Question 541620: on monday, the produce manager stocked his store's display case with 80 heads of lettuce. By the end of the day some heads of the lettuce had been sold. On Tusday, the manager counted the number of heads of a lettuce that were left and decided to add an equal number of heads of lettuce, thereby doubling the leftovers. By the end of the day he had sold the some 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. He sold the same number of heads of lettuce that day, too. At the end of the day, thought, there were no heads of lettuce left. How many were sold each day? Fing an equation that models this scenario. To do this you must first you must first define what it is you are trying to find. Finally, solve the equation.
Answer by KMST(5328)   (Show Source):
You can put this solution on YOUR website! Life is easy for that manager; he sells the same number of heads of lettuce every day.
Let the number of heads of lettuce sold daily be x, because that is the only mystery here, and is the unknown we are trying to find.
The store opens Monday withe 80 heads of lettuce on display.
As x sell on Monday, the day ends with  on display.
On Tuesday, before opening, the manager orders doubling the number of heads of lettuce that are on display. The store opens with  on display.
At the end on the day, the number of heads of lettuce that are on display is  .
On Wednesday, before opening, the manager orders tripling the number of heads of lettuce that are on display. The store opens with  on display. Some were starting to look a little wilted on the outside, but that astute store manager, knew that he would still sell them. He sold x that day as usual.
At the end of the day the number of head of lettuce on display was
 and it was also zero.
That manager knew his sale patterns well, and he was so good at algebra too.
 is the equation to solve.
You'll go through  to 
|
|
|
