The Ever-Evolving World of Science – Numerical Problems with Stepwise Solutions
CBSE Board Examinations – Systematic Numerical Practice
These NCERT-based Numerical Problems with Stepwise Solutions for Class 7 Science help in strong concept building and exam readiness for:
- CBSE Class 7 Annual / Term-End Examinations
- Class Tests, Periodic Tests and Internal Assessments
- Olympiads and Science Talent Exams (Class 7 level)
- Foundation for higher classes (Class 8–10)
Content Bank – Important Formulas from the Chapter
- Speed = Distance ÷ Time
- Time = Distance ÷ Speed
- Distance = Speed × Time
- Average = Sum of observations ÷ Number of observations
- Percentage = (Part ÷ Whole) × 100
- Energy consumed (Wh) = Power (W) × Time (hours)
- Energy consumed (kWh or unit) = Power (kW) × Time (hours)
- Temperature difference = Final temperature – Initial temperature
- Unit conversions: 1 km = 1000 m, 1 h = 60 min, 1 min = 60 s, 1 kL = 1000 L
Chapter 1 – "The Ever-Evolving World of Science" introduces the nature and scope of science and shows how scientific ideas are used in everyday life. It also builds basic understanding of scientific inquiry through observation, measurement and reasoning.
Below are 20 topic-wise Numerical Problems with clear, step-by-step solutions, designed strictly as per the NCERT syllabus for Class 7 Science.
Topic 1: Measurement and Speed in Everyday Life
Problem 1. Riya walks 3 km from her home to the library in 45 minutes. Find her walking speed in km/h.
- Given distance = 3 km, time = 45 minutes.
- Convert time in minutes to hours: \(45 \text{ min} = \frac{45}{60} = 0.75 \text{ h}\).
- Use the formula: Speed = Distance ÷ Time.
- So, speed = \( \frac{3}{0.75} = 4 \text{ km/h}\).
Final Answer: Riya’s walking speed = 4 km/h.
Problem 2. A student goes to school on a bicycle at a speed of 12 km/h and reaches in 1.5 hours. How far is his school from his home?
- Given speed = 12 km/h, time = 1.5 h.
- Use the formula: Distance = Speed × Time.
- Distance = \(12 \times 1.5 = 18 \text{ km}\).
Final Answer: The school is 18 km away from his home.
Problem 3. A boy takes 20 minutes to reach his tuition centre by walking at a speed of 4.5 km/h. Find the distance between his home and the tuition centre.
- Given speed = 4.5 km/h, time = 20 minutes.
- Convert time to hours: \(20 \text{ min} = \frac{20}{60} = \frac{1}{3} \text{ h}\).
- Use the formula: Distance = Speed × Time.
- Distance = \(4.5 \times \frac{1}{3} = 1.5 \text{ km}\).
Final Answer: The distance is 1.5 km.
Problem 4. A car travels at 60 km/h for 2 hours and then at 40 km/h for 1 hour. Calculate:
- (a) The total distance travelled.
- (b) The average speed for the whole journey.
- First part: speed = 60 km/h, time = 2 h.
- Distance in first part = \(60 \times 2 = 120 \text{ km}\).
- Second part: speed = 40 km/h, time = 1 h.
- Distance in second part = \(40 \times 1 = 40 \text{ km}\).
- (a) Total distance = \(120 + 40 = 160 \text{ km}\).
- Total time = \(2 + 1 = 3 \text{ h}\).
- (b) Average speed = Total distance ÷ Total time = \( \frac{160}{3} \approx 53.33 \text{ km/h}\).
Final Answer: (a) Total distance = 160 km; (b) Average speed ≈ 53.3 km/h.
Topic 2: Electricity, Energy Use and Everyday Calculations
Problem 5. A 60 W bulb is used for 5 hours in a day. Calculate the electrical energy consumed by the bulb in Wh and in kWh.
- Given power = 60 W, time = 5 h.
- Use the formula: Energy (Wh) = Power (W) × Time (h).
- Energy = \(60 \times 5 = 300 \text{ Wh}\).
