Solution:

Suppose that we have B buses at present and S refers to the space before we expanded. This implies that
S = B - 12 ----- [1]

When we expanded, we came up with more space. It can be equated as
S + 0.4 x S = B + 12 ------ [2]

Rewriting the equation we get
1.45 = B + 12
Also, B = 1.45 - 12 ----- [3]

The first equation can also be written as
B = S + 12 ----- [4]

If we equate the third and fourth equation, we will have
1.4 S - 12 = S + 12

Let us subtract S from both the sides
0.4 S - 12 = 12

Adding 12 to both sides
0.4 S = 24

Multiplying by ten and dividing by four on both sides
S = 60

If we insert this value in the fourth equation, we will have
B = 72

Thus there are 72 buses.

Initially we had room for 60 buses after expanding the garage space by 40 percent, we will be having enough room to have 84 buses out of which twelve is for future.

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:

There is no reason whatever why the customer's original deposit of Rs.100 should equal the total of the balances left after each withdrawal.
The total of withdrawals in the left-hand colum may equal Rs.100, but is is purely coincidence that the total of the right-hand column is close to Rs.100.
Let us show another example, but starting with Rs.200 in the bank:
Withdrawals Balance left
Rs.50 Rs.150
Rs.25 Rs.125
Rs.10 Rs.115
Rs.8 Rs.107
Rs.5 Rs.102
Rs.2 Rs.100
Rs.100 Rs.699
Moral of this story? Don't Total Balances

Solution:

Either your read from start or from ends, it will be same

Solution:

17 mins

The initial solution most people will think of is to use the fastest person as an usher to guide everyone across. How long would that take? 10 + 1 + 7 + 1 + 2 = 21 mins. Is that it? No. That would make this question too simple even as a warm up question.

Let’s brainstorm a little further. To reduce the amount of time, we should find a way for 10 and 7 to go together. If they cross together, then we need one of them to come back to get the others. That would not be ideal. How do we get around that? Maybe we can have 1 waiting on the other side to bring the torch back. Ahaa, we are getting closer. The fastest way to get 1 across and be back is to use 2 to usher 1 across. So let’s put all this together.

1 and 2 go cross
2 comes back
7 and 10 go across
1 comes back
1 and 2 go across (done)

Total time = 2 + 2 + 10 + 1 + 2 = 17 mins

Solution:

4 6 5 3 (from magician perspective)
3 5 6 4 (viewer perspective)

Detail:

From user perspective:
3 5 6 4 => 5 6 4 3 => 6 5 4 3

For magician perspective
3 4 5 6 (ascending order as)

Solution:

He askes the god to beat him half to death.

Solution:

There are 3 fruits (1 grapes, 1 apples, and 1 oranges).

Solution:

19

Explanation:
The children can purchase 19 chocolates in return.

Out of 77 wrappers, 75 will be used to buy 15 chocolates and two will be left spare.
The 15 chocolates will create 15 empty wrappers that can be exchanged to get three chocolates.
Three chocolates will return three wrappers which will help them buy another chocolate.
Now the wrapper from this chocolate and the two spare that were left earlier will get them another chocolate.
15 + 3 + 1 = 19

Solution:

Old password : altered

New Password : distinct

Solution:

23rd day

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17 + 18 + 19 + 20 + 21 + 22 + 23 = 276

Thus our anniversary falls on the 23rd day of the month.

You can apply other formulas to shorten the process but that is the simplest way to do it.