Help me figure out some math puzzles!

Andrew Lynch

Andrew Lynch

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Feb 6, 2000
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A (beautiful) friend of mine asked me to help her figure out these puzzles she needs for some strange college math class. Help make me look smart, LOL :D

Two days ago I was twenty. Later next year I'll be twenty three. How can that be?

I woman buys a horse for $500, sells it for $600, buys it for $700 and sells it for $800. How much did she gain or lose on these transactions?

There are ten identical barrels holding balls of some type. Nine of the barrels have 1 oz balls; one of the barrels has 2oz balls. How can you figure out which barrel has 2oz balls with just one weigh-in?


Thanks guys!! Oh, and please hurry, she needs them quick.
 
That's odd -- the second two are easy but I can't get the first one at all....

2) She paid $500 plus $700 for the horse = $1200
sold it for $600 plus $800 = $1400
profit = $200

3) The barrel with the 2 oz balls is lighter than the others, because larger balls don't pack as well.

I'll mull at the first one some more ...
 
first one:

Today is January 1st, 2005
Two days ago would be Dec 30, 2004
His birthday is Dec 31

So, he was 20 on the 30th.
21 on the 31st, 2004
22 on 12/31/2005
23 on 12/31/2006 - technically speaking...."next year"
 
Another way of looking at that second question: it doesn't matter that it was the same horse she bought and sold twice; it could just as well have been two different horses. So she bought and sold a horse for $100 profit, and then she bought and sold another horse for $100 profit = $200 :cool:
 
Yes, that is what I thought on the second one, but it seemed too easy.

As far as the second one is concerned, you still won't know which one is the light one without weighing them all. The riddle is how to find out which one is the lightest (or heaviest, depending on how you argue it) with just one weigh-in.

Thanks for the help guys!!
 
the problem with the barrel/balls riddle is that we're missing information.

Can you look inside the barrel?
can you swap any of the balls?
Are they all the same size, but different weights?
Or, are the 1oz exactly half as big as the 2oz...


And, what does "weigh in" mean....?

A similar riddle to this involves putting two on a balance scale (vs. just a single scale)

Another involves removing one ball from each barrel....



...see what I mean?
 
The barrels one is complicated. You first have to label all the barrels 0-9. Then take zero balls from barrel 0, 1 ball from barrel 1, 2 from barrell 2...etc. Weigh this subset of the balls. Then use this table to find the heavy barrel:

Total weight...heavy barrel
45 ounces..........0
46 ounces..........1
47 ounces..........2
48 ounces..........3
49 ounces..........4
50 ounces..........5
51 ounces..........6
52 ounces..........7
53 ounces..........8
54 ounces..........9

This works because there are 45 balls total. If they were all one ounce, the total would be 45 ounces. So if all the balls weigh 45 ounces, barrel zero has the heavy ones since its not represented. If its one ounce over 45, there is one heavy ball, so its barrel one. If its two ounces over, there are two heavy balls so the culprit is barrel 2.
 
The barrel one is easy. 1 weigh in.

Put all barrels on the scale. Take 1 by one off and see which barrel takes off more/less than the others. And that's the odd one out. Kind of cheating though.

The answer that murnax gave is overcomplicated and doesn't make sense. If you're taking out balls from the barrel wouldn't you know which ball from which barrel is heavier while you're taking it out?

Practically speaking though i'd just kick down all the barrels and just choose the one which feels different.

I think it needs more information. As Daniel already mentioned.
 
The barrel one is a slightly mangled version of an older riddle which makes more sense. Suppose you have ten bags of gold coins, except one of the bags is counterfeit. The real coins are an ounce each, the counterfeit are .9 ounce each. Take one from the first, two from the second, and so on. Weigh them all at once. The number of tenths of ounce that you're short tells you which bag is bad.
 
These are old and good math riddles, I think the one about the barrels is specially interesting because people tend to think that you have to use a balance for weighing, comparing balls from different barrels and it seems obvious that you would need to weigh more than once. It is sometimes worded as: How many weighings do you need to discover which barrel contains the heavier balls ?.

