# PROBLEMS ON TRAINS (PART-II)

## PROBLEMS ON TRAINS (PART-II)

#### QUERY 11

**A train got accidental after the journey of 50 km. Therefore its speed become 3/4th, so it took 35 more minutes to complete its destination. If accident was taken place before 24 km, then time taken to complete destination was 60 minutes more. What is the speed of the train?**

A) 19.2 km/h

B) 19 km/h

C) 20 km/h

D) 20.6 km/h

**MAHA GUPTA**

Let the speed of the train = x km/h

and the total distance = d km

CASE-I

The trains covers the whole distance in two ways; first 50 km at its normal speed and the rest i.e. (d – 50) km at ¾ of its normal speed i.e. 3x/4 km/h.

Therefore, time taken for the whole journey = ^{50}⁄_{x} + ^{d – 50}⁄_{3x/4}

= ^{50}⁄_{x} + (d – 50) × ^{4}⁄_{3x}

But it equals to 35 minutes more than the usual time

Hence, ^{50}⁄_{x} + (d – 50) × ^{4}⁄_{3x} = ^{d}⁄_{x} + 7/12 (35 minutes = 35/60 hour = 7/12 hour)

=> ^{150+4d–200}⁄_{3x} = ^{12d + 7x}⁄_{12x}

=> 4d – 7x = 200 —–(i)

CASE-II

Now the accident takes 24 km before, means the trains covers first 50 – 24 = 26 km at its normal speed and the rest (d – 26) at ¾ of its normal speed i.e. 3x/4 km/h

Therefore, time taken for the whole journey = ^{26}⁄_{x} + ^{d–26}⁄_{3x/4}

= ^{26}⁄_{x} + (d – 26) × ^{4}⁄_{3x}

But it equals to 60 minutes i.e. 1 hour more than the usual time

Hence, ^{26}⁄_{x} + (d – 26) × ^{4}⁄_{3x} = ^{d}⁄_{x} + 1

=> ^{78+4d–104}⁄_{3x} = ^{d+x}⁄_{x}

=> d – 3x = 26 —-(ii)

Solving (i) & (ii)

x = 96/5 = 19.2 km/h (option ‘A’)

SHORT

**SUMER SINGH CHOUHAN**

The difference in time in the two cases = (60 – 35) minutes = 25/60 hr = 5/12 hr

Obviously this difference in time is only because of that 24 km as all other conditions are same.

In the first case the trains runs this 24 km at its original speed and in the second case it runs this 24 km at 75% i.e. 3/4 of its original speed;

Now let the original speed of the train = x km/hr

We know that TIME = DISTANCE/SPEED

Hence, Time taken to run 24 km in the first case therefore = 24/x hr

And time taken to run this 24 km in the second case = ^{24}⁄_{3x/4} = 24 × ^{4}⁄_{3x} hr i.e. 32/x hr

Therefore Time taken in the second case – Time taken in the first case = Difference in time

= 32/x – 24/x = 5/12

=> x = 19.2 km/h (option ‘A’)

#### QUERY 12

**A train travelling got accidental 50 km after the journey and therefore proceeds at 3/4 of its former speed and arrives at its destination 35 minutes late. Had the accident occurred 24 km further, it would have reached the destination only 15 minutes late. Normal speed of the train?**

A) 19.2 km/h

B) 19 km/h

C) 20 km/h

D) 24 km/h

**SUMER SINGH CHOUHAN**

The difference in time in the two cases = (35 – 15) minutes = 20/60 hr = 1/3 hr

Obviously this difference in time is only because of that 24 km as all other conditions are same.

Obviously, in the first case the trains runs this 24 km at 3/4 of its original speed speed and in the second case it runs this 24 km at its original speed;

Now let the original speed of the train = x km/hr

We know that TIME = DISTANCE/SPEED

Hence, Time taken to run 24 km in the first case therefore = ^{24}⁄_{3x/4} = 24 × ^{4}⁄_{3x} hr i.e. 32/x hr

And time taken to run this 24 km in the second case = 24/x hr

Therefore, Time taken in the first case – Time taken in the second case = Difference in time

=> 32/x – 24/x = 1/3

=> x = 24 km/h (option ‘D’)

#### QUERY 12 (A)

**A train meets with an accident after travelling 30 km, after which it moves with 4/5 of its original speed and arrives at destination 45 minutes late. Had the accident happened 18 km further on, it would have reached 9 minutes before. Find the distance of the journey and original speed.**

A) 120 km, 25km/h

B) 125 km, 25 km/h

C) 130km, 30km/h

D) 120 km, 30km/h

**MAHA GUPTA**

Let the speed of the train = x km/h

and the total distance = d km

CASE-I

The trains covers the whole distance in two ways; first 30 km at its normal speed and the rest i.e. (d – 30) km at 4/5 of its normal speed i.e. 4x/5 km/h.

