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# family feud sample questions

From Wikipedia

Brain teaser

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.

One of the earliest known brain teaser enthusiasts was the GreekmathematicianArchimedes. He devised mathematical problems for his contemporaries to solve.

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.

## Example

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...)

## Intuition

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.

Answers:1.Chess 2. With A Different Answer 3. Ray's Dad Frank Barone on Everybody Loves Raymond (real name is peter boyle) 4. 70 5. wrinkles 6. alzheimers 7. hue hefner 8. speak louder 9. old hag 10. Hard of Hearing 11. Their Childhood 12. back pain 13. They're slow 14. steal their purse or wallet. 15. Forks Hope I helped! =]] Sorry, they might not be the best.

Question:

Answers:No not that I know of. It would be great know though how many are around that has. Let us keep asking until we find someone.

Question:Expenditure on one category of food is estimated on the basis of a family expenditure survey. The amounts spent are surpassed to vary around an unknown mean, with a standard deviation of 6 dollars. For a sample of 80 families, the average expenditure is 43 dollars. a) Compute a 95% confidence interval for the mean family expenditure? b) How large of a sample is required to estimated the mean expenditure to within 50 cents? c) Do we need to assume that the individual expenditure follow a normal distribution?

Answers:In this case n = 80, x-bar = 43, s = 6 Standard error of the mean = 6 / 80 = 0.67 The 95% confidence interval = ( - 1.96 x SE) - ( + 1.96 x SE), and 1.96 x 0.67 = 1.32 Hence (43 - 1.32) << (43 + 1.32), i.e. 41.68 << 44.32 b. (This corrects an error in my original answer to this question.) To get an estimate to within 50 cents with a 95% confidence interval it is necessary to apply the z-value as determined above to the standard error of the mean. In other words, we require that 1.96 x 6 / n = 0.5 So n = (1.96 x 6) / 0.5 = 23.52 or n = 553 So with the same sample mean, for example, this would give the confidence interval as 42.50 << 43.50 ------------------------------------------- c. No. The sample means are normally distributed in all cases, irrespective of the shape of the population distribution.

Question:Two solids, A and B, are located in the same family on the periodic table. A sample of each is placed in a beaker of HCl. Substance A produces a few bubbles that rise to the top of the liquid. Substance B bubbles vigorously. Based on this information, compare substances A and B. Be sure to describe the following: what the bubbles indicate where the substances are located in relation to one another on the periodic table and the activity series the name of a family the substances could belong to which substance will have a larger atomic radius which substance will have a larger first ionization energy

Answers:The alkali metals (Na, K, Rb...) react violently with water, producing enough heat to ignite the hydrogen that is produced. The alkaline earth metals, on the other hand, react with water to produce hydrogen gas, which bubbles to the surface, but do so much less violently. Clean Ca, Sr, and Ba will react vigorously with water to produce H2 gas, while clean Mg will barely react. A can be Mg B can be Ca Ca is below Mg on the PT, and Ca is above Mg in the activity series, indicating that Ca will reduce Mg2+, but Mg will not reduce Ca2+. Clearly, Ca is larger than Mg since Ca has an additional energy level. Mg has the greater first ionization energy since the outter electrons are closer to the nucleus.