Solution:

40

Cost of each alphabet is 5

Solution:

Steps:
1) Light fire on first bar from both the ends and second bar from one end.

2) After 30 seconds(when first bar gets completely burned out) , burn the second bar from second end as well.

3) When second run completely gets burned , you know its 45 seconds.

Solution:

If you see the picture, they formed a pentagonal shape and achieved the required.

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:

The code word is TIE.

Let us assume that the word is MOD. Now if you are told one character of this word, you would not be able to identify the number of vowels in the word. For example if you are told M, then the word can be MOD or AIM but they both have different number of vowels. If you are told about O, then there are two words MOD and OAT which again have different number of vowels. If you are told about D, there are two words again MOD and DUE which have different numbers of vowels again.

In such manner, all the words that have M, O or D in them can be excluded. Then only word that remains is TIE.
If you are told about T, there are two words OAT and TIE which both have two vowels.
If you are told about I, there are two words AIM and TIE which both have two vowels.
If you are told about E, there are two words DUE and TIE which both have two vowels.

Thus TIE is the code word.

Solution:

   9567
+ 1085
======
 10652

Solution:

Solution can be verified from the picture below.

Solution:

Since the door is not visible in this picture, the bus is moving in the left direction.

Solution:

Old password : altered

New Password : distinct

Solution:

1/3

Explanation
Assuming T1 is the Toy for brother1, T2 is the toy for brother2 and T3 is the toy for brother3.

Following are the possible cases for toys distribution:

Boy1 Boy2 Boy3
T1 T2 T3
T1 T3 T2
T2 T1 T3
T2 T3 T1 .... (A)
T3 T1 T2 .... (B)
T3 T2 T1

In both steps (A) & (B), no one gets the correct toy.
Therefore probability that none brother can get the own toy is 2/6 = 1/3

Solution:

It is best if the person sits on the back of the train. After the stop, the train will be accelerating as it goes into the tunnel. Thus the train will be much master when the back of the train enters the tunnel. In this way, he will have to spend much less time in the tunnel.