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a/g

4 days

8*1024+3*64+3
8192+192+3
8387
1000/0011/1000/0111
So ie 7. 1s here

Rs.265.80

10m

4/52 * 3/51

5 is the answer….
Remaining all are the multiples of 3 expect 5

Circular Track sis of 11 km

Speed of Mens are

4 , 5.5 , 8 km/hr

Time at Which they are at starting Points again

11/4 , 11/5 , 11/8

11/4 , 2.2 , 11/8

We need to find LCM of these

to find at what time they Meet again at starting point

LCM 11 * 2 = 22

After 22 Hrs they will meet at starting point

12 sec

X-Y=-1

Pfull = 4 Hrs; Qfull = 5 Hrs
Assume velocity Vp = x then Vq = 4/5 x –> avg velocity = (5/5 + 4/5) / 2 = 0.9
If fastest time = 4hrs then the time it would take to fill up tank by alternating is 4/0.9 = 4.44 Hrs

– While the train is moving, the jogger will also be running in the same direction.
– for the head(engine) of the train to get to the current position of the jogger 240m away, it will take:
45km/hr => 12.5m/s => 240/12.5 = 19.2 seconds.

– But in the same period of time, the jogger will still be running and will have moved to a new location by: 9km/hr => 2.5m/s => 2.5 * 19.2 = 48m

To get to the new location at the speed of 12.5m/s will take the train:
48/12.5 = 3.84sec

In this additional time, the jogger will move forward by:
3.84 * 2.5 = 9.6m

at a speed of 12.5m/s, it will take the train less than a second to cover the additional 9.6m

If we add the distance the jogger will cover in 1 second to 9.6, it is still less than what the train can cover per second. let us see (9.6 + 2.5 = 12.1)

Therefore, the head of the train will pass the runner at approximately: 19.2 + 3.84 + 1 => 24.04 seconds.

For the train to completely pass the runner, it will need its whole length of 120m to be in front of the runner.

This will take an additional (9.6 + 2) seconds.

Therefore for the length of the train to be ahead of the runner it will take approx. 35.65 (24.04 + 9.6 + 2) seconds