Solution:

2418

Explanation:
The rule that is being followed here is

P ~ Q ~ R ~ S = [first digit of (P * S)] [last digit of (Q * R)] [first digit of (Q * R)] [last digit of (P * S)]

Using the same rule in the incomplete equation,
7 ~ 2 ~ 7 ~ 4 = [first digit of (7 * 4)] [last digit of (2 * 7)] [first digit of (2 * 7)] [last digit of (7 * 4)]
= 2418

Thus 2418 is the missing number.

Solution:

The number is 281.

Solution:

18

Explanation:
Number of days with snow in morning and clear afternoon = A
Number of days with clear morning and snow in afternoon = B
Number of days with no snowfall in morning or afternoon = C

Now, number of days with snowfall = A + B = 13
Number of days with clear mornings = B + C = 11
Number of days with clear afternoons = A + C = 12

Adding these three equations, we get

2 * (A + B + C) = 36
Or, A + B + C = 18

Therefore, they spent eighteen days in total at their vacation.

Solution:

11550

Explanation:
There is just one way to find it out and that is the generic computing method.

The number of squares and rectangles formed = (Summation of column numbers)
= (1 + 2 + 3 +......+ 19 + 20) * (1 + 2 + 3 +......+ 9 + 10)
= 210 * 55
= 11550

Solution:

Assuming that John has A number of apples and Jacobs has B number of apples.

As per the question,
A + 1 = B - 1
=> B - A = 2 ------ (p)

Also, A - 1 = (B + 1) / 2
=> 2A - B = 3 ------ (q)

Solving (p) and (q)
A = 5 and B = 7

Therefore, A has 5 chocolates and B has 7 chocolates right now.

Solution:

The three numbers are:
-3, -1 and 1.

Solution:

119

Explanation:
Let the minimum number of chocolates = C
When divided by 2, remainder = 1
When divided by 3, remainder = 2
When divided by 4, remainder = 3
When divided by 5, remainder = 4
When divided by 6, remainder = 5
When divided by 7, remainder = 0

Therefore, C is divisible by 7.

Whenever C is divided by any number less than 7, the remainder is 1 less than the divisor.
This implies that, C + 1 is the LCM of 2, 3, 4, 5 and 6.

Now LCM of 2, 3, 4, 5 and 6 = 60
But 60 - 1 = 59 is not divisible by 7

60 * 2 = 120
120 - 1 = 119 which is divisible by 7

Therefore, C + 1 = 120 or C = 119

Number of minimum number of chocolates in the box = 119

Solution:

Girl

Explanation:
In the House P:
There can be three possibilities.
Older Younger
Boy Girl
Girl Boy
Boy Boy
Since a boy is mentioned, we can't have a Girl-Girl option. Now all these options are equally possible and thus we will take them all into account. Two of the outcome have girls as an option and thus the probability here will be 2/3.

In the House Q:
We know that the elder child is a boy and thus there can be only two possibilities:
Older Younger
Boy Girl
Boy Boy

Here, the probability of having a girl child will be 1/2.

Thus you must choose the house P as we have better probability of having a girl in that house.

Solution:

18

Explanation:
Let us suppose that there were (3 * G) garlic bread sticks originally.

On dividing equally, the number of bread sticks with each of the friend = G

Each one of them ate four. The remaining sticks with each one of them = (G - 4)

Now, the total number of breadsticks left = 3 * (G - 4)
This number is equal to the breadsticks each one of them had after division.

=> 3 * (G - 4) = G
(3 * G) - 12 = G
2 * G = 12
G = 6.

Therefore, the number of breadsticks originally = 3 * G = 3 * 6 = 18.

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.