Time and Distance Questions

Let the speed of Kaushik be Sk and speed of Piyush be Sp and let the time taken by Piyush be t second then,

 SkSp = 800t+20600t = 23since given ratio

t = 20 s

Speed of Piyush = Sp = 600/20 = 30 m/s

and    Speed of Kaushik = 800/40 = 20 m/s

Given that,

1st train speed = 62 km/hr 

2nd train speed  = 44 km/hr

Hence the relative speed of two trains is : 62-44 = 18 km/hr = 18×518 m/s = 5 m/s.

There fore, the first train crosses the second train i.e (85+75)mts of distance with a speed of 5m/s in, 

time = distancespeed = 85+755 = 32 sec.

Total running distance in four weeks = (24 x 240) + (4 x 400)

= 5760 + 1600

= 7360 meters

= 7360/1000

=> 7.36 kms

Average speed=total distance/total time
Let distance to store be  K
then, total time =(K/20)+(K/30)=K/12
and, total time =(2K)
so average speed= 2K / (K/12) = 24kmph.

Let 'd' be the distance between A and B
K -time = d/10 + d/9 = 19d/90 hours
L -time = 2d/12 = d/6 hours
We know that, 10 min = 1/6 hours
Thus, time difference between K and L is given as 10 minutes.
=> 19d/90 - d/6 = 1/6
=> (19d-15d)/90 = 1/6
=> 4d/90 = 1/6
Thus,
d= 15/4 km = 3.75 km.

hence the distance between A and B is 3.75 km.

To find the minimum distance, we have to get the LCM of 75, 80, 85

Now, LCM of 75, 80, 85 = 5 x 15 x 16 x 17 = 20400

Hence, the minimum distance each should walk so that thay can cover the distance in complete steps = 20400 cms = 20400/100 = 204 mts.

Speed of tractor = 360/12 = 30 kmph

Speed of jeep = 250 x 30/100 = 75 kmph

But, given ratio of speed of car, jeep and tractor is 3 : 5 : 2

Speed of car = 3 x 75/5 = 45 kmph

Required, average speed of car and jeep = 75 + 45/2 = 60 kmph.

Let the actual distance be t kms

Now, According to the data,

$$\begin{gathered} 5t + 860 = 6t \\ => 560t + 860 = 6t \\ => 60t + 8 = 72t \\ => 8 = 12t \\ => t = 8/12 = 2/3 hrs \end{gathered}$$

Now, the total distance = 6t = 6 x 2/3 = 2 x 2 = 4 kms.