- To convert to kWh, divide by 1000: \(300 \text{ Wh} = \frac{300}{1000} = 0.3 \text{ kWh}\).
Final Answer: Energy consumed = 300 Wh = 0.3 kWh.
Problem 6. A ceiling fan of 100 W runs for 3 hours every day. Find the total energy used by the fan in 10 days (in kWh).
- Daily energy = Power × Time = \(100 \times 3 = 300 \text{ Wh}\).
- For 10 days, energy = \(300 \times 10 = 3000 \text{ Wh}\).
- Convert Wh to kWh: \(3000 \text{ Wh} = \frac{3000}{1000} = 3 \text{ kWh}\).
Final Answer: The fan uses 3 kWh of energy in 10 days.
Problem 7. A family uses 90 kWh (units) of electricity in a month. If the cost of 1 kWh is ₹6, calculate the total electricity bill for the month.
- Given energy used = 90 kWh, cost per kWh = ₹6.
- Use the formula: Total cost = Energy used × Cost per unit.
- Total cost = \(90 \times 6 = 540\).
Final Answer: Electricity bill for the month = ₹540.
Problem 8. A 40 W bulb and a 10 W LED lamp are used for 4 hours daily for 30 days.
- (a) Find the energy used by each (in kWh).
- (b) How much energy is saved by using the LED instead of the bulb?
- Time used in 30 days = \(4 \times 30 = 120 \text{ h}\).
- Bulb (40 W): Energy = \(40 \times 120 = 4800 \text{ Wh} = 4.8 \text{ kWh}\).
- LED (10 W): Energy = \(10 \times 120 = 1200 \text{ Wh} = 1.2 \text{ kWh}\).
- Energy saved = \(4.8 - 1.2 = 3.6 \text{ kWh}\).
Final Answer: (a) Bulb: 4.8 kWh, LED: 1.2 kWh; (b) Energy saved = 3.6 kWh.
Topic 3: Heat, Temperature and Averages
Problem 9. The temperature of a city on four days is 25°C, 27°C, 26°C and 28°C. Find the average temperature of these four days.
- Add all temperatures: \(25 + 27 + 26 + 28 = 106\).
- Number of days = 4.
- Use the formula: Average = Sum ÷ Number of observations.
- Average temperature = \( \frac{106}{4} = 26.5^\circ\text{C}\).
Final Answer: Average temperature = 26.5°C.
Problem 10. The temperature of water in a beaker increases from 24°C to 32°C in 8 minutes when heated. Find the average rise in temperature per minute.
- Initial temperature = 24°C, final temperature = 32°C.
- Temperature rise = \(32 - 24 = 8^\circ\text{C}\).
- Time taken = 8 minutes.
- Rise per minute = \( \frac{8^\circ\text{C}}{8 \text{ min}} = 1^\circ\text{C per minute}\).
Final Answer: Average rise in temperature = 1°C per minute.
Problem 11. 500 mL of water at 20°C is mixed with 500 mL of water at 40°C in a well-insulated container (no heat is lost). What will be the final temperature of the mixture?
- The amounts of water are equal (500 mL each), so we can take the average of the two temperatures.
- Final temperature = \(\frac{20^\circ\text{C} + 40^\circ\text{C}}{2} = \frac{60}{2} = 30^\circ\text{C}\).
Final Answer: Final temperature of the mixture = 30°C.
Problem 12. The temperature inside a refrigerator is 5°C while the room temperature is 35°C.
- (a) Find the temperature difference between the room and the refrigerator.
- (b) If the power goes off and the temperature inside rises by 2°C every 10 minutes, how long will it take for the inside temperature to reach 15°C?
- (a) Temperature difference = \(35 - 5 = 30^\circ\text{C}\).
- (b) Initial inside temperature = 5°C, final = 15°C.
- Rise needed = \(15 - 5 = 10^\circ\text{C}\).
- Given rise = 2°C in 10 minutes.
- To rise 10°C, number of steps = \( \frac{10}{2} = 5\).
- Total time = \(5 \times 10 = 50 \text{ minutes}\).
Final Answer: (a) Temperature difference = 30°C; (b) Time taken = 50 minutes.
