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Download Time Distance and Speed Questions for TISSNET PDF â€“ TISSNET Time Distance and Speed questions PDF by Cracku. Practice TISSNET solved Time Distance and Speed Questions paper tests, and these are the practice question to have a firm topic in the Time Distance and Speed exam. Top 20 very Important Time Distance and Speed Questions for TISSNET based on asked questions in previous exam papers. Â The TISSNET question papers contain actual questions asked with answers and solutions.

Question 1:Â A person crosses a 900-metre long street in 6 minutes.The speed of the person in km/h is:

a)Â 9

b)Â 10

c)Â 15

d)Â 12

Solution:

Distance travelled = 900 m =Â $\frac{900}{1000}=0.9\ km$

Time taken = 6 minutes =Â $\frac{6}{60}=\frac{1}{10\ }=0.1\ hr$

Speed of person (km/hr) =Â $\frac{Dis\tan ce}{time\ }=\frac{0.9}{0.1}$

= 9 km/hr

Hence, Option A is correct.

Question 2:Â A train can travel 40% faster then a car.Both the train and the car start from point A at the same time and reach point B, which is 70km away point from A,at the same time.On the way, however,the train lost about 15 minutes while stopping at stations. The speed of the car in km/h is:

a)Â 120

b)Â 80

c)Â 90

d)Â 100

Solution:

let the speed of car be x.

then the speed of train =Â $x\left(1+\frac{40}{100}\right)=1.4x$

Time taken by car to cover 70km =Â $\frac{70}{x}$

Time takenÂ  by train to cover 70km =Â $\frac{70}{1.4x}=\frac{50}{x}$

According to question,

$\therefore\ \frac{70}{x}-\frac{50}{x}=\frac{15}{60}$

$\therefore\ \frac{20}{x}=\frac{1}{4}$

so, Speed of car =Â x = 80 km/hr

Hence, Option B is correct

Question 3:Â The speed of a motor boat in still water is 20 km/h.It travels 150 km downstream and then returns to the starting point. If the round trip takes a total of 16 hours, what is the speed (in km/h) of the flow of river?

a)Â 8

b)Â 6

c)Â 5

d)Â 4

Solution:

Let the speed of the flow of the river = s

The speed of motor boat in still water(m)Â = 20 km/h

Downstream speed = 20 + s

Upstream speed = 20 – s

Time taken for boat to travel 150 km downstream = $\frac{150}{20+s}$

Time taken for boat to travel 150 km upstream = $\frac{150}{20-s}$

Total time taken = 16 hours

$\frac{150}{20+s}$ +Â $\frac{150}{20-s}$ = 16

150[$\frac{20-s+20+s}{(20+s)(20-s)}$] = 16

$\frac{150\times40}{16}$ =Â (20+s)(20-s)

375 = 400 – s$^2$

s$^2$ = 25

s = 5 km/h

The speed of the flow of the river = s = 5 km/h

Hence, the correct answer is Option C

Question 4:Â A train is to cover 370 km at a uniform speed. After running 100 km,the train could run at a speed 5 km/h less than its normal speed due to some technical fault. The train got delayed by 36 minutes. What is the normal speed of the train, in km/h?

a)Â 48

b)Â 45

c)Â 40

d)Â 50

Solution:

Let the normal speed of the train = s km/h

According to the problem,

$\frac{100}{s}+\frac{270}{s-5}=\frac{370}{s}+\frac{36}{60}$

$\frac{270}{s-5}-\frac{270}{s}=\frac{3}{5}$

$90\left[\frac{s-s+5}{\left(s-5\right)s}=\frac{1}{5}\right]$

$s^2-5s-2250=0$

s = 50 or s = -45

‘s’ cannot be negative.

