ICSE Class 6 Mathematics · Selina Solutions Concise Mathematics Class 6 ICSE · 34 solved questions
A train covers 51 km in 3 hours. Calculate its speed. How far does the train go in 30 minutes?
Given,
Distance covered = 51 km and time taken = 3 hours.
Speed = $\frac{\text{Distance}}{\text{Time}} = \frac{51}{3} = 17$ km/h.
Converting 30 minutes into hours :
30 minutes = $\frac{30}{60}$ h = $\frac{1}{2}$ h.
Distance covered in 30 minutes = Speed × Time = 17 × $\frac{1}{2}$ = 8.5 km.
Hence, the speed of the train = 17 km/h and it goes 8.5 km in 30 minutes.
A motorist travelled the distance between two towns, which is 65 km, in 2 hours and 10 minutes. Find his speed in metre per minute.
Given,
Distance = 65 km and time = 2 hours 10 minutes.
As the speed is required in metre per minute, convert the distance into metres and the time into minutes.
Distance = 65 km
= 65 × 1000 m
= 65000 m.
Time = 2 hours 10 minutes
= (2 × 60 + 10) minutes
= 130 minutes.
Speed = $\frac{\text{Distance}}{\text{Time}} = \frac{65000}{130} = 500$ m/min.
Hence, the speed of the motorist = 500 m/min.
A train travels 700 metres in 35 seconds. What is its speed in km h-1?
Given,
Distance = 700 metres and time = 35 seconds.
Speed = $\frac{\text{Distance}}{\text{Time}} = \frac{700}{35} = 20$ m/s.
Converting the speed into km/h (multiply by $\frac{18}{5}$) :
Speed = 20 × $\frac{18}{5} = \frac{360}{5} = 72$ km/h.
Hence, the speed of the train = 72 km/h.
A racing car covered 600 km in 3 hours 20 minutes. Find its speed in metre per second. How much distance will the car cover in 50 sec?
Given,
Distance = 600 km and time = 3 hours 20 minutes.
As the speed is required in metre per second, convert the distance into metres and the time into seconds.
Distance = 600 km
= 600 × 1000 m
= 600000 m.
Time = 3 hours 20 minutes
= (3 × 3600 + 20 × 60) seconds
= (10800 + 1200) seconds
= 12000 seconds.
Speed = $\frac{\text{Distance}}{\text{Time}} = \frac{600000}{12000} = 50$ m/s.
Distance covered in 50 seconds = Speed × Time = 50 × 50 = 2500 m.
Hence, the speed of the car = 50 m/s and it covers 2500 m in 50 seconds.
Rohit goes 350 km in 5 hours. Find:
(i) his speed
(ii) the distance covered by Rohit in 6.2 hours
(iii) the time taken by him to cover 210 km.
[Assume that throughout the journey, the speed of Rohit remains uniform].
Given,
Distance = 350 km and time = 5 hours.
(i) Speed = $\frac{\text{Distance}}{\text{Time}} = \frac{350}{5} = 70$ km/h.
Hence, Rohit's speed = 70 km/h.
(ii) Distance covered in 6.2 hours = Speed × Time = 70 × 6.2 = 434 km.
Hence, the distance covered by Rohit in 6.2 hours = 434 km.
(iii) Time taken to cover 210 km = $\frac{\text{Distance}}{\text{Speed}} = \frac{210}{70} = 3$ hours.
Hence, the time taken to cover 210 km = 3 hours.
A boy drives his scooter with a uniform speed of 45 km h-1. Find:
(i) the distance covered by him in 1 hour 20 min.
(ii) the time taken by him to cover 108 km.
(iii) the time taken to cover 900 m.
Given,
Uniform speed of the scooter = 45 km/h.
(i) Time = 1 hour 20 min
= $1\frac{20}{60} \text{ h} = 1\frac{1}{3}\text{ h} = \frac{4}{3}\text{ h}$.
Distance covered = Speed × Time
= 45 × $\frac{4}{3} = \frac{180}{3}$
= 60 km.
