Question 56503: LETTUCE FOR SALE: On Monday, the produce manager stocked the display case with 80 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 that were 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. He sold the same number of head of lettuce that day too. However, at the end of that day there were no heads of lettuce left.
* How many heads of lettuce were sold each day?
* How many heads of lettuce were left over at the end of each day?
Make sure to define any variables, explain your equations and show your work. Don't forget to show that you checked your work.
Answer by AgusKwan(7)   (Show Source):
You can put this solution on YOUR website! Let the number of lettuce has been sold as n
End of Monday, the lettuce would be 80-n
Double the leftovers = 2(80-n) = 160-2n
End of tuesday (160-2n) - n = 160-3n
Triple the leftovers = 3(160-3n) = 480 - 9n
He sold the same number of head of lettuce that day too.
However, at the end of that day there were no heads of lettuce left.
This means the 480-9n=n --> 480 = 10n --> n=48
* How many heads of lettuce were sold each day? --> 48
* How many heads of lettuce were left over at the end of each day?
Left overs on:
Monday = 80-48 = 32
Tuesday = 32*2-48=16
Wednesday = 16*3 - 48 = 0
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