Another math problem

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 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?

1 for $0

Originally posted by XanthosNZ
1 for $0

The answer is not $0,

"Sell" them for free.

EDIT: Too late.

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?

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?

Buy 1 apple for $5 and get all the rest of the employee's apples for free

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?

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.

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.

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

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

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

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

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 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.)

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.)

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.

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.