Aptitude
1. Speed =  Time, Time = Speed , 
   Distance  =  (Speed *  Time)                          
2. x km / hr =  x  *  5/18                          
3. x  m/sec  = (x * 18/5) km /hr          
4. If the ratio of the speeds of A and B is a:b ,
 then the ratio of the times taken by them to cover the same distance is  1/1 : 1/b
5. Suppose a man covers a certain distance at x km/ hr and 
an equal distance at y km / hr . Then , the average speed during
 the whole journey is   2xy/x + y km/ hr.

1. How many minutes does Aditya take to cover a distance of 400 m, if he runs at a speed of 20 km/hr?

Aditya’s speed = 20 km/hr  = {20 * 5/18} m/sec  =   50/9 m/sec
Time taken to cover 400 m= { 400 * 9/50} sec =72 sec = 1 12/60  min 1 1/5min.                                                                                               

2. A cyclist covers a distnce of 750 m in 2 min 30 sec. What is the speed in km/hr of the cyclist?

Speed = { 750/150 } m/sec  =5 m/sec  = { 5  *  18/5 } km/hr =18km/hr                                                   

3. A dog takes 4 leaps for every 5 leaps of a hare but 3 leaps of a dog are equal to 4 leaps of the hare. Compare their speeds?

Let the distance covered in 1 leap of the dog be x 
and that covered in 1 leap of the hare by y.
Then , 3x = 4y => x = 4/3y  =>  4x = 16/3  y.
Ratio(speeds of dog and hare) = Ratio(distances covered by them)
=> 4x : 5y => 16/3 y : 5y  => 16/3  : 5  => 16:15

4.While covering a distance of 24 km, a man noticed that after walking for 1 hour and 40 minutes, the distance covered by him was 5/7 of the remaining distance. What was his speed in metres per second?

 Let the speed be x km/hr.
 Then, distance covered in 1 hr. 40 min. i.e., 1  2/3 hrs  = 5x/3  km
 Remaining distance = { 24 – 5x/3 } km.

          5x/3  =  5/7 {  24 -  5x/3  } 
      =>  5x/3  =  5/7 {  (72-5x)/3  }  =>  7x  = 72 – 5x                                            
                                        => 12x = 72  
                                        =>  x = 6
  Hence speed = 6 km/hr ={ 6 * 5/18 } m/sec  =  5/3  m/sec = 1 2/3

5.Peter can cover a certain distance in 1 hr. 24 min. by covering two-third of the distance at 4 kmph and the rest at 5 kmph. Find the total distance.

Let the total distance be x km . Then,
            (2/3)x / 4  + (1/3)x / 5 = 7/5
         => x/6 + x/15 = 7/5
         => 7x = 42
         => x = 6

6.A man traveled from the village to the post-office at the rate of 25 kmph and walked back at the rate of 4 kmph. If the whole journey took 5 hours 48 minutes, find the distance of the post-office from the village.

 Average speed   = 2xy/(x+y) km/hr  = 2*25*4/(25+4) km/hr  = 200  km/hr
 Distance traveled in 5 hours 48 minutes i.e., 5(4/5)  hrs.  =  (200/29)*(29/5) km  = 40km
 Distance of the post-office from the village = 40 / 2    = 20 km

7.An aeroplane files along the four sides of a square at the speeds of 200,400,600 and 800km/hr.Find the average speed of the plane around the field.

Let each side of the square be x km and let the average speed of the plane around the field by y km per hour then ,
 x/200+x/400+x/600+x/800=4x/yó25x/2500ó4x/yóy=(2400*4/25)=384
hence average speed =384 km/hr

8.Walking at 5/7 of its usual speed, a train is 10 minutes too late. Find its usual time to cover the journey.

New speed =5/6 of the usual speed
New time taken=6/5 of the usual time
So,( 6/5 of the usual time )-( usual time)=10 minutes.
=>1/5 of the usual time=10 minutes.
=>  usual time=10 minutes

9.If a man walks at the rate of 5 kmph, he misses a train by 7 minutes. 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.

Let the required distance be x km
Difference in the time taken at two speeds=1 min =1/2 hr
Hence x/5-x/6=1/5<=>6x-5x=6
=> x=6
Hence, the required distance is 6 km

10. A and B are two stations 390 km apart. A train starts from A at 10 a.m. and travels towards B at 65 kmph. Another train starts from B at 11 a.m. and travels towards A at 35 kmph. At what time do they meet?

Suppose they meet x hours after 10 a.m. Then,
   (Distance moved by first in x hrs) + [Distance moved by second in (x-1) hrs]=390.                                                                                                     
   65x + 35(x-1) = 390  => 100x = 425  => x = 17/4
So, they meet 4 hrs.15 min. after 10 a.m i.e., at 2.15 p.m.

11. A goods train leaves a station at a certain time and at a fixed speed. After ^hours, an express train leaves the same station and moves in the same direction at a uniform speed of 90 kmph. This train catches up the goods train in 4 hours. Find the speed of the goods train.

Let the speed of the goods train be x kmph.
     Distance covered by goods train in 10 hours= Distance covered by express train in 4 hours
     10x = 4 x 90 or x =36.
 So, speed of goods train = 36kmph.

12. A thief is spotted by a policeman from a distance of 100 metres. When the policeman starts the chase, the thief also starts running. If the speed of the thief be 8km/hr and that of the policeman 10 km/hr, how far the thief will have run before he is overtaken?

Relative speed of the policeman = (10-8) km/hr =2 km/hr.
Time taken by police man to cover 100m  (100/1000 x 1/2) hr = 1/20  hr.
In 1/20  hrs, the thief covers a distance of 8  x  1/20  km = 2/5  km  = 400 m

13. I walk a certain distance and ride back taking a total time of 37 minutes. I could walk both ways in 55 minutes. How long would it take me to ride both ways?

Sol. Let the distance be x km. Then, 
        ( Time taken to walk x km) + (time taken to ride x km) =37 min.
        ( Time taken to walk 2x km ) + ( time taken to ride 2x km )= 74 min.
 But, the time taken to walk 2x km = 55 min.
 Time taken to ride 2x km = (74-55)min =19 min.

www.000webhost.com