✅ SOLUTIONS
Class 5
Chapter 5: Far and Near
Math Mela · NCERT Solutions
Class 5Math Mela (NCERT)Chapter 5
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📏 Let Us Find — Choosing Units
Identify appropriate units (m or cm) for each quantity.
Solution
| Quantity | Unit |
|---|---|
| Height of India Gate | Metre (m) |
| Length of a handkerchief | Centimetre (cm) |
| Depth of a well | Metre (m) |
| Length of a mobile phone | Centimetre (cm) |
| Length of an elephant's trunk | Metre (m) |
| Distance between two buttons on a shirt | Centimetre (cm) |
🧵 Different Units but Same Measure
Match the measures that represent the same sari/stole length.
Solution
Convert each mixed measurement to pure cm, then match:
| In cm | Equals | Matches |
|---|---|---|
| 204 cm | 200 cm + 4 cm | 2 m 4 cm |
| 540 cm | 500 cm + 40 cm | 5 m 40 cm |
| 750 cm | 600 cm + 150 cm | 6 m 150 cm |
| 240 cm | 200 cm + 40 cm | 2 m 40 cm |
| 404 cm | 200 cm + 204 cm | 2 m 204 cm |
Double number line: 100cm=1m, 200cm=2m, 400cm=4m, 500cm=5m, 700cm=7m
⚖️ Let Us Compare
Q1. Compare using <, =, >
Solution
| Part | Comparison | Working | Answer |
|---|---|---|---|
| (a) | 456 cm ___ 5 m | 5m=500cm | 456 cm < 5 m |
| (b) | 55cm+200cm ___ 200cm+54cm | 255cm vs 254cm | 255 cm > 254 cm |
| (c) | 6m5cm ___ 6m50cm | 605cm vs 650cm | 6m5cm < 6m50cm |
| (d) | 2m150cm ___ 3m50cm | 350cm vs 350cm | 2m150cm = 3m50cm |
| (e) | 238cm ___ 138cm+1m | 238cm vs 238cm | 238 cm = 138cm+1m |
Q2. World's tallest statues — Statue of Unity (182m), Spring Temple Buddha (128m), Guanyin of Nanshan (108m), Statue of Liberty (93m), Motherland Calls (91m), Christ the Redeemer (38m).
Solution
(a) Difference: Statue of Unity − Statue of Liberty = 182 − 93 = 89 m
(b) Least difference: Sorted heights: 38,91,93,108,128,182. Closest pair: Motherland Calls(91) & Statue of Liberty(93) → difference = 2 m
(c) Largest difference: Statue of Unity(182) & Christ the Redeemer(38) → difference = 182−38 = 144 m
(d) Half of 182 = 91 → Motherland Calls, Russia (91 m) — doubled, it equals the Statue of Unity's height!
🏃 Kilometre Race (3 km)
Q1. Water stations every 500 m — how many, and at what positions?
Solution
3 km = 3,000 m. Positions every 500m: 500, 1000, 1500, 2000, 2500, 3000
6 water stations needed, at 500m, 1,000m, 1,500m, 2,000m, 2,500m, and 3,000m (finish)
Q2. Children at intervals of 300 m — how many, and where?
Solution
3,000 ÷ 300 = 10
10 children needed, at positions 300m, 600m, 900m, 1,200m, 1,500m, 1,800m, 2,100m, 2,400m, 2,700m, 3,000m
Q3. Red and blue flags alternately every 50 m till finish — how many of each?
Solution
Number of flag positions = 3,000 ÷ 50 = 60 positions
Alternating colours: 30 red flags and 30 blue flags
🚂 Longest Train Journey — Vivek Express
Answer the questions based on the station-distance table.
Solution
| Q | Working | Answer |
|---|---|---|
| 1. Total route length | Distance to Kanniyakumari (last row) | 4,187 km |
| 2. Vijayawada − Jalpaiguri Road | 2,800 − 983 | 1,817 km |
| 3. Vijayawada − Visakhapatnam | 2,800 − 2,450 | 350 km |
| 4. Farther apart pair? | Guwahati−Dimapur=556−306=250km; Bhubaneswar−Jalpaiguri=2007−983=1,024km | Bhubaneswar & Jalpaiguri Road (1,024 km apart) |
| 5. Guwahati − Coimbatore JN | 3,675 − 556 | 3,119 km |
🔢 Unit Conversion — Let Us Do
Q1. Fill in the double number lines.
