Solution:

library

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:

The old lady of course.
After helping the old lady into the car, I give the car keys to my best friend.
And then i go ib the bus with child love <3

Solution:

Birbal drew another line that was larger than the pre-drawn line. Now the pre drawn line was smaller.

Solution:

A Question

Solution:

If you are thinking to hold one chocolate from each box in hand and then balancing weight in bare hands, you are thinking all wrong.

Let us begin by labelling boxes as 1, 2, 3 and so on till 10.

Now pick one chocolate from box 1, two chocolates from box 2, three from box 3 and so on. In total, you will have 55 chocolates now. (1 + 2 + 3 + ..... + 10)

The ideal weight of the chocolates should be 55 * 20 = 1100. However, somewhere in there are the defected chocolate/s.

You can judge that clearly by noting down the result of 1100 - total weight of chocolates. If the weight is less than 1 gram, the defected box is box 1, if the weight is less than 2 grams, the defected box is box 2 and so on.

Solution:

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

Solution:

'five'

Its the count of alphabet

Solution:

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

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:

Costa

Explanation:
c + 1 characters-> d
o + 2 characters-> q
s + 3 characters-> v
t + 4 characters-> x
a + 5 characters-> f

=> dqvxf = costa