Hence, the distance covered in 1 hour 20 min = 60 km.
(ii) Time taken to cover 108 km = $\frac{\text{Distance}}{\text{Speed}}$
= $\frac{108}{45}$
= $\frac{12}{5}$
= 2.4 hours.
0.4 hour = 0.4 × 60 minutes = 24 minutes.
Hence, the time taken to cover 108 km = 2.4 hours (i.e. 2 hours 24 minutes).
(iii) 900 m = $\frac{900}{1000}$ km = 0.9 km.
Time taken to cover 900 m = $\frac{\text{Distance}}{\text{Speed}} = \frac{0.9}{45} = 0.02$ h.
0.02 h = 0.02 × 3600 seconds = 72 seconds.
Hence, the time taken to cover 900 m = 72 seconds.
I travel a distance of 10 km and come back in $2\frac{1}{2}$ hours. What is my speed?
Given,
Distance travelled one way = 10 km, so total distance = 10 km + 10 km = 20 km.
Total time taken = $2\frac{1}{2}$ hours = $\frac{5}{2}$ hours.
$$\begin{aligned}\text{Speed} = \frac{\text{Distance}}{\text{Time}} \\[1em] = \frac{20}{\frac{5}{2}} \\[1em] = \frac{40}{5} \\[1em] = 8 \text{ km/h}.\end{aligned}$$
Hence, my speed = 8 km/h.
A man walks a distance of 5 km in 2 hours. Then he goes in a bus to a nearby town, which is 40 km, in further 2 hours. From there, he goes to his office in an autorickshaw, a distance of 5 km, in $\frac{1}{2}$ hour. What was his average speed during the whole journey?
Given,
Distance walked = 5 km in 2 hours.
Distance by bus = 40 km in 2 hours.
Distance by autorickshaw = 5 km in $\frac{1}{2}$ hour.
Total distance covered = 5 km + 40 km + 5 km = 50 km.
Total time taken = 2 h + 2 h + $\frac{1}{2}$ h = $\frac{9}{2}$ h.
$$\begin{aligned}\text{Average speed} = \frac{\text{Total distance covered}}{\text{Total time taken}} \\[1em] = \frac{50}{\frac{9}{2}} \\[1em] = 50 \times \frac{2}{9} \\[1em] = \frac{100}{9} \\[1em] = 11\frac{1}{9} \text{ km/h}.\end{aligned}$$
Hence, the man's average speed during the whole journey = $11\frac{1}{9}$ km/h.
Jagan went to another town such that he covered 240 km by a car going at 60 km h-1. Then he covered 80 km by a train going at 100 km h-1, and the rest 200 km he covered by a bus going at 50 km h-1. What was his average speed during the whole journey?
Given,
Distance by car = 240 km at 60 km/h.
Distance by train = 80 km at 100 km/h.
Distance by bus = 200 km at 50 km/h.
Time taken by car = $\frac{240}{60} = 4$ hours.
Time taken by train = $\frac{80}{100} = \frac{4}{5} = 0.8$ hour.
Time taken by bus = $\frac{200}{50} = 4$ hours.
Total distance covered = 240 km + 80 km + 200 km = 520 km.
Total time taken = 4 h + 0.8 h + 4 h = 8.8 h = $\frac{44}{5}$ h.
$$\begin{aligned}\text{Average speed} = \frac{\text{Total distance covered}}{\text{Total time taken}} \\[1em] = \frac{520}{\frac{44}{5}} \\[1em] = 520 \times \frac{5}{44} \\[1em] = \frac{2600}{44} \\[1em] = \frac{650}{11} \\[1em] = 59\frac{1}{11} \text{ km/h}.\end{aligned}$$
Hence, Jagan's average speed during the whole journey = $59\frac{1}{11}$ km/h.
The speed of sound in air is about 330 m s-1. Express this speed in km h-1. How long will the sound take to travel 99 km?
Given,
Speed of sound = 330 m/s.
