Solution:

444 + 444 + 44 + 44 + 4 + 4 + 4 + 4 + 4 + 4 = 1000

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.

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

Solution:

He simply suggested that one of the brothers will be allowed to divide the land but he must take into consideration that it will be the other one who will have a choice of selecting his land between the two halves first.

Thus both will be happy in such manner.

Solution:

The figure tells that B is located to the North-East of E.

Solution:

5 leopard and one sheep

since all animal are rational , once 1st leopard eats the sheep all the rest of leopard would know about this.

Solution:

28 =>1+2+4+7+14

cannot be 1 (as number cannot be included)

Solution:

111111111

Explanation:
It's a maths magical square root series as :
Square root of number 121 is 11
Square root of number 12321 is 111
Square root of number 1234321 is 1111
Square root of number 123454321 is 11111
Square root of number 12345654321 is 111111
Square root of number 1234567654321 is 1111111
Square root of number 123456787654321 is 11111111
Square root of number 12345678987654321 is 111111111 (answer)

Solution:

One Minute building

building 62 => Sixty second building => One minute building

Solution:

Set the first switches on for abt 10min, and then switch on the second switch and then enter the room.
Three cases are possible
1.Bulb is on => second switch is the ans
2.Bulb is off and on touching bulb , you will find bulb to be warm
=>1st switch is the ans.
3.Bulb is off and on touching second bulb , you will find bulb to be normal(not warm)
=>3rd bulb is the ans.

Solution:

At each draw, the number of strawberry candies are either decreasing by 2 or not decreasing at all. In the case of orange candies, at each draw, they are either increasing by 1 or decreasing by 1.

Thus on an assumed outset with at least one candy in the jar to begin with, if the number of strawberry candies are 0 or are even in numbers, they will finish off leaving an orange candy at the end. If otherwise, the remaining candy will be a strawberry one.