Solution:

John starts after 3 days, thus Peter has already covered 60 km by then (travelling at 20km a day)
The distance between Peter and John will be increased to 60 + 5 + 1 = 66 km in the next two days.

Since John is following an arithmetic progression in the speed daily, the distance between him and peter will keep on decreasing from the 6th day. It will decrease by 3 km on the first day, 7 km on the next day and so on.

Now John will catch up with Peter after n + 5 days from the start of Peter
Thus
66 = (n/2)*(2*3 + (n-1)*4)

If we solve for n in the above equation, we will get n = 5.5

Thus John will catch up with Peter on the eights day when he starts running after Peter.

Solution:

The ant climbs 6 meters in one minute and then falls down 3 meters in the second minute, the distance that it covers in 2 minutes is 3 meters.

Thus for climbing 54 meters, the ant will require time = 52*2/3 = 36 minutes.

We have counted only for 54 meters as it will climb the remaining six meters in one minute.

Thus the total time required to climb the 60 meters wall = 36 + 1 = 37 minutes.

Solution:

First time = 20 ft
Second time = 10ft down + 10 ft up
Third time = 5ft up + 5ft down
It will go on like this the object stops

Total distance traveled = 20 + (10 + 10) + (5 + 5) + (2.5 + 2.5) + . . .
= 20 + (10 + 5 + 2.5 + ...) + (10 + 5 + 2.5 + ...)
Assuming the object comes to rest at infinity,

Total distance traveled = 20 + [10/ (1 - 0.5)] + [10/ (1 - 0.5)]
= 20 + 20 + 20 = 60

Thus option d is correct.

Solution:

1595

13
44 = 13 + 31
88 = 44 + 44
176 = 88 + 88
847 = 176 + 671
1595 = 847 + 748

Solution:

The last marble will be yellow.

Since marbles can only be taken out in pairs and you started off with an odd number of yellows there is always going to be one yellow left over that you'll keep putting back in the box until it's left on it's own.