Converting the speed into km/h (multiply by $\frac{18}{5}$) :
Speed = 330 × $\frac{18}{5} = \frac{5940}{5} = 1188$ km/h.
Time taken to travel 99 km = $\frac{\text{Distance}}{\text{Speed}} = \frac{99}{1188} = \frac{1}{12}$ h.
$\frac{1}{12}$ h = $\frac{1}{12}$ × 60 minutes = 5 minutes.
Hence, the speed of sound = 1188 km/h and it takes 5 minutes to travel 99 km.
A train 180 m long is running at a speed of 90 km/h. How long will it take to pass a railway signal?
Given,
Length of the train = 180 m and speed = 90 km/h.
To pass a railway signal (a stationary object), the distance to be covered = length of the train = 180 m.
Converting the speed into m/s (multiply by $\frac{5}{18}$) :
Speed = 90 × $\frac{5}{18} = \frac{450}{18} = 25$ m/s.
Time taken = $\frac{\text{Distance}}{\text{Speed}} = \frac{180}{25} = 7.2$ seconds.
Hence, the train takes 7.2 seconds to pass the railway signal.
A train whose length is 150 m, passes a telegraph pole in 10 s. Find the speed of the train in km/h.
Given,
Length of the train = 150 m and time taken to pass the telegraph pole = 10 s.
To pass a telegraph pole (a stationary object), the distance covered = length of the train = 150 m.
Speed = $\frac{\text{Distance}}{\text{Time}} = \frac{150}{10} = 15$ m/s.
Converting the speed into km/h (multiply by $\frac{18}{5}$) :
Speed = 15 × $\frac{18}{5} = \frac{270}{5} = 54$ km/h.
Hence, the speed of the train = 54 km/h.
A train 120 m long passes a railway platform 160 m long in 14 s. How long will it take to pass another platform which is 100 m long?
Given,
Length of the train = 120 m.
When the train passes the first platform (160 m long) :
Distance covered = length of the train + length of the platform = 120 m + 160 m = 280 m in 14 s.
Speed = $\frac{\text{Distance}}{\text{Time}} = \frac{280}{14} = 20$ m/s.
When the train passes the second platform (100 m long) :
Distance to be covered = length of the train + length of the platform = 120 m + 100 m = 220 m.
Time taken = $\frac{\text{Distance}}{\text{Speed}} = \frac{220}{20} = 11$ seconds.
Hence, the train takes 11 seconds to pass the platform which is 100 m long.
Mr. Amit can walk 8 km in 1 hour 20 minutes.
(a) How far does he go in:
(i) 10 minutes?
(ii) 30 seconds?
(b) How long will it take him to walk:
(i) 2500 m?
(ii) 6.5 km?
Given,
Mr. Amit walks 8 km in 1 hour 20 minutes.
Distance = 8 km = 8000 m.
Time = 1 hour 20 minutes = 80 minutes.
Speed = $\frac{\text{Distance}}{\text{Time}} = \frac{8000}{80} = 100$ m/min.
(a) (i) Distance covered in 10 minutes = Speed × Time = 100 × 10 = 1000 m = 1 km.
(ii) Speed in m/s = $\frac{100}{60} = \frac{5}{3}$ m/s.
Distance covered in 30 seconds = Speed × Time = $\frac{5}{3}$ × 30 = 50 m.
Hence, he goes 1 km in 10 minutes and 50 m in 30 seconds.
(b) (i) Time taken to walk 2500 m = $\frac{\text{Distance}}{\text{Speed}} = \frac{2500}{100} = 25$ minutes.
(ii) 6.5 km = 6.5 × 1000 m = 6500 m.
Time taken to walk 6.5 km = $\frac{\text{Distance}}{\text{Speed}} = \frac{6500}{100} = 65$ minutes = 1 hour 5 minutes.
Hence, he takes 25 minutes to walk 2500 m and 1 hour 5 minutes (65 minutes) to walk 6.5 km.
Which is greater: a speed of 45 km/h or a speed of 12.25 m/sec? How much is the distance travelled by each in 2 seconds?
To compare the two speeds, convert both into the same unit (m/s).