Therefore, time taken for the whole journey = ^{30}⁄_{x} + ^{d–30}⁄_{4x/5}

= ^{30}⁄_{x} + (d – 30) × ^{5}⁄_{4x}

But it equals to 45 minutes more than the usual time

Hence, ^{30}⁄_{x} + (d – 30) × ^{5}⁄_{4x} = ^{d}⁄_{x} + ^{3}⁄_{4} (45 minutes = 45/60 hour = 3/4 hour)

=> ^{120+5d–150}⁄_{4x} = ^{4d+3x}⁄_{4x}

=> d – 3x = 30 —–(i)

CASE-II

Now the accident takes 18 km further, means the trains covers first 30 + 18 = 48 km at its normal speed and the rest (d – 48) at 4/5 of its normal speed i.e. 4x/5 km/h

Therefore, time taken for the whole journey = ^{48}⁄_{x} + ^{d–48}⁄_{4x/5}

= ^{48}⁄_{x} + (d – 48) × ^{5}⁄_{4x}

But it equals to 9 minutes before the time taken earlier, means 45 – 9 = 36 minutes later than the usual time Hence, ^{48}⁄_{x} + (d – 48) × ^{5}⁄_{4x} = ^{d}⁄_{x} + ^{3}⁄_{5} (36 minutes = 36/60 hour = 3/5 hour)

=> ^{192+5d–240}⁄_{4x} = ^{5d+3x}⁄_{5x}

=> 5d – 12x = 240 —-(ii)

Solving (i) & (ii)

d = 120, x = 30 km/h (option ‘D’)

#### QUERY 13

**A train met with an accident 3 hours after starting, which detains it for an hour, after which it proceeds at 75% of its original speed and it arrives the destination 4 hours late. Had the accident taken place 150 km further along the railway line, the train would have arrived only 3½ hours late. The original speed of the train is?**

A) 50 km/hr

B) 75 km/hr

C) 60 km/hr

D) 100 km/hr

**SUMER SINGH CHOUHAN**

The difference in time in the two cases = 4 – 3½ = 1/2 hr

Obviously this difference in time is only because of that 150 km as all other conditions are same.

In the first case the trains runs this 150 km at 75% i.e. 3/4 of its original speed; and in the second case it runs this 150 km at its original speed.

Now let the original speed of the train = x km/hr

We know that TIME = DISTANCE/SPEED

Hence, Time taken to run 150 km in the first case therefore = ^{150}⁄_{3x/4} = 150 × ^{4}⁄_{3x} hr i.e. 200/x hr

And time taken to run this 150 km in the second case = 150/x

Therefore Time taken in first case – Time taken in the second case = Difference in time

= ^{200}⁄_{x} – ^{150}⁄_{x} = 1/2

=> x = 100 (option ‘D’)

QUERY 14

**Two trains, one from station A to station B, and the other from B to A start simultaneously. After meeting, the trains reach their destination after 9 hours and 16 hours respectively. The ratio of their speeds is?**

A) 2:3

B) 4:3

C) 6:7

D) 9:16

**MAHA GUPTA**

Suppose both the trains meet at R

And let first train’s speed be x km/h and the second train’s speed be y km/h

Then AR = 9x kilometers and BR = 16y kilometers

Now the time taken by the first train = ^{16y}⁄_{x} —-Distance covered by the second train/Speed of the first train

and the time taken by the other train = ^{9x}⁄_{y} —-Distance covered by the first train/Speed of the second train

It’s obvious that these two times are equal

Therefore ^{16y}⁄_{x }= ^{9x}⁄_{y}

=> 16y² = 9x²

=> ^{x²}⁄_{y²} = 16/9

=> ^{x}⁄_{y} = 4/3

=> x : y = 4 : 3 (option ‘B’)

QUERY 15

**Two guns were fired from the same place at the difference of 26 minutes. A person sitting in a train coming towards the same direction hears the second sound after 25 minutes. If speed of the sound is 350 m/s, find the speed of the train.**

