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.