One of my favorites (I hope I can put it clear in English):

Two trains are traveling toward each other on the same railroad so that they will collide, they both travel at 50 miles per hour. A fly is standing on the tip of one of them, the speed of the fly when flying is a constant 100 miles per hour (assume constant speed to simplify problem), when the distance between the trains is exactly 100 miles the fly departs from its position and flies straight toward the other train, as soon as it touches the tip of the train it turns around instantly and flies back straight to the first train, it keeps going back and forth between trains until they collide and it is smashed between them.

What distance did the fly travel ?

Luis
 
Yes 100 miles, this one is an example of how taking the right approach can make things easy, many people assume that they have to calculate the distance traveled in every turn and add them together, it seems like a complicated calculus problem, but if you just consider total time of flight it´s simple.

Luis
 
She paid $500 plus $700 for the horse = $1200
sold it for $600 plus $800 = $1400
profit = $200
Click to expand...

But if she starts by buying a horse for $500 then she is straight away -$500
Sells it for $600 dollars +$100
Buys another for $700 -$100
Sells it for $800 +$100 again.
So in total she makes $100 over all the transactions.?
Am I missing something here?
 
Musashi

Don't split that way, add them all together. Actually when buying the second horse your figure is off.

TLM
 
Point44 said:
The barrel one is easy. 1 weigh in.

Put all barrels on the scale. Take 1 by one off and see which barrel takes off more/less than the others. And that's the odd one out. Kind of cheating though.

The answer that murnax gave is overcomplicated and doesn't make sense. If you're taking out balls from the barrel wouldn't you know which ball from which barrel is heavier while you're taking it out?

Practically speaking though i'd just kick down all the barrels and just choose the one which feels different.

I think it needs more information. As Daniel already mentioned.
Click to expand...

His answer is correct. By taking no ball from the first one, one from the second one etc.. you have the following equation:

x1 + 2*x2 + ... + 9*x9 = N

there is only one j where xj=2, all the other xi = 1, i=0..9; i<>j

for each j from 0 to 9, you get:

j
0 N = 1+2+3+4+5+6+7+8+9 = 10 * (10-1) / 2 = 45
1 N = 2+2+3+4+5+6+7+8+9 = 46
2 N = 1+2*2+3+4+5+6+7+8+9 = 47
.....
9 N = 1+2+2+3+4+5+6+7+8+9*2 = 54

Just by measuring the total weight of the balls you took out and by doing a bit of math you have the answer.
 
Musashi: The gimmick (every math puzzle has a gimmick) is you're supposed to be fooled into thinking she lost $100 the second time she bought the horse. To clear up the confusion, suppose they're two different horses and suppose she bought both horses at the same time. Then it's easy: She buys two horses for $500 and $700, sells them for $600 and $800. :cool:
 
musashi said:
But if she starts by buying a horse for $500 then she is straight away -$500
Sells it for $600 dollars +$100
Buys another for $700 -$100
Sells it for $800 +$100 again.
So in total she makes $100 over all the transactions.?
Am I missing something here?
Click to expand...

Cougar's right, I did the same thing when I first looked at the puzzle. There's a little trouble with your math. After she sells the first horse, she has +$100.00. Buying the second horse, or buying back the same horse, puts her at -$600.00 (100-700). The successive increase of $100.00 at each step is distracting and can be confusing.

Jeremy
 
Cougar Allen said:
Musashi: The gimmick (every math puzzle has a gimmick) is you're supposed to be fooled into thinking she lost $100 the second time she bought the horse. To clear up the confusion, suppose they're two different horses and suppose she bought both horses at the same time. Then it's easy: She buys two horses for $500 and $700, sells them for $600 and $800. :cool:
Click to expand...

Yeah but she can write off the capital gain on one horse because it's for her own tranportation. If she has two horses than she's a broker and has to pay.......Oh forget it! :D
 
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