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

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

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

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

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

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

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

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

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

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

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

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

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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.

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

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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 $\displaystyle 12 \text{ km/hr }$?

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

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

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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?

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

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

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

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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?

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

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

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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?

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

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

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

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

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

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

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

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

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

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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.

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

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

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.

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

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

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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.

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

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

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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.

$\displaystyle \text{Let } x$ 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?

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.

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.

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

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

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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.

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.

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.

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.

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?

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.?

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

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

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

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

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

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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.

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

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

$\displaystyle \text{ Solving for } x = 20 \text{ km }$ (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?

$\displaystyle \text{ Let the train meet at a distance } = x \text{ km }$ 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.
$\displaystyle \text{ Time taken before them meet } = \frac{1000}{60} = 16 \frac{2}{3} \text{ hr }$