Riddle #992

logicmath

Camel and Banana

The owner of a banana plantation has a camel. He wants to transport his 3000 bananas to the market, which is located after the desert. The distance between his banana plantation and the market is about 1000 kilometer. So he decided to take his camel to carry the bananas. The camel can carry at the maximum of 1000 bananas at a time, and it eats one banana for every kilometer it travels. What is the most bananas you can bring over to your destination?
First of all, the brute-force approach does not work. If the Camel starts by picking up the 1000 bananas and try to reach point B, then he will eat up all the 1000 bananas on the way and there will be no bananas left for him to return to point A. So we have to take an approach that the Camel drops the bananas in between and then returns to point A to pick up bananas again. Since there are 3000 bananas and the Camel can only carry 1000 bananas, he will have to make 3 trips to carry them all to any point in between. When bananas are reduced to 2000 then the Camel can shift them to another point in 2 trips and when the number of bananas left are <= 1000, then he should not return and only move forward. In the first part, P1, to shift the bananas by 1Km, the Camel will have to Move forward with 1000 bananas – Will eat up 1 banana in the way forward Leave 998 banana after 1 km and return with 1 banana – will eat up 1 banana in the way back Pick up the next 1000 bananas and move forward – Will eat up 1 banana in the way forward Leave 998 banana after 1 km and return with 1 banana – will eat up 1 banana in the way back Will carry the last 1000 bananas from point a and move forward – will eat up 1 banana Note: After point 5 the Camel does not need to return to point A again. So to shift 3000 bananas by 1km, the Camel will eat up 5 bananas. After moving to 200 km the Camel would have eaten up 1000 bananas and is now left with 2000 bananas. Now in the Part P2, the Camel needs to do the following to shift the Bananas by 1km. Move forward with 1000 bananas – Will eat up 1 banana in the way forward Leave 998 banana after 1 km and return with 1 banana – will eat up this 1 banana in the way back Pick up the next 1000 bananas and move forward – Will eat up 1 banana in the way forward Note: After point 3 the Camel does not need to return to the starting point of P2. So to shift 2000 bananas by 1km, the Camel will eat up 3 bananas. After moving to 333 km the camel would have eaten up 1000 bananas and is now left with the last 1000 bananas. The Camel will actually be able to cover 333.33 km, I have ignored the decimal part because it will not make a difference in this example. Hence the length of part P2 is 333 Km. Now, for the last part, P3, the Camel only has to move forward. He has already covered 533 (200+333) out of 1000 km in Parts P1 & P2. Now he has to cover only 467 km and he has 1000 bananas. He will eat up 467 bananas on the way forward, and at point B the Camel will be left with only 533 Bananas.
83.88 %
60 votes

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How long was he walking

Every day, Jack arrives at the train station from work at 5 pm. His wife leaves home in her car to meet him there at exactly 5 pm, and drives him home. One day, Jack gets to the station an hour early, and starts walking home, until his wife meets him on the road. They get home 30 minutes earlier than usual. How long was he walking? Distances are unspecified. Speeds are unspecified, but constant. Give a number which represents the answer in minutes.
The best way to think about this problem is to consider it from the perspective of the wife. Her round trip was decreased by 30 minutes, which means each leg of her trip was decreased by 15 minutes. Jack must have been walking for 45 minutes.
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You are somewhere on Earth. You walk due south 1 mile, then due east 1 mile, then due north 1 mile. When you finish this 3-mile walk, you are back exactly where you started. It turns out there are an infinite number of different points on earth where you might be. Can you describe them all? It's important to note that this set of points should contain both an infinite number of different latitudes, and an infinite number of different longitudes (though the same latitudes and longitudes can be repeated multiple times); if it doesn't, you haven't thought of all the points.
One of the points is the North Pole. If you go south one mile, and then east one mile, you're still exactly one mile south of the North Pole, so you'll be back where you started when you go north one mile. To think of the next set of points, imagine the latitude slighty north of the South Pole, where the length of the longitudinal line around the Earth is exactly one mile (put another way, imagine the latitude slightly north of the South Pole where if you were to walk due east one mile, you would end up exactly where you started). Any point exactly one mile north of this latitude is another one of the points you could be at, because you would walk south one mile, then walk east a mile around and end up where you started the eastward walk, and then walk back north one mile to your starting point. So this adds an infinite number of other points we could be at. However, we have not yet met the requirement that our set of points has an infinite number of different latitudes. To meet this requirement and see the rest of the points you might be at, we just generalize the previous set of points. Imagine the latitude slightly north of the South Pole that is 1/2 mile in distance. Also imagine the latitudes in this area that are 1/3 miles in distance, 1/4 miles in distance, 1/5 miles, 1/6 miles, and so on. If you are at any of these latitudes and you walk exactly one mile east, you will end up exactly where you started. Thus, any point that is one mile north of ANY of these latitudes is another one of the points you might have started at, since you'll walk one mile south, then one mile east and end up where you started your eastward walk, and finally, one mile north back to where you started.
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How to measure exactly 4 gallon of water from 3 gallon and 5 gallon jars, given, you have unlimited water supply from a running tap.
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87.77 %
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Equation riddle

