Solution:

Let us assume,
the radius of the river is R.
the speed of beast is S.
the speed of duck is S/4.
Circumference = 2 * Pi * R.

Now, if duck swims R/4 distance from the center of the river and then begins to swim across the river in center implies both duck and beast can take the round trip in same time.
Explanation :
time by beast : 2 * Pi* R * S
time by duck : 2 * Pi* R/4 * S/4 => 2 * Pi* R * S


Now, the duck can move slowly inward toward the center of the river, and begin swimming around the center in a circle from this distance. It is now going around a very slightly smaller circle than it was a moment ago, and thus will be able to swim around this circle FASTER than the beast can run around the shore.

The duck can keep swimming around this way, pulling further away each second, until finally it is on the opposite side of its inner circle from where the beast is on the shore. At this point, the duck aims directly toward the closest shore and begins swimming that way. At this point, the duck has to swim [0.75R feet + 1 millimeter] to get to shore. Meanwhile, the beast will have to run R*pi feet (half the circumference of the river) to get to where the duck is headed.
The beast runs four times as fast as the duck, but you can see that it has more than four times as far to run:

[0.75R feet + 1 millimeter] * 4 < R*pi

[This math could actually be incorrect if R were very small, but, in that case, we could just say the duck swam inward even less than a millimeter and make the math work out correctly.]

Because the duck has less than a fourth of the distance to travel as the beast, it will reach the shore before the beast reaches where it is and successfully escape.

Solution:

It will take the man 58 minutes to reach the top.

Since he is climbing 3 feet and falling 2 feet back every passing minute, he is actually climbing 1 feet every minute.
Therefore, he would have climbed 57 feet in 57 minutes. The remaining distance is 3 feet. He will take one more minute to climb it and reach the surface. Since he has reached the top surface, he won't slip back.

Solution:

Yes it is possible in Roman Numerals. In Roman Numerals,
9 = IX
10 = X
11 = XI

Thus by removing one from 9, you are getting 10 and by removing one from 11, you are getting 10 again.

Solution:

Step 5 is invalid, because we are dividing by (x-y), and since x=y, we are thus dividing by 0. This is an invalid mathematical operation (division by 0), and so by not following basic mathematical rules

Solution:

The two numbers are 4 and 13 whose product is 52 and sum is 17.

Solution:

If we have to measure precisely, it is impossible. Because we are asked to measure odd number of liters whereas all the jars we have can contain only even liters of water.

* Tricky question asked in ias/infosys/infoedge/ibibo interview

Solution:

The mules is moving in circular path and hence outside legs travels more distance than inner legs.

Solution:

White should move the rook from
A1 to D1 and hence complete the castling move.

Solution:

4 6 5 3 (from magician perspective)
3 5 6 4 (viewer perspective)

Detail:

From user perspective:
3 5 6 4 => 5 6 4 3 => 6 5 4 3

For magician perspective
3 4 5 6 (ascending order as)

Solution:

It can be done in 8 moves as shown in the animated image below.

Solution:

Think deeply about the door number 56, people will visit it for every divisor it has. So 56 has 1 & 56, 2 & 28, 4 & 14, 7 & 8. So on pass 1, the 1st person will open the door; pass 2, 2nd one will close it; pass 4, open; pass 7, close; pass 8, open; pass 14, close; pass 28, open; pass 56, close.
Thus we can say that the door will just end up back in its original state for each pair of divisor. But what about the cases in which the pair of divisor has analogous number for example door number 16? 16 has the divisors 1 & 16, 2 & 8, 4&4. But 4 is recurrent because 16 is a perfect square, so you will only visit door number 16, on pass 1, 2, 4, 8 and 16… leaving it open at the end. So only perfect square doors will remain open at the end.