Solution:

We can calculate out how many days there will be for the 13th on each month if we count from the beginning of the year (January 1). Then we divide total days by 7 to get the remainders. We also need to consider the leap year. Through the whole year we had all kinds of remainders, from 0 to 6. The minimum of occurence for all the unique remainders was 1. It means that we have at least one Friday on 13th. In a regular year, the best chance you can get 3 Fridays on 13th, which are in February, March and December because the remainders of these 3 months are 2. In a leap year, the best chance you also can get 3 Fridays on 13th, which are in January, April and July because the remainders of these 3 months are 6.

Solution:

Wise you are, wise you be, I see you are too wise for Me

Solution:

I am 72 years old.

Let m be my age in years. If s is my son's age in years, then my son is 52s weeks old. If g is my grandson's age in years, then my grandson is 365g days old. Thus,
365g = 52s.
Since my grandson is 12g months old,
12g = m.
Since my grandson, my son and I together are 120 years,
g + s + m = 120.
The above system of 3 equations in 3 unknowns (g, s and m) can be solved as follows.
m / 12 + 365 m / (52 x 12) + m = 120 or
52 m + 365 m + 624 m = 624 x 120 or
m = 624 x 120 / 1041 = 72.
So, I am 72 years old.

Solution:

From left to right
#1 Mr. Reed Richard – Blue Marbles
#2 Ms. Sue Storm – Red Marbles
#3 Mr. Johnny Storm – White Marbles
#4 Mr. Ben Grimm – Green Marbles.

Let us label the houses from left to right as #1, #2, #3, #4. Now let us split the hints in different category.
a) Mr. Reed Richard’s house is to the left of the green marbles one
b) The third one is made of white marbles
c) Ms. Sue Storm owns a red marbles house
d) Mr. Johnny Storm does not live on either ends
e) Mr. Johnny Storm lives right of the blue marbles house
f) Mr. Ben Grimm lives in the fourth house
g) The first house is not made up of red marbles

By seeking all the (g) and (b), we are sure that the #1 is not made up of red marbles and nor is the #3. Now with (f) we know that Mr. Ben Grimm lives in #4 made up of green marbles and keeping in mind (c) #2 must be made of Red Marbles and Ms. Sue Storm lives there. Now moving to the hint (d) Mr. Johnny Storm must be living in #3 and adding up with hint (b), it must be made from white marbles. Now by hint (a) #4 must be made up of green marbles (or Mr. Reed Richard’s house can’t be on its left). Also by (f) Mr. Ben Grimm lives there. All of it leaves Mr. Reed Richards living in #1 with blue marbles construction.

Solution:

Arc + tick + circle.
= Artic Circle

Solution:

987,654

Let us assume with a smaller number and then increase the number by extrapolating. If there are two citizens, the number of hairs on their heads will be zero and one. Therefore for ‘n’ number of citizens, the hair distribution will range from 0 to (n-1).

If you take up a case of a number greater than 987,654 let’s say 987,655; and consider the fact that none of the citizens have 987,654 hairs and they must be having different number of hairs, one must be having more than 987,654 hairs which is bluntly violating the fact of the town that there are more citizens than the hairs on any individual.

Solution:

Let the speed of red ant be R.
Speed of Black ant be B.
Time at which they cross their paths be T.

Now we will apply the basic formula of distance
Distance = Speed * Time

The Red ant travels R * T distance before meeting and R * 20 after meeting.
The Black ant travels B * T distance before meeting and B * 5 after meeting.

Now as per the question,
The distance traveled by red ant before they both meet will be equal to the distance covered by black ant after they meet. We can say the same for the vice versa case as well.

Thus
R * T = B * 5 and B * T = R * 20

Using both the quations
T = 10

Thus the red ant will require 10 + 20 = 30 seconds to travel the distance.
And the black ant will take 10 + 5 = 15 seconds to travel the distance.

Solution:

We know that there may be many possibilities and if you find one it is great. For reference we have given one solution below.

(1 + (3 + 4*5)*6)*2

Solution:

Suppose if there is just one unfaithful husband, then his wife will kill her husband as she must be knowing that all other men are faithful but the queen declared that one of them all is unfaithful.

If there are two unfaithful husbands, then both their wives will believe there is just one unfaithful husband, that too the other one whom she knows. Thus both of them will expect the above case to happen at midnight. But when no one is shot, they will realize that there are two unfaithful husbands and since they know about everyone, they will know that the other unfaithful man is their own husband on the next day.

If there are three unfaithful husbands, each of their wives will be knowing about two other unfaithful men so they will be expecting the above case and will be waiting for the gunshots on the second day. When that does not happens, they will realize that there are more than two and since they all will know about the rest, they will realize that their own husband is unfaithful and they will kill him on the third day.

Thus if we talk about general terms and suppose that there are n unfaithful husbands, their wives will believe that there is n-1 unfaithful husbands and will expect the gunshot on n-1 day. When they don’t hear that, they will realize their own husband is the nth.

Solution:

You can put one precious stone in any one of the bag and all the rest in the other bag. If you do it, you will get just a shade under ¾ probability of picking up a precious stone.