by John P. Pratt
last updated 31 Mar 2007
All Puzzles Menu
This is sort of a trick question, but it usually takes people a surprisingly long time to get it. The trick is the word "plus". In reality, the $2 should be subtracted from (not added to) the $27 to get $25 paid for the room. There is no missing dollar.
You would have 2 dimes. You might also have 40 pennies and 8 nickels, or you might have 45 pennies, a quarter and 2 nickels.
No, the answer is not 90 mph, and it is also not 120 mph. The average time is the total distance divided by the total time. 60 mph is 1 mile per minute. To average one mile per minute around a 4 mile track, it must finish the race in four minutes. It spent one minute on the first leg, another on the second, and 2 minutes on the third. Thus it has used up all four of the minutes needed! Thus, no matter how fast it goes, it cannot average 60 mph all the way around.
This is similar to the last problem, in that you cannot simply average speeds and get 4 mph as the correct answer. The answer is easiest to see if you pick any distance at all for the length of the trail, say 6 miles. Going up takes 3 hours and coming down takes one hour. That is 12 miles in 4 hours, so the average is 3 mph.
If 6 can eat 6 in 6 minutes, then those same 6 can eat 12 in 12 minutes, 24 in 24 minutes or 100 in 100 minutes. You can think of it as all 6 as working on one ant in one minute and then going to the next ant.
As for the chickens and eggs, you can come up with a formula, which I did in my last posted solution, but it gives no insight into how to do the problem or what are the principles at work. One method I use to solve problems is to begin with something I know, which is whole numbers of chickens. Suppose 2 chickens could lay 2 eggs in 2 days. Let's try varying some of those numbers to see what happens. If 2 chickens lay 2 eggs in 2 days, then it takes each chicken two days to lay an egg. So 1 chicken would lay 1 egg in 2 days, and 47 chickens would lay 47 eggs in 2 days. So we learn Rule 1: We can multiply the first two numbers by the same number and keep the third the same. Now let's keep the middle number fixed. How long would it take for 1 chicken to lay 2 eggs? Half as many chickens working would take twice as long to produce the same output. So 1 chicken could lay 2 eggs in 4 days, and 4 chickens could produce 2 eggs in 1 day. So we have Rule 2: If we multiply the number of workers by N and keep the output fixed, then we must divide the time by N. Now suppose we keep the number of chickens fixed. How long would it take 2 chickens to produce 4 eggs? It would take 4 days. And 2 chickens could produce 6 eggs in 6 days. So we have Rule 3: We can keep the first number fixed and multiply the last two by the any number and it will be true. Now let's solve the problem using these rules.
Method 1: Use two of the rules to change the numbers to the desired ones. If 7/4 chickens can lay 7/4 eggs in 7/4 days, how many chickens would it take to lay 12 eggs in 7 days? We can use either Rule 2 or Rule 3 to change the number of days to 7. Using rule 3, we multiply the last two numbers by 4 and get that 7/4 chickens can produce 7 eggs in 7 days. Now use Rule 1 to change the number of eggs while leaving the days unchanged. We multiply the first two numbers by 12/7 (to get a dozen eggs) and get the answer that 3 chickens can lay a dozen eggs in a week. (Exercise for student: Get same result by using Rule 2.)
Method 2: This problem was submitted by my friend Moray King who prefers the method of always beginning by applying Rule 1 first to find the output of one worker. Multiplying the first two numbers by 4/7 means that 1 chicken can produce 1 egg in 7/4 days. Now we can apply Rule 3 and multiply by 4 to get one week: 1 chicken can produce 4 eggs in 7 days. And now Rule 1 again and multiply by three to get the desired number of eggs: 3 chickens can lay 12 eggs in 7 days. That took an extra step, but it is easier to remember and conceptually very easy. Which of the Rules would have solved the anteater problem in one step?
