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A brain teaser is a form of puzzle that requires thought to solve. It often requires thinking in unconventional ways with given constraints in mind; sometimes it also involves lateral thinking. Logic puzzles and riddles are specific types of brain teasers.
A brain teaser may also refer to an object. A brain teaser might be formed from wood, rope, metal, plastic, foam, or rubber. Brain teaser objects utilize many different problem solving skills. There are several categories of puzzles which are not limited to mathematical, kinesthetic, logical reasoning, and visual puzzles.
Metal, iron, wire, tavern, and rope puzzles are often referred to as disentanglement brain teasers. There are several types of wood brain teasers, such as interlocking, jigsaw, dyecut, and sequential.
- Q: If threehens lay three eggs in three days, how many eggs does a (statistical) hen lay in one day?
- A1: One third. (Note: 3 hens = 3 eggs / 3 days â†’ 3 hens = (3 / 3) (eggs / days) â†’ 1 hen = (1 / 3) (egg / days))
- A2: Zero or one (it's hard to lay a third of an egg).
One can argue about the answers of many brain teasers; in the given example with hens, one might claim that all the eggs in the question were laid in the first day, so the answer would be three.
- Q: Mary's father has five daughters: 1. Nana, 2. Nene, 3. Nini, 4. Nono. What is the name of the fifth daughter?
- A: Mary. The first four daughters all have names with the first 4 vowels, so if someone does not think about the question, they may say the name with the fifth vowel, Nunu. The answer was given at the beginning of the question (ie.'Mary's father has five...)
The difficulty of many brain teasers relies on a certain degree of fallacy in human intuitiveness. This is most common in brain teasers relating to conditional probability, because the casual human mind tends to consider absolute probability instead. As a result, controversial discussions emerge from such problems, the most famous probably being the Monty Hall problem. Another (simpler) example of such a brain teaser is given here:
- If we encounter someone with exactly two children, given that at least one of them is a boy, what is the probability that both of her children are boys?
(For simplicity, assume that boys and girls are born with equal probability.) The common intuitive way of thinking is that the births of the two children are independent of each other, and so the answer must be the absolute probability of one child being a boy, 1/2. However, the correct answer is 1/3 as shown by the following argument:
- For a single birth, there are two possibilities (a boy or a girl) with equal probability.
- Therefore, for two births, there are four possibilities: 1) two boys, 2) two girls, 3) first a boy, then a girl, and 4) first a girl, then a boy; all of them have equal probability.
- We are given that one of the children is a boy. Thus, only one of the four possibilitiesâ€”two daughtersâ€”is eliminated. Three possibilities with equal probabilities (1/3) remain.
- Out of those three, only oneâ€”two sonsâ€”is what we are looking for. Hence, the answer is 1/3.
Alternatively, one can see that in any sample of families with two children, 3/4 of them will have at least one son, and 1/4 will have two sons. The probability is thus (1/4)/(3/4) = 1/3. The common intuitive way of thinking is equivalent to considering families in which a particular child (e.g. the first-born, or the one that comes first in the alphabet, etc.) is a son (which is only 1/2 of the sample, not 3/4) and seeing how many of them have two sons.
One might formulate the above as
- If someone has two children, and one of them is a son, what is the probability that the other is also a son?
but that would be (more) ambiguous, since it could mean that we chose a person at random, and learnt that at least one of their two children was a son (in which case we get 1/3), or it could mean that we chose a person at random, and met one of their children, which turned out to be a son. This would then be a particular child, so the probability of the other being a son is 1/2.
The difference lies in the specific choice of words: The first example is considering the probability of a family having two sons in a row, if at least one of them is a son already (as shown in the proof). The second example might be understood to only ask for the sex of the second child, which is, given an even distribution of children born to each gender, one half or 1/2 either way.
