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:

If you are thinking 5 times powerful, then you are in a completely wrong direction. The answer will be the square root of ten i.e. 3.16.

Thus if it was half a point higher on the Richter scale, the earthquake will be 3.16 times more powerful.

Solution:

There is only one possible shape i.e. a cone.

Solution:

This quite an easy question and the answer is Rachel.

Since she is in the room, she is also a primate and being human, she is the smartest of all.

Solution:

Let us assume that all of them are standing on the X shore and they have to travel to the Y shore.

X = mother, first daughter, second daughter, father, first son, second son, maid and dog
Y = None

Maid and dog go first and maid returns back
X = mother, first daughter, second daughter, father, first son, second son and maid
Y = dog

Maid and first son go and maid returns with dog
X = mother, first daughter, second daughter, father, second son, maid and dog
Y = first son

Father and second son go and father returns back
X = mother, first daughter, second daughter, father, maid and dog
Y = first son and second son

Mother and father go and mother comes back
X = mother, first daughter, second daughter, maid and dog
Y = father, first son and second son

Maid and Dog go and Father returns back
X = mother, first daughter, second daughter and father
Y = first son, second son, maid and dog

Father and Mother go and mother comes back
X = mother, first daughter and second daughter
Y = father, first son, second son, maid and dog

Mother and first daughter go and maid comes back with dog
X = second daughter, maid and dog
Y = mother, first daughter, father, first son and second son

Maid and second daughter go and maid comes back
X = Maid and dog
Y = mother, first daughter, second daughter, father, first son and second son

Maid and dog go
X= none
Y = mother, first daughter, second daughter, father, first son, second son, maid and dog

Thus everyone has reached the other end.

Solution:

13 8

prime number and then number of letter in prime number

examples
2 (two - 3 letters) => 2,3
3 (three - 5 letters) =>3,5

Solution:

The girl was born on 29 February. Thus her birthday came once in four years only while her sister was born on a normal day and celebrated her birthday every year. Thus the girl had a chance of depositing money only 10 times in 4 years through which she collected $2500 while her sister took $50 from the piggy bank every year making the total amount to be $2000. Thus when the piggy bank was opened, it had just $500.

Solution:

25 Fruits

Remember we told you that Birbal is a witty trader. So his sole motive is to get rid of the sacks as fast as he can.

For the first sack:
He must be able to fill fruits from one sack to other two sacks. Assume that he is able to do that after M check points. Now to find M,

(Space in first sack) M + (Space in second sack) M = (Remaining fruits in Third Sack) 30 – M
M = 10

Thus after 10 checkpoints, Birbal will be left with only 2 sacks containing 30 fruits each.

Now he must get rid of the second sack.
For that, he must fill the fruits from second sack to the first sack. Assume that he manages to do that after N checkpoints.

(Space in First Sack) N = (Remaining fruits in second sack) 30 – N
N = 15

Thus after he has crossed 25 checkpoints, he will be left be one sack with 30 fruits in it. He has to pass five more checkpoints where he will have to give five fruits and he will be left with twenty five fruits once he has crossed all thirty check points.

Solution:

1:1 (approx)

Explanation:
We know that the probability of having a boy or having a girl is same and thus, half of the couples will stop after having a boy child. Half of the others will have a girl and will have a baby again. Out of those half of the couples, half will have a boy and will stop and half will have a girl again. This will keep on going like this.

Now, if there are X number of couples, there will be X boys.

1/2 have a boy and stop: 0 girls
1/4 have a girl, then a boy: X/4 girls
1/8 have 2 girls, then a boy: 2*X/8 girls
1/16 have 3 girls, then a boy: 3*X/16 girls
1/32 have 4 girls, then a boy: 4*X/32 girls
…
Total: X boys and
1X 2X 3X 4X
– + – + – + — +… = ~X

Therefore, the proportion of boys to girls will be extremely close to 1:1

Solution:

It is not as tough as it seems. The strategy is quite simple. You just have to pick up a card out of first three whenever you find a 9 or higher number.

Let us offer you a strong valid reason for the strategy. The probability that the four card kept below are below the number nine is 32*31*30*29 / 52*51*50*49 = 0.133
Thus in such case, you will lose $2 to $9 with equal probability = 0.133/8 = 0.0166

Let us now calculate the probability that one of the four cards is nine or even higher.
1 – 0.133 = 0.867
Now you will surely stop at the first sight of 9 or any value above 9 as per the strategy. Thus you have a chance of winning -1, 0, 1, 2, 3 with an equal probability = 0.867/5 = 0.173

In such case you stand a chance of winning overall amount of 0.14 every game you play.

Let us consider the other strategies. If you want to stop at the sight of 10 or above 10, the expected amount you will win per game will be 0.09 which is way below the present strategy. If you want to stop at above ten only, you will have an expected winning in negative.