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Chapter 12

Speed, Distance & Time - Exercise 12(A)

Class - 7 RS Aggarwal Mathematics Solutions



Exercise 12(A)

Question 1

Convert each of the following speeds into m/sec:

(i) 90 km/hr

(ii) 27 km/hr

(iii) 315 km/hr

(iv) 16.2 km/hr

Answer

(i) 90 km/hr

To convert km/hr to m/sec, multiply by 518\dfrac{5}{18}:

∴90 km/hr=(90×518) m/sec=(5×5) m/sec=25 m/sec\therefore 90 \text { km/hr} =\Big(90 \times \dfrac{5}{18}\Big)\text{ m/sec} \\[1em] = (5 \times 5) \text{ m/sec} \\[1em] = 25 \text{ m/sec}

Hence, the answer is 25 m/sec.

(ii) 27 km/hr

To convert km/hr to m/sec, multiply by 518\dfrac{5}{18}:

∴27 km/hr=(27×518) m/sec=(3×52) m/sec=152 m/sec=7.5 m/sec\therefore 27 \text { km/hr} = \Big(27 \times \dfrac{5}{18}\Big)\text{ m/sec} \\[1em] = \Big(3 \times \dfrac{5}{2}\Big)\text{ m/sec} \\[1em] = \dfrac{15}{2}\text{ m/sec} \\[1em] = 7.5\text{ m/sec}

Hence, the answer is 7.5 m/sec.

(iii) 315 km/hr

To convert km/hr to m/sec, multiply by 518\dfrac{5}{18}:

∴315 km/hr=(315×518) m/sec=157518 m/sec=87.5 m/sec\therefore 315 \text { km/hr} = \Big(315 \times \dfrac{5}{18}\Big)\text{ m/sec} \\[1em] = \dfrac{1575}{18}\text{ m/sec} \\[1em] = 87.5\text{ m/sec}

Hence, the answer is 87.5 m/sec.

(iv) 16.2 km/hr

To convert km/hr to m/sec, multiply by 518\dfrac{5}{18}:

∴16.2 km/hr=(16.2×518) m/sec=8118 m/sec=4.5 m/sec\therefore 16.2 \text { km/hr} = \Big(16.2 \times \dfrac{5}{18}\Big)\text{ m/sec} \\[1em] = \dfrac{81}{18}\text{ m/sec} \\[1em] = 4.5\text{ m/sec}

Hence, the answer is 4.5 m/sec.

Question 2

Convert each of the following speeds into km/hr:

(i) 35 m/sec

(ii) 14 m/sec

(iii) 60 m/sec

(iv) 3.5 m/sec

Answer

(i) 35 m/sec

To convert m/sec to km/hr, multiply by 185\dfrac{18}{5}:

∴35 m/sec=(35×185) km/hr=(7×18) km/hr=126 km/hr\therefore 35 \text { m/sec} = \Big(35 \times \dfrac{18}{5}\Big)\text{ km/hr} \\[1em] = (7 \times 18) \text{ km/hr} \\[1em] = 126 \text{ km/hr}

Hence, the answer is 126 km/hr.

(ii) 14 m/sec

To convert m/sec to km/hr, multiply by 185\dfrac{18}{5}:

∴14 m/sec=(14×185) km/hr=2525 km/hr=50.4 km/hr\therefore 14 \text { m/sec} = \Big(14 \times \dfrac{18}{5}\Big)\text{ km/hr} \\[1em] = \dfrac{252}{5}\text{ km/hr} \\[1em] = 50.4\text{ km/hr}

Hence, the answer is 50.4 km/hr.

(iii) 60 m/sec

To convert m/sec to km/hr, multiply by 185\dfrac{18}{5}:

∴60 m/sec=(60×185) km/hr=(12×18) km/hr=216 km/hr\therefore 60 \text { m/sec} =\Big(60 \times \dfrac{18}{5}\Big)\text{ km/hr} \\[1em] = (12 \times 18) \text{ km/hr} \\[1em] = 216 \text{ km/hr}

Hence, the answer is 216 km/hr.

