Time and Distance Questions for SSC MTS

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Question 1: The time taken by a boat to travel 13 km downstream is the same as time taken by it to travel 7 km upstream. If the speed of the stream is 3 km/h, then how much time (in hours) will it take to travel a distance of 44.8 km in still water?

a) $4\frac{12}{25}$

b) $5\frac{3}{5}$

c) $5\frac{2}{5}$

d) $4\frac{13}{25}$

1) Answer (A)

Solution:

Speed of the stream = 3 km/h

Let the speed of the boat be v.

Speed in downstream = 3 + v

Speed in upstream = v – 3

Time = distance/speed

According to the question,

The time taken by a boat to travel 13 km downstream = time taken by boat to travel 7 km upstream

13/(3 + v) = 7/(v – 3)

13v – 39 = 21 + 7v

6v = 60

v = 10 km/h

Speed of boat = 10 km/h

The time taken by a boat to travel a distance of 44.8 km in still water = 44.8/10 = 112/25 = $ 4 \frac{12}{25}$ hr

Question 2: 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) 120

b) 80

c) 90

d) 100

2) Answer (B)

Solution:

let the speed of car be x.

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

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

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

According to question,

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

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

so, Speed of car = x = 80 km/hr

Hence, Option B is correct

Question 3: 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) 3

b) 2.5

c) 2

d) 3.2

3) Answer (A)

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 4: 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) 380 km

b) 385 km

c) 375 km

d) 285 km

4) Answer (B)

Solution:

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

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

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

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

Hence, the correct answer is Option B

Question 5: 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) 12 hours

b) 14 hours

c) 10 hours

d) 8 hours

5) Answer (C)

Solution:

Given, the car covered 150 km in 5 hours

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

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

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

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Question 6: 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) 35 km/h

b) 20 km/h

c) 25 km/h

d) 30 km/h

6) Answer (D)

Solution:

Let the speed of Rahul = s

$\Rightarrow$ Speed of Mithun = 70 – s

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

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 = $\frac{21}{10}$ hours

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

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

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

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

$\Rightarrow$  $70s-s^2=1000$

$\Rightarrow$  $s^2-70s+1000=0$

$\Rightarrow$  $s^2-50s-20s+1000=0$

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

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

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

$\Rightarrow$  s = 50 km/h or   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 7: 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) 214 km

b) 212 km

c) 216 km

d) 218 km

7) Answer (C)

Solution:

Let the total distance be 3d

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

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

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

$\Rightarrow$  $\frac{13d}{72}=13$

$\Rightarrow$   d = 72 km

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

Hence, the correct answer is Option C

Question 8: 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) 130 km

b) 120 km

c) 140 km

d) 100 km

8) Answer (B)

Solution:

Let the total distance covered by Mohan = d

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

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

Total time for the journey = 5 hours

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

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

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

$\Rightarrow$  d = 120 km

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

Hence, the correct answer is Option B

Question 9: 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) 8 km

b) 10 km

c) 15 km

d) 12 km

9) Answer (A)

Solution:

Let the distance travelled on foot = d

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

Speed of the man on foot = 4 km/h

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

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

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

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

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

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

$\Rightarrow$  d = 8 km

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

Hence, the correct answer is Option A

Question 10: A person covers 700 m distance in 6 minutes. What is his speed in km/h?

a) 3.45 km/h

b) 7 km/h

c) 6 km/h

d) 6.23 km/h

10) Answer (B)

Solution:

Given,

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

Time = 6 minutes = $\frac{6}{60}$ 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 11: Two cars start from the same place at the same time at right angles to each other. Their speeds are 54 km/hr and 72 km/hr, respectively. After 20 seconds the distance between them will be:

a) 720 m

b) 480 m

c) 500 m

d) 540 m

11) Answer (C)

Solution:

Let the cars start at a point A

Speed of first car = 54 km/hr =  $54\times\frac{5}{18}$ m/s = 15 m/s

Distance travelled by first car in 20 seconds = $15\times20$ = 300 m

Speed of second car = 72 km/hr = $72\times\frac{5}{18}$ = 20 m/s

Distance travelled by second car in 20 seconds = $20\times20$ = 400 m

Two cars move at right angle to each other

Let the position of the cars are B and C after 20 seconds respectively as shown in figure

