Applications: Paths, Tanks & Cost

🎯 Learning Objective

Solve mixed exam problems on paths and roads, tank capacity, and the cost of painting, flooring or fencing.

💡 Concept

• Path/road area = (outer area) − (inner area)
• Two crossing roads inside a field: area = road₁ + road₂ − overlap (subtract the square once)
• Cost = area (or length) × rate per unit
• Capacity of a tank: 1 m³ = 1000 litres
• Walls of a room = 2h(l + b); add the ceiling (l × b) if it is also painted

🧮 Key Formulas

Path area = outer area − inner area

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Crossing roads area = l×w + b×w − w²

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Cost = area × rate

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1 m³ = 1000 litres

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4 walls area = 2h(l + b)

✏️ Easy Example

Q. Find the cost of fencing a square field of side 25 m at ₹40 per metre.

1. Perimeter = 4 × side = 4 × 25 = 100 m
2. Cost = length × rate = 100 × 40

Answer: ₹4,000

🇮🇳 Real-Life Example

Estimating the paint for a railway waiting hall, or the concrete for a platform road, is exactly these path-and-cost sums done at scale.

📝 Exam-Level Example

Q. A rectangular garden 50 m × 30 m has a 2 m wide path around it on the outside. Find the area of the path and the cost of paving it at ₹20 per m².

1. Outer dimensions = (50 + 2 + 2) × (30 + 2 + 2) = 54 × 34 = 1836
2. Inner (garden) area = 50 × 30 = 1500
3. Path area = 1836 − 1500 = 336 m²
4. Cost = 336 × 20

Answer: Path = 336 m², Cost = ₹6,720

📝 Exam-Level Example

Q. A park is 60 m × 40 m. Two roads, each 3 m wide, run through the middle — one parallel to the length and one parallel to the breadth. Find the total road area and cost of construction at ₹50 per m².

1. Road along length = 60 × 3 = 180
2. Road along breadth = 40 × 3 = 120
3. Overlap counted twice = 3 × 3 = 9, subtract once
4. Total road area = 180 + 120 − 9 = 291 m²
5. Cost = 291 × 50

Answer: Road area = 291 m², Cost = ₹14,550

📝 Exam-Level Example

Q. A cuboidal water tank is 3 m long, 2 m wide and 1.5 m deep. How many litres of water can it hold?

1. Volume = l × b × h = 3 × 2 × 1.5 = 9 m³
2. 1 m³ = 1000 litres, so 9 × 1000

Answer: 9,000 litres

📝 Exam-Level Example

Q. A room is 6 m long, 5 m wide and 4 m high. Find the cost of painting its four walls at ₹25 per m².

1. Area of 4 walls = 2h(l + b) = 2 × 4 × (6 + 5)
2. = 8 × 11 = 88 m²
3. Cost = 88 × 25

Answer: ₹2,200

🪄 Memory Trick

Read carefully whether a path is inside or outside — outside adds twice the width to each dimension, inside subtracts it. That single word decides the whole answer.

⚠️ Common Mistakes

• ❌ Adding the path width only once instead of on both sides
• ❌ Forgetting to subtract the overlapping square in crossing-roads problems
• ❌ Using m³ directly as litres without the ×1000 conversion

🏆 Exam Tips

• ✅ Draw a quick rough figure — it instantly shows inside vs outside paths
• ✅ Convert every measurement to the same unit before multiplying

📌 Summary

• Path area = outer area − inner area
• Crossing roads = l×w + b×w − w² (remove double-counted square)
• Cost = area or length × rate
• 1 m³ = 1000 litres; 4 walls = 2h(l+b)