**Nishu is standing on a railway bridge which is 180 m long. He finds that a train crosses the bridge in 20 seconds but himself in 8 seconds. Find the speed of the train?**

A.35 kmph

B.54 kmph

C.62 kmph

D.70 kmph

E.None of these

**Solution**

**Answer – B (54 kmph)**

**Explanation** – Let the length of the train be x metres.

Then, the train covers x metres in 8 seconds (train is actually covering itself because length of man is very less compartable to train)and (x + 180) metres in 20 seconds.

equate speed in both case,s=d/t

∴ x/8 =( x + 180) / 20 ⇔ 20x = 8(x + 180) ⇔ x = 120.

∴ Length of the train = 120 m.

Speed of the train = [120/8]m/sec (convert to km/hr i.e x 18/5)

=[120/8] x 18/5kmph = 54 kmph.

**Two trains are running at 40 km/hr and 20 km/hr respectively in the same direction. Fast train completely passes a man sitting in the slower train in 5 seconds. What is the length of the fast train?**

A.27 7/9 m

B.28 m

C.29 m

D.30 2/7 m

E.None of these

**Solution**

**Answer – A (27 7/9 m)**

**Explanation **– When SAME direction- MINUS

Relative speed = (40-20) km/hr = [20 x 5/18] m/sec = [50/9] m/sec.

Length of faster train = sxt=[50/9 x 5] m = 250/9 m = 27 7/9 m.

**Two train travel in opposite directions at 36 kmph and 45 kmph and a man sitting in slower train passes the faster train in 8 seconds. Then length of the faster train is:**

A.120 m

B.140 m

C.160 m

D.180 m

E.None of these

**Solution**

**Answer – D (180 m)**

**Explanation **– When OPP. direction-PLUS

Relative speed = (36 + 45) km/hr

= [81 x 5/18] m/sec = [45/2] m/sec.

Length of train = [45/2 x 8] m = 180 m.

**Two trains of equal lengths take 10 seconds and 15 seconds respectively to cross a telegraph post. If the length of each train be 120 metres, in what time (in seconds) will they cross each other travelling in opposite direction?**

A.12 sec

B.18 sec

C.14 sec

D.25 sec

E.None of these

**Solution**

**Answer – A (12 sec)**

**Explanation **– Speed of the first train = [120 / 10] m/sec = 12 m/sec.

Speed of the second train = [120 / 15] m/sec = 8 m/sec.

Relative speed = (12 + 8) = m/sec = 20 m/sec.

∴ Required time = (120 + 120) / 20 secc = 12 sec

**Two trains are running in opposite directions with the same speed. If the length of each train is 120 metres and they cross each other in 12 seconds, then the speed of each train (in km/hr) is:**

A.12 kmph

B.24 kmph

C.36 kmph

D.48 kmph

E.None of these

**Solution**

**Answer – C (36 kmph)**

**Explanation** – Let the speed of each train be x m/sec.

Then, relative speed of the two trains = 2x m/sec.

So, 2x = (120 + 120)/12 ⇔ 2x = 20 ⇔ x = 10.

∴ Speed of each train = 10 m/sec = [10 x 18/5] km/hr = 36 km/hr

**Excluding stoppages, the speed of a bus is 54 kmph and including stoppages, it is 45 kmph. For how many minutes does the bus stop per hour?**

A.10 min

B.12 min

C.18 min

D.15 min

E.None of these

**Solution**

**Answer – A (10 min)**

**Explanation** – Due to stoppages, it covers 9 km less.Time taken to cover 9 km = (9/54 x 60) min = 10 min

**A motor car starts with the speed of 70 km/hr with its speed increasing every two hours by 10 kmph. In how many hours will it cover 345 kms?**

A.4 hrs

B.4 hrs 5 mins

C.4 1/2 hrs

D.2 hrs

E.None of these

**Solution**

**Answer – C (4 1/2 hrs)**

**Explanation **– Distance covered in first 2 hours = (70 x 2) km = 140 km

Distance covered in next 2 hours = (80 x 2) km = 160 km

Remaining distance = 345 – (140 + 160) = 45 km.

Speed in the fifth hour = 90 km/hr

Time taken to cover 45 km = as speed is 90km/hr means it covers 90 km in 1 hour

so,if 90km………..1 hr

45km……….?

