How much dirt would be in a hole 6 feet deep and 6 feet wide that has been dug with a square edged shovel?
None.
No matter how big a hole is, it's still a hole: the absence of dirt.
And those of you who said 36 cubic feet are wrong for another reason, too.
You would have needed the length measurement too.
So you don't even know how much air is in the hole.
A farmer lived in a small village. He had three sons. One day he gave $100 dollars to his sons and told them to go to market. The three sons should buy 100 animals for $100 dollars. In the market there were chickens, hens and goats. Cost of a goat is $10, cost of a hen is $5 and cost of a chicken is $0.50.
There should be at least one animal from each group. The farmer’s sons should spend all the money on buying animals. There should be 100 animals, not a single animal more or less! What do the sons buy?
They purchased 100 animals for 100 dollars.
$10 spent to purchase 1 goat.
$45 spent to purchase 9 hens.
$45 spent to purchase 90 chickens.
You have 25 horses. When they race, each horse runs at a different, constant pace. A horse will always run at the same pace no matter how many times it races.
You want to figure out which are your 3 fastest horses. You are allowed to race at most 5 horses against each other at a time. You don't have a stopwatch so all you can learn from each race is which order the horses finish in.
What is the least number of races you can conduct to figure out which 3 horses are fastest?
You need to conduct 7 races.
First, separate the horses into 5 groups of 5 horses each, and race the horses in each of these groups. Let's call these groups A, B, C, D and E, and within each group let's label them in the order they finished. So for example, in group A, A1 finished 1st, A2 finished 2nd, A3 finished 3rd, and so on.
We can rule out the bottom two finishers in each race (A4 and A5, B4 and B5, C4 and C5, D4 and D5, and E4 and E5), since we know of at least 3 horses that are faster than them (specifically, the horses that beat them in their respective races).
This table shows our remaining horses:
A1 B1 C1 D1 E1
A2 B2 C2 D2 E2
A3 B3 C3 D3 E3
For our 6th race, let's race the top finishers in each group: A1, B1, C1, D1 and E1. Let's assume that the order of finishers is: A1, B1, C1, D1, E1 (so A1 finished first, E1 finished last).
We now know that horse D1 cannot be in the top 3, because it is slower than C1, B1 and A1 (it lost to them in the 6th race). Thus, D2 and D3 can also not be in the to 3 (since they are slower than D1).
Similarly, E1, E2 and E3 cannot be in the top 3 because they are all slower than D1 (which we already know isn't in the top 3).
Let's look at our updated table, having removed these horses that can't be in the top 3:
A1 B1 C1
A2 B2 C2
A3 B3 C3
We can actually rule out a few more horses. C2 and C3 cannot be in the top 3 because they are both slower than C1 (and thus are also slower than B1 and A1). And B3 also can't be in the top 3 because it is slower than B2 and B1 (and thus is also slower than A1). So let's further update our table:
A1 B1 C1
A2 B2
A3
We actually already know that A1 is our fastest horse (since it directly or indirectly beat all the remaining horses). So now we just need to find the other two fastest horses out of A2, A3, B1, B2 and C1. So for our 7th race, we simply race these 5 horses, and the top two finishers, plus A1, are our 3 fastest horses.
An infinite number of mathematicians are standing behind a bar. The first asks the barman for half a pint of beer, the second for a quarter pint, the third an eighth, and so on. How many pints of beer will the barman need to fulfill all mathematicians' wishes?
