Solution:

Let the number of people required originally = X
10*X = Work --- eq (1)
(10+10)*(X-10) = Work --- eq (2)
Equating eq (1) and eq (2)
10X=200
=>X=20
Therefore originally 20 people were tasked at the work
Thus option d is the correct answer.

Solution:

Four

As the window was closed it is obvious that the six candles that were not extinguished were lit till they melted away entirely. Thus only four candles that were extinguished remain at the end.

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:

5 leopard and one sheep

since all animal are rational , once 1st leopard eats the sheep all the rest of leopard would know about this.

Solution:

Step 5 is invalid, because we are dividing by (x-y), and since x=y, we are thus dividing by 0. This is an invalid mathematical operation (division by 0), and so by not following basic mathematical rules

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:

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:

The answer is 2. First, divide the coins into 3 equal piles. Place a pile on each side of the scale, leaving the remaining pile of 3 coins off the scale. If the scale does not tip, you know that the 6 coins on the scale are legitimate, and the counterfeit is in the pile in front of you. If the scale does tip, you know the counterfeit is in the pile on the side of the scale that raised up. Either way, put the 6 legitimate coins aside. Having only 3 coins left, put a coin on each side of the scale, leaving the third in front of you. The same process of elimination will find the counterfeit coin.

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.

Solution:

For this answer is 3^0, 3^1, 3^2... That is 1,3,9,27,81,243 and 729.

Solution:

Solution can be verified from the picture below.