If, Fernando + Alonso + McLaren = 6 Fernando x Alonso = 2 Alonso x McLaren = 6 Then, McLaren x Fernando =?
3 Explanation: Fernando + Alonso + McLaren = 6 Fernando x Alonso = 2 Alonso x McLaren = 6 Given. Fernando x Alonso = 2 Alonso x McLaren = 6 Rewriting Both equations in terms of Alonso, Fernando = 2/Alonso McLaren = 6/Alonso Replacing above values in equation "Fernando + Alonso + McLaren = 6" 2/Alonso + Alonso + 6/Alonso =6 (2 + Alonso^2 + 6)/Alonso = 6 8 + Alonso^2 = 6Alonso Alonso^2 - 6Alonso + 8 = 0 (Alonso - 4) (Alonso - 2) = 0 Therefore; Alonso = 4 or 2 Let's take value of Alonso as 2 Fernando = 2/2 = 1 McLaren = 6/2 = 3 Therefore; McLaren x Fernando = 3 x 1 = 3
87.71 %
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Tiling Without Corners

You can easily "tile" an 8x8 chessboard with 32 2x1 tiles, meaning that you can place these 32 tiles on the board and cover every square. But if you take away two opposite corners from the chessboard, it becomes impossible to tile this new 62-square board. Can you explain why tiling this board isn't possible?
Color in the chessboard, alternating with red and blue tiles. Then color all of your tiles half red and half blue. Whenever you place a tile down, you can always make it so that the red part of the tile is on a red square and the blue part of the tile is on the blue square. Since you'll need to place 31 tiles on the board (to cover the 62 squares), you would have to be able to cover 31 red squares and 31 blue squares. But when you took away the two corners, you can see that you are taking away two red spaces, leaving 30 red squares and 32 blue squares. There is no way to cover 30 red squares and 32 blue squares with the 31 tiles, since these tiles can only cover 31 red squares and 31 blue squares, and thus, tiling this board is not possible.
87.71 %
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Half of two plus two

Is half of two plus two equal to two or three?
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2 Player and N Coin – Strategy Puzzle

There are n coins in a line. (Assume n is even). Two players take turns to take a coin from one of the ends of the line until there are no more coins left. The player with the larger amount of money wins. Would you rather go first or second? Does it matter? Assume that you go first, describe an algorithm to compute the maximum amount of money you can win. Note that the strategy to pick maximum of two corners may not work. In the following example, first player looses the game when he/she uses strategy to pick maximum of two corners. Example 18 20 15 30 10 14 First Player picks 18, now row of coins is 20 15 30 10 14 Second player picks 20, now row of coins is 15 30 10 14 First Player picks 15, now row of coins is 30 10 14 Second player picks 30, now row of coins is 10 14 First Player picks 14, now row of coins is 10 Second player picks 10, game over. The total value collected by second player is more (20 + 30 + 10) compared to first player (18 + 15 + 14). So the second player wins.
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You are standing in a pitch-dark room. A friend walks up and hands you a normal deck of 52 cards. He tells you that 13 of the 52 cards are face-up, the rest are face-down. These face-up cards are distributed randomly throughout the deck. Your task is to split up the deck into two piles, using all the cards, such that each pile has the same number of face-up cards. The room is pitch-dark, so you can't see the deck as you do this. How can you accomplish this seemingly impossible task?
Take the first 13 cards off the top of the deck and flip them over. This is the first pile. The second pile is just the remaining 39 cards as they started. This works because if there are N face-up cards in within the first 13 cards, then there will be (13 - N) face up cards in the remaining 39 cards. When you flip those first 13 cards, N of which are face-up, there will now be N cards face-down, and therefore (13 - N) cards face-up, which, as stated, is the same number of face-up cards in the second pile.
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