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Question

A train runs at a speed of 90 km/h in the first 10 minutes and 60 km/h in next 25 minutes and 15 km/h in last 4 minutes. What is the average speed of the train?

The correct answer is \(63{1 \over13}\)km/h

Calculating Train Average Speed

The question asks for the average speed of a train that travels at different speeds over different time intervals. To find the average speed, we need to calculate the total distance covered by the train and divide it by the total time taken for the journey.

The formula for average speed is:

$$ \text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} $$

The journey is divided into three segments with different speeds and durations.

Step 1: Calculate the distance covered in each segment

We know that Distance = Speed $\times$ Time. We must ensure that units are consistent. The speeds are given in km/h, and the times are given in minutes. Let's convert the time for each segment into hours.

Segment 1:

  • Speed = 90 km/h
  • Time = 10 minutes
  • Converting time to hours: $$ 10 \text{ minutes} = 10 \times \frac{1}{60} \text{ hours} = \frac{1}{6} \text{ hours} $$
  • Distance in Segment 1: $$ \text{Distance}_1 = 90 \text{ km/h} \times \frac{1}{6} \text{ h} = \frac{90}{6} \text{ km} = 15 \text{ km} $$

Segment 2:

  • Speed = 60 km/h
  • Time = 25 minutes
  • Converting time to hours: $$ 25 \text{ minutes} = 25 \times \frac{1}{60} \text{ hours} = \frac{25}{60} \text{ hours} = \frac{5}{12} \text{ hours} $$
  • Distance in Segment 2: $$ \text{Distance}_2 = 60 \text{ km/h} \times \frac{5}{12} \text{ h} = \frac{60 \times 5}{12} \text{ km} = 5 \times 5 \text{ km} = 25 \text{ km} $$

Segment 3:

  • Speed = 15 km/h
  • Time = 4 minutes
  • Converting time to hours: $$ 4 \text{ minutes} = 4 \times \frac{1}{60} \text{ hours} = \frac{4}{60} \text{ hours} = \frac{1}{15} \text{ hours} $$
  • Distance in Segment 3: $$ \text{Distance}_3 = 15 \text{ km/h} \times \frac{1}{15} \text{ h} = \frac{15}{15} \text{ km} = 1 \text{ km} $$

Step 2: Calculate the total distance

The total distance is the sum of the distances covered in each segment.

$$ \text{Total Distance} = \text{Distance}_1 + \text{Distance}_2 + \text{Distance}_3 $$

$$ \text{Total Distance} = 15 \text{ km} + 25 \text{ km} + 1 \text{ km} = 41 \text{ km} $$

Step 3: Calculate the total time

The total time is the sum of the time taken for each segment. It's easier to sum in minutes first and then convert to hours for the final calculation.

Total time in minutes = 10 minutes + 25 minutes + 4 minutes = 39 minutes.

Converting total time to hours:

$$ \text{Total Time} = 39 \text{ minutes} = 39 \times \frac{1}{60} \text{ hours} = \frac{39}{60} \text{ hours} = \frac{13}{20} \text{ hours} $$

Step 4: Calculate the average speed

Now, use the formula for average speed with the total distance and total time.

$$ \text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{41 \text{ km}}{\frac{13}{20} \text{ h}} $$

$$ \text{Average Speed} = 41 \times \frac{20}{13} \text{ km/h} = \frac{820}{13} \text{ km/h} $$

Step 5: Convert the average speed to a mixed fraction

To convert $\frac{820}{13}$ to a mixed fraction, divide 820 by 13.

$$ 820 \div 13 $$

$$ 820 = 13 \times 63 + 1 $$

The quotient is 63 and the remainder is 1. So, the mixed fraction is $63 \frac{1}{13}$.

Therefore, the average speed of the train is $63 \frac{1}{13}$ km/h.

Segment Speed (km/h) Time (minutes) Time (hours) Distance (km)
1 90 10 $10/60 = 1/6$ $90 \times 1/6 = 15$
2 60 25 $25/60 = 5/12$ $60 \times 5/12 = 25$
3 15 4 $4/60 = 1/15$ $15 \times 1/15 = 1$
Total - 39 $39/60 = 13/20$ $15 + 25 + 1 = 41$

Average Speed = Total Distance / Total Time = $41 \text{ km} / (13/20) \text{ h} = 41 \times 20/13 \text{ km/h} = 820/13 \text{ km/h}$

$$ \frac{820}{13} = 63 \frac{1}{13} $$

The average speed of the train is $63 \frac{1}{13}$ km/h.

Revision Table: Average Speed Calculation

Concept Formula/Definition Application in this problem
Average Speed Total Distance / Total Time Used to find the overall speed considering varying speeds.
Distance Speed $\times$ Time Calculated for each segment of the journey.
Unit Conversion Minutes to Hours (divide by 60) Essential for consistency when speed is in km/h.
Total Distance Sum of distances of all segments Added $15 \text{ km} + 25 \text{ km} + 1 \text{ km} = 41 \text{ km}$.
Total Time Sum of time for all segments Added $10 + 25 + 4 = 39$ minutes, converted to $13/20$ hours.

Additional Information: Speed, Distance, and Time

Speed, distance, and time are fundamental concepts in physics and mathematics, often related by the formula: $$ \text{Speed} = \frac{\text{Distance}}{\text{Time}} $$ This can be rearranged to find distance ($\text{Distance} = \text{Speed} \times \text{Time}$) or time ($\text{Time} = \frac{\text{Distance}}{\text{Speed}}$).

When an object travels at different speeds over different parts of its journey, the average speed is not simply the average of the speeds. It must be calculated using the total distance and the total time.

  • Constant Speed: If speed is constant, the relationship is straightforward.
  • Varying Speed: When speed varies, the journey is often broken down into segments. Calculate distance for each segment and sum them up. Sum the time taken for each segment. Then, use the total distance and total time to find the average speed.
  • Units: Always pay close attention to units (e.g., km/h, m/s, minutes, hours) and ensure they are consistent throughout the calculation. Convert units if necessary.
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Important Questions from Average Speed

  1. A car covers the first 41 km of its journey in 45 min and covers the remaining 23 km in 35 min. What is the average speed (in m/sec) of the car?

  2. Akhil drives a car from his home to office at an average speed of 50 km/h and reaches office 10 minutes early. But one day due to some problem with the car, he could drive at an average speed of 30 km/h only and reached office 10 minutes late. How far is his office from home ?

  3. During a flight of 900 km, an aircraft was slowed down due to bad weather. Its average speed for the trip was reduced by 300 km/h and the time of flight increased by 30 min. The original duration of the flight was :

  4. If a man travels from A to B at a speed of 50 km/h and returns by increasing his speed by 40%, then find his average speed (to 2 decimal places) for both the trips.

  5. A man travels a distance of 420 km by train which moves at the speed of 75 km/h and returns back by car at the speed of 50 km/h. Find his average speed for the whole journey.

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