15 Tricky Math Questions
1) There are 8 Apples on the table, you take 3. How many do you have?
2) 10 Birds in a field. 2 were shot, how many were left?
3) Take away the first letter, take away the last letter, then take away all the other letters. What do you have left?
4) If you have 4 melons in one hand, and 7 apples in the other - What do you have?
5) A box has nine ears of corn in it. A Squirrel carries out three ears a day, and yet it takes him nine days to carry out all the corn. Explain?
6) Why do white sheep eat more than black sheep?
7) A man wanted to plant 4 trees, but all 4 had to be equal distances from each other. How did he do it?
8) I have 2 coins in my hand that add up to 60c. One of the coins isn't a 50c piece. What are the coins?
9) A fisherman was asked how long was the fish he had caught. He said "it is 30cms plus half its length" How long was the fish?
10) A Hammer and a Nail cost $31. If the Hammer cost $30 more than the Nail, what is the cost of each?
11) It takes 7 men 2 hours to build a wall. How long does it take 3 men to build the same wall?
12) "I will bet you $1" said Fred, "that if you give me $2, I will give you $3 in return."
"Done," replied Tom. Was he clever?
13) "How much will one cost?"
"25 cents"
"How much will fifteen cost?"
"50 cents"
"OK then, I'll take one hundred and sixteen"
"Thank you, that will be 75 cents please"
Explain.
14) What comes next in the following sequence ?
1, 4, 5, 6, 7, 9, 11,...
15) In a scientific context, what could the following phrase mean?
How I want a drink, alcoholic of course, after the heavy chapters involving quantum mechanics
Answers:
1) If you take 3 then you have 3.
2) 2 - the others flew away.
3) The Mailman.
4) Big hands.
5) He has 2 of his own ears, so he carries out only 1 ear of corn per day.
6) There are more white sheep than black sheep.
7) He planted 3 trees at the corners of an equilateral triangle. He built a mound in the middle and planted the 4th on the top of the mound so that it was the same distance from the other 3 trees. (on the points of a tetrahedron.)
8) 50c and a 10c (The coin that isn't a 50c piece is a 10c. The other coin is the 50c).
9) 60 cms.
10) Hammer $30.50, Nail $0.50
11) No need to bother, the 7 men have already built it.
12) Tom accepts the bet, gives Fred $2. Fred does not give Tom $3 so loses the bet and has to pay Tom $1. Result Fred gains $1.
13) He is buying house numbers, each separate digit costs 25c, so 116 is three digits so 3 x 25c = 75c.
14) 100 (The next number that doesn't contain a "T" in the spelling).
15) The number of letters in each word refers to pi to 14 decimal places, i.e. 3.14159265358979
The Old Woman and the Airplane:
There is an airplane with 100 seats, and there are 100 passengers each of whom has an assigned seat on the plane. The passengers line up in a random order to get on the plane. The first woman on line is a confused old lady who doesn't know how to find her proper seat. She just sits in a random seat. When each person after her gets on the plane, they look to see if their assigned seat is available. If it is, they sit in it. If it is not (i.e., if the old lady or someone else before them has sat in it), they sit in a random seat. What are the odds that the 100th person sits in his/her proper assigned seat?
There are two very different ways to do this. The brute force long equation way, and the cute, sneaky, but simple way. Hopefully we'll get both on here.
Here are the proofs that the correct answer is 50%:
1. (cute shortcut). Model this process as a decision tree with 100 steps, with step i representing the decision of the ith person to sit down. At step i, there are two ways the process can terminate -- either the ith person sits in seat 100, guaranteeing person 100 is out of luck, or the ith person sits in the lady's seat, guaranteeing that people (i+1) thru 100 all get the correct seats. If the ith person makes any other choice, we have to continue down the tree to another decision.
All we care about are the stopping points where we learn person 100's fate. Note that at each step, no matter what has happened before, there is always an equal chance of stopping with 100 in the right seat and of stopping with 100 in the wrong seat. Thus, there is a 50%-50% chance -- the # of ways to get person 100 in the right seat always equals the # of ways to guarantee person 100 gets the wrong seat.
2. We can also write out the equation. Let n=100 be the number of passengers. We know the 100th passenger will get to sit in the correct seat if and only if his/her seat is left open through the 99 previous rounds. Thus, the odds the old lady takes his seat are 1/n=1/100. The odds the second person takes seat 100 are 1/(n-1)*(n-2)/n -- if the old lady chose seat 100 (1/n) or her own seat (1/n) then the second person will take his own seat, and if he has to pick a new seat, he has 1/(n-1) chance of picking seat 100.
Continuing this analysis, we get the total chance of passenger 100 getting the right seat as:
1/n + 1/(n-1)*(n-2)/n + 1/(n-2)*(n-3)/(n-1)*(n-2)/n
+ 1/(n-3) * (n-4)/(n-2)*(n-3)/(n-1)*(n-2)/n) ...
= 1/n + (n-2)/n(n-1) + (n-3)/n(n-1) + (n-4)/n(n-1) + ... 1/n(n-1) [by cancelling like terms]
= [(n-1) + (n-2) + (n-3) + ... + 1]/(n(n-1))
= [1/2 * n(n-1)] / n(n-1) [by taking the sum of the arithmetic series in brackets]
= 1/2.