Solution:

$6 for red sponges = 1 sponges
$87 for yellow sponges = 29 sponges
$7 for blue sponges = 70 sponges

Explanation:
a = red sponges
b = yellow sponges
c = blue sponges

Now using a simple expression:
A + b + c = 100
6a + 3b + 0.1c = 100

Now if we multiply the first equation by 6 and then 3 and then subtracting the two equations, we will be able to get two more equations:

6a + 6b + 6c = 600
3a + 3b + 3c = 300

3b = 500 - 5.9c
3a = 2.9c - 200

Now if you are wondering how will we solve three variables using just two equations, we have to tell you that a and b are non-negative integers. Thus, if 3b >/= 0, then 500 - 5.9c >/= 0.
Using this, c Also, if 3a >/= 0, then 2.9c - 200 > 0. This implies that c >/= 68.97.

But buying the blue sponges is the only way to spend in fraction thus the number of blue sponges that we buy must cost an even dollar amount. Now if you consider the expressions, there can be only two numbers we can buy between 68.97 and 84.75 that satisfies the given condition i.e. 70 and 80.

Let's say that we replace c by 80 in the last two expressions that we have formed above, a will equal to 10.67 and b will equal to 9.33. However, the values of a and b must be an integer and not fraction. Therefore, c can't be 80.

Let us now replace c with 70. After solving the expressions, we will obtain a as 1 and b as 29.

Thus here are the amounts that must be used to buy:
$6 for red sponges = 1 sponges
$87 for yellow sponges = 29 sponges
$7 for blue sponges = 70 sponges

Solution:

Zouma is 3 and has two footballs.
Ake is 1 and has one football.
Kalas is 4 and has three footballs.
Aina is 2 and has four footballs.

Solution:

If you are thinking to hold one chocolate from each box in hand and then balancing weight in bare hands, you are thinking all wrong.

Let us begin by labelling boxes as 1, 2, 3 and so on till 10.

Now pick one chocolate from box 1, two chocolates from box 2, three from box 3 and so on. In total, you will have 55 chocolates now. (1 + 2 + 3 + ..... + 10)

The ideal weight of the chocolates should be 55 * 20 = 1100. However, somewhere in there are the defected chocolate/s.

You can judge that clearly by noting down the result of 1100 - total weight of chocolates. If the weight is less than 1 gram, the defected box is box 1, if the weight is less than 2 grams, the defected box is box 2 and so on.

Solution:

They are matched as shown in the picture below.

Solution:

The matchsticks that are marked with red are to be removed and the final figure will have just two squares. You can remove either pair from the possible four ways.

Solution:

The first step that they will take will be a leap of logic. What it means is that they will deduce that every color must appear twice at least. Why? Because the master logician has assured them that the puzzle will not be impossible for anyone of them. And if a color appears only once in the circle, the person wearing it will have no clue about that color which will not be fair for him.

Then the logicians will follow the same and look for all the colors of hats in the circle. If one of them sees a color just once, he can safely assume that he is also wearing the hat of the same color as by leap of logic, no color can appear just once. Thus when the bell is rung, he will leave.

In the similar fashion, if anyone sees another color just once, he can determine that he is wearing the hat of the same color and will leave when the bell rings or will be disqualified and sent home. Unvaryingly, if a color is seen twice, they will be eliminated after the first bell. Hence, there must be at least three hats of any of the remaining color.

Assume that you are sitting in the circle and you don’t see a color once but see it twice. Then if they were the only two hats of the same color, the logicians must have left at the first bell already. But they did not. Which means that there are three hats of that color and you are wearing one. Thus you will leave after the second bell.

Solution:

Since the door is not visible in this picture, the bus is moving in the left direction.

Solution:

Adam and Eve were the only people without navel. As they were not born out of a mother, they never had umbilical cords and thus never had navels.

Solution:

step 3 is wrong

Solution:

It is best if the person sits on the back of the train. After the stop, the train will be accelerating as it goes into the tunnel. Thus the train will be much master when the back of the train enters the tunnel. In this way, he will have to spend much less time in the tunnel.

Solution:

Let m be initial number of flowers and n be the flowers offered at each temple by the man.
Initial flowers = m
After stepping into the first temple = 2m
Flowers offered at the first temple = n
Remaining flowers = 2m – n

After stepping into the second temple = 4m – 2n
Flowers offered at the second temple = n
Remaining flowers = 2m – n

After stepping into the third temple = 8m – 6n
Flowers offered at the third temple = n
Remaining flowers = 8m – 7n

Now as per the question (8m – 7n) must be equal to 0.
i.e. 8m – 7 n = 0
or 8m = 7y

Now the least numbers required to solve this equation are 7 and 8 for m and n respectively.

Thus m = 7; n = 8
Which means, the man had 7 flowers in the beginning and he offered 8 flowers at each temple.