(iv) 3.5 m/sec

To convert m/sec to km/hr, multiply by 185\dfrac{18}{5}:

∴3.5 m/sec=(3.5×185) km/hr=635 km/hr=12.6 km/hr\therefore 3.5 \text { m/sec} = \Big(3.5 \times \dfrac{18}{5}\Big)\text{ km/hr} \\[1em] = \dfrac{63}{5}\text{ km/hr} \\[1em] = 12.6\text{ km/hr}

Hence, the answer is 12.6 km/hr.

Question 3

A car travels a distance of 258 km in 3 hours 35 minutes. What is the speed of the car?

Answer

Given:

Distance = 258 km

Time = 3 hours 35 mins

Let's convert Time to hours:

1 hour = 60 mins

∴ 3 hours 35 mins = 33560 hr=3712 hr=4312 hr3\dfrac{35}{60} \text{ hr} = 3\dfrac{7}{12} \text{ hr} = \dfrac{43}{12} \text{ hr}

And

Speed=DistanceTime=(2584312) km/hr=(258×1243) km/hr=6×12 km/hr=72 km/hr\text {Speed} = \dfrac{\text{Distance}}{\text{Time}} \\[1em] = \left(\dfrac{258}{\dfrac{43}{12}}\right)\text{ km/hr} \\[1em] = \Big(258 \times \dfrac{12}{43}\Big)\text{ km/hr} \\[1em] = 6 \times 12\text{ km/hr} \\[1em] = 72 \text{ km/hr}

The speed of the car is 72 km/hr.

Question 4

A train covers 14 km in 8 minutes. What is the speed of the train?

Answer

Given:

Distance = 14 km

Time = 8 minutes

Let's convert Time to hours:

1 hour = 60 mins

∴ 8 minutes = 860 hr=215 hr\dfrac{8}{60} \text{ hr} = \dfrac{2}{15} \text{ hr}

And

Speed=DistanceTime=(14215) km/hr=(14×152) km/hr=7×15 km/hr=105 km/hr\text {Speed} = \dfrac{\text{Distance}}{\text{Time}} \\[1em] = \left(\dfrac{14}{\dfrac{2}{15}}\right)\text{ km/hr} \\[1em] = \Big(14 \times \dfrac{15}{2}\Big)\text{ km/hr} \\[1em] = 7 \times 15\text{ km/hr} \\[1em] = 105 \text{ km/hr}

The speed of the train is 105 km/hr.

Question 5

A man can walk 8 km in 1 hour 15 minutes. Find his speed.

Answer

Given:

Distance = 8 km

Time = 1 hour 15 minutes

Let's convert Time to hours

1 hour = 60 mins

∴ 1 hour 15 minutes = 11560 hr=114 hr=54 hr1\dfrac{15}{60} \text{ hr} = 1\dfrac{1}{4} \text{ hr} = \dfrac{5}{4} \text{ hr}

And

Speed=DistanceTime=(854) km/hr=(8×45) km/hr=325 km/hr=6.4 km/hr\text {Speed} = \dfrac{\text{Distance}}{\text{Time}} \\[1em] = \left(\dfrac{8}{\dfrac{5}{4}}\right)\text{ km/hr} \\[1em] = \Big(8 \times \dfrac{4}{5}\Big)\text{ km/hr} \\[1em] = \dfrac{32}{5}\text{ km/hr} \\[1em] = 6.4 \text{ km/hr}

The speed of the man is 6.4 km/hr.

Question 6

Rahul cycles at 12 km per hour. How much distance does he cover in 36 minutes?

Answer

Given:

Speed = 12 km/hr

Time = 36 minutes

Let's convert Time to hours:

1 hour = 60 mins

∴ 36 minutes = 3660 hr=35 hr\dfrac{36}{60} \text{ hr} = \dfrac{3}{5} \text{ hr}

And

Distance = Speed x Time

Substituting the values in above, we get:

Distance=(12×35) km=365 km=7.2 km\text {Distance} = \Big(12 \times \dfrac{3}{5}\Big)\text{ km} \\[1em] = \dfrac{36}{5}\text{ km} \\[1em] = 7.2 \text{ km}

The distance covered by Rahul is 7.2 km.

Question 7

Ravi walks at a uniform speed of 4 km per hour. How much time does he take to cover 600 metres?

