Question 1: Express the following speeds in meter/sec:

\displaystyle \text{i)  }72 \text{ km/hr } \displaystyle \text{ii)  }117 \text{ km/hr } \displaystyle \text{iii)  }5.4 \text{ km/hr } \displaystyle \text{iv)  }12.6 \text{ km/hr }

Answer:

\displaystyle \text{i)  }72\hspace{2pt} \frac{\text{km}}{\text{hr}} = \frac{72\times 1000 \ m}{3600 \ sec} = 20\hspace{2pt} \frac{\text{m}}{\text{s}}  

\displaystyle \text{ii)  }117\hspace{2pt} \frac{\text{km}}{\text{hr}} = \frac{117\times 1000 \ m}{3600 \ sec} = 32.5\hspace{2pt} \frac{\text{m}}{\text{s}}  

\displaystyle \text{iii)  }5.4\hspace{2pt} \frac{\text{km}}{\text{hr}} = \frac{5.4\times 1000 \ m}{3600 \ sec} = 1.5\hspace{2pt} \frac{\text{m}}{\text{s}}  

\displaystyle \text{iv)  }12.6\hspace{2pt} \frac{\text{km}}{\text{hr}} = \frac{12.6 \times 1000 \ m}{3600 \ sec} = 3.5\hspace{2pt} \frac{\text{m}}{\text{s}}  

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Question 2: Express the following speeds in km/hr:

\displaystyle \text{i)  }18 \text{ m/sec } \displaystyle \text{ii)  }2 \text{ m/sec } \displaystyle \text{iii)  }3  \frac{1}{3} \text{ m/sec } \displaystyle \text{iv)  }12.5 \text{ m/sec }

Answer:

\displaystyle \text{i)  }18\hspace{2pt} \frac{\text{m}}{\text{s}} = \frac{18 \ km \times 3600}{1000 \ hr} = 64.8\hspace{2pt} \frac{\text{km}}{\text{hr}}  

\displaystyle \text{ii)  }2\hspace{2pt} \frac{\text{m}}{\text{s}}= \frac{2 \ km \times 3600}{1000 \ hr} = 7.2\hspace{2pt} \frac{\text{km}}{\text{hr}}  

\displaystyle \text{iii)  }3\frac{1}{3}\hspace{2pt} \frac{\text{m}}{\text{s}} = \frac{10 \ km \times 3600}{3 \times 1000 \ hr} = 12\hspace{2pt} \frac{\text{km}}{\text{hr}}  

\displaystyle \text{iv)  }12.5\hspace{2pt} \frac{\text{m}}{\text{s}} = \frac{12.5 \ km \times 3600}{1000 \ hr} = 45\hspace{2pt} \frac{\text{km}}{\text{hr}}  

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Question 3: An athlete covers a distance of \displaystyle 1200 meters in \displaystyle 4 minutes \displaystyle 48 seconds. Find his speed in km/hr.

Answer:

\displaystyle Speed = \frac{1200 \ m}{4 min \ 48 \ sec} = \frac{1200 \ km \times 3600}{1000 \times 288 \ hr} = 15\hspace{2pt}\frac{\text{km}}{\text{hr}}

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Question 4: Walking at the rate of \displaystyle 4 \text{ km/hr } a man covers a certain distance in \displaystyle 2\frac{1}{2} hours. How much time will be taken by the man to cover the same distance, if he cycles at \displaystyle 12 \text{ km/hr } ?

Answer:

\displaystyle \text{Distance covered walking    } = 4 \times 2.5 = 10 \text{ km } 

\displaystyle \text{Time to cover 10 km on cycle    } = \frac{10 \ km \times hr}{12 \ km} = \frac{5}{6} \text{ hr or 50 minutes }   

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Question 5: A car can finish a certain journey in \displaystyle 10 hours at a speed of \displaystyle 48 \text{ km/hr } . By how much the speed of car must be increased to cover the same distance in \displaystyle 8 hours?

