Loading...
If \(\frac{1}{2}\) of the air in a tank is removed with each stroke of a vacuum pump, what fraction of the original amount of air has been removed after 4 strokes ?
Let's break down what this problem is asking in plain English:
The key insight here is to think about what stays versus what gets removed. If half is removed each time, then half remains each time.
Process Skill: TRANSLATE - Converting "removed" vs "remaining" language into mathematical understandingLet's imagine we start with 1 unit of air (representing 100% of the original air). Now let's see what happens after each stroke:
After Stroke 1: Half is removed, so half remains
Remaining = \(\frac{1}{2}\) of original
After Stroke 2: Half of what's left is removed
Remaining = \(\frac{1}{2}\) of \(\left(\frac{1}{2}\right) = \frac{1}{4}\) of original
After Stroke 3: Half of what's left is removed
Remaining = \(\frac{1}{2}\) of \(\left(\frac{1}{4}\right) = \frac{1}{8}\) of original
After Stroke 4: Half of what's left is removed
Remaining = \(\frac{1}{2}\) of \(\left(\frac{1}{8}\right) = \frac{1}{16}\) of original
So after 4 strokes, only \(\frac{1}{16}\) of the original air remains in the tank.
Now we need to answer the actual question: what fraction has been removed?
If \(\frac{1}{16}\) of the original air remains in the tank, then the rest must have been removed.
Fraction removed = Original amount - Amount remaining
Fraction removed = \(1 - \frac{1}{16} = \frac{16}{16} - \frac{1}{16} = \frac{15}{16}\)
This means \(\frac{15}{16}\) of the original air has been removed from the tank.
Looking at our answer choices:
Final Answer: A. \(\frac{15}{16}\)
We can verify this makes sense: after 4 strokes of removing half each time, we'd expect most of the air to be gone, and \(\frac{15}{16}\) represents \(93.75\%\) of the air removed, which seems reasonable.
1. Misunderstanding what "\(\frac{1}{2}\) of the air is removed" means: Students might think that exactly \(\frac{1}{2}\) of the original amount is removed with each stroke, rather than understanding that \(\frac{1}{2}\) of whatever air currently remains in the tank is removed. This leads to incorrect calculations like: after stroke 1, \(\frac{1}{2}\) removed; after stroke 2, another \(\frac{1}{2}\) removed, so total removed = 1. This fundamental misinterpretation derails the entire solution.
2. Focusing on tracking removed air instead of remaining air: Students might attempt to directly calculate how much is removed at each stroke rather than tracking what remains. This approach becomes mathematically complex because the amount removed changes with each stroke (\(\frac{1}{2}\), then \(\frac{1}{4}\), then \(\frac{1}{8}\), etc.), making it easier to make errors compared to the simpler pattern of tracking what remains (\(\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}\)).
3. Confusing the final question requirement: Students may set up their approach to find how much air remains in the tank instead of how much has been removed, especially since the calculation naturally leads to finding the remaining fraction first. This confusion in approach setup leads to selecting \(\frac{1}{16}\) instead of \(\frac{15}{16}\) as the final answer.
1. Arithmetic errors in fraction multiplication: When calculating what remains after each stroke, students may make errors in multiplying fractions. For example, calculating \(\frac{1}{2} \times \frac{1}{4}\) incorrectly as \(\frac{1}{6}\) instead of \(\frac{1}{8}\), or \(\frac{1}{2} \times \frac{1}{8}\) as \(\frac{1}{10}\) instead of \(\frac{1}{16}\). These multiplication errors compound and lead to incorrect final fractions.
2. Errors in subtraction when converting from remaining to removed: Students may make mistakes when calculating \(1 - \frac{1}{16}\). Common errors include: getting the common denominator wrong (using 15 instead of 16), or making arithmetic mistakes like \(\frac{16}{16} - \frac{1}{16} = \frac{14}{16}\) instead of \(\frac{15}{16}\).
1. Selecting the fraction that remains instead of the fraction removed: After correctly calculating that \(\frac{1}{16}\) remains in the tank, students may mistakenly select answer choice E (\(\frac{1}{16}\)) because this is the number they calculated, forgetting that the question asks for the fraction removed, not the fraction remaining.
2. Second-guessing the reasonableness of the answer: Students might doubt that \(\frac{15}{16}\) (which is \(93.75\%\)) could be removed in just 4 strokes and switch to a smaller fraction like \(\frac{7}{8}\), not fully grasping how powerful the exponential decay effect is when repeatedly removing half of what remains.
Step 1: Choose a convenient starting amount
Let's say the tank initially contains 16 units of air. We choose 16 because it's a power of 2, which works well when repeatedly dividing by 2.
Step 2: Track the air remaining after each stroke
Step 3: Calculate how much air has been removed
Amount removed = Initial amount - Remaining amount = \(16 - 1 = 15\) units
Step 4: Find the fraction removed
Fraction removed = Amount removed ÷ Initial amount = \(15 \div 16 = \frac{15}{16}\)
Step 5: Verify with answer choices
Our result of \(\frac{15}{16}\) matches answer choice A.
Why this smart number works: By choosing 16 as our starting value, we can easily track the halving process through 4 strokes without dealing with messy fractions during calculation. The final fraction \(\frac{15}{16}\) represents the general solution regardless of the actual starting amount.