Topic 4: Data Handling, Percentages and Scientific Surveys
Problem 13. In a class survey of 40 students, 18 students said that Science is their favourite subject. Find the percentage of students who like Science the most.
- Total students = 40, students liking Science = 18.
- Use the formula: Percentage = (Part ÷ Whole) × 100.
- Percentage = \(\frac{18}{40} \times 100 = 45\%\).
Final Answer: 45% of the students like Science the most.
Problem 14. The rainfall (in mm) recorded in a town for 7 days is: 4, 0, 12, 0, 8, 10, 6. Find the average rainfall for these 7 days.
- Add all rainfall amounts: \(4 + 0 + 12 + 0 + 8 + 10 + 6 = 40 \text{ mm}\).
- Number of days = 7.
- Average rainfall = \( \frac{40}{7} \approx 5.71 \text{ mm}\).
Final Answer: Average rainfall ≈ 5.7 mm (correct to one decimal place).
Problem 15. In a school, 120 students of Class 7 were asked if they visited the Science exhibition. 72 students said “Yes”.
- (a) What fraction of students visited the exhibition?
- (b) What percentage of students visited the exhibition?
- (a) Fraction = \(\frac{72}{120}\).
- Simplify: divide numerator and denominator by 24: \(\frac{72 \div 24}{120 \div 24} = \frac{3}{5}\).
- (b) Percentage = \( \frac{72}{120} \times 100 = 60\%\).
Final Answer: (a) Fraction = 3/5; (b) Percentage = 60%.
Problem 16. Heights of 5 plants grown in the school garden are 12 cm, 15 cm, 13 cm, 10 cm and 20 cm. Find the average height of the plants.
- Add all heights: \(12 + 15 + 13 + 10 + 20 = 70 \text{ cm}\).
- Number of plants = 5.
- Average height = \( \frac{70}{5} = 14 \text{ cm}\).
Final Answer: Average height of the plants = 14 cm.
Topic 5: Estimation and Unit Conversions in Science
Problem 17. A science documentary lasts for 2.5 hours. Convert this time into:
- (a) Minutes
- (b) Seconds
- 2.5 hours = 2 hours + 0.5 hour.
- (a) Convert to minutes: \(2.5 \times 60 = 150 \text{ minutes}\).
- (b) Convert minutes to seconds: \(150 \times 60 = 9000 \text{ seconds}\).
Final Answer: (a) 150 minutes; (b) 9000 seconds.
Problem 18. A scientist measures the length of a small road as 3.6 km. Express this distance in metres.
- Use the conversion: 1 km = 1000 m.
- 3.6 km = \(3.6 \times 1000 = 3600 \text{ m}\).
Final Answer: The distance is 3600 m.
Problem 19. A car travels 300 km in 5 hours at a constant speed.
- (a) Find the speed of the car in km/h.
- (b) How long will it take to travel 90 km at the same speed?
- (a) Use Speed = Distance ÷ Time = \( \frac{300}{5} = 60 \text{ km/h}\).
- (b) Distance to travel = 90 km, speed = 60 km/h.
- Time = Distance ÷ Speed = \( \frac{90}{60} = 1.5 \text{ h}\).
- Convert 1.5 h to hours and minutes: 1 h + 0.5 h = 1 h 30 min.
Final Answer: (a) Speed = 60 km/h; (b) Time = 1 hour 30 minutes.
Problem 20. A city uses 25,000 kL (kilolitres) of water per day.
- (a) Convert this amount into litres.
- (b) If the population of the city is 1,00,000, how many litres of water are available per person per day (on average)?
- (a) Use the conversion: 1 kL = 1000 L.
- Total water in litres = \(25{,}000 \times 1000 = 25{,}000{,}000 \text{ L}\) (25 million litres).
- (b) Population = 1,00,000 people.
- Water per person per day = Total water ÷ Population.
- Water per person = \( \frac{25{,}000{,}000}{100{,}000} = 250 \text{ L}\).
Final Answer: (a) Total water = 25,000,000 L; (b) Water per person per day = 250 L.