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:

This statement is true but only in a specific case i.e. if both of the siblings continue to live till the day the younger is as old as the elder one was during the time of their birth.

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:

The brain teaser can be solved as shown in the image below.

Solution:

36 meters.

100/v1 = 80/v2
100/v2 = 80/v3

20+(20/v2)*v3=20+16 = 36

Solution:

Fifteen Minutes.

After frying for five minutes, you can take out one steak and put the third one inside while turning the other one inside. After five more minutes, one steak is fried from both sides and you can take it out. Now put in the steak that we removed and turn the other one inside. After five more minutes, they both will also be fried from both sides.

Solution:

You better switch!

Your first choice has a 1/3 chance of having the car, and that does not change.<br>The other two doors HAD a combined chance of 2/3, but now a Goat has ben revealed behind one, all the 2/3 chance is with the other door.

Solution:

Solution can be verified from the picture below.

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:

1653

Explanation:
For an ideal case, the batsman will hit a six on each ball. But if he hits six on the last ball of the over, the strike will change in the next over. Thus, the best he can do in the last ball is run 3 runs so that he retains the strike even in the next over. Thus the total runs that he can score in each over:
6 * 5 + 3 = 33
But this will have to go like it only till the 49th over. In the last over, he can hit a six in the last ball as well as that will be the last ball of the match.
Thus runs for the last over will be 6 * 6 = 36.
Hence the maximum runs = 33 * 49 + 36 = 1653

Solution:

She is near the trunk of the tree as shown below.