Solution:

If you remember the formulas, you will find out that the first option cannot exist.
Side A^2 + Side B^2 = Side C^2

Thus the only possible option here is the second one and of course, it will have a greater area then

Solution:

The women is a of short stature. She can't reach the upper elevator buttons, but can push is with his umbrella.

Solution:

This puzzle has no answer if you are considering this figure in a 2d domain. However if you follow it on a three-dimensional plane, you can do it like shown in the figure given.

Solution:

C7 => B8
Exchange pawn for a knight

Check-Mate

GameOver

Solution:

Jeff Killed Mike

1. Jack is not the murderer, because he is the brother of the murderer.
2. Dan can't be the murderer since he ran a marathon, and the murderer recently had his leg amputated, and wouldn't be running a marathon of any magnitude that quickly.
3. Ben is not the murderer if he just met Jack, since Jack and the murderer grew up together.
4. This leaves Jeff and Mike.

Since Jeff is still alive (he wants to install a new computer next week, present tense) he must be the murderer. Mike also didn't grow up with Jack. It has been determined that Jack, Dan and Jeff are all alive. Ben must also be alive since Jeff plans to install Ben's computer next week. This means that Jeff killed Mike.

Solution:

Q = 1
U = 9
I = 6
C = 8
K = 3.

Explanation:
Replace values in( Q + U + I + C + K)^3 = QUICK
(1+9+8+6+3)^3 = 19863
27^3 = 19863
19863 = 19863

Solution:

Yes, some people are going get really tiny pieces but you were allowed for that right. Also, there is another way of doing it. Can you think of it?

Solution:

Let us assume that all of them are standing on the X shore and they have to travel to the Y shore.

X = mother, first daughter, second daughter, father, first son, second son, maid and dog
Y = None

Maid and dog go first and maid returns back
X = mother, first daughter, second daughter, father, first son, second son and maid
Y = dog

Maid and first son go and maid returns with dog
X = mother, first daughter, second daughter, father, second son, maid and dog
Y = first son

Father and second son go and father returns back
X = mother, first daughter, second daughter, father, maid and dog
Y = first son and second son

Mother and father go and mother comes back
X = mother, first daughter, second daughter, maid and dog
Y = father, first son and second son

Maid and Dog go and Father returns back
X = mother, first daughter, second daughter and father
Y = first son, second son, maid and dog

Father and Mother go and mother comes back
X = mother, first daughter and second daughter
Y = father, first son, second son, maid and dog

Mother and first daughter go and maid comes back with dog
X = second daughter, maid and dog
Y = mother, first daughter, father, first son and second son

Maid and second daughter go and maid comes back
X = Maid and dog
Y = mother, first daughter, second daughter, father, first son and second son

Maid and dog go
X= none
Y = mother, first daughter, second daughter, father, first son, second son, maid and dog

Thus everyone has reached the other end.

Solution:

There can be two possibilities for the given situation.

One is if I run on the perimeter. Then, the lion will eventually catch me as the lion can follow my radius and then his trajectory will be half of the circle and not a spiral. Therefore, the lion will subsequently catch me in a finite amount of time.

Secondly if I don’t run on the perimeter, then clearly, I have an infinite amount of time before the lion catches me.

Solution:

25 Fruits

Remember we told you that Birbal is a witty trader. So his sole motive is to get rid of the sacks as fast as he can.

For the first sack:
He must be able to fill fruits from one sack to other two sacks. Assume that he is able to do that after M check points. Now to find M,

(Space in first sack) M + (Space in second sack) M = (Remaining fruits in Third Sack) 30 – M
M = 10

Thus after 10 checkpoints, Birbal will be left with only 2 sacks containing 30 fruits each.

Now he must get rid of the second sack.
For that, he must fill the fruits from second sack to the first sack. Assume that he manages to do that after N checkpoints.

(Space in First Sack) N = (Remaining fruits in second sack) 30 – N
N = 15

Thus after he has crossed 25 checkpoints, he will be left be one sack with 30 fruits in it. He has to pass five more checkpoints where he will have to give five fruits and he will be left with twenty five fruits once he has crossed all thirty check points.

Solution:

10,15,9,20,12

Explanation:
The score obtained on the first throw is 10.
The score obtained on the second throw is 15.
The score obtained on the third throw is 9.
The score obtained on the fourth throw is 20.
The score obtained on the fifth throw is 12.