Records for Sale

The owner of a record store hires a pimply faced, high-school kid to work on Saturdays. He says to the kid, “You know what to do. But I got one little extra thing for you. I’ve got two piles of used 45-PM records that I’m selling for my friend Sam. Each pile consists of 30 records.

“The records in the first pile are 2 for \$1. The other pile is 3 for \$1. I don’t want you to put the money in the register. I’ve got to give it to Sam. Put the money into the cigar box under the counter.”

At the end of the first day, the owner comes back to the store, finds that all of the records have been sold and there is \$25 in the cigar box. The two-for-a-dollar records sold for a total of \$15, and the three-for-a-dollar records sold for a total for ten bucks.

Encouraged by the rapid sales, the next week Sam shows up with 60 more records. The owner gives the kid the same instructions. This time, the kid says, “I noticed last week that people were taking two records from one pile and three records from another pile, so I decided that this week I’m going to sell 5 for \$2.”

The fellow who owns the record store says, “Seems like a good idea.” At the end of the day, though, the owner opens the cigar box and there’s \$24 in there. He says, “You’re missing a dollar!” The kid says, “No, I sold all the records.” And that’s the question. Where’s the missing dollar?

5 guesses to Records for Sale

Warning: Use of undefined constant bfa_comments - assumed 'bfa_comments' (this will throw an Error in a future version of PHP) in /home/customer/www/riddledude.com/public_html/wp-content/themes/atahualpa/comments.php on line 132
• det

There’s no missing dollar. 60/5*2\$=24\$

• Roolstar

The owner should’ve paid a bit more attention before giving the go ahead on the not so good idea of the salesman.

At first glace it would seem that nothing changed 2 Disks for 1\$ and 3 Disks for 1\$ means that 5 Disks would go for \$5. Well, not really so.

1- Logical way to see what’s wrong is to simulate the sales:
DAY 1, 15 customers bought 15 disks of Pile 1 (2 for \$1) and 10 cutomers bought 15 disks of Pile 2(3 for \$1)

DAY 2, 12 customers bought all 60 disks (5 for \$2)
=> Number of disks sold for 2 for 1 = 2 x 12 = 24 disks (Pile 1)
=> Number of disks sold for 3 for 1 = 3 x 12 = 36 disks (Pile 2)
=> Less Disks in Pile 1 (expensive) and more disks in Pile 2 (cheep)

2- Mathematical way to see this:
Salesman idea suggests that 1/2 + 1/3 = 2/5. Obivously mistaken.

PS: He mislead the owner by saying that the first day, customers would buy 2 disks from Pile 1 and 3 from Pile 2. And that’s not totally correct. In fact, if the 1st 10 customers would do this, Pile 2 would be completely sold out, and the next 5 customers can only get Disks from the 1st pile!

And yes, I still like to complicate stuff :)

• Roolstar

Correction:

* At first glace it would seem that nothing changed 2 Disks for 1\$ and 3 Disks for 1\$ means that 5 Disks would go for \$2. Well, not really so.

• Paul

But you guys seem to be missing the main point, as pile one consisted of the sweet tunes from the 60’s, while pile two consists of disco music from 1977. The only reason they made more money the first day was because music connoisseurs flocked around the table, arguing over pile one and the value of each 45’s B side, causing the musically deficient to purchase the remaining disco records. On day two the connoisseurs avoided the five for \$2, because they didn’t want to be seen doling out good money for three disco records in order to get two sweet stylings from the 60’s. You have to have musical standards for crying out loud.

• Det, you’re correct that there is no missing dollar. Here’s how we see the discrepancy:

In the record sales during week #1, the records in the pile marked 2 for \$1 were priced at 50 cents each. And the records in the pile marked 3 for \$1 were priced at 33.33 cents each. In the record sales during week #2, the records were all marked 5 for \$2, which averages out to 40 cents per record. And that average price is the problem, because if you averaged the price of the records for sale in Week #1, the true average is 41.67 cents (50 cents + 33.33 cents = 83.33 cents /2 = 41.67 cents per record).

(To make \$25 in the record sales in Week #2, the clerk should have priced the records at 5 for \$2.08 to reach a total sales figure of \$25).

So there is no missing dollar. Rool, you were a little more elaborate on why this occurred so kudos to both of you.