if you have a jar filled with candy and every time you take one candy out you put the wrapper back into the jar. the wrapper takes 10% of the volume of a piece of candy. How many times will you have to fill the jar before half the jar is filled with candy wrappers?
To solve this problem, we need to determine the number of times you would have to fill the jar before half of it is filled with candy wrappers.
First, let's establish the initial volume of the jar filled with candies. Let's assume the jar is completely filled with candies, so the volume of candies is 100%.
Each time you remove a candy and put the candy wrapper back into the jar, the volume of the candy is reduced by 10%. This means that after each cycle, the volume of candies is 90% of its original volume, and the volume of wrappers is 10% of the original volume.
Based on this information, we can calculate how the volume changes after each filling cycle:
1st cycle: The jar is filled with 90% candies and 10% wrappers.
2nd cycle: The volume of candies becomes 90% of 90%, which is (90% * 90%) = 81% of the jar's volume. The volume of wrappers becomes 10% of 90%, which is (10% * 90%) = 9% of the jar's volume.
3rd cycle: The volume of candies becomes 90% of 81%, which is (90% * 81%) = 72.9% of the jar's volume. The volume of wrappers becomes 10% of 81%, which is (10% * 81%) = 8.1% of the jar's volume.
As you can see, after each cycle, the volume of candies decreases, while the volume of wrappers increases. So, we need to continue this process until the volume of wrappers reaches exactly 50% of the jar's volume.
Let's calculate the number of cycles necessary to reach this point:
Let's represent X as the number of cycles needed.
The equation to find the volume of wrappers after X cycles is: 10% * (90%)^X.
We need to solve the following equation: 10% * (90%)^X = 50%.
To solve this equation, we can take the logarithm of both sides. Assuming a base-10 logarithm:
log(10% * (90%)^X) = log(50%).
log(10%) + log((90%)^X) = log(50%).
-log(10) + X*log(90%) = log(50%).
Simplifying further, we have:
X*log(90%) = log(50%) + log(10).
X*log(9/10) = log(50%) + log(10).
X*log(9) = log(50%) + log(10).
X = (log(50%) + log(10)) / log(9).
Using a calculator, we can evaluate the right side of the equation to find the value of X:
X ≈ (−0.301 − 1) / 0.954 = −1.301 / 0.954 ≈ −1.363 ≈ 1.363.
Since X represents the number of cycles, it cannot be a fraction or negative. Therefore, we take the ceiling of X, which gives us the smallest integer greater than or equal to X:
Ceiling(X) = 2.
Thus, you would need to fill the jar approximately 2 times before half of the jar is filled with candy wrappers.