Maths Riddles

Every girl has seven bags comprising of seven adult cats each with seven little cats per adult cat.
Thus for one girl:
1 girl + 49 adult cats + 343 little cats
Since there are 7 girls:
7 girls + 343 adult cats + 2410 little cats

Now calculating the legs:
14 legs of girls + 10976 legs of cats (There are total 2744 cats with 4 legs each).
Thus total legs = 10990

Now I climbed aboard the bus as well, thus total number of legs = 10990 + 2 = 10992 legs
If we add two legs of the driver as well,
Total legs = 10992 + 2 = 10994

512
A layer will mean to remove from all sides of the cube. Thus it would reduce the dimension be two not one. In this way only a cube of 8x8x8 will remain. This totals to be 512.

3^2 = 4 + 5
(^ is used for squaring the term, it is not a mathematical symbol.)

5832/17496
This fraction will result in 1/3.

If we place a decimal between the numbers it will become 5.9 which is what we needed.

This question can be a bit tricky. But if you pay attention, you can solve it. When fifteen minutes have passed after three, the hour hand must have moved exactly one fourth of an hour. Thus both the hands will make an angle of 7.5 degree between them.

10

In such type of questions we can come to a conclusion by working our way backwards.

He is left with one apple in the end. He gave half of what he had plus one to Diana. If we remove the extra one, the he must have had two apples. Since now both of them have half of the remaining ones, it must be 2 * 2 = 4 apples.

Again move back. Adam gave half of what he had plus one extra to Eve. Thus if we remove that extra one and give it back to Adam, then Adam must have 5 apples. Since he gave half to Eve, it means he must have had 5 * 2 = 10 apples in the beginning.

Thus Adam had 10 apples in the beginning.

This can be solved in a generic and complex way but let us not go into all that.

There can be four ways through which the pair of dice results in a sum of 5. There can be six ways through which the pair of dice can result in a sum of 7.

Now, we want the probability of the pair of dice resulting in a sum of 5 before a sum of 7. Thus probability = 4/(4+6) = 4/10 or 2/5.

60%

Explanation:
This type of questions call in for the application of conditional probability as the question is asking you to find out the probability that the second test was passed with the given situation that the first one was passed.

The formula for conditional probability:
P (Second/First) = P (First and Second)/ P (First)
= 0.25/0.42
=0.60
=60%

37

The number is 37.
The missing guess can be either 24 or 50 and the missing difference is 8.