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View Full Version : Help with MATH... IM STUMPED!


vonedaddy
01-02-2004, 01:43 PM
this is much better is it is said outloud to someone, (and yes i DO know the answer i am not that stupid.)

here goes:

three people go to get a hotel room. the room cost $30. each person pays $10. after they go to the room, the clerk realizes he overcharged them. the room should have been $25.
The clerk gives $5 in singles to the bell-hop and tells him to give it back to the people in the room evenly. as he is walking to the room, he cant figure out how to give it back to them evenly and decides to pocket $2.00
he gives each person back $1.00 each. ($3 to each of them and he pockes $2...this is $5)

(this is where i need help)
all the money is accounted for, but if you do the math, it doesnt work out.

by recieving $1.00 back from their original $10, this means that each person realy only payed $9.00...right?

ok...now 3 X $9.00 = $27.
the bell-hop pocketed only $2.00
add thoes together and you have $29


WTF HAPPENED TO THE EXTRA DOLLAR???

In Odder Words
01-03-2004, 02:36 AM
The buck stops... HERE!

The three roommates originally paid THIRTY dollars totaled (ten bucks each).

When the bellboy returns one dollar apiece, that means the roommates instead paid TWENTY-SEVEN dollars totaled (nine bucks apiece.) This is accounted for by totalin' the twenty-five kept by the clerk added to the two dollar "tip" appropriated by the bellboy.

The confusion arises out of mentally hangin' on to the prior payment of "thirty dollars" rather than reassessin' what they paid to be "twenty-seven."

Of course, once the roommates discover they shoulda been charged twenty-five instead of twenty-seven and that the hotel hires thievin' bellboys, there would likely be some kind of... CHANGE... Never let a busboy do a... number... on you!


At least, that's how I... figure it... Were you usin' NEW math, or OLD?

;)

BorgHunter
01-03-2004, 09:53 PM
That's it? Pfffffft. I was hoping for at least some algebra. Odder already gave you the answer, so I shall throw this one out for all the other math buffs on AFN. (I do know the answer to this one.)

What is (x-a)(x-b)(x-c)...(x-z)?

es347fan
01-03-2004, 10:51 PM
Well, I know it's not X26+ABCD... and I do know both the answer & the explanation.

In Odder Words
01-04-2004, 03:04 AM
Evidently luck ain't a... factor... in this, 'cuz no matter how hard I

try ta GUESS the answer...



...NUTHIN' comes ta mind!

;)