Math Problem

Where did the money go?

There were three friends having dinner together, when the waiter brought their bill, which was $30. Each paid $10.
After collecting the money the waiter realized that t he bill should have been $26 insted, dua to a special discount.
Unsure on how to divide $4 equally among three people, he put $1 in his pocket ang gave each person $1. This means that each person paid $9 for their dinner.
It happens that 9 x 3 + 1 = 28. What happened to the other $2?

sorry about the typos... =P *(due, and)

There are no two other dollars.  $9 x 3 = $27.  Minus the $1 the waiter took, and you have the $26.

Another way to think of it is that the $9 each person paid is postive becuase they go towards paying for the bill.  So +27.  But the dollar the waiter took doesn't go towards paying for the bill, so it's negative.  Thus +27 -1 = +26.

It doesn't make sense to think of it in the way the teaser suggests.

The other $2 is in the pocket of one Malthus Poswilly of Towel Outlet, Alabama.

I love this question because people try to answer it. The fact of the matter is, you can't. You can't go backwards in math. The simple answer is, "we don't know" another example is: If I drive 40 mph for the first mile of my trip and 60  mph for the second mile of my trip, how fast must I go on the third mile to average 50 mph? It's impossible to know.

Also, let's try it this way. 2. Where did the 2 come from? was it 4-2 was it f'(x) = 2x ? same thing

The problem arises in the problem when we assume that each person "paid 9 dollars"

Huh, >>5?

Less drugs, more mathbooks, man.

>>5
If you drive 40mph for the first mile, and 60mph, you would need to drive 50mph on the third mile to average 50mph for the three miles.

>>7

Actually, in order to have an average speed of 50 mph, you would have to travel 54 6/11 mph for the third mile.

Think about it: If you travel at 40 mph for the first mile, it will take you 90 seconds to travel that mile. Likewise, it will take you 60 seconds to travel the second mile at 60 mph. 50 mph equals one mile per 72 seconds. In order to have an average speed of 50 mph over the three miles, you must cover 3 miles in 3*72=216 seconds in order to have a total distance/total time equal to 50 mph. You have already used 60+90=150 seconds; therefore, you must travel the remaining mile in 216-150=66 seconds. One mile per 66 seconds times 3600 seconds in an hour equals 54 and 6/11 miles per hour.

>>5
I think you are thinking of problems more like "If it takes a man 40 minutes to dig half a hole, how long does it take him to dig a hole?" This problem truly has no solution, because really, how does one dig half of a hole? Any hole you dig will be a whole hole.

>>8
Not if he's digging through something, such as bedrock or whatever.  Also, if, for instance, he has to dig the hole for constructin, and it needs to be x feet deep, and he digs x/2 feet into the ground, that's "half" of one of the x feet deep holes he needs.  You're just thinking too hard ;)

actually it would be 50 mph.  If you think of an hour as 60 minutes, then 40mph become 4/6 mpm, 60 mph become 1 or 6/6 mpm, and 50 mph becomes 5/6 mpm, so we say(4/6+6/6+x/6)/3 (since it is an average of three speeds)=5/6 (50 mph).  Multiply 5/6 by three and get 15/6=(4/6+6/6+x/6), simplify...15/6=(10/6+x/6), subtract 10/6 from both sides and you get x/6=5/6 otherwise known as 50 mph

i dont get this at all, is >>3 right or wrong?

>>3 is right. To simplify it:

Each customer paid $9, making a total of $27, but the bill was only $26 so the waiter kept the extra one.

The wording is misleading. It happens that 9 x 3 - 1 = 26. The extra $1 was pinched, it wasn't refunded or put towards paying the bill. It certainly wasn't refunded AND put towards paying the bill, which the question was implying ($28).

Think of it this way:
There are 26 dollars with the bill
There is one dollar in the waiter's pocket
There are three dollars with the respective three people
That adds up to 30, no?
The question is asking you where the rest of the money is, when it's really tricking you by including the three people's three dollars to the bill, plus the waiter's one dollar. In other words, it's asking you how much has been paid, but fakes that out by asking you where the rest of the change is.

Going into even simpler terms, it says that there are 28 dollars total, correct? In other words, it's saying that they paid 28 dollars. But the fact is that the three people paid twenty-seven, and that extra one dollar FROM that twenty-seven that goes into the waiter's pocket is NOT added to the twenty-seven. That doesn't make sense, see! The rest should play out easily in your head.

Yes, now you can flame me for bumping a four-day-old topic.

>>2
"Math" was another typo. It's "Maths".

I love math threads.  They remind me every time that 4chan is full of idiots and makes me feel like a fucking genius.  The ones on /b/ are the best.

Cash flow timeline.

Fa0, Fb0, Fc0, W0
Fa-10, Fb-10, Fc-10, W30
Fa-9, Fb-9, Fc-9, W26+1

T= till

F30, W0, T0
F0, W30, T0
F0, W4, T26
F3, W1, T26

So the correct answer is "What $2?".

