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:

720

(0!+0!+0!+0!+0!+0!)!
=(1+1+1+1+1+1)!
=(5)!
=6! is 6*5*4*3*2*1=720

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:

129.

These are the first 10 prime numbers (2, 3, 5...) prefixed with a 1

Solution:

SQRT(.2 ^ -2)

Take .2 and raise it to the power of -2 and then take the square root.

Solution:

1) 8/(3-(8/3))
= 8/(1/3)
= 24

2)((8 x 3!)/3)+8
= ((8 × 3 × 2 × 1)/3)+8
= (48/3)+8
= (16)+8
= 24

3)(3!/)*8

4)(8-3)!/(8-3)
( × )!
( + )!
?(8×8×3×3)
8+(8×(3!/3))
((?(8+8) × (3/3))!
?(8+8) × (3+3)
(log base(3!/3) of 8) × 8
((log base(3!/3) of (8+8))!

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:

1/6227020800

Explanation
This can be solved through a simple mathematical equation.

N cards have n! permutations.
Therefore, the probability of obtaining a specific one by chance is 1/n!

Now, in this particular case, n = 13
Therefore, the required probability is 1/6227020800

Solution:

Let us assume,
the radius of the river is R.
the speed of beast is S.
the speed of duck is S/4.
Circumference = 2 * Pi * R.

Now, if duck swims R/4 distance from the center of the river and then begins to swim across the river in center implies both duck and beast can take the round trip in same time.
Explanation :
time by beast : 2 * Pi* R * S
time by duck : 2 * Pi* R/4 * S/4 => 2 * Pi* R * S


Now, the duck can move slowly inward toward the center of the river, and begin swimming around the center in a circle from this distance. It is now going around a very slightly smaller circle than it was a moment ago, and thus will be able to swim around this circle FASTER than the beast can run around the shore.

The duck can keep swimming around this way, pulling further away each second, until finally it is on the opposite side of its inner circle from where the beast is on the shore. At this point, the duck aims directly toward the closest shore and begins swimming that way. At this point, the duck has to swim [0.75R feet + 1 millimeter] to get to shore. Meanwhile, the beast will have to run R*pi feet (half the circumference of the river) to get to where the duck is headed.
The beast runs four times as fast as the duck, but you can see that it has more than four times as far to run:

[0.75R feet + 1 millimeter] * 4 < R*pi

[This math could actually be incorrect if R were very small, but, in that case, we could just say the duck swam inward even less than a millimeter and make the math work out correctly.]

Because the duck has less than a fourth of the distance to travel as the beast, it will reach the shore before the beast reaches where it is and successfully escape.

Solution:

If all other conditions are fair, then the toss will generate a 52:48 distribution in the favor of heads. The toss boas factor will then be 52/48.

Using the same approach, if the coin will generate a 49:51 distribution in the favor of tails, the coin bias factor for heads will be 49/51.

Using the results, we can say that the combined bias factor is (52 * 49) / (51 * 48) = 2548 / 2448 which can be canceled if the floor bias factor is 2448 / 2548. The floor will generate a distribution factor of 2448 / (2558 + 2448) : 2548 + 2448) in favor of tails. If you calculate roughly, it will be 51.00080064% tails.