Answer

Given:

Speed = 4 km/hr

Distance = 600 metres

Let's convert Distance to km:

1 km = 1000 m

∴ 600 m = 6001000 km=0.6 km\dfrac{600}{1000} \text{ km} = 0.6 \text{ km}

Time = DistanceSpeed\dfrac{\text{Distance}}{\text{Speed}}

Substituting the values in above, we get:

Time = 0.64\dfrac{0.6}{4} hours = 0.15 hours

Let's convert time to minutes:

1 hour = 60 minutes

∴ 0.15 hours = 0.15 x 60 minutes = 9 minutes

The time taken by Ravi is 9 minutes.

Question 8

A train is running at 45 km per hour. How far does it go in 8 seconds?

Answer

Given:

Speed = 45 km/hr

Let's convert speed into m/sec:

45 km/hr=(45×518) m/sec=22518 m/sec=12.5 m/sec45 \text { km/hr} = \Big(45 \times \dfrac{5}{18}\Big)\text{ m/sec} \\[1em] = \dfrac{225}{18}\text{ m/sec} \\[1em] = 12.5\text{ m/sec}

Time = 8 seconds

Distance = Speed x Time

Substituting the values in above, we get:

Distance = 12.5 x 8 m = 100 m

The distance covered by train is 100 m.

Question 9

Which is more : a speed of 36 km/hr or a speed of 10.5 m/sec?

Answer

Let's convert 36 km/hr into m/sec:

36 km/hr=(36×518) m/sec=(2×5) m/sec=10 m/sec36 \text { km/hr} = \Big(36 \times \dfrac{5}{18}\Big)\text{ m/sec} \\[1em] = (2 \times 5)\text{ m/sec} \\[1em] = 10 \text{ m/sec}

Now, compare 10 m/sec and 10.5 m/sec:

10.5 m/sec > 10 m/sec

10.5 m/sec is more.

Question 10

A car, running at 45 km/hr takes 6 hours to cover a journey. At what speed must it travel to complete the journey in 5 hours?

Answer

Given:

Speed = 45 km/hr

Time = 6 hours

Distance = Speed x Time

Substituting the values in above, we get:

Distance = 45 x 6 km = 270 km

Now, let's find the required speed for the new time:

Distance = 270 km, New Time = 5 hours

New Speed = DistanceNew Time\dfrac{\text{Distance}}{\text{New Time}}

Substituting the values in above, we get:

New Speed = 2705\dfrac{270}{5} km/hr

New Speed = 54 km/hr

Speed of the car to complete the journey in 5 hours is 54 km/hr.

Question 11

A car covers a certain distance in 50 minutes if it runs at 45 km per hour. How much time will it take to cover the same distance at a speed of 30 km per hour?

Answer

Given:

Speed = 45 km/hr

Time = 50 minutes

Let's convert time to hour:

1 hour = 60 minutes

∴ 50 minutes = 5060 hr=56 hr\dfrac{50}{60}\text{ hr} = \dfrac{5}{6}\text{ hr}

Distance = Speed x Time

Substituting the values in above, we get:

Distance = 45×5645 \times \dfrac{5}{6} km = 2256\dfrac{225}{6} km

Let's find the time for the new speed:

Distance = 2256\dfrac{225}{6} km, New Speed = 30 km/hr

Time = DistanceNew Speed\dfrac{\text{Distance}}{\text{New Speed}}

Substituting the values in above, we get:

Time = 225630 hours\dfrac{\dfrac{225}{6}}{30}\text{ hours}

=(2256×130) hours=(456×16) hours=4536 hours=1.25 hours= \Big(\dfrac{225}{6} \times \dfrac{1}{30}\Big)\text{ hours} \\[1em] = \Big(\dfrac{45}{6} \times \dfrac{1}{6}\Big)\text{ hours} \\[1em] = \dfrac{45}{36} \text{ hours} \\[1em] = 1.25 \text{ hours}

1.25 hours = 1 hour + (0.25 x 60) minutes = 1 hour 15 minutes

Time taken by the car is 1 hour 15 minutes.

Question 12

A bus covers 100 km in first 2 hours, 64 km in the next 1 hour and 72 km in the next 2 hours. Find the average speed of the bus during the whole journey.