>>15
We study mathematics, not mathsematics.

x_1 = 10
x_2 = 10
x_3 = 10
waiter = 0
business = 0
total = 30

pay for dinner
x_1 = 0
x_2 = 0
x_3 = 0
waiter = 0
business = 30
total = 30

refund 4, 1 to each friend, 1 for waiter.
x_1 = 1
x_2 = 1
x_3 = 1
waiter = 1
business = 26
total = 30

>>1

it's simple order of operations. you don't do 9 * 3 + 1. you do (9+1) * 3. problem solved.

>>10
>>8 is right.

look at this:
Vt is total average velocity.
Tt is total time.
Dt is total distance.

we know that Vt= Dt/Tt

Dt we know is 3 miles. Tt we don't know, but Vt we do: 50 mph. So Tt=Dt/Vt, or 3/50 of an hour (3.6 minutes). However, we just want the speed for the last mile. In order to find that, let's look at the first two miles. To find the average velocity for the first 2 miles, we use the velocity equation V = d/t. The only thing we know is d, which is 2 miles. We must break it down further and look at just the first mile and second mile separately.

Now, we know the V for the first mile, 40 mph. D is 1 mile. So t = 1/40 of an hour. V for the second mile is 60, so t = 1/60 of an hour. So total time for the first two miles must be 1/40 + 1/60, or 1/24. Now let's go back up and and consider the average velocity for the first two miles. V = 2 miles/(1/24)hours, or 48 mph.

Ok, now for the final mile. The first two miles took 1/24 of an hour. The total trip took 3/50 of an hour. Therefore the final mile must have taken 11/600 of an hour, or .01833333. Taking the average velocity equation one last time, we find that V = 1 mile/(11/600) hours, or 54.5454545454 mph.

[4:25] Rational numbers as a countable infinity
1 Name: MathMajor : 2004-12-27 13:40

       This past semester I took the first halves of Introduction to Analysis and Modern (Abstract) Algebra in which, naturally, we covered the basics of the cardinalities of infinities.  I understand the concepts of what makes something countable verses uncountable.  I simply have an intuitive problem with the fact that the rational numbers are countable.  The other basics countable infinite sets (Integers, Evens, Odds, Natural Numbers, etc.) all sit well with me.  For some reason, I just have a problem accepting that the Rationals are truly countable.  I understand that a definition is not arguable, but it just does not feel right.
       Part of my discontent lies in the fact that like the continueum, there is no smallest positive rational number.  This is not true for the previously mentioned sets.  I feel that the rationals are somehow set apart from this set.  If I recall correctly, it was proven that "It cannot be proven nor disproven that there exists an infinite set whose cardinality is greater than the cardinality of the countable infinite sets and less than the cardinality of the continueum (the reals)."

       If anyone knows where I may be able to find more information on this, or even has an answer to my dilemma, please let me know.  Thanks!

    -Will

The 5 newest replies are shown below.
Read this thread from the beginning
21 Name: Anonymous : 2008-01-04 18:15

    FLOOD

22 Name: Anonymous : 2008-01-05 01:18

    But how are they less than the infinite complex numbers.

23 Name: Anonymous : 2008-01-05 01:59

    old thread is old

24 Name: Anonymous : 2008-01-05 10:56

    They are both countable, just the rationals are not well order with the operator (<)


    Imagine it this way, if we define every even integer n to have "size" 1/n, and then ordered to integers according to this definition we have this list:


    ....10, 8, 6, 4, 2, 1, 3, 5, 7, 9.......

    there is no least, and no greatest number, very much resembling the rationals.


    There are ways of ordering the rationals such that there is a "least member" just that the ordering isn't using the operator (<)


    The fact that they have equal cardinality is obvious consider the injection f from Q to Z

    (Post truncated.)

25 Name: Anonymous : 2008-01-05 12:12

    butt sex

Name: Anonymous : 2008-01-05 17:20
[4:25] Rational numbers as a countable infinity
1 Name: MathMajor : 2004-12-27 13:40

       This past semester I took the first halves of Introduction to Analysis and Modern (Abstract) Algebra in which, naturally, we covered the basics of the cardinalities of infinities.  I understand the concepts of what makes something countable verses uncountable.  I simply have an intuitive problem with the fact that the rational numbers are countable.  The other basics countable infinite sets (Integers, Evens, Odds, Natural Numbers, etc.) all sit well with me.  For some reason, I just have a problem accepting that the Rationals are truly countable.  I understand that a definition is not arguable, but it just does not feel right.
       Part of my discontent lies in the fact that like the continueum, there is no smallest positive rational number.  This is not true for the previously mentioned sets.  I feel that the rationals are somehow set apart from this set.  If I recall correctly, it was proven that "It cannot be proven nor disproven that there exists an infinite set whose cardinality is greater than the cardinality of the countable infinite sets and less than the cardinality of the continueum (the reals)."

       If anyone knows where I may be able to find more information on this, or even has an answer to my dilemma, please let me know.  Thanks!

    -Will

The 5 newest replies are shown below.
Read this thread from the beginning
21 Name: Anonymous : 2008-01-04 18:15

    FLOOD

22 Name: Anonymous : 2008-01-05 01:18

    But how are they less than the infinite complex numbers.

