Solution:

Since it is a 3x3 grid, the area of the grid = 9
We have to arrange 3 white marbles in 9 places, the possible ways = 9c3 = 84

Now you don't have to arrange the blue marbles and thus the possible ways will be 84 only.

Also if you would have started with the blue marbles, the number of ways would have been the same i.e. 84 (9c6)

Solution:

You must put labels on the tablets as A and B before using. In that case, if you pour tablets together, you will get 3A, 2A 1B, 1A 2B or 3B. If they are from the same bottles you can take one from another bottle and save the remaining two for another day. If you get two from same and one from other, you can draw one from another bottle and you will have two pairs of which you can eat one and save the other.

Solution:

If you see the picture, they formed a pentagonal shape and achieved the required.

Solution:

There can be two possibilities for the given situation.

One is if I run on the perimeter. Then, the lion will eventually catch me as the lion can follow my radius and then his trajectory will be half of the circle and not a spiral. Therefore, the lion will subsequently catch me in a finite amount of time.

Secondly if I don’t run on the perimeter, then clearly, I have an infinite amount of time before the lion catches me.

Solution:

1 2 3 4 5 6 and 7 8 6

Since summation of numbers 1-9(1 2 3 4 5 6 7 8 9) is 45 , so its impossible to divide into two equal parts.
Since we cant flip over 6 to make it 9 but we can flip 9 to make it 6 and hence the solution

Solution:

It can be down as shown below.

Solution:

At each draw, the number of strawberry candies are either decreasing by 2 or not decreasing at all. In the case of orange candies, at each draw, they are either increasing by 1 or decreasing by 1.

Thus on an assumed outset with at least one candy in the jar to begin with, if the number of strawberry candies are 0 or are even in numbers, they will finish off leaving an orange candy at the end. If otherwise, the remaining candy will be a strawberry one.

Solution:

What must be kept in mind before you start calculating through any method is that the weights can be put on both the pans.

So the weights - 1, 3, 7, 12, 43, 76, 102

The oldest man that can be weighed using these weights is a man aged 122 years

Solution:

A) Fill 5 ml gallon ( 5mlGallon - 5, 3mlGallon - 0)
B) Transfer to 3 ml gallon (5mlGallon - 2, 3mlGallon - 3)
C) Empty 3 ml gallon ( 5mlGallon - 2, 3mlGallon - 0)
D) Transfer 2 ml from 5 ml gallon to 3 ml gallon (5mlGallon - 0, 3mlGallon - 2)
E) Fill 5 ml gallon(5mlGallon - 5, 3mlGallon - 2)
F) Transfer 1 ml from 5 ml gallon to 3 ml gallons(5mlGallon - 4, 3mlGallon - 3)

Solution:

1. Fill 5 liters jar ( 5l-jar:5, 3l-jar:0)
2. Transfer to 3 liters pail (5l-jar:2, 3l-jar:3)
3. Empty 3 liters jar ( 5l-jar:2, 3l-jar:0)
4. Transfer 2q from 5 pail to 3 pail (5l-jar:0, 3l-jar:2)
5. Fill 5 liters pail(5l-jar:5, 3l-jar:2)
6. Transfer 1q from 5 pail to 3 pail(5l-jar:4, 3l-jar:3)

Solution:

This problem can be solved with algebra as well as logical reasoning. We are going to tell you how it is logically possible.

Be it any quantity that was present in the beginning and any liquid as well. We know that we have taken a spoonful from one glass and put it into another. Then, we have taken a spoonful from the other glass and put it into the first glass. So, at the end of it, the quantity of liquid in either glass remains same as it was in the beginning.

Therefore, we can conclude the fact that any amount of coke that is missing in the glass with coke will be present in the lemonade glass. Also, the same quantity of lemonade will be missing from the lemonade glass and will be present in the coke glass.