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Q1. Express the following speeds in meter/sec:

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

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

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

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

Q2. Express the following speeds in km/hr:

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

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

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

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

Q3. An athlete covers a distance of 1200 meters in 4 min.48 sec. Find his speed in km/hr.

Answer:

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

Q4. Walking at the rate of 4 km/hr a man covers a certain distance in $2\frac{1}{2}$ hours. How much time will be taken by the man to cover the same distance, if he cycles at 12 km/hr?

Answer:

Distance covered walking  $= 4 \times 2.5 = 10 km$

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

Q5. A car can finish a certain journey in 10 hours at a speed of 48 km/hr. By how much the speed of car must be increased to cover the same distance in 8 hours?

Answer:

Distance covered by the car  $= 48 \times 10 = 480 km$

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

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

Q6. A bus covers a certain distance in 50 minutes, if it runs at a speed of 54 km/hr. What must be the speed of the bus in order to reduce the time of journey to 40 minutes?

Answer:

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

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

Q7. A motor car starts with the speed of 70 km/hr with its speed increasing every two hours by 10 km/hr. In how much time will it cover a distance of 345 km?

Answer:

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

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

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

Distance covered in 3rd and the 4th hour $= 160 km$

Total distance covered by end of 4th hours $= 140 + 160 = 300 km$

Distance left to be covered after end of 4th hour $= 345 - 300 = 45 km$

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

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

Hence the total time to cover  $345 km = 4.5 hr$

Q8. A man takes 150 steps in walking 75 meters. If he takes 3 steps in 1 second, find his speed in (i) m/sec (ii) km/hr.

Answer:

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

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

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

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

Q9. A man walks at 5 km/hr for 6 hours and at 4 km/hr for 12 hours. Find his average speed.

Answer:

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

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

Q10. A man covers a distance of 144 km at the speed of 36 km/hr and another 256 km at the speed of 64 km/hr. Find his average speed for the whole journey.

Answer:

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

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

Q11. Two buses travel to a place at 45 km/hr and 60 km/hr respectively. If the second bus takes  $5\frac{1}{2} hr$ hours less than the first for the same journey, find the length of the journey.

Answer:

Let $x$ be the distance of the journey

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

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

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

Solving for $x = 990 km$

Q12. A boy goes to school from his village at 3km/hr and returns back at 2km/hr. If he takes 5 hours in all, what is the distance between the village and the school?

Answer:

Let $x$ be the distance of the journey

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

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

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

Solving for $x = 6 km$

Q13. A bus completes a journey of 420 km in $6\frac{1}{2} hr$. The first ¾ part of the journey performed at 63 km/hr.  Calculate the speed of the rest of the journey.

Answer:

Let $x$ be the speed of the rest of the journey

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

Solving for $x = 70 \frac{km}{hr}$

Q14. A man drives 150 km from town A to town B in 3 hours 20 min and returns back to town A from town B in 4 hours 10 min. Find his average speed for the whole journey.

Answer:

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

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

Q15. A car completed a journey in 7 hours.  One-third of the journey was performed at 20 km/hr and the rest at 30 km/hr. Find the total length of the journey.

Answer:

Let $x$ be the distance of the journey.

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

Solving for $x = 180 km$

Q16. A cyclist covered a certain distance in  $3\frac{1}{2}$ hours. The speed for first half of the distance was 15 km/hr and for the second half it was 20 km/hr. Find the total distance covered by him.

Answer:

Let $x$ be the distance of the journey.

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

Solving for $x = 60 km$

Q17. A person travels equal distances with speeds of 3 km/hr, 4 km/hr and 5 km/hr and takes a total time of 47 minutes. Find the total distance.

Answer:

Let $x$ be the distance of each leg

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

Solving for $x = 1 km$ and the total distance is  $3 km$

Q18. A farmer traveled a distance of 61 km in 9 hours. He traveled partly on foot at 4 km/hr and partly on bicycle at 9 km/hr. Find the distance traveled by him on foot.

Answer:

Let $x$ be the distance traveled by foot

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

Solving for $x = 16 km$

Q19. Rohan cycles to his office at the rate of $12\frac{1}{2}$km/hr and is late by 3 minutes. However, if he travels at 15 km/hr, he reaches 5 minutes earlier than the usual time. What is the distance of his office from his residence?

Answer:

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

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

Solving for$x=10 km$ (Distance of his office from his residency)

Q20. Robert is travelling on his cycle and has calculated to reach point A at 2 p.m. if he travels at 10 km/hr. However, he will reach there at 12 noon if he travels at 15 km/hr. At what speed must he travel to reach A at 1 p.m.?

Answer:

Let the distance traveled $= x km$

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

Solving for$x=60 km$

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

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

Q21. If a train runs at 40 km/hr, it reaches its destination late by 11 minutes, but if it runs at 50 km/hr, it is late by 5 minutes only. Find the correct time for the train to complete its journey.

Answer:

Let the distance traveled by the train $= x km$

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

Solving for $x = 20 km$   (Distance of his office from his residency)

Time taken to cover 20 km=30 minutes

Therefore the correct time is 19 minutes

Q22. The distance between Delhi and Hyderabad is 1800 km. A train leaves Delhi and proceeds towards Hyderabad at a uniform speed of 60 km/hr. Another train leaves Hyderabad at the same time and proceeds towards Delhi at a uniform speed of 48 km/hr. When and where will they meet?

Answer:

Let the train meet at a distance $= x km$ from Delhi

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

Solving for $x = 1000 km$. The trains will meet 1000 km from Delhi.

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

ICSE Board: Suggested Books     ICSE Board:  Foundation Mathematics
Class 8: Reference Books               Class 8: NTSE Preparation
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