Carmen received some money for her birthday on Friday. That night, she took her friend out to dinner. She spent half of her check for dinner. Then she tipped the waiter $6. The next evening she went to dinner alone, spent half of the money she had left, and tipped the waiter $2. The next night she went to dinner again, spent half of the money she had left, and tipped the waiter $1. At this time she had $6 left. How much money did she originally have?
To find out how much money Carmen originally had, we need to work backwards from what we know.
Let's represent the amount of money Carmen had originally as "X". We know that the following things happened:
1. Carmen spent half of her money on the first dinner, so she had X/2 left.
2. Carmen tipped the waiter $6, so she had X/2 - $6 left.
3. The next evening, Carmen spent half of the money she had left, which is (X/2 - $6)/2.
4. Carmen tipped the waiter $2, so she had (X/2 - $6)/2 - $2 left.
5. On the third night, Carmen spent half of the money she had left, which is ((X/2 - $6)/2 - $2)/2.
6. Carmen tipped the waiter $1, so she had ((X/2 - $6)/2 - $2)/2 - $1 left, which equals $6.
Now we can write an equation to solve for X:
((X/2 - $6)/2 - $2)/2 - $1 = $6
To simplify the equation, let's remove $ signs and multiply both sides by 2:
((X/2 - 6)/2 - 2)/2 - 1 = 6
(X/2 - 6)/2 - 2 = 12
(X/2 - 6)/2 = 12 + 2
(X/2 - 6)/2 = 14
X/2 - 6 = 14 * 2
X/2 - 6 = 28
X/2 = 28 + 6
X/2 = 34
To solve for X, we multiply both sides by 2:
X = 34 * 2
X = 68
Therefore, Carmen originally had $68.
First Trip
Check = X dollars.
Spent: X/2 + 6,
BAL = X - (X/2 + 6) = X/2 - 6.
SECOND TRIP
Spent: 1/2(X/2 - 6) + 2 = X/4 - 1.
BAL = (X/2 - 6) - (X/4 - 1) = X/4 - 5.
THIRD TRIP
Spent: 1/2(X/4 - 5) = X/8 - 5/2.
BAL=(X/4 - 5) - (X/8 - 5/2) = X/8 - 5/2
X/8 - 5/2 = 6,
Multiply both sides by 8:
X - 20 = 48,
X=48 + 20 = $68 = The original amount.