I don't see a way to solve this by inspection because the numbers are too close. One way to solve it is by beginning with my Rule 2 (which keeps production the same) from the last problem. Thus, to see how many of the latter mixture would be needed to produce the same milk in 5 days (5/4 the time), we multiply the cows by 4/5. That means that 2.4 black cows and 4 brown cows produce the same milk in 5 days as do four black and three brown. That means that 1 brown cow is producing as much as 1.6 black cows, so the brown is more.
Now let's solve it with algebra. The rate of production of each cow times the time equals the amount of milk. Let rB be the rate for Black cows and rb be for brown cows. Then (4*rB + 3*rb)* 5 days = (3*rB + 5*rb)* 4 days. Solving this equation for the ratio of rates rB/rb yields 5/8, which is the same answer. So the brown cow produces 8/5 as much!
No, it is not 24. Look at a clock. The hands align at 12:00, then 1:06, 2:12, ... 10:54, but that is all. There is no 11:60 because that is the same ast 12:00. So they align 11 times per revolution of the hour hand, or 22 times per day.
E L E V E N - T H R E E E I G H T
This is my all time favorite problem. To do it, first write down all the real possibilities that the number on that building might have been. Assuming integer ages you get the following which equal 36 when multiplied:
Ages = 1,1,36 (sum = 38)
Ages = 1,2,18 (sum = 21)
Ages = 1,3,12 (sum = 16)
Ages = 1,4,9 (sum = 14)
Ages = 1,6,6 (sum = 13)
Ages = 2,2,9 (sum = 13)
Ages = 2,3,6 (sum = 11)
Ages = 3,3,4 (sum = 10)
The big clue is that the second professor DID NOT KNOW after having been told the sum equalled the number on the house. Why didn't he know? The only reason would be that the number was thirteen, in which case there are two possible answers. For any other number, the answer is unique and the professor would have known after the second clue. So he asked for a third clue. The clue that the oldest had red hair is really just saying that there is an "oldest", meaning that the older two are not twins. Hence, the answer is that the redhead is 9 years old, and the younger two are both two years old.
11 pieces of the nickel candy, 19 of the 2 cent, and 70 of the 10/penny.
The odds favor changing your answer. This problem is famous because Marilyn vos Savant, the "smartest person in the world," answered this question in her column in Parade, in one sentence like I just did. Then several famous math professors challenged her, saying that it made no difference, that the chance became 1/2 for each after the other door was opened. Marilyn turned out to be right and gave a nice, short proof. I'll give a different proof which my mother used.When I asked my non-mathmatical ninety-year-old Mother this puzzle, I was astounded that she got the right answer in seconds, reasoning as follows. She said, "the chance is 1 in 3 that it is behind the door she chose, and zero chance that it is behind the one that was opened, so there must be a 2 in 3 chance that it is behind the other one." That is exactly right. To see why, pretend that there were 100 doors instead of three. Suppose they kept opening all 98 other doors, so that only the original door chosen was left and one other. Now would you change doors? There is only a 1/100 chance that the first door was right, and a huge chance (99/100) that the other is right.
There are equal amounts in each. I'll come back and show why some day.
No the answer is not 42 (going straight up or down and around). The only way I know to do this problem is to drawn pictures of every possible way to cut the room apart and have the spider walk in a straight line between the points. You have to know the Pythagorean formula that sum of the squares of the legs of a right triangle equals the square of the hypotenuse. You'll know when you have the answer because it comes out in round numbers.
You need higher math for this one, but I put it in because of how nifty the answer is. It turns out that the probability of n errors per page is 1/(en!) where e = 2.71828 and n! means n x (n-1) x (n-2) ... 1.
For example, the probabily it zero errors per page is 1/e or 1/2.7, which is also the probabily of 1 error per page. The chance of 2 errors per page is half that and of 3 is 1(3x2) or 1/6 of the chance of 1. If you add up all of the possible probabilities of up to all 600 errors being on one page, the all add up to 1 (or very close to one. Technically, if you all them all the way to infinity, they add to one.)