Converting 45 km/h into m/s (multiply by $\frac{5}{18}$) :
45 km/h = 45 × $\frac{5}{18} = \frac{225}{18} = 12.5$ m/s.
Since 12.5 m/s > 12.25 m/s, a speed of 45 km/h is greater.
Distance travelled in 2 seconds :
At 12.5 m/s : Distance = Speed × Time = 12.5 × 2 = 25 m.
At 12.25 m/s : Distance = Speed × Time = 12.25 × 2 = 24.5 m.
Hence, a speed of 45 km/h is greater. In 2 seconds, an object moving at 45 km/h covers 25 m, whereas an object moving at 12.25 m/s covers 24.5 m.
A and B start from the same point and at the same time with speeds 15 km/h and 12 km/h respectively. Find the distance between A and B after 6 hours if both move in:
(i) same direction
(ii) the opposite directions.
Given,
Speed of A = 15 km/h, speed of B = 12 km/h and time = 6 hours.
Distance covered by A in 6 hours = Speed × Time = 15 × 6 = 90 km.
Distance covered by B in 6 hours = Speed × Time = 12 × 6 = 72 km.
(i) When both move in the same direction, the distance between them = difference between the distances covered.
Distance between A and B = 90 km − 72 km = 18 km.
Hence, in the same direction the distance between A and B = 18 km.
(ii) When both move in opposite directions, the distance between them = sum of the distances covered.
Distance between A and B = 90 km + 72 km = 162 km.
Hence, in opposite directions the distance between A and B = 162 km.
A and B start from the same place, in the same direction and at the same time with speeds 6 km/h and 2 m/sec respectively. After 5 hours who will be ahead and by how much?
Given,
Speed of A = 6 km/h, speed of B = 2 m/sec and time = 5 hours.
Converting B's speed into km/h (multiply by $\frac{18}{5}$) :
Speed of B = 2 × $\frac{18}{5} = \frac{36}{5} = 7.2$ km/h.
Distance covered by A in 5 hours = Speed × Time = 6 × 5 = 30 km.
Distance covered by B in 5 hours = Speed × Time = 7.2 × 5 = 36 km.
As both move in the same direction, the distance between them = difference between the distances covered = 36 km − 30 km = 6 km.
Since B covers the greater distance, B is ahead.
Hence, B will be ahead of A by 6 km.
Mohit covers a certain distance in 6 hours by his scooter at a speed of 40 km h-1.
(i) Find the time taken by Manjoor to cover the same distance by his car at the speed of 60 km h-1.
(ii) Find the speed of Joseph, if he takes 8 hours to complete the same distance.
Given,
Mohit covers the distance in 6 hours at a speed of 40 km/h.
Distance = Speed × Time = 40 × 6 = 240 km.
(i) Time taken by Manjoor at 60 km/h = $\frac{\text{Distance}}{\text{Speed}} = \frac{240}{60} = 4$ hours.
Hence, Manjoor takes 4 hours to cover the same distance.
(ii) Speed of Joseph (who takes 8 hours) = $\frac{\text{Distance}}{\text{Time}} = \frac{240}{8} = 30$ km/h.
Hence, the speed of Joseph = 30 km/h.
A boy swims 200 m in still water and then returns back to the point of start in total 10 minutes. Find the speed of his swim in (i) m s-1 (ii) km h-1.
Given,
The boy swims 200 m and then returns, so total distance = 200 m + 200 m = 400 m.
Total time taken = 10 minutes
= 10 × 60 seconds
= 600 seconds.
(i) Speed = $\frac{\text{Total distance}}{\text{Total time}} = \frac{400}{600} = \frac{2}{3}$ m/s.
Hence, his swimming speed = $\frac{2}{3}$ m/s.
(ii) Converting the speed into km/h (multiply by $\frac{18}{5}$) :
Speed = $\frac{2}{3}$ × $\frac{18}{5} = \frac{36}{15} = \frac{12}{5} = 2.4$ km/h.
Hence, his swimming speed = 2.4 km/h.
