Sol: Total length of ribbon = 96 m
Number of equal pieces = 12
Length of each piece = 96 ÷ 12 = 8 m
Each piece is 8 m long.
Q13: A man walks 3 km every day. How many kilometers will he walk in 15 days?
Q14: A train travels 125 km in 1 hour. How much distance will it cover in 8 hours?
Sol: Distance covered in 1 hour = 125 km
Time = 8 hours
Total distance = 125 × 8 = 1000 km
The train will cover 1000 km in 8 hours.
Q15: A road is 420 m long. It is divided into 7 equal sections. What is the length of each section?
Sol: Total length of road = 420 m
Number of equal sections = 7
Length of each section = 420 ÷ 7 = 60 m
Each section is 60 m long.
The document Worksheet Solutions: Far and Near | Mathematics (Maths Mela) Class 5 - New NCERT is a part of the Class 5 Course Mathematics (Maths Mela) Class 5 - New NCERT.
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FAQs on Worksheet Solutions: Far and Near - Mathematics (Maths Mela) Class 5 - New NCERT
| 1. What is the main focus of the "Far and Near" topic in Class 8? |  |
Ans. The "Far and Near" topic in Class 8 primarily focuses on the concepts of distance and proximity in various contexts, including geography, physics, and daily life. It explores how different objects or places are perceived in terms of their distance from a point of reference, and the implications of these distances on vision, travel, and interaction.
| 2. How do we define the terms "far" and "near"? | |
Ans. "Far" refers to a considerable distance away from a reference point, making objects or locations less accessible or visible. In contrast, "near" indicates proximity to a reference point, allowing for easier access and visibility. These terms are relative and depend on the context and the observer’s perspective.
| 3. Can you explain the significance of understanding "far" and "near" in everyday life? | |
Ans. Understanding "far" and "near" is significant in everyday life as it influences our navigation, planning, and decision-making. For instance, recognizing how far a location is can affect travel time and costs. Additionally, it aids in spatial awareness, helping individuals interact effectively with their environment, whether it’s in finding directions or estimating distances for activities.
| 4. What are some examples of activities that illustrate the concepts of "far" and "near"? | |
Ans. Examples of activities that illustrate "far" and "near" include measuring distances while walking or driving, playing sports where distance affects performance (like basketball), and even in technology, such as using GPS to determine how far away a destination is. Additionally, educational activities, such as drawing objects at varying distances, can help reinforce these concepts.
| 5. How can teachers effectively teach students about the concepts of "far" and "near"? | |
Ans. Teachers can effectively teach students about "far" and "near" by using interactive activities such as mapping exercises, distance estimation games, and visual aids like pictures showing various distances. Incorporating technology, like virtual tours or distance-measuring apps, can also enhance understanding. Engaging students in discussions about their own experiences with distance can further solidify their grasp of these concepts.
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