23 Name: Anonymous : 2008-01-05 01:59

    old thread is old

24 Name: Anonymous : 2008-01-05 10:56

    They are both countable, just the rationals are not well order with the operator (<)


    Imagine it this way, if we define every even integer n to have "size" 1/n, and then ordered to integers according to this definition we have this list:


    ....10, 8, 6, 4, 2, 1, 3, 5, 7, 9.......

    there is no least, and no greatest number, very much resembling the rationals.


    There are ways of ordering the rationals such that there is a "least member" just that the ordering isn't using the operator (<)


    The fact that they have equal cardinality is obvious consider the injection f from Q to Z

    (Post truncated.)

25 Name: Anonymous : 2008-01-05 12:12

    butt sex

Name: Anonymous : 2008-01-06 02:01
Name: Anonymous : 2008-01-05 17:20
[4:25] Rational numbers as a countable infinity
1 Name: MathMajor : 2004-12-27 13:40

       This past semester I took the first halves of Introduction to Analysis and Modern (Abstract) Algebra in which, naturally, we covered the basics of the cardinalities of infinities.  I understand the concepts of what makes something countable verses uncountable.  I simply have an intuitive problem with the fact that the rational numbers are countable.  The other basics countable infinite sets (Integers, Evens, Odds, Natural Numbers, etc.) all sit well with me.  For some reason, I just have a problem accepting that the Rationals are truly countable.  I understand that a definition is not arguable, but it just does not feel right.
       Part of my discontent lies in the fact that like the continueum, there is no smallest positive rational number.  This is not true for the previously mentioned sets.  I feel that the rationals are somehow set apart from this set.  If I recall correctly, it was proven that "It cannot be proven nor disproven that there exists an infinite set whose cardinality is greater than the cardinality of the countable infinite sets and less than the cardinality of the continueum (the reals)."

       If anyone knows where I may be able to find more information on this, or even has an answer to my dilemma, please let me know.  Thanks!

    -Will

The 5 newest replies are shown below.
Read this thread from the beginning
21 Name: Anonymous : 2008-01-04 18:15

    FLOOD

22 Name: Anonymous : 2008-01-05 01:18

    But how are they less than the infinite complex numbers.

23 Name: Anonymous : 2008-01-05 01:59

    old thread is old

24 Name: Anonymous : 2008-01-05 10:56

    They are both countable, just the rationals are not well order with the operator (<)


    Imagine it this way, if we define every even integer n to have "size" 1/n, and then ordered to integers according to this definition we have this list:


    ....10, 8, 6, 4, 2, 1, 3, 5, 7, 9.......

    there is no least, and no greatest number, very much resembling the rationals.


    There are ways of ordering the rationals such that there is a "least member" just that the ordering isn't using the operator (<)


    The fact that they have equal cardinality is obvious consider the injection f from Q to Z

    (Post truncated.)

25 Name: Anonymous : 2008-01-05 12:12

    butt sex

Math Problem
25 Name: Anonymous : 2008-01-06 19:33
Name: Anonymous : 2008-01-06 02:01
Name: Anonymous : 2008-01-05 17:20
[4:25] Rational numbers as a countable infinity
1 Name: MathMajor : 2004-12-27 13:40

       This past semester I took the first halves of Introduction to Analysis and Modern (Abstract) Algebra in which, naturally, we covered the basics of the cardinalities of infinities.  I understand the concepts of what makes something countable verses uncountable.  I simply have an intuitive problem with the fact that the rational numbers are countable.  The other basics countable infinite sets (Integers, Evens, Odds, Natural Numbers, etc.) all sit well with me.  For some reason, I just have a problem accepting that the Rationals are truly countable.  I understand that a definition is not arguable, but it just does not feel right.
       Part of my discontent lies in the fact that like the continueum, there is no smallest positive rational number.  This is not true for the previously mentioned sets.  I feel that the rationals are somehow set apart from this set.  If I recall correctly, it was proven that "It cannot be proven nor disproven that there exists an infinite set whose cardinality is greater than the cardinality of the countable infinite sets and less than the cardinality of the continueum (the reals)."

       If anyone knows where I may be able to find more information on this, or even has an answer to my dilemma, please let me know.  Thanks!

    -Will

The 5 newest replies are shown below.
Read this thread from the beginning
21 Name: Anonymous : 2008-01-04 18:15

    FLOOD

22 Name: Anonymous : 2008-01-05 01:18

    But how are they less than the infinite complex numbers.

23 Name: Anonymous : 2008-01-05 01:59

    old thread is old

24 Name: Anonymous : 2008-01-05 10:56

    They are both countable, just the rationals are not well order with the operator (<)


    Imagine it this way, if we define every even integer n to have "size" 1/n, and then ordered to integers according to this definition we have this list:


    ....10, 8, 6, 4, 2, 1, 3, 5, 7, 9.......

    there is no least, and no greatest number, very much resembling the rationals.


    There are ways of ordering the rationals such that there is a "least member" just that the ordering isn't using the operator (<)


    The fact that they have equal cardinality is obvious consider the injection f from Q to Z

    (Post truncated.)

25 Name: Anonymous : 2008-01-05 12:12

    butt sex