A distance of 14.4 km is covered in 2 hours 40 minutes. Find the speed in m s-1. With this speed Sakshi goes to her school, 240 m away from her house and then returns back. How much time, in all, will Sakshi take?
Given,
Distance = 14.4 km and time = 2 hours 40 minutes.
Converting into metres and seconds :
Distance = 14.4 km
= 14.4 × 1000 m
= 14400 m.
Time = 2 hours 40 minutes
= (2 × 3600 + 40 × 60) seconds
= (7200 + 2400) seconds
= 9600 seconds.
Speed = $\frac{\text{Distance}}{\text{Time}} = \frac{14400}{9600} = 1.5$ m/s.
Sakshi goes 240 m to school and returns, so total distance = 240 m + 240 m = 480 m.
Total time taken = $\frac{\text{Distance}}{\text{Speed}} = \frac{480}{1.5} = 320$ seconds.
Total time taken = 5 minutes 20 seconds.
Hence, the speed = 1.5 m/s and Sakshi takes 320 seconds (5 minutes 20 seconds) in all.
36 km/hr in m/s is:
Converting 36 km/hr into m/s (multiply by $\frac{5}{18}$) :
36 × $\frac{5}{18} = \frac{180}{18} = 10$ m/s.
Hence, option 2 is the correct option.
20 m/s in km/h is:
Converting 20 m/s into km/h (multiply by $\frac{18}{5}$) :
20 × $\frac{18}{5} = \frac{360}{5} = 72$ km/h.
Hence, option 3 is the correct option.
A car covers 120 km in 3 hrs. Distance covered by it, with the same speed, in 5 hrs is:
Speed of the car = $\frac{\text{Distance}}{\text{Time}} = \frac{120}{3} = 40$ km/h.
Distance covered in 5 hours = Speed × Time
= 40 × 5 = 200 km.
Hence, option 1 is the correct option.
You are travelling at a speed of 30 km/h. The time you will take to cover 750 km is:
Time taken = $\frac{\text{Distance}}{\text{Speed}} = \frac{750}{30} = 25$ hours.
Hence, option 2 is the correct option.
A bus covers 80 km in 2 hrs, next 20 km in 1 hr and last 50 km in 2 hrs. Its average speed is:
Total distance covered = 80 km + 20 km + 50 km = 150 km.
Total time taken = 2 h + 1 h + 2 h = 5 h.
Average speed = $\frac{\text{Total distance covered}}{\text{Total time taken}} = \frac{150}{5} = 30$ km/h.
Hence, option 3 is the correct option.
A girl covers 6 km in 30 minutes. Her average speed is:
Time = 30 minutes = $\frac{30}{60}$ h = $\frac{1}{2}$ h.
$$\begin{aligned}\text{Average speed} = \frac{\text{Distance}}{\text{Time}} \\[1em] = \frac{6}{\frac{1}{2}} \\[1em] = 6 \times 2 \\[1em] = 12 \text{ km/h}.\end{aligned}$$
Hence, option 2 is the correct option.
Rohan covers a distance of 24 km and returns back in 3 hrs. His average speed is:
Since Rohan returns back, total distance = 24 km + 24 km = 48 km.
Total time taken = 3 hours.
Average speed = $\frac{\text{Total distance}}{\text{Total time}} = \frac{48}{3} = 16$ km/h.
Hence, option 2 is the correct option.
Sangeeta covers first 60 km of her journey at 20 km/h and next 60 km in 2 hrs. Her average speed is:
Time taken for the first 60 km = $\frac{\text{Distance}}{\text{Speed}} = \frac{60}{20} = 3$ hours.
Time taken for the next 60 km = 2 hours.
Total distance covered = 60 km + 60 km = 120 km.
Total time taken = 3 h + 2 h = 5 h.
Average speed = $\frac{\text{Total distance covered}}{\text{Total time taken}} = \frac{120}{5} = 24$ km/h.
Hence, option 2 is the correct option.
