+19 Question:
On Monday, the produce manager stocked the display case with eighty heads of lettuce. By the end of the day some of the lettuce had been sold.
On Tuesday, the manager surveyed the display case and counted the number of heads of lettuce left. He decided to add an equal number of heads of lettuce. (He doubled the leftovers.) By the end of the day he had sold the same number of heads of lettuce as on Monday.
On Wednesday, the manager decided to triple the number of heads of lettuce that had been left in the case Tuesday night. He sold the same number of heads of lettuce that day, too. However, at the end of the day there were no heads of lettuce left.
How many heads of lettuce were sold each day? At least one method for solving this problem must be algebraic.
Quoting CPhill's Answer
"On Monday, the produce manager stocked the display case with eighty heads of lettuce. By the end of the day some of the lettuce had been sold.
On Tuesday, the manager surveyed the display case and counted the number of heads of lettuce left. He decided to add an equal number of heads of lettuce. (He doubled the leftovers.) By the end of the day he had sold the same number of heads of lettuce as on Monday.
On Wednesday, the manager decided to triple the number of heads of lettuce that had been left in the case Tuesday night. He sold the same number of heads of lettuce that day, too. However, at the end of the day there were no heads of lettuce left. How many heads of lettuce were sold each day?
I only know one method
Call N the number sold on Monday.......so...80 - N were left
On Tuesday (80 - N ) were added to the (80-N) and N were sold...so 2(80 - N) - N = 160 - 2N - N = (160 - 3N) were left
On Wed 2(160 - 3N) were added to (160 - N) and N were sold....and none were left ...so...
3(160 - 3N) - N = 0
480 - 9N - N = 0
480 -10N = 0
480 = 10N
N = 48 = number sold each day"