Answer:

\displaystyle \text{Distance covered by the car    } = 48 \times 10 = 480 \text{ km } 

\displaystyle \text{Speed to cover the same distance in 8 hr    } = \frac{480 \ km}{8 \ hr} = 60 \frac{\text{km}}{\text{hr}}  

\displaystyle \text{Hence the speed must be increased by    } = 60 -48 = 12\hspace{2pt} \frac{\text{km}}{\text{hr}}

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Question 6: A bus covers a certain distance in \displaystyle 50 minutes, if it runs at a speed of \displaystyle 54 \text{ km/hr } . What must be the speed of the bus in order to reduce the time of journey to \displaystyle 40 minutes?

Answer:

\displaystyle \text{Distance covered by the bus    } = \frac{50}{60} \times 54 = 45 \frac{\text{km}}{\text{hr}}

\displaystyle \text{Speed to cover the same distance in 40 min    } = \frac{45 \ km \times 60}{40 \ hr } = 67.5 \frac{\text{km}}{\text{hr}}

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Question 7: A motor car starts with the speed of \displaystyle 70 \text{ km/hr } with its speed increasing every two hours by \displaystyle 10 \text{ km/hr } . In how much time will it cover a distance of \displaystyle 345 km?

Answer:

\displaystyle \text{Speed at the start    } = 70\frac{\text{km}}{\text{hr}}

\displaystyle \text{Distance covered in first 2 hours    } = 140 \frac{\text{km}}{\text{hr}}

\displaystyle \text{Speed After 2 hours    } = 80 \frac{\text{km}}{\text{hr}}

\displaystyle \text{Distance covered in 3rd and the 4th hour    } = 160 \text{ km } 

\displaystyle \text{Total distance covered by end of 4th hours    } = 140 + 160 = 300 \text{ km } 

\displaystyle \text{Distance left to be covered after end of 4th hour    } = 345 - 300 = 45 \text{ km } 

\displaystyle \text{Speed After 4 hours    } = 90 \frac{\text{km}}{\text{hr}}

\displaystyle \text{Time taken to cover 45 km at speed of    } 90 \frac{\text{km}}{\text{hr}} = 0.5 \text{ hr } 

\displaystyle \text{Hence the total time to cover    } 345 km = 4.5 \text{ hr } 

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Question 8: A man takes \displaystyle 150 steps in walking \displaystyle 75 meters. If he takes \displaystyle 3 steps in \displaystyle 1 second, find his speed in (i) m/sec (ii) km/hr.

Answer:

\displaystyle \text{Length of one step    } = \frac{75}{150} = 0.5 \text{m}

\displaystyle \text{Distance covered in 1 sec    } = 0.5\times3 = 1.5 \frac{\text{m}}{\text{s}}

i) \displaystyle \text{Speed in    } \frac{\text{m}}{\text{s}} = 1.5 \frac{\text{m}}{\text{s}}

ii) \displaystyle \text{Speed in    } \frac{\text{km}}{\text{hr}} = \frac{1.5 km \times 3600}{1000\times hr} = 5.4 \frac{\text{km}}{\text{hr}}

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Question 9: A man walks at \displaystyle 5 \text{ km/hr } for \displaystyle 6 hours and at \displaystyle 4 \text{ km/hr } for \displaystyle 12 hours. Find his average speed.

Answer:

\displaystyle \text{Average Speed    } = \frac{Distance \ Covered}{Total \ time \ taken \ to \ cover \ the distance}  

\displaystyle \text{Average Speed    } = \frac{5 \times 6 + 4 \times 12}{6 + 12} = 4 \frac{1}{3} \frac{\text{km}}{\text{hr}}  

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Question 10: A man covers a distance of \displaystyle 144 km at the speed of \displaystyle 36 \text{ km/hr } and another \displaystyle 256 km at the speed of \displaystyle 64 \text{ km/hr } . Find his average speed for the whole journey.

