Thread: How Good Are You At Reasoning?

Okay, here's the answers. They've pretty much all been done now, though there are simpler options:

Spoiler for Answers:

1. Coin n gets flipped for every number that divides n except 1, so coins with an odd number of divisors will be heads, and an even number will give tails.

For each number p that divides n, you get a second number q such that p*q = n, for instance, for n = 12

1*12
2*6
3*4,

so the only way a number can have an odd number of divisors is if for one of the divisors the second number is the same, e.g. for 36

1*36
2*18
3*12
4*9
6*6.

In other words, p = q, i.e. n = p*q = p^2, i.e. n is a square number.


2. A line can intersect another straight line at most once. Each time your cut hits another cut, or hits the edge of the pizza, you add another piece. So you get 1 piece for 0 cuts, 1 + 1 = 2 pieces for 1 cut, 2 + 2 = 4 pieces for 2 cuts, 4 + 3 = 7 pieces for 3 cuts, ... , 22 pieces for 6 cuts.

Technically all we have proved is that 22 is the maximum; it could be that it's impossible to intersect each line with a previous cut. But we can draw a diagram to show that it is possible:



And thus the maximum number of pieces for 6 cuts is 22.

It seems 'obvious' that you can always cut the previous lines no matter how many there are, but can you prove it..?


4. Simple math; total people claiming to be tree-huggers is 80 + 950 = 1030, total people claiming to be tree-huggers who are tree-huggers is 80, thus the fraction is 80/1030.

This was not a hard question, so why did I include it? Because it's actually exactly the same as question 3. The probability that somebody is a tree hugger is 100/10,000 = 1%, the probability that you find out they are when they are is 80/100 = 80%, and the probability that you think falsely that they are when they aren't is 950/9900 = 9.6%.


3. Like all of these questions, you don't actually need to know any maths. You just need to know that the probability of something being A when there are n things which are A and m things in total is n/m, which is obvious.

Draw a rectangle representing 10,000 average people. Split it into two groups, those with cancer and those without, and work out the number of people in each group. Then further split each group according to positive and negative test results, so you have four groups, (those with cancer who test positive, those with cancer who test negative, those without cancer who test positive, those without cancer who test negative). Work out the numbers in each and you can then easily work out the ratio of the number people who have cancer and positive tests to the total number of people who have positive tests.

The thing about this question is that if you are clear headed and define exactly what the question means, you easily see that it is the same question as 3, which is just simple arithmetic. Amazingly only 15% of doctors were able to do the question, but far more are able to do it when you change it into a problem about numbers of people rather than probabilities.


5. We obviously need to check that the card with an even number is red on the other side. We do not need to check the red card; if it was odd on the other side then that wouldn't contradict even cards being red (and for the same reason you don't need to turn over the card with a 1). We do however need to check the blue card, because if it were even on the other side then it would contradict even cards being red.

Again hardly anybody gets this, yet they do get the next question, which is exactly the same but with different words.


6. Obviously you need to check that the 16 year old isn't drinking alcohol and that the person drinking beer isn't under 18.


7. If my claim were false, we could have a group of n people where no two of them had the same number of friends, i.e. n people all with a different number of friends.

There are only n options (0 friends, 1 friend, ... , n - 1 friends), thus we must have one person with 0 friends, another person with 1 friend, ... , one person with n - 1 friends, i.e. friends with everybody.

However, if we have somebody who is friends with everybody, we can't have somebody who is friends with nobody, as they would have to be friends with at least this person.

Originally Posted by tommo

I'm just gonna say 9 people.

Given n people, you need to work out the probability that none of them shares a birthday, i.e. n people have n different birthdays, i.e. person 2 doesn't have person 1's birthday, person 3 doesn't have person 2 or person 1's birthday, etc.

As for the other questions you asked for, 1 is wrong because the assertion that all coins get flipped to tails isn't right (just consider coin 4), 2 is correct and for the right reason, 3 is definitely wrong because it would mean testing positive for cancer doesn't make it more likely that you have cancer, 4 I'm guessing was an arithmetic slip, 5 is wrong because a red card having an odd number on the back doesn't contradict my claim that all cards with even numbers on the back are red, for 6 you need to check the drink of the 16 year old, the reason for 7 is wrong (why don't you have to check it for groups of size 3 or more; my claim is about groups of any size?). For more detail check the spoiler.