**Light as a feather, nothing is there in it.
A strong man can’t hold it more than a minute.
what is it?**

## Rearrange the letters

**Rearrange the letters
in new door to make one word.**

## River crossing – the harder one

Now the difficulties are : if the dog is left with anyone and the maid isn’t there to control him, he’ll bite. The dad can’t be left with any of the daughters when the mom isn’t there. Likewise, the mom can’t be trusted alone with either of the sons when the dad isn’t there.

Remember! only an adult can operate the boat, and the boat can’t drive itself.See Solution : River crossing - the harder one

Say

West shore is {W} and East shore is {E}

Lets give the nick names to each family member and dog 😉

Mother – {m}, Father – {F}, Daughters – {d1, d2}, Sons – {s1, s2}

House maid – {h}, Dog – {d}

Initially,

W = {m, d1, d2, f, s1, s2, h, d}

E = {…}

let’s move everyone, over…

housemaid and dog go east, and the housemaid comes back:

W = {m, d1, d2, f, s1, s2, h}

E = {d}

housemaid and s1 go east, h and d come back:

W = {m, d1, d2, f, s2, h, d}

E = {s1}

father and s2 go east, father comes back:

W = {m, d1, d2, f, h, d}

E = {s1, s2}

mother and father go east, mother comes back:

W = {m, d1, d2, h, d}

E = {f, s1, s2}

h and d go east, father comes back:

W = {m, d1, d2, f}

E = {s1, s2, h, d}

father and mother go east, mother comes back:

W = {m, d1, d2}

E = {f, s1, s2, h, d}

mother and d1 go east, housemaid and d come back:

W = {d2, h, d}

E = {m, d1, f, s1, s2}

h and d2 go east, h comes back

W = {h, d}

E = {m, d1, d2, f, s1, s2}

h and d go east

W = {}

E = {m, d1, d2, f, s1, s2, h, d}

And we are done.

## Daughter’s ages

**The first grad says to the second**: “how have you been?”

**Second**: “great! i got married and i have three daughters now”

**First**: “really? how old are they?”

**Second**: “well, the product of their ages is 72, and the sum of their ages is the same as the number on that building over there..”

**First**: “right, ok.. oh wait.. hmm, i still don’t know”

**Second**: “oh sorry, the oldest one just started to play the piano”

**First**: “wonderful! my oldest is the same age!”

How old are the daughters ?

See Solution : Daughter's agesWe know that there are 3 daughters whose ages multiply to 72. Let’s look at the possibilities…

Ages:Sum of ages: 1 1 72 74 1 2 36 39 1 3 24 28 1 4 18 23 1 6 12 19 1 8 9 18 2 2 18 22 2 3 12 17 2 4 9 15 2 6 6 14 3 3 8 14 3 4 6 13

## Ants on a triangle

There are three ants on a triangle, one at each corner. At a given moment in time, they all set off for a different corner at random. What is the probability that they don’t collide ?

See Solution : Ants on a triangleSolution 1:

Let the three ants are a, b, c.

So probability that they will not collide is 2/(2+6) i.e. 1/4

Solution 2 :

Total no. of movements: 8

A->B, B->C, C->A; A->B, B->A, C->A; A->B, B->A, C->B; A->B, B->C, C->B; A->C, B->C, C->A; A->C, B->A, C->A; A->C, B->A, C->B; A->C, B->C, C->B

Non-colliding movements: 2

A->B, B->C, C->A; A->C, B->A, C->B

(i.e. the all ants move either in the clockwise or anti-clockwise direction at the same time)

So probability of not colliding = 2/8 = 1/4

## Calendar cube

A man has two cubes on his desk. Every day he arranges both cubes so that the front faces show the current day of the month. what numbers are on the faces of the cubes to allow this?

See Solution : Calendar cubeSo that means we need to have 0 on both the cubes, but that makes it 13, but there are only 12 faces. Thats the trick in this question, we can place cubes upside down too, now which is the number we can use both the ways, yes its 6, it can be used as 9 and then we can have all the possible dates.

so first cube : 0, 1, 2, 3, 4, 5

second cube : 0, 1, 2, 6, 7, 8

## Cube Puzzle

You have a normal six sided cube. I give you six different colors that you can paint each side of the cube with (one color to each side). How many different cubes can you make?

Different means that the cubes can not be rotated so that they look the same. This is important! If you give me two cubes and i can rotate them so that they appear identical in color, they are the same cube.

See Solution : Cube puzzleLet X be the number of “different” cubes (using the same definition as in the problem). Let Y be the number of ways you can “align” a given cube in space such that one face is pointed north, one is east, one is south, one is west, one is up, and one is down. (We’re on the equator.) Then the total number of possibilities is X * Y. Each of these possibilities “looks” different, because if you could take a cube painted one way, and align it a certain way to make it look the same as a differently painted cube aligned a certain way, then those would not really be different cubes. Also note that if you start with an aligned cube and paint it however you want, you will always arrive at one of those X * Y possibilities.

How many ways can you paint a cube that is already “aligned” (as defined above)? You have six options for the north side, five options for the east side, etc. So the total number is 6! (that’s six factorial, or 6 * 5 * 4 * 3 * 2 * 1). Note that each way you do it makes the cube “look” different (in the same way the word is used above). So 6! = X * Y.

How many ways can you align a given cube? Choose one face, and point it north; you have six options here. Now choose one to point east. There are only four sides that can point east, because the side opposite the one you chose to point north is already pointing south. There are no further options for alignment, so the total number of ways you can align the cube is 6 * 4.

Remember, Y is defined as the number of ways you can align the cube, so Y = 6 * 4. This gives us 6! = X * 6 * 4, so X = 5 * 3 * 2 * 1 = 30.

Reference : http://www.techinterview.org/Solutions/fog0000000128.html

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