1. From height of 8 mts a ball fell down and each time it bounces half the distance back. What will be the distance travelled
Ans.: 24
Sol. 8+4+4+2+2+1+1+0.5+0.5+ and etc .. =24
Hence ,the required distance is 6 km
2. Two trains 200mts and 150mts are running on the parallel rails at this rate of 40km/hr and 45km/hr. In how much time will they cross each other if they are running in the same direction?
Ans: 252sec
Sol: Relative speed=45-40=5km/hr=25/18 mt/sec
Total distance covered =sum of lengths of trains =350mts.
So, time taken =350*18/25=252sec.
3.Vivek travelled 1200km by air which formed 2/5 of his trip.One third of the whole trip , he travelled by car and the rest of the journey he performed by train. The distance traveled by train was ?
Ans.800km
Sol: Let the total trip be x km.
Then 2x/5=1200
x=1200*5/2=3000km
Distance travelled by car =1/3*3000=1000km
Journey by train =[3000-(1200+1000)]=800km.
4. The diameter of the driving wheel of a bus is 140cm. How many revolutions per minute must the wheel make in order to keep a speed of 66 kmph?
Ans. 250
Sol. Distance to be covered in 1 min=(66*1000)/60 m=1100m
Circumference of the wheel =(2*22/7*0.70)m=4.4m.
So, Number of revolutions per min=1100/4.4=250.
5. Walking 5/6 of its usual speed, a train is 10min late. Find the usual time to cover the journey?
Ans:50 min
Sol: New speed = 5/6 of usual speed
New time = 6/5 of usual time
Therefore, (6/5 of usual time) – usual time = 10min
Therefore Usual time = 50min
6. A man is standing on a railway bridge which is 180m long. He finds that a train crosses the bridge in 20seconds but himself in 8 seconds. Find the length of the train and its speed.
Ans: length of train=120m
Speed of train=54kmph
Sol: Let the length of the train be x meters
Then, the train covers x meters in 8 seconds and (x + 180) meters in 20 seconds.
Therefore x/8 = (x+180)/20 ó 20x = 8(x+180) ó x = 120
Therefore Length of the train = 120m
Speed of the train = 120/8 m/sec = 15 m/sec =15 * 18/5 kmph = 54kmph
7. A train running at 54 kmph takes 20 seconds to pass a platform. Next it takes 12 seconds to pass a man walking at 6 kmph in the same direction in which the train is going. Find the length of the train and the length of the platform.
Ans. length of the train=160m
length of the platform=140 m.
Sol: Let the length of the train be x meters and length of the platform be y meters.
Speed of the train relative to man=(54-6) kmph =48 kmph.
=(48*5/18) m/sec =40/3 m/sec.
In passing a man, the train covers its own length with relative speed.
Therefore, length of the train=(Relative speed *Time)
=(40/3 * 12) m =160 m.
Also, speed of the train=(54 * 5/18) m/sec=15 m/sec.
Therefore, x+y/2xy=20 or x+y=300 or y=(300-160 m=140 m.
Therefore, Length of the platform=140 m.
8. If a man walks at the rate of 5kmph, he misses a train by only 7min. However if he walks at the rate of 6 kmph he reaches the station 5 minutes before the arrival of the train. Find the distance covered by him to reach the station.
Ans:6km.
Sol: Let the required distance be x km.
Difference in the times taken at two speeds=12mins=1/5 hr.
Therefore x/5-x/6=1/5 or 6x-5x=6 or x=6km.
Submitted by subbaiah