So s = 50 km/h

Normal speed of the train = s = 50 km/h

Hence, the correct answer is Option D

Question 5:Â A train running at $40\frac{1}{2}$ km/h takes 24 seconds to cross a pole. How much time (in seconds) will it take to pass a 450 m long bridge?

a)Â 56

b)Â 52

c)Â 60

d)Â 64

Solution:

Let the length of the train = L

Speed of the train =Â $40\frac{1}{2}$ km/h =Â $\frac{81}{2}$ km/h =Â $\frac{81}{2}\times\frac{5}{18}$ m/sec = $\frac{45}{4}$ m/sec

Train crosses a pole in 24 seconds.

$\frac{L}{\frac{45}{4}}$ = 24

$\frac{4L}{45}$ = 24

L = 270 m

Length of the train = L = 270 m

Time required for train to passÂ a 450 m long bridge = $\frac{L+450}{\frac{45}{4}}$

=Â $\frac{270+450}{\frac{45}{4}}$

=Â $\frac{720\times4}{45}$

= 64 sec

Hence, the correct answer is Option D

Question 6:Â A train covers 450 km at a uniform speed. If the speed had been 5 km/h more,Â it would have taken 1 hour less to cover the same distance. How much time will it take to cover 315 km at its usual speed?

a)Â 7h 52m

b)Â 6h 30m

c)Â 6h 18m

d)Â 7h

Solution:

Let the uniform speed of train = s

According to the problem,

$\frac{450}{s+5}=\frac{450}{s}-1$

$\frac{450}{s}-\frac{450}{s+5}=1$

$450\left(\frac{s+5-s}{s\left(s+5\right)}\right)=1$

$450\left(\frac{5}{s^2+5s}\right)=1$

$s^2+5s-2250=0$

$\left(s+50\right)\left(s-45\right)=0$

s = -50 or s = 45

s cannot be negative, so s = 45 km/h

The uniform speed of train =Â 45 km/h

Time taken by train to cover 315 km at its usual speed =Â $\frac{315}{45}$ = 7 hours

Hence, the correct answer is Option D

Question 7:Â A train running at 48 km/h crosses a man going with the speed of 12 km/h, in the same direction, in 18 seconds and passes a woman coming from the opposite direction in 12 seconds. The speed (in km/h) of the woman is :

a)Â 6

b)Â 8

c)Â 9

d)Â 10

Solution:

Let the length of the train = L

Relative speed between train and man = 48 – 12 = 36 km/h

= 36 x $\frac{5}{18}$ m/sec

= 10 m/sec

Time taken by train to cross the man = 18 seconds

$\frac{L}{10}$ = 18

L = 180 m

Length of the train = 180 m

Let the speed of the woman = s km/h

RelativeÂ speed between train and woman = (48 + s) km/h

=Â (48 + s)Â x $\frac{5}{18}$ m/sec

Time taken by train to cross the woman = 12 seconds

$\frac{L}{\left(48+s\right)\times\frac{5}{18}}=12$

$\frac{180}{\left(48+s\right)\times\frac{5}{18}}=12$

$\frac{180\times18}{\left(48+s\right)5}=12$

48 + s = 54

s = 6

Speed of the woman = 6 km/h

Hence, the correct answer is Option A

Question 8:Â A car can cover a distance of 144 km in 1.8 hours. In what time(in hours) will it cover double the distance when its speed is increased by 20% ?

a)Â 3

b)Â 2.5

c)Â 2

d)Â 3.2

Solution:

Speed of the car = $\frac{144}{1.8}$ = 80 km/hr

Speed of the car when increased by 20% = $\frac{120}{100}\times$80 = 96 km/hr

Required time = $\frac{288}{96}$

= 3 hours

Hence, the correct answer is Option A

Question 9:Â A boat goes 30 km upstream in 3 hours and downstream in 1 hour. How much time (in hours) will this boat take to cover 60 km in still water?

a)Â 6

b)Â 3

c)Â 2

d)Â 5

Solution:

Let the speed of the boat in still water = m

Speed of the stream = s

Upstream speed = m – s

$\frac{30}{3}$ = m – s

m – s = 10…………(1)

