# Time, Speed and Distance Questions for RRB NTPC Set-2

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## Time, Speed and Distance Q & A for RRB NTPC Set-2

Download RRB NTPC Time, Speed and Distance Q & A Set-2 PDF. Top 15 RRB NTPC Maths questions based on asked questions in previous exam papers very important for the Railway NTPC exam.

Question 1:Â A train â€˜Aâ€™ of 180 metres is running at the rate of 72 km/hr. Another train â€˜Bâ€™ of 120 metres is coming from opposite direction is the rate of 108 km/hr. How long will they take to cross one another ?

a)Â 24 sec

b)Â 12 sec

c)Â 6 sec

d)Â 30 sec

Question 2:Â A man went to his office on cycle at the rate of 10 km/hr and reached late by 6 minutes. When he increased the speed by 2 km/hr, he reached 6 minutes before time. What is the distance between his office and his departure point ?

a)Â 6 km

b)Â 7 km

c)Â 12 km

d)Â 16 km

Question 3:Â 100 kms is the distance between the stations A and B. One train departed from A towards B at the rate of 50 km/hr and from B towards A at the rate of 75 km/hr. Both the trains departed simultaneously. At what distance from station A, both the trains will cross one -another ?

a)Â 40 kms

b)Â 20 kms

c)Â 30 kms

d)Â None of these

Question 4:Â What will be the time taken for a 400 meter long train speeding at 80 km/h to cross a pole?

a)Â 18 seconds

b)Â 12 seconds

c)Â 11 seconds

d)Â 10 seonds

Question 5:Â Two cars A and B started a race of 150 km with the constant speeds of 30 km/h and 50 km/h. After 2 hours A increases the speed so that they both finish the race together. What was the increased speed of the car A?

a)Â 90 kmph

b)Â 80 kmph

c)Â 60 kmph

d)Â 40 kmph

Question 6:Â Starting from his house Arun moves 5 km to the north east and then 4 km towards south He now moves 3 km towards east and then move 8 km towards north to reach his friendâ€™s hourse. Then his friendâ€™s house is

a)Â 5 km north-east of Arunâ€™s house

b)Â 10 km south-east of Arunâ€™s house

c)Â 10 km north-east of Arunâ€™s house

d)Â 10 km north-west of Arunâ€™s house

d)Â Reptiles

Question 7:Â A man swim at 5 km per hour velocity in still water He takes 75 minutes to swim from position A to the position B and back in a river when it is flowing at 1 km per hour The distance between A and B is

a)Â 6 km

b)Â 5 km

c)Â 2.5 km

d)Â 3 km

Question 8:Â A train of length 600 m is travelling at a speed of 54 kmph. A person is running on a platform in the direction opposite to that of the train at a speed of 5 m/s. Find the time taken by the train to cross the person?

a)Â 20 seconds

b)Â 30 seconds

c)Â 40 seconds

d)Â 1 minute

Question 9:Â A man travelled to his hometown at 5 kmph more than his usual speed and reached 2 hours early. On another occasion, he travelled at 2 kmph less than his usual speed and reached the destination 1.5 hours later than usual. What is the usual speed of the man?

a)Â 10 kmph

b)Â 12 kmph

c)Â 15 kmph

d)Â 20 kmph

Question 10:Â 2 runners start running from the opposite sides of a track AB. The speed of the runner who starts running from A towards B is 30 kmph and the speed of the runner who starts running from B to A is 40 kmph. If the length of the track is 200 m, at what distance from A will the two runners meet?

a)Â 400/7 m

b)Â 800/7 m

c)Â 500/ 7 m

d)Â 600/7 m

Speed of the train A = 72 * 5 /18 = 20 m/s

Speed of train B = 108 * 5/18 = 30 m/s

Total time to cross each other = total distance / total speed

= 300/50 = 6 seconds

Let the distance be d.

D/10 = T+ 1/10

D/12 = TÂ – 1/10

So, D = 12 kilometers.

Distance from A = 100 * 50 / (50 + 75) = 40 kms

Time taken to cross the pole will be = $\frac{400}{400/18}$ = 18 seconds (As 80 km/h = 80*(5/18) Meter/seconds)
Hence, answer will be 18 seconds

B will finish the race in 150/50 = 3 hours
In 2 hours, distance travelled by A = 30*2 = 60 km
As they both finished together, in 1 hour distance travelled by A = 90 km
Hence, increased speed of car B = 90 km/h

The total distance of his friend’s house from his house is 10 kilometers north-east. This can be obtained by using Pythagoras theorem where the distance is the sum of the squared distance of 3 and 4 which is 5 kilometers.

Let the distance be x

Relative speed when travelling upstream = 5 – 1 = 4 kmph

Relative speed when travelling downstream = 5 + 1 = 6 kmph

Time taken = 75 minutes = 1.25 hours

So, x/4 + x/6 = 1.25

So, x = 3 kms

Relative distance to be covered = length of the train = 600 m
Relative speed = Speed of the train + Speed of the person (opposite direction)
= 54 * 5/18 m/s + 5 m/s = 15 + 5 = 20 m/s
So, time taken = 600/20 = 30 seconds

Let the speed of the man be â€˜sâ€™ and the distance be â€˜dâ€™.
So, d/(s+5) = d/s – 2
d/(s-2) = d/s + 1.5

By checking the options, we see that if s = 10 kmph, then d = 60 km.