Solution
(a) cm ↔ mm (×10 conversion):
cm: 1, 7, 15, 25, 32, 45, 50 | mm: 10, 70, 150, 250, 320, 450, 500 (=0.5 m)
(b) m ↔ cm (×100 conversion):
m: 1, 4, 10, 12, 14, 17, 21 | cm: 100, 400, 1000, 1200, 1400, 1700, 2100
(c) km ↔ m (×1000 conversion):
km: 1, 9, 25, 41, 55, 67, 82 | m: 1000, 9000, 25000, 41000, 55000, 67000, 82000
Q2. Fill in the unit conversion blanks.
Solution
| Problem | Answer |
|---|---|
| (a) 4 cm 5 mm = ___ mm | 45 mm |
| (b) 89 mm = ___ cm ___ mm | 8 cm 9 mm(given) |
| (c) 234 cm = ___ mm | 2,340 mm |
| (d) 514 mm = ___ cm ___ mm | 51 cm 4 mm |
| (e) 6 m 34 cm = ___ cm | 634 cm |
| (f) 20 m 12 cm = ___ cm | 2,012 cm |
| (g) 397 m = ___ cm | 39,700 cm |
| (h) 5,792 cm = ___ m ___ cm | 57 m 92 cm(given) |
| (i) 9,108 cm = ___ m ___ cm | 91 m 8 cm |
| (j) 34 km = ___ m | 34,000 m |
| (k) 6,870 m = ___ km ___ m | 6 km 870 m |
| (l) 10,552 m = ___ km ___ m | 10 km 552 m |
| (m) 29 km 30 m = ___ m | 29,030 m |
| (n) 32 km 359 m = ___ m | 32,359 m |
➕➖ Adding & Subtracting Lengths
Q1. Rani has ribbons 3m75cm and 2m25cm. Total?
Solution
375 cm + 225 cm = 600 cm = 6 m exactly!
Q2. Bhopal-Sanchi = 48km700m. Waterfall is 17km900m from Bhopal. Waterfall to Sanchi distance?
Solution
48,700 m − 17,900 m = 30,800 m = 30 km 800 m
Q3. Gulmarg Gondola: section1=2km300m, section2=2km650m. Total distance?
Solution
2km300m + 2km650m = 4km + 950m = 4 km 950 m
Q4. Circle the bigger length and find the difference.
Solution
| Part | Compare | Bigger | Difference |
|---|---|---|---|
| (a) | 11mm vs 1cm(10mm) | 11 mm | 1 mm |
| (b) | 26mm vs 2cm(20mm) | 26 mm | 6 mm |
| (c) | 20cm(200mm) vs 201mm | 201 mm | 1 mm |
| (d) | 1020mm vs 1m(1000mm) | 1,020 mm | 20 mm |
| (e) | 2m(200cm) vs 245cm | 245 cm | 45 cm |
| (f) | 5678m vs 6km(6000m) | 6 km | 322 m |
| (g) | 6km1480m(7480m) vs 7km479m(7479m) | 6km 1480m | 1 m |
✖️➗ Multiplying & Dividing Lengths
Q2. Cloth costs ₹100 for 5m. Fill the double number line for cost/length.
Solution
Rate: ₹100 ÷ 5m = ₹20 per metre
5m→₹100 | 10m→₹200 | 20m→₹400 | 40m→₹800 | 100m→₹2,000
Q3. Anita needs 1m thread per 50cm sari border. Thread needed for 5m border? Cost at ₹50/m?
Solution
5 m border = 500 cm. Number of 50cm sections = 500 ÷ 50 = 10
Thread needed = 10 × 1 m = 10 m thread
Cost = 10 × ₹50 = ₹500
Q4. 12km600m road laid equally over 6 days. Length per day?
Solution
12 km 600 m = 12,600 m. 12,600 ÷ 6 = 2,100 m
Road laid per day = 2 km 100 m
🧩 Try This — Operation Grid Puzzle
Fill +,−,×,÷ so each row, column, AND each 2×2 box has no repeats.
Solution
Given clues: (1,1)=÷, (1,4)=+, (2,2)=×, (3,3)=+, (4,4)=−
| ÷ | − | × | + |
| + | × | − | ÷ |
| − | ÷ | + | × |
| × | + | ÷ | − |
✅ Every row has all 4 signs, every column has all 4 signs, and each of the four 2×2 corner boxes also contains all 4 signs exactly once!
💡 Method: Start from the given clues and use the "no repeat in row/column" rule to eliminate possibilities one cell at a time — just like solving a mini-Sudoku!