Downstream speed = m + s

$\frac{30}{1}$ = m + s

m + s = 30………..(2)

2m = 40

m = 20

Speed of the boat in still water = 20 km/h

Time required for theÂ boat to cover 60 km in still water = $\frac{60}{20}$ = 3 hours

Hence, the correct answer is Option B

Question 10:Â The speed of a train is 220% of the speed of a car. The car covers a distance of 950 km in 19 hours. How much distance will the train cover in $3 \frac{1}{2}$ hours?

a)Â 380 km

b)Â 385 km

c)Â 375 km

d)Â 285 km

Solution:

Speed of the car =Â $\frac{950}{19}$ = 50 km/h

Given, speed of the train is 220% of the speed of the car

$\Rightarrow$Â  Speed of the train =Â $\frac{220}{100}\times50$ =Â  110 km/h

$\therefore\$Distance covered by the train inÂ $3 \frac{1}{2}$ hours =Â $110\times3\frac{1}{2}$ =Â $110\times\frac{7}{2}$ = 385 km

Hence, the correct answer is Option B

Question 11:Â A car covered 150 km in 5 hours. If it travels at one-third its usual speed, then how much more time will it take to cover the same distance?

a)Â 12 hours

b)Â 14 hours

c)Â 10 hours

d)Â 8 hours

Solution:

Given, the car coveredÂ 150 km in 5 hours

Speed of the car =Â $\frac{150}{5}$ = 30 km/h

One third of the speed =Â $\frac{1}{3}\times30$ = 10 km/h

Time required for the car to cover 150 km with one third speed =Â $\frac{\text{Distance}\\ }{\text{Speed}}$ =Â $\frac{150}{10}$ = 15 hours

$\therefore\$Extra time required to cover the distance with one third speed = 15 – 5 = 10 hours

Hence, the correct answer is Option C

Question 12:Â Rahul and Mithun travel a distance of 30 km. The sum of their speeds is 70 km/h and the total time taken by both to travel the distance is 2 hours 6 minutes. The difference between their speeds is:

a)Â 35 km/h

b)Â 20 km/h

c)Â 25 km/h

d)Â 30 km/h

Solution:

Let the speed of Rahul = s

$\Rightarrow$ Speed of Mithun = 70 – s

Time taken by Rahul to cover 30 km distance =Â $\frac{30}{s}$

Time taken by Mithun to cover 30 km distance = $\frac{30}{70-s}$

Given, total time = 2 hours 6 minutes = 2 + $\frac{6}{60}$ hours = 2 + $\frac{1}{10}$ hours =Â $\frac{21}{10}$ hours

$\Rightarrow$ Â $\frac{30}{s}+\frac{30}{70-s}=\frac{21}{10}$

$\Rightarrow$ Â $\frac{1}{s}+\frac{1}{70-s}=\frac{7}{100}$

$\Rightarrow$ Â $\frac{70-s+s}{s\left(70-s\right)}=\frac{7}{100}$

$\Rightarrow$ Â $\frac{70}{70s-s^2}=\frac{7}{100}$

$\Rightarrow$ Â $70s-s^2=1000$

$\Rightarrow$ Â $s^2-70s+1000=0$

$\Rightarrow$ Â $s^2-50s-20s+1000=0$

$\Rightarrow$ Â $s\left(s-50\right)-20\left(s-50\right)=0$

$\Rightarrow$ Â $\left(s-50\right)\left(s-20\right)=0$

$\Rightarrow$ Â $s-50=0$ Â  or Â  $s-20=0$

$\Rightarrow$Â  s = 50 km/h or Â  s = 20 km/h

When speed of Rahul = 50 km/h, speed of mithun = 20 km/h

When speed of Rahul = 20 km/h, speed of mithun = 50 km/h

$\therefore\$Difference between their speeds = 30 km/h

Hence, the correct answer is Option D

Question 13:Â Mohan travels three equal distances at speeds of 12 km/h, 18 km/h and 24 km/h.If he takes a total of 13 hours, then what is the total distance covered?