From the figure,

$AC^2+AB^2=BC^2$

$=$>  $400^2+300^2=BC^2$

$=$>  $1600+900=BC^2$

$=$>  $BC^2=2500$

$=$>  $BC=500$ m

$\therefore\ $Distance between the cars after 20 seconds = BC = 500 m

Hence, the correct answer is Option C

Question 12: A person walks a distance from point A to B at 15 km/h, and from point B to A at 30 km/h. If he takes 3 hours to complete the journey, then what is the distance from point A to B?

a) 15 km

b) 30 km

c) 25 km

d) 10 km

12) Answer (B)

Solution:

Let the distance between point A and B = ‘d’ km

Speed of the person from point A to B = 15 km/h

Time taken by the person to walk from point A to B = $\frac{d}{15}$ hours

Speed of the person from point B to A = 30 km/h

Time taken by the person to walk from point B to A = $\frac{d}{30}$ hours

Total time taken to complete the journey = 3 hours

$=$>  $\frac{d}{15}+\frac{d}{30}=3$

$=$>  $\frac{2d+d}{30}=3$

$=$>  $\frac{3d}{30}=3$

$=$>  $d=30$ km

$\therefore\ $Distance from point A to B = 30 km

Hence, the correct answer is Option B

Question 13: The distance covered by a train in $(x + 1)$ hours is $( x^3 + 1)$ km. What is the speed of the train?

a) $(x+ 1)$ km/h

b) $(x^2 + x + 1) km/h$

c) $(x^2 – x  + 1) km/h$

d) $(x^3 – 1) km/h$

13) Answer (C)

Solution:

Given,

Distance covered by the train = $( x^3 + 1)$ km

Time taken by the train =  $(x+1)$ hours

Speed of the train = $\frac{\text{Distance covered}}{\text{Time taken}}$

$=\frac{\left(x^3+1\right)}{\left(x+1\right)}$

$=\frac{\left(x+1\right)\left(x^2-x+1\right)}{\left(x+1\right)}$

$=\left(x^2-x+1\right) km/h$

Hence, the correct answer is Option C

Question 14: A train, 150 m long is running at 90 km/h. How long (in seconds) will it take to clear a platform that is 300 m long?

a) 6

b) 12

c) 50

d) 18

14) Answer (D)

Solution:

Given,

Length of the train (L) = 150 m

Length of the platform (l) = 300 m

Speed of the train (s) = 90 km/h = $\text{90}\times\frac{5}{18}$ m/s = $\text{25}$ m/s

Time taken by the train to cross the platform = $\frac{L+l}{s}=\frac{150+300}{25}=\frac{450}{25}= \text{18}$ seconds

Hence, the correct answer is Option D

Question 15: Tap A can fill a tank in 6 hours and tap B can empty the same tank in 10 hours. If both taps are opened together, then how much time (in hours) will be taken to fill the tank?

a) 20

b) 15

c) 18

d) 16

15) Answer (B)

Solution:

Let the Volume of the tank = $V$

Given,

Tap A can fill the tank in 6 hours

$=$>  Volume filled by Tap A in 1 hour = $\frac{V}{6}$

Tap B can empty the tank in 10 hours

$=$>  Volume emptied by Tap B in 1 hour = $\frac{V}{10}$

$=$>  Volume filled by both the taps in 1 hour = $\frac{V}{6}-\frac{V}{10}=\frac{5V-3V}{30}=\frac{V}{15}$

$\therefore\ $Number of hours required for both the taps to fill the tank = $\frac{V}{\frac{V}{15}}=15$ hours

Hence, the correct answer is Option B

Question 16: Mohan covers a distance of 2.5 km by scooter at the rate of 30 km/h. The time taken by Mohan to cover the given distance in minutes is:

a) 10

b) 6

c) 8

d) 5

16) Answer (D)

Solution:

Given,

Distance travelled = 2.5 km

Speed = 30 km/hr

$\text{Time taken}=\ \frac{\text{Distance travelled}}{\text{Speed}}=\frac{2.5}{30}hr=\frac{25}{30\times10}hr=\frac{25}{30\times10}\times60\min=5\min$

Hence, The correct answer is Option D

Question 17: Ramu works 4 times as fast as Somu. If Somu can complete a work in 20 days independently, then the number of days in which Ramu and Somu together can complete the work is:

a) 4 days

b) 5 days

c) 3 days

d) 6 days

17) Answer (A)

