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:

10

Explanation:
Equation 1 + 9 + 11 = 1 can be derived from
One (o) + nine (n) + eleven (e) = one => 1

Similarly for equation,
12 + 11 + 9
Twelve (t) + eleven (e) + nine (n) => ten (10)

Solution:

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

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:

He askes the god to beat him half to death.

Solution:

111111111

Explanation:
It's a maths magical square root series as :
Square root of number 121 is 11
Square root of number 12321 is 111
Square root of number 1234321 is 1111
Square root of number 123454321 is 11111
Square root of number 12345654321 is 111111
Square root of number 1234567654321 is 1111111
Square root of number 123456787654321 is 11111111
Square root of number 12345678987654321 is 111111111 (answer)

Solution:

charles put the ring into the box, secure it with one of your locks, and send the box to the girl firend.
She should then attach one of his own locks and return it. When you receive it again, remove your lock and send it back with her lock. Now she can unlock his own lock and retrieve the object.

Solution:

Yes, some people are going get really tiny pieces but you were allowed for that right. Also, there is another way of doing it. Can you think of it?

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:

146900

Explanation:
Whenever you face such type of questions, it is wise to begin from the last thing. Here in this question the last thing will be the 9th temple. He climbed down 100 steps and thus you know, he had Rs. 100 before beginning climbing down. Thus, he must have offered Rs. 100 to the god in that temple too (he offered half of the total amount). Also, he must have dropped Rs. 100 while climbing the steps of the ninth temple. This means that he had Rs. 300 before he begand climbing the steps of the ninth temple.

Now, we will calculate in the similar manner for each of the temples backwards.
Before the devotee climbed the eight temple: (300+100)*2 + 100 = 900
Before the devotee climbed the seventh temple: (900+100)*2 + 100 = 2100
Before the devotee climbed the Sixth temple: (2100+100)*2 + 100 = 4300
Before the devotee climbed the fifth temple: (4300+100)*2 + 100 = 8900
Before the devotee climbed the fourth temple: (8900+100)*2 + 100 = 18100
Before the devotee climbed the third temple: (18100+100)*2 + 100 = 36,500
Before the devotee climbed the second temple: (36500+100)*2 + 100 = 73300
Before the devotee climbed the first temple: (73300+100)*2 + 100 = 146900

Therefore, the devotee had Rs. 146900 with him initially.

Solution:

Blue could only be wearing white or red and we know that there is already someone else wearing the white shirt so Mr. Blue could only be wearing the red shirt.

Mr. White could have only been wearing a blue or a red shirt, and red is already taken, so Mr. White is wearing a blue shirt.

Mr. Red now has to be wearing a white shirt.