a)Â 214 km

b)Â 212 km

c)Â 216 km

d)Â 218 km

Solution:

Let the total distance be 3d

Time taken to cover the distance at speed 12 km/h =Â $\frac{d}{12}$ hour

Time taken to cover the distance at speed 18 km/h = $\frac{d}{18}$ hour

Time taken to cover the distance at speed 24 km/h = $\frac{d}{24}$ hour

Given, total time = 13 hours

$\Rightarrow$ Â $\frac{d}{12}+\frac{d}{18}+\frac{d}{24}=13$

$\Rightarrow$ Â $\frac{6d+4d+3d}{72}=13$

$\Rightarrow$ Â $\frac{13d}{72}=13$

$\Rightarrow$Â Â  d = 72 km

$\therefore\$Total distance = 3d = 3 x 72 = 216 km

Hence, the correct answer is Option C

Question 14:Â The diameter of a wheel is 49 cm. The number of revolutions in which it will have to cover a distance of 770 m, is:

a)Â 400

b)Â 500

c)Â 700

d)Â 600

Solution:

Given, diameter of the wheel = 49 cm

Perimeter of the wheel = 49$\pi =49\times\frac{22}{7}=154$cm

Distance to be covered = 770 m = 770$\times$100 cm

$\therefore\$The number of revolutions in which it will have to cover a distance of 770 m = $\frac{770\times100}{154}=500$

Hence, the correct answer is Option B

Question 15:Â Mohan finishes a journey by scooter in 5 hours. He travels the first half of the journey at 30 km/h and the second half of the journey at 20 km/h. The distance covered by him is:

a)Â 130 km

b)Â 120 km

c)Â 140 km

d)Â 100 km

Solution:

Let the total distance covered by Mohan = d

Time taken by Mohan to cover the first half =Â $\frac{\frac{d}{2}}{30}$ =Â $\frac{d}{60}$ hours

Time taken by Mohan to cover the second half =Â $\frac{\frac{d}{2}}{20}$ =Â $\frac{d}{40}$ hours

Total time for the journey = 5 hours

$\Rightarrow$ $\frac{d}{60}+\frac{d}{40}=5$

$\Rightarrow$ $\frac{2d+3d}{120}=5$

$\Rightarrow$ $\frac{5d}{120}=5$

$\Rightarrow$Â  d = 120 km

$\therefore\$Distance covered by Mohan = 120 km

Hence, the correct answer is Option B

Question 16:Â A car covers 15 km, 20 km, 30 km and 12 km at speeds of 20 km/h, 30 km/h, 40 km/h and 30 km/h,respectively. The average speed of the car for the total journey is :

a)Â 88 km/h

b)Â 60 km/h

c)Â 30 km/h

d)Â 40 km/h

Solution:

Time taken by the car to cover 15 km distance at speed of 20 km/h =Â $\frac{15}{20}$ h

Time taken by the car to cover 20 km distance at speed of 30 km/h = $\frac{20}{30}$ h

Time taken by the car to cover 30 km distance at speed of 40 km/h = $\frac{30}{40}$ h

Time taken by the car to cover 12 km distance at speed of 30 km/h = $\frac{12}{30}$ h

Total Time =Â $\frac{15}{20}+\frac{20}{30}+\frac{30}{40}+\frac{12}{30}$ =Â $\frac{90+80+90+48}{120}$ =Â $\frac{308}{120}$ =Â $\frac{77}{30}$ h

$\therefore\$Average speed of the car for the total journey = $\frac{\text{Total distance}}{\text{Total Time}}$ = $\frac{15+20+3+12}{\frac{77}{30}}$ = $\frac{77\times30}{77}$ = $30$ km/h