A) 14 m/s

B) 60 m/s

C) 80 m/s

D) 40 m/s

**MAHA GUPTA
**Let the speed of the train be x m/m

Obviously, the distance traveled by train in 25 minutes = distance traveled by sound in 26 minutes – 25 minutes i.e. 1 minute

Therefore according to above; x*25 = 350*1

or x = 350/25 = 14 (option ‘A’)

QUERY 16

**Two trains A and B are 110 km apart on a straight line. One train starts at 8 am and travels toward the train B at 40 km/h. Another train starts at 10 am and travels towards the train A at 50 km/h. At what time will they meet?**

A) 10:30 am

B) 11:00 am

C) 10:20 am

D) 10:40 am

**MAHA GUPTA**

Train A travels alone for 2 hours

So distance travelled by it in 2 hours at 40 km/h = 2*40 = 80 km

Thus the remaining distance = 110 – 80 = 30 km

Obviously this distance of 30 km has to be travelled by both the trains running together in the opposite direction

Here you can do the sum by finding the relative speed of both the trains

The relative speed here = 40 + 50 = 90 km/h

Time taken to meet = DISTANCE/SPEED = 30/90 hours = 20 minutes

So the time at which both the trains will meet = 10:20 am (option ‘C’)

QUERY 17

**Two trains A and B start from Howrah and Patna towards Patna and Howrah respectively at the same time. After passing each other they take 4h 48m and 3h 20m to reach Patna & Howrah respectively. If the train from Howrah is moving at 45 km/hr, then the speed of the other train is?**

A) 45 km/h

B) 54 km/h

C) 40 km/h

D) 60 km/h

**MAHA GUPTA**

Suppose both the trains meet at R and let the second train’s speed be y km/h

Then AR = (24/5)45 = 216 kilometers and BR = (10/3)y kilometers [4 hours 48 minutes = 24/5 hours; and 3 hours 20 minutes = 10/3 hours]

Now time taken by the first train = [(10/3)y]/45 —-Distance covered by the second train/speed of the first train

and time taken by the other train = 216/y —-Distance covered by the first train/speed of the second train

It’s obvious that these two times are equal

Therefore [(10/3)y]/45 = 216/y

=> 2y/27 = 216/y

=> y = (108*27)/y

=> y² = 108*27

=> y² = 2*2*27*27

=> y = 54

Hence the speed of the second train = 54 km/h (option ‘B’)

QUERY 18

**Without stoppages a train travels a certain distance with an average speed of 60 km/h and with stoppage it covers the same distance with the average speed of 40 km/h. On an average how many minutes per hour does the train stops during the journey?**

A) 20 minutes

B) 18 minutes

C) 25 minutes

D) 15 minutes

**MAHA GUPTA**

Due to stoppages, the train covers 20 km/h less

Therefore time taken to cover that 20 km in minutes = (20/60)60 = 20 minutes (option ‘A’)

QUERY 19

**Two trains start running from the same place on parallel lines in the same direction at speed of 45 km/h and 40 km/h respectively. Find the distance between them after 45 minutes.**

A) 10/3 km

B) 15/4 km

C) 12/5 km

D) 15/7 km

**MAHA GUPTA**

Distance covered by the first train in 60 minutes = 45 km

Therefore distance covered by it in 45 minutes = (45/60)45 = 135/4 km

Now distance covered by the other train in 60 minutes = 40 km

Therefore distance covered by it in 45 minutes = (40/60)45 = 30 km

Hence the distance between them = 135/4 – 30 = 15/4 km (answer)

TRICK

To find the distance between them, find their relative speed and do

Now their relative speed = 45 – 40 = 5 km/h

Difference between them in 60 minutes = 5 km

Hence the distance between them in 45 minutes = (5/60)45 = 15/4 km (option ‘B’)

QUERY 20

**A train crosses a 80 meter long platform in 20 seconds and a man standing on the platform crosses it in 12 seconds. Find the length of the train.**

A) 100 meter

B) 110 meter

C) 120 meter

D) 200 meter

**MAHA GUPTA**

To remember:

To cross a pole or a standing man, a train is required to cover distance equal to its own length, and to cross a platform or any other stationary object of some length it requires to cover distance equal to the sum of its own length and the length of the platform or the object.

Now, let the length of the train = x meters

We know that SPEED = DISTANCE/TIME

Therefore the speed of train in case when it crosses the man = x/12 seconds per meter

And the speed in case when it crosses the platform = (x + 80)/20

As the speed in both the cases must be equal, therefore

x/12 = (x + 80)/20

=> x = 120

So the length of the train = 120 meters (option ‘C’)