?=45/90hr=1/2hr

Total time taken = 2 + 2 +1/2=4 (1/2)hrs

**A person travels from P to Q at a speed of 40 kmph and returns by increasing his speed by 50%. What is his average speed for the both the trips?**

A.35 kmph

B.4o kmph

C.48 kmph

D.55 kmph

E.None of these

**Solution**

**Answer – C (48 kmph)**

**Explanation **– Speed on return trip = 150% of 40 = 60 kmph

Average speed =

[2 x 40 x 60] / [40 + 6]km/hr =4800/100km/hr = 48 km/hr.

**Three trains are running from a place A to another place B. Their speeds are in the ratio of 4 : 3 : 5. The time ratio to reach B by trains will be**

A. 4 : 3 : 5

B. 5 : 3 : 4

C. 15 : 9 : 20

D. 15 : 20 : 12

**Solution**

**Answer: D(15 : 20 : 12)**

**Explanation:** Ratio of speeds = 4 : 3 : 5

Therefore Ratio of times taken [t=d/s or t indirectly proportional to s when distance is same]= (1/4) : (1/3) : (1/5) = 15 : 20 : 12

**A man in a train notices that he can count 21 telephone posts in one minute. If they are known to be 50 meters apart, then at what speed is the train travelling?**

A.50 kmph

B.54 kmph

C.60 kmph

D.62 kmph

E.None of these

**Solution**

**Answer – C (60 kmph)**

**Explanation **– Number of gaps between 21 telephone posts = 20 Distance traveled in 1 minute = (50 x 20) m = 1000 m = 1 km

Speed = 60 km/hr

**Two trains starting at the same time from two stations 200 km apart and going in opposite directions cross each other at a distance of 110 km from one of the stations. What is the ratio of their speeds?**

A.9 : 28

B.11 : 9

C.11 : 8

D.9 : 22

E.None of these

**Solution**

**Answer – B (11 : 9)**

**Explanation **– In the same time, they cover 110 km and 90 km respectively.Ratio of their speeds = 110 : 90 = 11 : 9

**A and B walk around a circular track. They start at 8 a.m. from the same point in the opposite directions. A and B walk at a speed of 2 rounds per hour and 3 rounds per hour respectively. How many times shall they cross each other before 9.30 a.m.?**

A.15

B.8

C.7

D.10

E.None of these

**Solution**

**Answer – C (7)**

**Explanation** – Relative speed = (2 + 3) = 5 rounds per hourSo, they cross each other 5 times in an hour and 2 times in half an hourHence, they cross each other 7 times before 9.30 a.m.

**The distance between two cities A and B is 330 km. A train starts from A at 8 a.m. and travels towards B at 60 km/hr. Another train starts from B at 9 a.m. and travels towards A at 75 km/hr. At what time do they meet?**

A.10:30 am

B.10:45 am

C.11 am

D.11:25 am

E.None of these

**Solution**

**Answer – C (11 am)**

**Explanation **– Suppose they meet x hrs after 8 a.m.

Then, (Distance moved by first in x hrs) + [Distance moved by second in (x-1) hrs] = 330

60x + 75(x – 1) = 330

x = 3

So, they meet at (8 + 3), i.e. 11 a.m

**The speed of a car increases by 2 kms after every one hour. If the distance travelled in the first one hour was 35 kms, what was the total distance traveled in 12 hours?**

A.456 kms

B.482 kms

C.552 kms

D.556 kms

E.None of these

**Solution**

**Answer – C (552 kms)**

**Explanation** – Total distance travelled in 12 hours = (35 + 37 + 39 + …… upto 12 terms) This is an A.P. with first term,

a = 35, number of terms, n = 12, common difference, d =2.

Required distance =12/2 (2 x 35 + (12 – 1) x 2) = 6(70 + 22) = 552 kms

**A thief is noticed by a policeman from a distance of 200 m. The thief starts running and the policeman chases him. The thief and the policeman run at the rate of 10 km and 11 km per hour respectively. What is the distance between them after 6 minutes?**

A.100 m

B.150 m

C.190 m

D.200 m

E.None of these

**Solution**

**Answer- A(100m)**

**Explanation:**

Relative speed of the thief and policeman = (11 – 10) km/hr = 1 km/hr

Distance covered in 6 minutes =[(1/60)*6] km = (1/10)km = 100 m.

Distance between the thief and policeman = (200 – 100) m = 100 m.