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

Solution:

Let us denote the number of boys as 'p' and the number of girls as 'q'.

The first statement says that each boy has equal numbers of brothers and sisters
(p - 1) = q
=> p - q = 1 ------- (a)

The second statement says that every girl has twice as many brothers as that of sisters
2 (q - 1) = p
=> 2q - p = 2 ------- (b)

By solving (a) and (b)
We get q as 3 and p as 4.

Therefore, there are 4 boys and 3 girls in the family.

Solution:

On the back, since bullet train is accelerating, therefore,the person at the back will spend less time than one in front.

Solution:

In the corners of the square, tighten two ropes. Tighten another rope between them by putting rings on each rope. That rope should be tightened on the sheep's neck on the other end.

Solution:

3,3,8

Lets break it down. The product of their ages is 72. So what are the possible choices?

2, 2, 18 sum(2, 2, 18) = 22
2, 4, 9 sum(2, 4, 9) = 15
2, 6, 6 sum(2, 6, 6) = 14
2, 3, 12 sum(2, 3, 12) = 17
3, 4, 6 sum(3, 4, 6) = 13
3, 3, 8 sum(3, 3, 8 ) = 14
1, 8, 9 sum(1,8,9) = 18
1, 3, 24 sum(1, 3, 24) = 28
1, 4, 18 sum(1, 4, 18) = 23
1, 2, 36 sum(1, 2, 36) = 39
1, 6, 12 sum(1, 6, 12) = 19

The sum of their ages is the same as your birth date. That could be anything from 1 to 31 but the fact that Torres was unable to find out the ages, it means there are two or more combinations with the same sum. From the choices above, only two of them are possible now.

2, 6, 6 sum(2, 6, 6) = 14
3, 3, 8 sum(3, 3, 8 ) = 14

Since the eldest kid is taking piano lessons, we can eliminate combination 1 since there are two eldest ones. The answer is 3, 3 and 8.

Solution:

The receptionist was pretty clever she played with the mute button of the phone while talking with her mother. She muted everything except the word emergency, call and help. So while talking, she sounded like 'Emergency… Call… Help' to her mother and thus she called the police.

Solution:

See the colored sticks in the picture. We have moved those sticks in the position given in the second image corresponding with the color.

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:

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.

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