# PROBLEMS ON TRAINS (PART-I)

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## PROBLEMS ON TRAINS (PART-I)

#### QUERY 1

**A train 105 mts long moving at a speed of 54 kmph crosses another train in 6 sec. Then which of the following is true?**

A) trains are moving in the opposite direction

B) trains are moving in the same direction

C) the other train is not moving

**[email protected]**

As the train crosses the other train in a time given in seconds, we assume the speed per second

Let length and speed of the second train be x m and s m/s respectively

Speed of the first train in meter per second = 54*(5/18) m/s = 15 m/s

Now there may be three cases

1) when trains are moving in the same direction

105 + x = (15 – s)* 6

or x + 6s = -15

which is not possible (as the sum of ‘x’ and ‘s’ should be positive)

2) when second train is not moving

105 + x = (15 – 0)*6

or x = -15

which is not possible as ‘x’ is a length which can’t be negative

3) when trains are moving in opposite direction

105 + x = (15 + s)*6

or 6s – x = 15

which is possible; so option ‘A’ is correct

QUERY 2

** Two guns were fired from the same place at an interval of 10 minutes and 30 seconds, but a person in the train approaching the place hears the second shot 10 minutes after the first. The speed of the train(in km/hr), supposing that sound travels at 330 m per second, is?**

A) 19.8

B) 58.6

C) 59.4

D) 111.80

**[email protected]**

Let the speed of the train be x m/s

Obviously, the distance traveled by train in 10 minutes (as the person in the train approaching the place hears the second shot 10 minutes after the first) = distance traveled by sound in 10 minutes and 30 seconds – 10 minutes i.e. 30 seconds

therefore according to above; x*10*60 = 330*30 (10 minutes = 10*60 seconds)

or x = 16.5 m/s

= (16.5)*(18/5) km/h

= 59.4 km/h (option ‘C’)

QUERY 3

**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 5/3 hours and 27/5 hours respectively. If speed of the first train is 63 km/h, then find the speed of the second train.**

A) 35

B) 25

C) 45

D) 60

**MAHA GUPTA
**Let the second train’s speed be y km/h

Then distance covered by the first train after the point of meeting = (5/3)*63 = 105 kilometers and distance covered by the second train after the point of meeting = (27/5)*y kilometers

Now time taken by the first train before the point of meeting = Distance covered by the second train after meeting/speed of the first train = [(27/5)*y]/63

and time taken by the second train before the point of meeting = Distance covered by the first train after meeting/speed of the second train = 105/y

It’s obvious that these two times are equal

Therefore [(27/5)*y]/63 = 105/y

=> 3y/35 = 105/y

=> y/35 = 35/y

=> y² = 35*35

=> y = 35

Hence the speed of the second train = 35 km/h (option ‘A’)

#### QUERY 4

**A train travelling with a speed of 60 km/h catches another train travelling in the same direction and leaves it 120 meters behind in 18 seconds. The speed of the second train is?**

A) 26 km/h

B) 35 km/h

C) 36 km/h

D) 63 km/h

**Raj kumar**

Letting the first train is moving at x m/s and the second train at y m/s As both these trains are moving in the same direction their relative speed = (x – y) m’s i.e. 120/18 m/s

This speed in km/hr = 120/18*5/18 = 24

Now the speed of 2nd train = speed of the first train – relative speed i.e. 60 – 24 = 36 km/hr (option ‘C’)

QUERY 5

**Two goods train each 500 m long, are running in opposite directions on parallel tracks. Their speeds are 45 km/hr and 30 km/hr respectively. Find the time taken by the slower train to pass the driver of the faster one?**

A) 12 seconds

B) 20 seconds

C) 22 seconds

D) 24 seconds

**RONNIE BANSAL
**One needs to calculate the relative speed in such questions. When two bodies are moving in opposite directions x m/s and y m/s, their relative speed is (x + y) m/s.

Thus the relative speed of these two trains = (45 + 30) km/h = 75 km/h.

But as the distance given here is in meters, so better if we convert it into meters per second i.e. m/s

So the relative speed in m/s = 75*(5/18) = 125/6 m/s

Now we need to know the distance so that time taken is known. Here we have to know the distance covered by the slower train to pass the driver of the other train, so it will be only the length of the slower train. We don’t have to add to it the length of the other train we should remember. You can say it like this also:

distance covered = Length of the slower train.

Therefore, time taken by the slower train to pass the driver of the faster one = 500*(6/125) seconds = 24 seconds (option ‘D’) —- (Time = Distance/speed)

#### QUERY 6

**Train A leaves Meerut at 4 pm and arrives Ghaziabad at 5pm. Train B leaves Ghaziabad at 4 pm and arrives Meerut at 5.30 pm. At what time both the trains would meet each other?**

A) 4:00 pm

B) 4:36 pm

C) 4:25 pm

D) 4:50 pm

**JAYANTCHARAN CHARAN
**Let the distance be D km

Therefore the speed of the train leaving Meerut = D/1 kmph (It takes 1 hour from Meerut to Ghaziabad)

and the speed of the train leaving Ghaziabad = D/1.5 kmph (It takes 1.5 hour from Ghaziabad to Meerut)

As they are moving in the opposite direction their relative speed = (D/1 + D/1.5) km/h

So the time taken in meeting each other = D/(D/1 + D/1.5) —- (Distance/Speed)

= 3/5 hour = (3/5)*60 minutes = 36 minutes

Hence the time at which both the trains will meet each other = 4 pm + 36 minutes = 4:36 pm (option ‘B’)

QUERY 7

**A train leaves Meerut at 5 am and reaches Delhi at 9 am. An other train leaves Delhi at 7 am and reaches Meerut at 10.30 am. At what time will they meet?**

A) 7.46 am

B) 7.56 am

C) 8 am

D) 8.26 am

**RONNIE BANSAL
**Let the distance between Meerut and Delhi be x km & the train starting from Meerut be M and the one starting from Delhi be D.

