Solution:

Were you thinking of using the fastest person who takes one minute to travel to and fro till everyone has reached the other end? Yes that can be done but it will take 10 + 1 + 7 + 1 + 2 = 21 minutes in total.

If we pay attention towards finding a way to merge the one taking 7 minutes and the one taking 10 minutes, we will have to acknowledge the fact that one of them will also have to return back which will take much more time than 21 minutes.

What if we use the person taking 2 minutes to escort the one taking 1 minutes across? Let's come to the conclusion:

1 and 2 cross the bridge
2 returns with the torch
7 and 10 cross the bridge
1 comes back
1 and 2 cross the bridge

In such a manner, the total time taken will be
2 + 2 + 10 + 1 + 2 = 17 minutes.

Solution:

To solve this question, we will have to use the concept of relative speed.

Let us assume that the distance between the station A and B is D Miles
The relative speed at which the train are approaching each other is 20 + 15 = 35 mph
When the train collides with each other, the sum of the distance covered will be the distance between the stations i.e. D.
Now Speed = Distance/Time
Which means that the trains will collide in d/35 hours after their start

Another constant things we have here is the speed of the vulture i.e. 25 mph.
Thus by the time the trains collide, the vulture would have covered
25*(D/35) miles = 5D/7 Miles

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:

Let x be the distance of train from bridge, 2l be the length of the bridge and s be the speed of cow

Then according to the question
(x-2)/4s = ( l-5)/s
4l-a = 18 ----- (1)
Also
(a+2l-2)/4s = (l+5-2)/s
a-2l = 14 ----- (2)

On adding (1) and (2), we get
2l = 32

Thus the length of the bridge is 32 meters.

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:

1600m

Solution:

Both A and B

Solution:

6 miles/hour

Solution:

Let us denote the cats with A, B, C and D in a fashion that A is chasing B, B is chasing C, C is chasing D and D is chasing A. The cats are moving in symmetry and thus the only logical answer is that they will meet at the center of the square.

At any point of time, the Cat A is perpendicular to cat B and Cat B is perpendicular to Cat C and so on. Cat A runs for Cat B but Cat B does not run towards or away from Cat A for it is moving in a perpendicular direction to Cat A. Thus the distance that need to be covered by Cat A to grab Cat B will be exactly same as the original distance between them which in this case is one unit (as it is a unit square).

Thus the speed of each cat towards the cat it is chasing will be [1 + cos (x)] where x denotes the angle on each corner.

Thus speed of the cat = 1 + cos (90) = 1 + 0 = 1

The time needed = Distance / Speed = 1/1 = 1 unit