Answer:

\displaystyle \text{Average Speed    } = \frac{Distance \ Covered}{Total \ time \ taken \ to \ cover \ the distance}  

\displaystyle \text{Average Speed    } = \frac{144 +256}{\frac{144}{36} + \frac{256}{64}} = 50 \frac{\text{km}}{\text{hr}}

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Question 11: Two buses travel to a place at \displaystyle 45 \text{ km/hr } and \displaystyle 60 \text{ km/hr } respectively. If the second bus takes \displaystyle 5\frac{1}{2} \text{ hr }  hours less than the first for the same journey, find the length of the journey.

Answer:

\displaystyle \text{Let    } x  be the distance of the journey

\displaystyle \text{Time taken by the First bus    } = \frac{x}{45} \text{ hr } 

\displaystyle \text{Time taken by the Second bus    } = \frac{x}{60} \text{ hr } 

\displaystyle \text{Therefore    } \colon \frac{x}{45}- \frac{x}{60} = \frac{11}{2}  

\displaystyle \text{Solving for    } x = 990 \text{ km } 

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Question 12: A boy goes to school from his village at \displaystyle 3 \text{ km/hr } and returns back at \displaystyle 2 \text{ km/hr } . If he takes \displaystyle 5 hours in all, what is the distance between the village and the school?

Answer:

Let \displaystyle x be the distance of the journey

\displaystyle \text{Time taken to reach school    } = \frac{x}{3} \text{ hr } 

\displaystyle \text{Time taken return from    } = \frac{x}{2} \text{ hr } 

\displaystyle \text{Therefore    } \colon \frac{x}{3} + \frac{x}{2} = 5

\displaystyle \text{Solving for    } x = 6 \text{ km } 

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Question 13: A bus completes a journey of \displaystyle 420 km in \displaystyle 6\frac{1}{2} \text{ hr }  . The first \displaystyle \frac{3}{4} part of the journey performed at \displaystyle 63 \text{ km/hr } . Calculate the speed of the rest of the journey.

Answer:

Let \displaystyle x be the speed of the rest of the journey

\displaystyle \Rightarrow \frac{3}{4} \times 420 \times \frac{1}{63} + \frac{105}{x} = 6 \frac{1}{2}

\displaystyle \text{Solving for    } x = 70 \frac{\text{km}}{\text{hr}}

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Question 14: A man drives \displaystyle 150 km from town A to town B in \displaystyle 3 hours \displaystyle 20 min and returns back to town A from town B in \displaystyle 4 hours \displaystyle 10 min. Find his average speed for the whole journey.

Answer:

\displaystyle \text{Average Speed    } =  \frac{Distance \ Covered}{Total \ time \ taken \ to \ cover \ the distance}  

\displaystyle \text{Average Speed    } =  \frac{150 + 150}{3\frac{1}{3} + 4\frac{1}{6}} = 40 \frac{\text{km}}{\text{hr}}

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Question 15: A car completed a journey in \displaystyle 7 hours. One-third of the journey was performed at \displaystyle 20 \text{ km/hr } and the rest at \displaystyle 30 \text{ km/hr } . Find the total length of the journey.

Answer:

Let \displaystyle x be the distance of the journey.

\displaystyle \frac{\frac{x}{3}}{20} + \frac{\frac{2x}{3}}{30} = 7

\displaystyle \text{Solving for    } x = 180 \text{ km }

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Question 16: A cyclist covered a certain distance in \displaystyle 3 \displaystyle \frac{1}{2} hours. The speed for first half of the distance was \displaystyle 15 \text{ km/hr } and for the second half it was \displaystyle 20 \text{ km/hr } . Find the total distance covered by him.

Answer:

Let \displaystyle x be the distance of the journey.