Thus time taken by M to cover x km = 9 – 4 = 4 hours

and time taken by D to cover x km = 10.30 – 7 = 3.30 hours = 7/2 hours

Therefore speed of M = x/4 km/h & speed of D = 2x/7 km/h

Now let them meet at y hours after 5 a.m.

Now distance traveled by train M till it meets D = speed*time taken = x/4*y = xy/4

and distance traveled by train D till it meets M = speed*time taken = (2x/7)*(y -2)

(as train D starts after 2 hours that of M, we have deducted 2 from y)

As we know distance traveled by first train+ distance traveled by second train= total distance

Therefore xy/4 + (2x/7)*(y – 2) = x

=> 7xy + 8xy – 16x = 28x

=> 15xy = 44x

=> y = 44/15 hours

=> y = (44/15)*60 = 176 minutes = 2 hours 56 minutes

So they will meet at 5 + 2:56 = 7:56 am (option ‘B’)

SHORT TRICK

**JAYANTCHARAN CHARAN**

The first train travels half the distance in half the time till 7 am (till the starting time of second train).

Now let the total distance between Meerut and Delhi be ‘x’ km

Therefore speed of the first train = x/4 km/h

and speed of the second train = x/(7/2) = 2x/7 km/h (It takes 3.5 = 7/2 hours in the journey)

Thus their relative speed = (x/4 + 2x/7) km/h

Thus time taken in meeting each other = DISTANCE LEFT/RELATIVE SPEED = (x/2)/(x/4 + 2x/7) hour= 14/15 hour

= (14/15)*60 minutes = 56 minute

Hence the time at which both the trains will meet each other = 7 am + 56 minutes = 7:56 am (option ‘B’)

QUERY 8

**A speed train is travelling east at 200 km/hr. I am in the train walking west at 5 km/hr. What is my speed relative to the ground**?

A) 205 km/h

B) 200 km/h

C) 195 km/h

D) 190 km/h

**JAYANTCHARAN CHARAN**

East at 200 km/hr – walking west at 5 km/hr = East at 195 km/hr

So my speed relative to the ground = 195 km/hr (option ‘C’)

**RONNIE BANSAL**

Dekho agar koi aadmi train mein h aur train ki opposte direction mein chal rha h, aur train mein uski speed 5 km/h hai. Obviously wo aadmi 1 hour mein 5 km chalega. Lekin usi 1 hour mein train 200 km chalegi. Means wo train us aadmi ko us point se jahan uske 5 km khatam hue hain, 195 km east mein le jaayegi, To chala na wo admi 1 hour mein 200 – 5 = 195 km . (option ‘C’)

Suppose wo aadmi train ki direction mein hi chal rahaa hota, to pahunch jaata na wo us 1 hour mein 205 km (200 + 5). To opposite direction ki speed train ki speed mein se subtract hogi is question mein.

Yahan ye baat sochne ki h ki koi aadmi khadaa khadaa ya baithha baithha bhi actually ground ke relative chal rhaa hota h agar wo body move kar rhi ho jisme wo swaar h (us speed se jis speed se wo body chal rhi hoti h).

QUERY 9

**Two bullets are fired at a platform in the interval of 13 seconds. A passenger who is sitting in the train coming towards the platform heard the sound in the interval of 12.5 seconds. If speed of sound is 330 m/s, train’s speed is?**

A) 47^{13}⁄_{25} km/h

B) 50 km/h

C) 66/5 m/h

D) 45 km/h

**MAHA GUPTA**

Let the speed of the train be x m/s

Obviously, the distance traveled by train in 12.5 seconds = distance traveled by sound in 13 seconds – 12,5 seconds i.e. 0.5 seconds

therefore according to above; x*12.5 = 330*0.5

or x = 66/5 m/s = 47^{13}⁄_{25} km/h (option ‘A’)

QUERY 10

**Two trains start simultaneously from two stations Howrah and Delhi, respectively towards each other on the same track. The distance between the two stations is 560 km and the speeds of trains are 30 km/h and 40 km/h. Simultaneously with the trains, a sparrow sitting on the top of one of the train starts towards the other and reverses its direction on reaching the other train and so on. If the speed of sparrow is 80 km/h then the distance that the sparrow flies before being crushed between the train is.**

A) 500 km

B) 555 km

C) 600 km

D) 640 km

**MAHA GUPTA**

Here we need to know the relative speed of the two trains. If two objects are moving in opposite directions (either towards each other or away from each other) there relative speed is equal to the sum of their speeds.

So relative speed of the trains 30+40 = 70 km/h

Obviously the distance of 560 km has to be covered by the trains at 70 km/h

Therefore time taken by the trains to cover it = DISTANCE/SPEED = 560/70 = 8 hours

It also means the sparrow flies for 8 hours

Now, speed of the sparrow = 80 km/h

Hence the distance flown by the sparrow before being crushed = SPEED*TIME = 80*8 = 640 km (option ‘D’)