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A spherical balloon originally had a radius of \(\mathrm{x}\) centimeters. The balloon was partially deflated by letting out some air so that the volume of the balloon decreased by \(\mathrm{a}%\). If the balloon remained spherical while the air was being let out, which of the following expressions in \(\mathrm{x}\) and \(\mathrm{a}\) gives the radius, in centimeters, of the partially deflated balloon?
(Note: The volume of a sphere with radius \(\mathrm{r}\) is \(\frac{4}{3}\pi\mathrm{r}^3\).)
Let's break down what's happening in plain English first. We start with a spherical balloon that has radius x centimeters. When we let out some air, the volume decreases by a%.
Now, what does "decreased by a%" actually mean? If something decreases by a%, then what remains is (100-a)% of the original. For example, if the volume decreases by 20%, then 80% of the original volume remains.
So mathematically, if the original volume was V, then the new volume is:
New volume = \(\frac{100-a}{100} \times \text{Original volume} = \left(1 - \frac{a}{100}\right) \times V\)
This means the new volume is \(\left(1 - \frac{a}{100}\right)\) times the original volume.
Process Skill: TRANSLATE - Converting the percentage decrease language into a mathematical relationship
Now let's use the sphere volume formula. We know that for any sphere with radius r, the volume is \(\frac{4}{3}\pi r^3\).
Original sphere: radius = x, so volume = \(\frac{4}{3}\pi x^3\)
New sphere: radius = \(r_{\text{new}}\) (unknown), so volume = \(\frac{4}{3}\pi (r_{\text{new}})^3\)
From step 1, we know:
New volume = \(\left(1 - \frac{a}{100}\right) \times \text{Original volume}\)
Substituting our formulas:
\(\frac{4}{3}\pi (r_{\text{new}})^3 = \left(1 - \frac{a}{100}\right) \times \frac{4}{3}\pi x^3\)
We can cancel out the \(\frac{4}{3}\pi\) from both sides:
\((r_{\text{new}})^3 = \left(1 - \frac{a}{100}\right) \times x^3\)
This is where the key insight comes in. We have:
\((r_{\text{new}})^3 = \left(1 - \frac{a}{100}\right) \times x^3\)
To find \(r_{\text{new}}\), we need to take the cube root of both sides. Remember that when you take the cube root of a product, you can take the cube root of each factor separately:
\(r_{\text{new}} = \sqrt[3]{\left(1 - \frac{a}{100}\right) \times x^3}\)
\(r_{\text{new}} = \sqrt[3]{1 - \frac{a}{100}} \times \sqrt[3]{x^3}\)
\(r_{\text{new}} = \sqrt[3]{1 - \frac{a}{100}} \times x\)
Therefore: \(r_{\text{new}} = x\sqrt[3]{1 - \frac{a}{100}}\)
This makes intuitive sense: if the volume shrinks by a certain factor, the radius shrinks by the cube root of that factor because volume depends on the cube of the radius.
Process Skill: MANIPULATE - Using cube root properties to isolate the new radius
Looking at our result: \(r_{\text{new}} = x\sqrt[3]{1 - \frac{a}{100}}\)
Comparing with the answer choices:
The radius of the partially deflated balloon is \(x\sqrt[3]{1 - \frac{a}{100}}\) centimeters.
Answer choice D is correct.
Students often confuse "decreased by a%" with "decreased to a%". When volume decreases by a%, what remains is (100-a)% or (1-a/100) of the original. Many students incorrectly think the new volume is a/100 of the original volume, leading them toward answer choice E.
Students may attempt to directly apply the percentage decrease to the radius instead of the volume. Since the problem states that volume decreased by a%, they might think the radius also decreased by the same percentage, missing that volume depends on the cube of the radius.
When solving \((r_{\text{new}})^3 = \left(1 - \frac{a}{100}\right) \times x^3\), students often make the error of taking the cube root incorrectly. They might write \(r_{\text{new}} = \left(1 - \frac{a}{100}\right) \times x\) instead of \(r_{\text{new}} = \sqrt[3]{1 - \frac{a}{100}} \times x\), essentially "forgetting" to take the cube root of the scaling factor.
Students frequently mix up cubing and taking cube roots. When they see that volume scales by factor \(\left(1 - \frac{a}{100}\right)\), they might incorrectly conclude that radius scales by \(\left(1 - \frac{a}{100}\right)^3\), leading them to select answer choice B.
Even after correctly deriving that the answer involves a cube root, students might select answer choice C: \(x\sqrt[3]{100 - a}\) instead of the correct D: \(x\sqrt[3]{1 - \frac{a}{100}}\). This happens when they use (100-a) instead of properly converting the percentage to the decimal form \(\left(1 - \frac{a}{100}\right)\).
Step 1: Choose convenient values
Let's select smart numbers that make calculations clean:
• Original radius: x = 10 cm (easy to work with)
• Volume decrease: a = 19% (chosen because 100 - 19 = 81, and 81 = \(3^4\), making cube root calculations exact)
Step 2: Calculate original volume
Original volume = \(\frac{4}{3}\pi(10)^3 = \frac{4}{3}\pi(1000) = \frac{4000\pi}{3}\) cubic cm
Step 3: Find new volume after decrease
Volume decreased by 19%, so new volume is 81% of original:
New volume = \(0.81 \times \frac{4000\pi}{3} = \frac{81}{100} \times \frac{4000\pi}{3} = \frac{3240\pi}{3}\) cubic cm
Step 4: Set up equation for new radius
If new radius is r, then:
\(\frac{4}{3}\pi r^3 = \frac{3240\pi}{3}\)
\(r^3 = \frac{3240}{4} = 810\)
\(r^3 = 810\)
Step 5: Solve for new radius
\(r^3 = 810 = 81 \times 10 = 3^4 \times 10\)
\(r = \sqrt[3]{81 \times 10} = \sqrt[3]{81} \times \sqrt[3]{10}\)
Since \(81 = 3^4\), we have \(\sqrt[3]{81} = \sqrt[3]{3^4} = 3^{\frac{4}{3}}\)
But more directly: \(810 = (1000) \times \frac{81}{100} = 10^3 \times \frac{81}{100}\)
So \(r = \sqrt[3]{10^3 \times \frac{81}{100}} = 10 \times \sqrt[3]{\frac{81}{100}} = 10 \times \sqrt[3]{0.81}\)
Step 6: Connect to answer format
Since a = 19, we have \(\frac{100-a}{100} = \frac{81}{100} = 0.81 = 1 - \frac{a}{100}\)
Therefore: \(r = 10 \times \sqrt[3]{1 - \frac{19}{100}} = x\sqrt[3]{1 - \frac{a}{100}}\)
Step 7: Verify against answer choices
Our result \(r = x\sqrt[3]{1 - \frac{a}{100}}\) matches answer choice D exactly.