Solution:

Number of days required for Somu to complete work = 20 days

Ramu works 4 times as fast as Somu

$=$> Number of days required for Ramu to complete work = $\frac{20}{4}=5$ days

Let the Total Work = W

Work done by Ramu in 1 day = $\frac{W}{5}$

Work done by Somu in 1 day = $\frac{W}{20}$

Work done by Ramu and Somu in 1 day = $\frac{W}{5}+\frac{W}{20}=\frac{5W}{20}=\frac{W}{4}$

Number of days required for both Ramu and Somu together to complete work = $\frac{W}{\frac{W}{4}}=4$ days

Hence, the correct answer is Option A

Question 18: A boat takes 80 minutes to row 12 km upstream and 60 minutes to row 15 km downstream. How long will it take to row a distance of 36 km in still water?

a) 2 hours

b) 3 hours

c) 4 hours

d) 2.5 hours

18) Answer (B)

Solution:

Let’s assume the speed of the boat in still water and the speed of the stream are B and C respectively.

Speed of boat downstream = (B+C)

Speed of boat upstream = (B-C)

A boat takes 80 minutes to row 12 km upstream and 60 minutes to row 15 km downstream.

$\frac{12}{B-C}\ =\ \frac{80}{60}$

$\frac{12}{B-C}\ =\ \frac{4}{3}$

$\frac{3}{B-C}\ =\ \frac{1}{3}$

(B-C) = 9    Eq.(i)

$\frac{15}{B+C}\ =\ \frac{60}{60}$

(B+C) = 15    Eq.(ii)

Add Eq.(i) and Eq.(ii).

(B-C)+(B+C) = 9+15

2B = 24

B = 12 km/h

Time taken by the boat to row a distance of 36 km in still water = $\frac{36}{12}$

= 3 hours

Question 19: Two trains each having a length of 160 meters moving in opposite direction crossed each other in 9 seconds. If one train crossed a 200-metre-long platform in 18 seconds,then the ratio of their speeds is:

a) 2 : 3

b) 9 : 7

c) 5 : 8

d) 3 : 4

19) Answer (B)

Solution:

Two trains each having a length of 160 meters moving in the opposite directions crossed each other in 9 seconds.

Let’s assume the speed of trains are ‘a’ and ‘b’ respectively.

$\frac{\left(160+160\right)}{a+b}\ =\ 9$

$\frac{320}{9}\ =\ \left(a+b\right)$    Eq.(i)

If one train crossed a 200 metre long platform in 18 seconds.

speed of one train = a = $\frac{\left(160+200\right)}{18}$

= $\frac{\left(360\right)}{18}$

= 20 m/s    Eq.(ii)

Put Eq.(ii) in Eq.(i).

$\frac{320}{9}\ =\ \left(20+b\right)$
$\frac{320}{9} – 20 = b$

$\frac{320-180}{9}=b$

$\frac{140}{9} = b$    Eq.(iii)

Ratio of their speeds is: Eq.(ii) : Eq.(iii)

$20 : \frac{140}{9}$

$1 : \frac{7}{9}$

9 : 7

Question 20: A boat goes a distance of 4 km upstream in 2 hours and the same distance downstream in 20 minutes. How long will it take to go $10\frac{1}{2}$ km in still water?

a) $1\frac{1}{2}$ hours

b) 48 minutes

c) $1\frac{1}{4}$ hours

d) 1 hour

20) Answer (A)

Solution:

Let’s assume the speed of the boat in still water is B and the speed of the stream is C.

A boat goes a distance of 4 km upstream in 2 hours and the same distance downstream in 20 minutes.

$\frac{4}{B-C}\ =\ 2$

B-C = 2    Eq.(i)

$\frac{4}{B+C}\ =\ \frac{20}{60}$

$\frac{4}{B+C}\ =\ \frac{1}{3}$
B+C = 12    Eq.(ii)
Add Eq.(i) and Eq.(ii),
B-C+B+C = 2+12
2B  =14
B = 7 km/h

Time taken to go $10\frac{1}{2}$ km in still water = $\frac{10.5}{7}$

= 1.5 hours

= $1\frac{1}{2}$ hours

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