Hence, the correct answer is Option C

Question 17:Â A man travelled a distance of 35 km in 5 hours. He travelled partly on foot at the rate of 4 km/h and the rest on bicycle at the rate of 9 km/h. The distance travelled on foot is:

a)Â 8 km

b)Â 10 km

c)Â 15 km

d)Â 12 km

Solution:

Let the distance travelled on foot = d

$\Rightarrow$Â  Distance travelled on bicycle = 35 – d

Speed of the man on foot = 4 km/h

Time taken by the man on foot =Â $\frac{d}{4}$

Speed of the man on bicycle = 9 km/h

Time taken by the man on bicycle = $\frac{35-d}{9}$

Total time taken by the man = 5 hours

$\Rightarrow$ Â $\frac{d}{4}+\frac{35-d}{9}=5$

$\Rightarrow$ Â $\frac{d}{4}+\frac{35}{9}-\frac{d}{9}=5$

$\Rightarrow$ Â $\frac{9d-4d}{36}=5-\frac{35}{9}$

$\Rightarrow$ Â $\frac{5d}{36}=\frac{10}{9}$

$\Rightarrow$ Â d = 8 km

$\therefore\$Distance travelled on foot = 8 km

Hence, the correct answer is Option A

Question 18:Â The wheel of a car has 210 cm diameter. How many revolutions per minute must the wheel make so that the speed of the car is kept at 120 km/h ?

a)Â 245

b)Â 289

c)Â 326.42

d)Â 303.03

Solution:

Speed of the car = 120 km/h =Â $120\times\frac{1000\times100}{60}$ cm/min = 2,00,000 cm/min

$\Rightarrow$ Car has to travel 2,00,000 cm in 1 minute

Diameter of the wheel of the car (d) = 210 cm

Perimeter of the wheel of the car = $\pi\ d$ = $\frac{22}{7}\times210$ = 660 cm

For one revolution, distance travelled by the car = 660 cm

Number of revolutions made by the car to cover 2,00,000 cm =Â $\frac{200000}{660}$ = 303.03

$\therefore\$Number of revolutions made in a minute = 303.03

Hence, the correct answer is Option D

Question 19:Â A person covers 700 m distance in 6 minutes. What is his speed in km/h?

a)Â 3.45 km/h

b)Â 7 km/h

c)Â 6 km/h

d)Â 6.23 km/h

Solution:

Given,

Distance = 700 m =Â $\frac{700}{1000}$ km

Time = 6 minutes =Â $\frac{6}{60}$ hr

Speed = $\frac{\text{Distance}}{\text{Time}}$ = $\frac{\frac{700}{1000}}{\frac{6}{60}}$ = $\frac{700}{1000}\times\frac{60}{6}$ = 7 km/h

Hence, the correct answer is Option B

Question 20:Â Ravi starts for his school from his house on his cycle at 8:20 a.m. If he runs his cycle at a speed of 10 km/h, he reaches his school 8 minutes late, and if he drives the cycle at a speed of 16 km/h, he reaches his school 10 minutes early. The school starts at:

a)Â 9:40 a.m.

b)Â 8:40 a.m.

c)Â 8:50 a.m.

d)Â 9:00 a.m.

Solution:

Let the distance between Ravi’s house and his school = $d$

Time taken by him when he runs his cycle at a speed of 10km/h =Â $\frac{d}{10}$

Time taken by him when he drives his cycle at a speed of 16km/h = $\frac{d}{16}$

According to the problem,

$\frac{d}{10}-\frac{d}{16}=\frac{18}{60}$ hours

$=$> Â $\frac{8d-5d}{80}=\frac{18}{60}$

$=$> Â $\frac{3d}{80}=\frac{18}{60}$

$=$> Â $d=8$ km

Time taken by him when he runs his cycle at a speed of 10km/h = $\frac{8}{10}$ hour =Â $\frac{8}{10}\times60$ minutes = 48 minutes

The time at which school starts = 8:20 a.m+ 48 minutes – 8 minutes = 8:20 + 40 minutes = 9:00 a.m

Hence, the correct answer is Option D