Solve this simple logic/math problem

Three guys go to a motel to rent a room for the night. The clerk tells them the room cost $30.00. Each guy ponies up $10.00 and they go to their room.

Then the clerk realizes there’s actually a special going on and the room is only $25.00 so he summons the maid, gives her five $1.00 bills and tells her to give the guys a refund.

On the way to the room, the maid starts wondering how she’s supposed to divide five $1.00 bills among three people. Then she decides to just pocket 2 bucks and give each guy one dollar and they’ll never know what the discount really was.

She refunds each guy a buck, so they paid 9 bucks each for the room.

9 x 3 = 27.

27 + the two dollars the maid pocketed is 29.

Where is the 30th dollar?

It’s a riddle. Read it carefully. The second half doesn’t add up to 30. It’s using misdirection.
$30 = $1 (inside guy 1 pocket) + $1 (inside guy 2 pocket) + $1 (inside guy 3 pocket) + $2 (inside maid pocket) + $25 (in register). Fuck the cunt maid, stealing shit.

It’s at the front desk with the clerk. You have to watch that guy.

They each paid $9.333333 after the refund.

$30 - $5 = $25 + $1 refunded to each guy x 3 = $28. 28/3=9.333

Btw, it’s not 27+2, it’s 27-2.

Try reducing it to one guy who gives the $2 to the maid after receiving the $5 refund.

The 30th dollar is in the maids pocket along with the 29th dollar.

The 3 men paid a combined $28, +$2 in the pocket = $30.

They should have paid $25, plus the $3 change and the $2 stolen = $30.

Maid sounds like a cunt

Fellas sound like they party

No each guy paid $9 per room. They have a whole dollar after handing over a 10. So they paid $27, and not anything else.

The problem is you don’t add the $2 to the 27, you subtract it to get what’s left with the clerk.

Then they would have change in their pockets instead of a whole bill. Imagine they each had a single 10 and now they have $1. They obviously spent $9. 9x3=27, so they paid a combined $27.

Ah yes, too early for this shit.

All I know is I took the day off, dropped my kids at school, and just found the Rams won me $240 last night. Fuck this maid and her $2 lol

It’s tricky on purpose. It’s designed to confuse.

You have ten stacks of coins one stack has coins that is light by one ounce per coin. You have a scale that is only capable of performing one measurement. How can you determine which stack contains coins that are each light by one ounce with only one measurement? And not a trick question.

You have nine coins but one of them is an ounce lighter than the other right You have a balance scale. What is the minimum number of weighings required to guarantee you know which of the nine coins is the light one?