Answer

Given:

Distance : (D1) = 100 km, (D2) = 64 km, (D3) = 72 km

Time : (T1) = 2 hrs, (T2) = 1 hr, (T3) = 2 hrs

Total Distance = 100 km + 64 km + 72 km = 236 km

Total Time = 2 hrs + 1 hrs + 2 hrs = 5 hrs

Average Speed = Total DistanceTotal Time\dfrac{\text{Total Distance}}{\text{Total Time}}

Substituting the values in above, we get:

Average Speed = 2365\dfrac{236}{5} km/hr = 47.2 km/hr

Average speed of the bus is 47.2 km/hr.

Question 13

An aeroplane covers 2500 km, 1200 km and 500 km at the rate of 500 km/hr, 400 km/hr and 250 km/hr respectively. Find its average speed for the whole journey.

Answer

Given:

Distances: (D1) = 2500 km, D2 = 1200 km, D3 = 500 km

Speed: (S1) = 500 km/hr, (S2) = 400 km/hr, (S3) = 250 km/hr

Let's find time taken for each part:

Time = DistanceSpeed\dfrac{\text{Distance}}{\text{Speed}}

T1 = D1S1=2500500\dfrac{D1}{S1} = \dfrac{2500}{500} = 5 hrs

T2 = D2S2=1200400\dfrac{D2}{S2} = \dfrac{1200}{400} = 3 hrs

T3 = D3S3=500250\dfrac{D3}{S3} = \dfrac{500}{250} = 2 hrs

Total Distance = 2500 km + 1200 km + 500 km = 4200 km

Total Time = 5 hrs + 3 hrs + 2 hrs = 10 hrs

Average Speed = Total DistanceTotal Time\dfrac{\text{Total Distance}}{\text{Total Time}}

Substituting the values in above, we get:

Average Speed = 420010\dfrac{4200}{10} km/hr = 420 km/hr

Average speed of the aeroplane is 420 km/hr.

Question 14

A scooterist travels at 36 km/hr for 5 hours and at 45 km/hr for 4 hours. Find his average speed for the whole journey.

Answer

Given:

Speed: S1 = 36 km/hr, S2 = 45 km/hr

Time: T1 = 5 hrs, T2 = 4 hrs

Let's find distance covered in each part:

Distance = Speed x Time

D1 = S1 x T1 = 36 x 5 = 180 km

D2 = S2 x T2 = 45 x 4 = 180 km

Total Distance = 180 km + 180 km = 360 km

Total Time = 5 hrs + 4 hrs = 9 hrs

Average Speed = Total DistanceTotal Time\dfrac{\text{Total Distance}}{\text{Total Time}}

Substituting the values in above, we get:

Average Speed = 3609\dfrac{360}{9} km/hr = 40 km/hr

Average speed of the scooterist is 40 km/hr.

Question 15

Raju's school is 4 km away from his house. He walks from his house to school and comes back in 2122\dfrac{1}{2} hours. Find his average speed.

Answer

Given:

Distance (one way) = 4 km

Total Distance = 4 km (to school) + 4 km (back home) = 8 km

Total Time = 212 hours=2.5 hours2\dfrac{1}{2} \text{ hours} = 2.5 \text{ hours}

Average Speed = Total DistanceTotal Time\dfrac{\text{Total Distance}}{\text{Total Time}}

Substituting the values in above, we get:

Average Speed = 82.5\dfrac{8}{2.5} km/hr = 3.2 km/hr

Average speed of Raju is 3.2 km/hr.

Question 16

Find the time taken by a train 150 m long, running at 60 km/hr in crossing a man standing on the platform.

Answer

Given:

Distance = Length of train = 150 m

Speed = 60 km/hr

Let's convert speed to m/sec:

60 km/hr=(60×518) m/sec=(10×53) m/sec=503 m/sec60 \text { km/hr} = \Big(60 \times \dfrac{5}{18}\Big)\text{ m/sec} \\[1em] = \Big(10 \times \dfrac{5}{3}\Big)\text{ m/sec} \\[1em] = \dfrac{50}{3} \text{ m/sec}

Time = DistanceSpeed\dfrac{\text{Distance}}{\text{Speed}}

Substituting the values in above, we get:

Time=150503 seconds=150×350 seconds=3×31 seconds=3×3 seconds=9 seconds\text {Time} = \dfrac{150}{\dfrac{50}{3}} \text{ seconds} \\[1em] = 150 \times \dfrac{3}{50}\text{ seconds} \\[1em] = 3 \times \dfrac{3}{1}\text{ seconds} \\[1em] = 3 \times 3\text{ seconds} \\[1em] = 9 \text{ seconds}

Hence, the train will cross a man in 9 seconds.

