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Chapter 9: Problem 88

A train covers a certain distance moving at a speed of 60 \(\mathrm{km} / \mathrm{h}\). However if it were to halt for a fixed time every hour, its average speed comes out to be \(50 \mathrm{~km} / \mathrm{h}\). For how much tine does the train halt for every hour? (a) \(6 \mathrm{~min}\) (b) \(10 \mathrm{~min}\) (c) \(12 \mathrm{~min}\) (d) none of these

Short Answer

Expert verified
Answer: (b) 10 min

Step by step solution

01

Identify the given information

The train moves at a speed of 60 km/h, and its average speed including the halt time is 50 km/h.
02

Find the proportion of time the train is moving per hour

To find this proportion, we can divide the average speed by the speed of the moving train: \(\frac{50}{60} = \frac{5}{6}\). So, the train is moving for \(\frac{5}{6}\) of the hour.
03

Find the moving time per hour

Since the train is moving for \(\frac{5}{6}\) of the hour, the moving time per hour can be calculated by multiplying the total time of an hour with this fraction: \(60 \times \frac{5}{6} = 50\) minutes.
04

Calculate the halt time per hour

To find the halt time per hour, we can subtract the moving time per hour from the total time of an hour: \(60 - 50 = 10\) minutes. So, the train halts for 10 minutes every hour. The correct answer is (b) \(10 \mathrm{~min}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Average Speed
When solving problems like this, it's crucial to understand the concept of average speed. Average speed is not simply the arithmetic mean of speeds but involves the total distance and the total time taken. It is calculated using:
\[\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}\]Identifying average speed requires considering pauses or variations in speed over the journey. In our exercise, the train's average speed decreases due to halt times. The problem shows how halts impact the overall speed since the train's ideal speed is 60 km/h, but stops reduce the average to 50 km/h.
  • It's essential to always consider total time, including halts, when calculating average speed.
  • This real-life application helps in understanding how delays affect travel plans.
Train Problems
Train problems are a popular topic in distance-speed-time calculations. Here, the train travels a specific distance at different speeds due to halts.
Key to solving this is understanding how speed and time relate. Trains often involve calculations with steady speed interrupted by stops, impacting overall travel time.
  • Understand the difference between constant speed (movement) and average speed (including stops).
  • Translate time spent on halts into a practical understanding of journey length.
In the given exercise, we explore how stopping affects a journey, transforming a seemingly simple trip into a complex calculation of varied speeds and times. By determining how long the train stops, we adjust the puzzle pieces of travel, revealing the true dynamics of motion.
Ratios and Proportions
Ratios and proportions play a significant role in problems involving time, speed, and distance. In these calculations, ratios help compare different values and understand their relationship.
The proportion in our exercise is derived by comparing the average speed (50 km/h) with the constant speed (60 km/h). This gives us a proportion of:
\[\frac{50}{60} = \frac{5}{6}\]Highlighting how the train moves for \(\frac{5}{6}\) of an hour when at 60 km/h. This means using proportions to simplify and break complex time calculations into comprehensible segments.
  • Utilize ratios to zero in on chunks of time the train is moving versus halted.
  • Proportions allow for intuitive problem-solving by scaling parts of the journey against a full hour.
By understanding these core concepts, ratios can make intricate train travel scenarios much clearer and easier to approach.

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Most popular questions from this chapter

The distance between Lucknow and Delhi is \(700 \mathrm{~km}\). Rajdhani express starts from Delh? for the Lucknow at \(60 \mathrm{~km} / \mathrm{h} .50\) minutes later Lucknow express leaves Lucknow for Delhi on the parallel uracks at \(70 \mathrm{~km} / \mathrm{h}\). How far from Lucknow will they cross each other? (a) \(250 \mathrm{~km}\) (b) \(360 \mathrm{~km}\) (c) \(350 \mathrm{~km}\) (d) \(475 \mathrm{~km}\)

Pushpak express leaves Lucknow at 6 am and two hours later another train Bhopal express leaves Lucknow. Both trains arrive Bhopal at \(4 \mathrm{pm}\) on the same day. If the difference between their speeds be \(10 \mathrm{~km} / \mathrm{h}\), what is the average speeds of both the trains over entire route: (a) \(40 \mathrm{~km} / \mathrm{h}\) (b) \(44 \frac{4}{9} \mathrm{~km} / \mathrm{h}\) (c) \(42 \frac{3}{5} \mathrm{~km} / \mathrm{h}\) (d) none of these

Two trains lewe Meerut at the difference of 4 hours. The first train leaves of 8 am at \(40 \mathrm{~km} / \mathrm{h}\) and the faster train leaves later at \(60 \mathrm{kmh}\) in the same direction. When the faster train will overtake the slower train? (a) \(4 \mathrm{pm}\) (b) \(2 \mathrm{pm}\) (c) \(8 \mathrm{pm}\) (d) \(6: 30 \mathrm{pm}\)

A motor boat takes 12 hours to go downstream and it tabes 24 hours to return the same distance. What is the time taken by boat in still water? (a) \(15 \mathrm{~h}\) (b) \(16 \mathrm{~h}\) (c) \(8 \mathrm{~h}\) (d) \(20 \mathrm{~h}\)

In a 1000 metres race Ravi gives Vinod a start of \(40 \mathrm{~m}\) and beats him by 19 seconds. If Ravi gives a start of 30 seconds then Vinod beats Ravi by \(40 \mathrm{~m}\). What is the ratio of speed of Ravi to that of Vinod? (a) \(4: 5\) (b) \(6: 5\) (c) \(3: 8\) (d) \(5: 4\)
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