\displaystyle \frac{\frac{x}{2}}{15} + \frac{\frac{x}{2}}{20} = 3 \displaystyle \frac{1}{2}  

\displaystyle \text{Solving for    } x = 60 \text{ km }

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Question 17: A person travels equal distances with speeds of \displaystyle 3 \text{ km/hr } , \displaystyle 4 \text{ km/hr } and \displaystyle 5 \text{ km/hr } and takes a total time of \displaystyle 47 minutes. Find the total distance.

Answer:

Let \displaystyle x be the distance of each leg

\displaystyle \frac{x}{3} + \frac{x}{4} + \frac{x}{5} = \frac{47}{60}  

\displaystyle \text{Solving for    } x = 1 \ km  \text{ and the total distance is  } 3 \text{ km }

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Question 18: A farmer traveled a distance of \displaystyle 61 km in \displaystyle 9 hours. He traveled partly on foot at \displaystyle 4 \text{ km/hr } and partly on bicycle at \displaystyle 9 \text{ km/hr } . Find the distance traveled by him on foot.

Answer:

Let \displaystyle x be the distance traveled by foot

\displaystyle \frac{x}{4} + \frac{61 - x}{9} = 9

\displaystyle \text{Solving for    } x = 16 \text{ km }

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Question 19: Rohan cycles to his office at the rate of \displaystyle 12 \frac{1}{2} \text{ km/hr } and is late by \displaystyle 3 minutes. However, if he travels at \displaystyle 15 \text{ km/hr } , he reaches \displaystyle 5 minutes earlier than the usual time. What is the distance of his office from his residence?

Answer:

Let \displaystyle x be the distance to office. The difference of the time in the two cases is 8 minutes.

\displaystyle \frac{x}{12.5} - \frac{x}{15} = \frac{8}{60}  

Solving for \displaystyle x=10 \text{ km } (Distance of his office from his residency)

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Question 20: Robert is travelling on his cycle and has calculated to reach point A at 2 p.m. if he travels at \displaystyle 10 \text{ km/hr } . However, he will reach there at 12 noon if he travels at \displaystyle 15 \text{ km/hr } . At what speed must he travel to reach A at 1 p.m.?

Answer:

\displaystyle \text{Let the distance traveled    } = x \text{ km } 

\displaystyle \frac{x}{10} - \frac{x}{15} =2

\displaystyle \text{Solving for    } x=60 \text{ km } 

\displaystyle \frac{60}{10} - \frac{60}{s} =1

\displaystyle \text{Solving for    } s=12 \frac{\text{km}}{\text{hr}}

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Question 21: If a train runs at \displaystyle 40 \text{ km/hr } , it reaches its destination late by \displaystyle 11 minutes, but if it runs at \displaystyle 50 \text{ km/hr } , it is late by \displaystyle 5 minutes only. Find the correct time for the train to complete its journey.

Answer:

\displaystyle \text{ Let the distance traveled by the train  } = x \text{ km }

\displaystyle \frac{x}{40} - \frac{x}{50} = \frac{6}{60}  

\displaystyle \text{ Solving for  } x = 20 \text{ km } (Distance of his office from his residency)

Time taken to cover 20 km=30 minutes

Therefore the correct time is 19 minutes

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Question 22: The distance between Delhi and Hyderabad is \displaystyle 1800 km. A train leaves Delhi and proceeds towards Hyderabad at a uniform speed of \displaystyle 60 \text{ km/hr } . Another train leaves Hyderabad at the same time and proceeds towards Delhi at a uniform speed of \displaystyle 48 \text{ km/hr } . When and where will they meet?

Answer:

\displaystyle \text{ Let the train meet at a distance  } = x \text{ km } from Delhi

\displaystyle \frac{x}{60} = \frac{(1800-x)}{48}  

\displaystyle \text{Solving for    } x = 1000 \text{ km } . The trains will meet 1000 km from Delhi.

\displaystyle \text{ Time taken before them meet  } = \frac{1000}{60} = 16 \frac{2}{3} \text{ hr }