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Answers:I added some letters to the other spaces to help explain the solution. .............. 7 C ..... ------------ A B ) d e f g ........... h i .......... ------- ........... j k l ........... m 2 n .......... ------- ................. 0 First, you are subtracting a two digit number (hi) from a three digit number (def). That means that you have to borrow. Thus d must be 1. Also, the last subtraction results in zero, so k must be 2 and jkl = m2n. .............. 7 C ..... ------------ A B ) 1 e f g ........... h i .......... ------- ........... j 2 l ........... j 2 l .......... ------- ................. 0 Next you are multiplying 7 times a two digit number (AB) and getting a two digit number... the choices are: 7 x 10 = 70, 7 x 11 = 77, 7 x 12 = 84, 7 x 13 = 91, 7 x 14 = 98. Thus A must be 1. .............. 7 C ..... ------------ 1 B ) 1 e f g ........... h i .......... ------- ........... j 2 l ........... j 2 l .......... ------- ................. 0 For the second multiplication, C must be bigger than 7, because you get a *three* digit result. Our choices for AB are 10, 11, 12, 13 or 14. When you try 8 times these numbers you get: 8 x 10 = 80 8 x 11 = 88 8 x 12 = 96 8 x 13 = 104 8 x 14 = 112 None of these have a 2 in the middle, so C must be 9. .............. 7 9 ..... ------------ 1 B ) 1 e f g ........... h i .......... ------- ........... j 2 l ........... j 2 l .......... ------- ................. 0 Trying 9 times these numbers you get: 9 x 10 = 90 9 x 11 = 99 9 x 12 = 108 9 x 13 = 117 9 x 14 = 126 The only solution that works is 9 x 14 since it has a 2 in the middle. So B = 9. .............. 7 9 ..... ------------ 1 4 ) 1 e f g ........... h i .......... ------- ........... 1 2 6 ........... 1 2 6 .......... ------- ................. 0 Now you can figure out the rest easily. Here's the complete equation: .............. 7 9 ..... ------------ 1 4 ) 1 1 0 6 ........... 9 8 .......... ------- ........... 1 2 6 ........... 1 2 6 .......... ------- ................. 0 So 1106 / 14 = 79 A = 1, B = 4, C = 9
Answers:1. nothing: there is no oxygen to combust 2. Mt. Everest; it just wasn't discovered yet 3. holding a letter d in your left hand 4. your mother
Answers:1.What is as big as you are and yet does not weigh anything? 2.Two cannibals were chatting as they had their dinner. One complained that he really quite disliked his new mother-in-law. What was the advice given to him by his companion? 3. A cowboy rode into town on Friday, stayed three days, and rode out again on Friday. How did he do that? 4.Complete this sequence of letters: o, t, t, f, f, s, s, _, _, _.
Answers:Hi Jordan 1. Find the product: 3/a * b/4 = 3b/4a 2. If the area of the base of a pyramid is 2.5cm^2 and its height is 3cm, what is its volume? Volume of any solid tapering to a vertex = Area of base*Height/3 = 2.5 * 3/3 = 2.5 cm^3 3.If the edge lengths of the square were doubled, how would the perimeter be affected? The illustration shows a square whose sides are 2 cm long. perimter of square of edge 'a' = 4a = 4*2 = 8 If the edge length doubles the new edge A = 2a = 2*2 = 4 So new perimeter = 4A = 4*4 = 16 Th perimeter also doubles 4. Describe the smallest rectangular piece of contact paper that would cover a 2ft. by 1ft. by 0.5ft. box. This is the surface area of the box = 2(lw + wh + lh) wher l = length, w = width and h = height of the box The Area of my smallest rectangular piece of cnatct papaer = 2(2*1 + 1*0.5 + 2*0.5) = 2*3.5 = 7 ft^2 5. The measures of the sides of a triangle are 12cm, 11cm, and 15cm, is it a right triangle? 12^2 + 11^2 = 144 + 121 = 265 15^2 = 225 The sum of the squares of the two smaller sides > than square of longer side so i cannot be a right angled triangle. It will be an acute angled triangle (Note: if they are equal(=), then right angled triangle. If smaller (<), then obtuse angled triangle) 6. Two dice are rolled. Find the probability that a prime number is rolled on one and and a multiple of 5 on the other. Prime nos = 2,3 and 5 = 3 ways Multiples of 5 = 5 = 1 way So no. of ways of getting a prime number is rolled on one and and a multiple of 5 on the other = 2* (3) (1) = 6 Total no. of events = 6*6 = 36 So P = 6/36 = 1/6 7. If the edge of a cube is 4in, then what is the surface area? Surface Area of cube of edge 'a' = 6a^2 = 6*4^2 = 6*16 = 96 cm^2 8. Measure the length of your desk to the nearest centimeter. [Sorry I don't have a desk right now and am not sure what to do. Maybe you could help..?] My desk is 91.5 cm 9. A bag contains 2 red marbles, 3 white marbles, and 5 green marbles. What is the probability of selecting a white, then a green (without replacement)? P(white) = 3/10 then P(green) = 5/9 P(white AND then green) = 3/10 * 5/9 = 15/90 = 1/6 Shy