Math error ???

[hellah10]

So me and a few buddies are eating at IHOP and these 2 pretty sexy girls come up and start talking to us asking us where we going from here - they were sluts beyond sluts but it had us thinking, with todays folks being like this, do they really care about school or even math? I told my friend, Im a math genious -- which is a lie by the way....so he says "genious huhhhh, well figure this out then"

Im sure all of you have heard this before, but Im super stumped and he claims there is an equation for it but my other friends said its a loophole in math

3 guys get a hotel room - the hotel room cost $30 so each guy pays $10. The bell boy takes their shit up to the room and heads back to the lobby. His manager says that they were overcharged by $5 and tells the bell boy to give them the $5 back. So he goes up and he keeps $2 for himself considering they didnt tip him and nobody will ever know...so that leaves $3 left. He gives $1 to each guy....which meanssss...each guy now paid $9 for the room. So among the 3 guys thats a total of $27 + $2 that the bell boy kept = $29....where did the other dollar go :shrug:

I gave up - my head hurts

 

[Scott4USC]

Hotel Owner keeps $25

Bell Boy keeps $2

3 guys keep $1 each = $3


$25 + $2 + $3 = $30

 

[KMA]

The manager returned $5, so the cost of the room is $25. However, the busboy kept $2, so the room now costs $25 + $2 = $27.
Therefore, each guy pays $27 / 3 = $9.
$25 + $2 = $9 * 3
The $1 is not missing. It was never there.

 

[hellah10]

Hotel Owner keeps $25

Bell Boy keeps $2

3 guys keep $1 each = $3


$25 + $2 + $3 = $30

yea but scott, that means each guy paid $9 for the room...which adds up to $27

 

[caught up]

scott, what he was saying is--if you lookatit like each guy was paying $9 and add it together--thats only $9 times the three of them--but bellboy only kept 2--so again, wheres the other dollar go?


since it was five in question, one of the guys actually picks up the tab in paying more.

it breaks down as each is firstly slated to pay 10 each for a total of 30=easy enough.


for the bellboy to actually pay each guy the same amount back, he woulda had to kept 3 dollars--then making it 27 exactly which as we know is 3 multiples of nine.

not following
lets try this: lets say the bellboy keeps the entire $5, so now the guys are paying a total of $25 for the room--that unfortunately doesnt divide nicely by three. lets do the math anyways--and say each paid $8.33. 3 times 8.33=24.99 and if the bell boy has only five, then ONE guy actually paid 8.34 with the other two making out a WHOLE cent paying only 8.33. lets add the three dollars now that the bellboy returns. 8.33+8.33+8.34+3=28. the guys are actually paying 8.33, 8.33, and 8.34 for the room.
-now, if the bellboy kept
$3, $6, or $5.01 then each would be an equal amount--because, he then would be subtracting a multiple of 3 (the three guys) 3 into 3 one. three into 6 twice. three into 5.01, 1.67 times


i know it didnt--but hopefully that helps



...ok, that confuses even me, but KMA hasit right.

as far as the guys are concerned they think they paid $9 each for the room--as each got a buck back. (10-1=9) out of pocket expenses is actually $27>>>that goes to the hotel (dont add the two here)

as far as the hotel is concerned, they paid $25 (each guy tabbed at $8.3X)

..and as far as the bellboy is concerned--neither the hotel or the guests paid enough--so he keeps the two --which is only a discrepency between the 25 that the hotel actually keeps and the 27 that the guys think they pay. like KMA sez, the dollar isnt there to be lost--but i know for some reason my mind is still twizted around it.

kinda like that riddle where father and son get in an accident and only the son survives, to be rushed to the hospital to have live saving surgery, only to have the doctor say cant operate as the is the doctor's son. (<while that goes to speak more on society's views on sexism in the workplace) it hints at how the brain jumps and bundles numbers and ideas together quickly--even sometimes when they doesnt make sense. ok, i'll stop now

 

[hellah10]

 

[freelancc]

:lol:

 

[Sportsaholic]

take the $ left over and get a 3 dollar hooker!!!!!!