Question 17

The length of a train is 175 m. It is running at 63 km/hr. In how much time would it cross an electric pole?

Answer

Given:

Distance = Length of train = 175 m

Speed = 63 km/hr

Let's convert speed to m/sec:

63 km/hr=(63×518) m/sec=(7×52) m/sec=352 m/sec63 \text { km/hr} = \Big(63 \times \dfrac{5}{18}\Big)\text{ m/sec} \\[1em] = \Big(7 \times \dfrac{5}{2}\Big)\text{ m/sec} \\[1em] = \dfrac{35}{2} \text{ m/sec}

Time = DistanceSpeed\dfrac{\text{Distance}}{\text{Speed}}

Substituting the values in above, we get:

Time=175352=175×235 seconds=5×21 seconds=5×2 seconds=10 seconds\text {Time} = \dfrac{175}{\dfrac{35}{2}} \\[1em] = 175 \times \dfrac{2}{35}\text{ seconds} \\[1em] = 5 \times \dfrac{2}{1}\text{ seconds} \\[1em] = 5 \times 2\text{ seconds} \\[1em] = 10 \text{ seconds}

Hence, the train will cross an electric pole in 10 seconds.

Question 18

A train 150 m long crosses a tree in 6 seconds. What is the speed of the train in km per hour?

Answer

Given:

Distance = Length of train = 150 m

Time = 6 seconds

Speed = DistanceTime\dfrac{\text{Distance}}{\text{Time}}

Substituting the values in above, we get:

Speed = 1506\dfrac{150}{6} m/sec = 25 m/sec

Let's convert speed to km/hr:

∴25 m/sec=25×185 km/hr=5×181 km/hr=5×18 km/hr=90 km/hr\therefore 25 \text{ m/sec} = 25 \times \dfrac{18}{5} \text{ km/hr} \\[1em] = 5 \times \dfrac{18}{1} \text{ km/hr} \\[1em] = 5 \times 18 \text{ km/hr} \\[1em] = 90 \text{ km/hr}

Hence, speed of the train is 90 km/hr.

Question 19

A train 120 m long is running at 54 km per hour. In how much time will it pass a bridge 180 m long?

Answer

Given:

Distance (D1) = Length of train = 120 m

Distance (D2) = Length of bridge = 180 m

Total Distance = 120 m + 180 m = 300 m

Speed = 54 km/hr

Let's convert speed to m/sec:

54 km/hr=(54×518) m/sec=(3×51) m/sec=15 m/sec54 \text { km/hr} = \Big(54 \times \dfrac{5}{18}\Big) \text{ m/sec} \\[1em] = \Big(3 \times \dfrac{5}{1}\Big) \text{ m/sec} \\[1em] = 15 \text{ m/sec}

Time = DistanceSpeed\dfrac{\text{Distance}}{\text{Speed}}

Substituting the values in above, we get:

Time = 30015\dfrac{300}{15} seconds = 20 seconds

Hence, the train will pass a bridge in 20 seconds.

Question 20

A train 760 m long crosses a platform 440 m long in 40 seconds. Find the speed of the train in km per hour.

Answer

Given:

Distance (D1) = Length of train = 760 m

Distance (D2) = Length of platform = 440 m

Total Distance = 760 m + 440 m = 1200 m

Time = 40 seconds

Speed = DistanceTime\dfrac{\text{Distance}}{\text{Time}}

Substituting the values in above, we get:

Speed = 120040\dfrac{1200}{40} m/sec = 30 m/sec

Let's convert speed to km/hr:

∴30 m/sec=30×185 km/hr=6×181 km/hr=6×18 km/hr=108 km/hr\therefore 30\text{ m/sec} = 30 \times \dfrac{18}{5} \text{ km/hr} \\[1em] = 6 \times \dfrac{18}{1} \text{ km/hr} \\[1em] = 6 \times 18 \text{ km/hr} \\[1em] = 108 \text{ km/hr}

Hence, speed of the train is 108 km/hr.

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