Solution:
Algebraically: Solve for X
-->The First Person
We have three people, one woke up and ate 1/3 of the apples. So, 1- (1/3)= 2/3 left. So, we can write Y= (2/3)X, for X is the total number of apples in the bag and Y is the number of apples left in the bag.
-->The Second Person
Next, the second person woke up and ate 1/3 of what was left in the bag. So, we can write
Z = (2/3) Y, for Z is the number of apples left in the bag after the second person eats 1/3 of the previous remainder.
-->The Third Person
Then, the third person wakes up and eats 1/3 of the remaining apples, leaving 8 apples in the bag. So, we have (2/3)Z - 8 = 0.
-->Now Solve for Z.
solving for Z: (2/3)Z – 8 = 0, we get (2/3)Z = 8, where Z = 12.
-->Now Work Backwards
Working backwards, we can plug Z = 12 into (2/3)Y = Z. So, we get (2/3)Y = 12. Solving for Y, we get Y = 18. •Continuing to work backwards, we can plug Y = 18 into (2/3)X = Y. Solving for X, we get X = 27.
-->The Answer
The original number of apples in the bag was 27!
Algebraically: Solve for X
-->The First Person
We have three people, one woke up and ate 1/3 of the apples. So, 1- (1/3)= 2/3 left. So, we can write Y= (2/3)X, for X is the total number of apples in the bag and Y is the number of apples left in the bag.
-->The Second Person
Next, the second person woke up and ate 1/3 of what was left in the bag. So, we can write
Z = (2/3) Y, for Z is the number of apples left in the bag after the second person eats 1/3 of the previous remainder.
-->The Third Person
Then, the third person wakes up and eats 1/3 of the remaining apples, leaving 8 apples in the bag. So, we have (2/3)Z - 8 = 0.
-->Now Solve for Z.
solving for Z: (2/3)Z – 8 = 0, we get (2/3)Z = 8, where Z = 12.
-->Now Work Backwards
Working backwards, we can plug Z = 12 into (2/3)Y = Z. So, we get (2/3)Y = 12. Solving for Y, we get Y = 18. •Continuing to work backwards, we can plug Y = 18 into (2/3)X = Y. Solving for X, we get X = 27.
-->The Answer
The original number of apples in the bag was 27!