- 28 Jun '06 18:36I am the boss. and I have employee 1, employee 2, employee 3.

I give them 90 apples distributed in the following order:

Employee 1: 50 apples

Employee 2: 30 apples

Employee 3: 10 apples

My order is that they have to sell all apples at the same prize employee 1 sells them, if he sells them two for a dollar, then the others must do so, and they must sell all apples, an by doing this they must obtain the same amount of money, if employee one obtains 5 dollars for the 50 apples, then employee two must have obtained 5 dollars for his 30 apples, and employee one as well.

At what prize do they have to sell the apples? - 28 Jun '06 19:32

Buy 1 apple for $5 and get all the rest of the employee's apples for free*Originally posted by Superman***I am the boss. and I have employee 1, employee 2, employee 3.**

I give them 90 apples distributed in the following order:

Employee 1: 50 apples

Employee 2: 30 apples

Employee 3: 10 apples

My order is that they have to sell all apples at the same prize employee 1 sells them, if he sells them two for a dollar, then the others must do so, and they mus ...[text shortened]... s for his 30 apples, and employee one as well.

At what prize do they have to sell the apples? - 28 Jun '06 19:54

My english is not very good, I meant "price". but the problen has a logical solution and it is not a word game, there is a way to sell all apples an obtain the same amount of money for each employee following the rules.*Originally posted by aginis***of course they are free they are prizes.**

(50 apples)*price = 30 apples* price

50-30=20 apples* price = 0 => price = 0

perhaps you meant something else?

Hint: you can sell them in diferent packs. - 28 Jun '06 20:01

then any price will do*Originally posted by Superman***My english is not very good, I meant "price". but the problen has a logical solution and it is not a word game, there is a way to sell all apples an obtain the same amount of money for each employee following the rules.**

Hint: you can sell them in diferent packs.

1 sells 50 apples for $x

2 sells 30 apples for $x

3 sells 10 apples for $x

or

1 sells 25,10,5 for $x each = 2x,5x,10x respectively

2 sells 15,6,3 for $x = 2x, 5x,10x respectively

3 sells 5,2,1 for $x each = 2x,5x,10x respectively

and the price x is irrelevent - 28 Jun '06 20:07

sell them in packs so that the prize is the same, imagine that they went out two times to sell them, and by imposible reasons the price had, they could not sell them all the first time, so they regrouped, rearanged the price for the last ones and finished.*Originally posted by aginis***then any price will do**

1 sells 50 apples for $x

2 sells 30 apples for $x

3 sells 10 apples for $x

or

1 sells 25,10,5 for $x each = 2x,5x,10x respectively

2 sells 15,6,3 for $x = 2x, 5x,10x respectively

3 sells 5,2,1 for $x each = 2x,5x,10x respectively

and the price x is irrelevent

this hint is too much i think, I am going out now be back in a few hours - 28 Jun '06 20:42

ok, its not a hint if the question doesn't make sense. you have to set some sort of rules.*Originally posted by Superman***sell them in packs so that the prize is the same, imagine that they went out two times to sell them, and by imposible reasons the price had, they could not sell them all the first time, so they regrouped, rearanged the price for the last ones and finished.**

this hint is too much i think, I am going out now be back in a few hours

once again the price could be anything

day1 employee 1 sells 49 apples for 261 dollars a piece netting $12,789

employee's 2 and 3 have a slow day and sell no apples

day2 employee 1 is trying to sell his last apple for 441 dollars but fails

employee 2 sells 29 apples at 441 each netting 12,789

employee 3 has a slow day no sales

day 3 employee 1 is selling his last apple at 1421

he and employee 2 fail

but employee sells 9 apples and the brooklyn bridge for $12,789

day 4 blow out sale all three employees sell their last remaining apple for $1 each nets $12790

do me a favor don't become a math teacher (i hear McDonalds is hiring.) - 29 Jun '06 03:45

Yes, I guess you are right, you reached the solution, only in a different way than the one I have, but it is like you said, it doesn’t mater the amount, so this problem has as much solutions as numbers, unless you make a rule for that.*Originally posted by aginis***ok, its not a hint if the question doesn't make sense. you have to set some sort of rules.**

once again the price could be anything

day1 employee 1 sells 49 apples for 261 dollars a piece netting $12,789

employee's 2 and 3 have a slow day and sell no apples

day2 employee 1 is trying to sell his last apple for 441 dollars but fails

employee 2 sells 29 a ...[text shortened]... 1 each nets $12790

do me a favor don't become a math teacher (i hear McDonalds is hiring.)

Here is the solution I have.

Employee 1 started selling seven apples per dollar, so she sold 49 apples obtaining 7 dollars, and 1 apple left.

Employee 2 was obliged to sell the same so he sold 28 apples obtaining 4 dollars, and 2 apples left.

Employee 3 did the same and sold 7 apples obtaining 1 dollar and 3 apples left.

Because it was impossible to sell 7 apples for them, they regrouped and employee 1 set a new price:

Employee 1 sold his remaining apple for 3 dollars

Employee 2 did the same with his apples and obtained 6 dollars

Employee 3 did the same with his 3 apples and obtained 9 dollars

So employee 1 has 7 dollars the first sale and 3 the second equals 10

And employee 2 has 4 dollars the first sale and 6 the second equals 10

Employee 3 had 1 dollar from the first sale and 9 the second it equals 10

I guess the rule would be to sell the apples at a reasonable price, because who would buy an apple for 441 dollars?

I will take your advice, wont be a teacher.