Sadaf gave Tanu as many rupees as Tanu started out with.
Tanu then gave Sadaf back as much as Sadaf had left.
Sadaf then gave Tanu as back as many rupees as Tanu had left, which left Sadaf broke and gave Tanu a total of 80 pounds. How much did Sadaf and Tanu have at the beginning of their exchange?
== Mathematical Answer ==
Mathematical Answer = number of seconds in a minute * number of minute in an hour * number hour in a day * number of day in year
60 * 60 * 24 * 365 = 31536000
60 * 60 * 24 * 365.25 = 31557600 (if you want account for leap year)
== We are looking for trick answer (which is 24) ==
Every month contains only 2 second like
A) 2nd January ("second" January)
B) 22nd January (twenty"second" January)
There stand nine temples in a row in a holy place. All the nine temples have 100 steps climb. A fellow devotee comes to visit the temples. He drops a Re. 1 coin while climbing each of the 100 steps up. Then he offers half of the money he has in his pocket to the god. After that, he again drops Re. 1 coin while climbing down each of the 100 steps of the temple.
If he repeats the same process at each temple, he is left with no money after climbing down the ninth temple. Can you find out the total money he had with him initially?
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.
Birbal is a witty trader who trade of a mystical fruit grown far in north. He travels from one place to another with three sacks which can hold 30 fruits each. None of the sack can hold more than 30 fruits. On his way, he has to pass through thirty check points and at each check point, he has to give one fruit for each sack to the authorities.
How many mystical fruits remain after he goes through all the thirty check points?
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.
In front of you, there are 9 coins. They all look absolutely identical, but one of the coins is fake. However, you know that the fake coin is lighter than the rest, and in front of you is a balance scale. What is the least number of weightings you can use to find the counterfeit coin?
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.
You have to place all the digits from 1 to 9 without repeating any number in a tic tac toe board. The condition is that the numbers should add up to 15 whether you add the numbers in each row, column or diagonally.