A train of length 80 m, moving with uniform speed and passes a pole in 10 s. The speed of the train is:
To pass a pole (a stationary object), the distance covered = length of the train = 80 m.
Speed = $\frac{\text{Distance}}{\text{Time}} = \frac{80}{10} = 8$ m/s.
Hence, option 2 is the correct option.
A train of length 80 m passes a platform in 20 s with uniform speed of 10 m/s. The length of the platform is:
Distance covered in passing the platform = Speed of train × Time = 10 × 20 = 200 m.
Distance covered in passing the platform = length of the train + length of the platform.
⇒ 200 = 80 + length of the platform
⇒ length of the platform = 200 − 80 = 120 m.
Hence, option 1 is the correct option.
Statement 1 : A boy covers a distance of 15 km in 2 min and next 15 km in 3 min. The speed of the boy is uniform.
Statement 2 : Since total distance covered = 30 km and total time taken = 5 min ∴ Average speed of the boy = 6 km/min
Which of the following options is correct?
The boy covers the first 15 km in 2 min and the next 15 km in 3 min.
For a uniform speed, equal distances must be covered in equal intervals of time. Here equal distances (15 km each) are covered in different times (2 min and 3 min), so the speed is not uniform, it is variable.
So, Statement 1 is false.
Total distance covered = 15 km + 15 km = 30 km.
Total time taken = 2 min + 3 min = 5 min.
Average speed = $\frac{\text{Total distance covered}}{\text{Total time taken}} = \frac{30}{5} = 6$ km/min.
So, Statement 2 is true.
Hence, option 4 is the correct option.
Statement 1 : Mr. John covers a certain distance between two places in 2 hr 30 min with the uniform speed of 60 km/hr. The distance between the two places = 60 × 150 km.
Statement 2 : Distance covered = Speed × time = 60 × 150 km
Which of the following options is correct?
Given,
Uniform speed = 60 km/hr and time = 2 hr 30 min.
To use Distance = Speed × time, the time must be in hours.
Time = 2 hr 30 min = $2\frac{1}{2}$ h = $\frac{5}{2}$ h.
∴ Distance = Speed × time = 60 × $\frac{5}{2}$ = 150 km, and not 60 × 150 km.
In both statements the time 2 hr 30 min has been wrongly taken as 150 (minutes) instead of $\frac{5}{2}$ hours, which gives the incorrect expression 60 × 150 km.
So, Statement 1 is false.
The relation Distance covered = Speed × time is correct, but the substitution 60 × 150 km is wrong, so the value stated in Statement 2 is also incorrect.
So, Statement 2 is false.
Hence, option 2 is the correct option.
Assertion (A) : A cyclist covers 200 m in 40 sec, his speed is 18 km/h.
Reason (R) : Speed in km/h = (speed in m/s) × $\frac{18}{5}$
For the cyclist :
Speed = $\frac{\text{Distance}}{\text{Time}} = \frac{200}{40} = 5$ m/s.
Converting into km/h (multiply by $\frac{18}{5}$) :
Speed = 5 × $\frac{18}{5} = 18$ km/h.
So, Assertion (A) is true.
To convert a speed from m/s to km/h, we multiply by $\frac{18}{5}$, i.e. speed in km/h = (speed in m/s) × $\frac{18}{5}$, which is correct.
So, Reason (R) is true.
Hence, option 3 is the correct option.
Assertion (A) : A car moving with a speed of 60 km/h covers a certain distance in 3 hrs. The time taken by the car to cover the same distance with the speed of 50 km/h is $\frac{60 \times 3}{50}$ hrs.
Reason (R) : Distance covered = Speed × time
Distance covered by the car = Speed × time = 60 × 3 = 180 km.
Time taken to cover the same distance at 50 km/h
= $\frac{\text{Distance}}{\text{Speed}} = \frac{180}{50} = \frac{60 \times 3}{50}$ hrs.
So, Assertion (A) is true.
The distance was found using Distance covered = Speed × time (60 × 3 = 180 km).
So, Reason (R) is true.
Hence, option 3 is the correct option.
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