Direct and Inverse Proportions Worksheet 97 | Filo - Free Printable

Problem Set: Direct and Inverse Proportions

We will solve each problem step by step, explaining the reasoning and calculations.

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#### Problem 1:
The scale of a map is given as \(1 : 50000000\). Two cities are 5 cm apart in the map. Find the actual distance between them.

Solution:
- The scale \(1 : 50000000\) means that 1 cm on the map represents 50,000,000 cm in reality.
- Distance between the two cities on the map = 5 cm.
- Actual distance = \(5 \times 50000000\) cm.
- Convert cm to km: \(1 \text{ km} = 100000 \text{ cm}\).
- Actual distance in km = \(\frac{5 \times 50000000}{100000} = 2500 \text{ km}\).

Answer: \(\boxed{2500 \text{ km}}\)

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#### Problem 2:
135 members in a hostel consumed 2025 kg of cereals in a month. Find the quantity of cereals required for 102 members during the month.

Solution:
- Let the quantity of cereals consumed per member per month be \(x\) kg.
- For 135 members, total consumption = \(135x = 2025\) kg.
- Solving for \(x\): \(x = \frac{2025}{135} = 15\) kg/member/month.
- For 102 members, total consumption = \(102 \times 15 = 1530\) kg.

Answer: \(\boxed{1530 \text{ kg}}\)

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#### Problem 3:
A car travels 11 km in 12 minutes. If the speed remains the same, how far will it travel in 3 hours 48 minutes?

Solution:
- Speed of the car = \(\frac{\text{Distance}}{\text{Time}} = \frac{11 \text{ km}}{12 \text{ minutes}}\).
- Convert 3 hours 48 minutes to minutes: \(3 \times 60 + 48 = 228\) minutes.
- Distance traveled in 228 minutes = Speed \(\times\) Time = \(\frac{11}{12} \times 228\) km.
- Simplify: \(\frac{11 \times 228}{12} = \frac{11 \times 19}{1} = 209\) km.

Answer: \(\boxed{209 \text{ km}}\)

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#### Problem 4:
There are 1200 students in a hostel. Food provision for them is for 25 days. How long will these provisions last if 300 more students join the group?

Solution:
- Total food provision for 1200 students for 25 days.
- Total food units = \(1200 \times 25 = 30000\) student-days.
- After 300 more students join, total number of students = \(1200 + 300 = 1500\).
- Let the food last for \(x\) days.
- Total food units consumed = \(1500 \times x\).
- Equating the total food units: \(1500 \times x = 30000\).
- Solving for \(x\): \(x = \frac{30000}{1500} = 20\) days.

Answer: \(\boxed{20 \text{ days}}\)

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#### Problem 5:
12 labourers can complete a piece of work in 72 days. How many men will be required to do the same work in 54 days?

Solution:
- Total work units = Number of labourers \(\times\) Number of days = \(12 \times 72 = 864\) labourer-days.
- Let the number of men required to complete the work in 54 days be \(x\).
- Total work units = \(x \times 54\).
- Equating the total work units: \(x \times 54 = 864\).
- Solving for \(x\): \(x = \frac{864}{54} = 16\).

Answer: \(\boxed{16}\)

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#### Problem 6:
Subhash cycles to his school at an average speed of 12 km/h. It takes him 20 minutes to reach the school. If he wants to reach his school in 15 minutes, what should be his average speed?

Solution:
- Distance to school = Speed \(\times\) Time = \(12 \text{ km/h} \times \frac{20}{60} \text{ hours} = 12 \times \frac{1}{3} = 4\) km.
- If Subhash wants to reach in 15 minutes, time = \(\frac{15}{60} = \frac{1}{4}\) hours.
- Required speed = \(\frac{\text{Distance}}{\text{Time}} = \frac{4 \text{ km}}{\frac{1}{4} \text{ hours}} = 4 \times 4 = 16\) km/h.

Answer: \(\boxed{16 \text{ km/h}}\)

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Final Answers:

1. \(\boxed{2500 \text{ km}}\)
2. \(\boxed{1530 \text{ kg}}\)
3. \(\boxed{209 \text{ km}}\)
4. \(\boxed{20 \text{ days}}\)
5. \(\boxed{16}\)
6. \(\boxed{16 \text{ km/h}}\)