 

[KotysDad]

Just look at is as an accountant would. Everytime a transaction takes place, someone loses X dollars and someone gains X dollars. The transaction sums to 0 (-X +X = 0)

Customers give hotel $30. So..

customer A -10
customer B -10
customer C -10
Hotel +30

Sum = 0

Now the hotel gives the bell boy $5 to return.

customer A -10
customer B -10
customer C -10
Hotel +25
bell boy +5

Sum = 0

Bell boy gives back to each customer $1

Customer A -9
Customer B -9
Customer C -9
Hotel +25
Bellboy +2

Sum = 0 = -27 +27

All monies accounted for.
Zero sum game.

 

[BobbyBlueChip]

Love the T-Accounts, KotysDad.

Just think of it as there was never $30 and that the $2 that the bellhop got was part of the $27 that was paid.

Shame on the professors at USC :nono:

 

[DOGS THAT BARK]

Guy owns movie house.
He charges $5 for men $2 for women and 10 cents for children

On one particular night he had exactly 100 people pay and collected exactly $100.

How many of each men,women and children attended.

+++++++++++++++++++++++++++++++++++++++++++
What is the simplest way to calculate the addition of #'s from 1 to 100. 1+2+3+4---------+100

Should take a couple of seconds.

 

[KMA]

5M + 2W + .1C = 100
M + W + C = 100


3 variables / 2 formulae, not solvable???

 

[KMA]

9 men, 24 women, 70 children, (3 of the children were paid for by their adult escort. If there are other solutions, I dont have time to look for them.)

5M + 2W + .1C
P=100

5(9) + 24(2) + 0.1(70) = $100
9 + 24 + 67 = 100

(We know P=100, and that none of the variables can be zero,
Also that P does not equal total cash received.)

(Must be a animated Disney film)

 

[MonsterNco]

11 men 55$
19 women 38$
70 children 7$

 

[KotysDad]

There is actually a way to solve that problem without trial and error. Since the total money is in whole units, the number of children must be a multiple of 10. Also, m and w must be integers.

m + w + c = 100
5m + 2w + .1c = 100
c = 10k

where k is some unknown constant.

Now if you solve this system for m and w in terms of k, you get

m = (19k -100) / 3
w = (400-49k) / 3

Looking at the formula for m, you can see that k must be at least 6 (since anything less would give you less than 1 male) and cant be more than 9 or else you would have 100 children alone.

So, k must be 6, 7, 8, or 9. k=7 is the only solution that gives you an integer solution for m and w.

k = 7
c = 70
w = 19
m = 11

 

[DOGS THAT BARK]

11-19-70 is correct. Only way I could figure it was thru reasoning,knowing the children had to end in 0 and had to be a lot like 70 or 80--and knew all end #'s had to add to 0. Took a while to consider 7-8-5---not common #'s to consider.

on the adding all #'s from 1 to 100 the answer is 5050 which you can get in short order by reasoning. if instead adding each in order if you add in pairs 100+1 you get 101----99+2=101 ect. There are 50 pairs all adding to 101. 50X101=5050

 

[KotysDad]

DTB,

I didnt even see that second problem in your post. I saw the line break and thought it was a signature block and just skimmed right over it. :banghead:

 

[SimonSezs]

The HOTEL room costs $30, overcharged $5 so the HOTEL room costs $25. The Bell Boy does not give $5 back instead he gives $3 back keeping $2 for himself. Nothing changed here except the statement 'each guy now paid $9 for the room' is incorrect from the Managers perspective. The three guys were scammed by the Bus Boy.

The Bus Boy gives back $3 so the 3 guys think they paid $27 per room when in reality the manager charged them $25. So the Hotel has $25, the Bus Boy has $2, and each of the three guys has $1 which totals $3 (who were scammed by the bus boy who took the tip without them knowing it). 25 + 2 + 3 = 30.

 

[IntenseOperator]

The manager returned $5, so the cost of the room is $25. However, the busboy kept $2, so the room now costs $25 + $2 = $27.
Therefore, each guy pays $27 / 3 = $9.
$25 + $2 = $9 * 3
The $1 is not missing. It was never there.

KMA=stud