A spherical balloon originally had a radius of x centimeters. The balloon was partially deflated by letting out some air...

Strategic Approach

1. Translate the problem requirements: A sphere starts with radius x, its volume decreases by a%, and we need to find the new radius. The key insight is that "decreased by a%" means the new volume is (100-a)% of the original, or (1 - a/100) times the original volume.
2. Set up the volume relationship: Express both the original and new volumes using the sphere formula, then create an equation showing the new volume is (1 - a/100) times the original volume.
3. Solve for the radius relationship: Since volume is proportional to radius cubed, if volume changes by a factor, radius changes by the cube root of that factor.
4. Verify against answer choices: Confirm our expression matches one of the given options and check the mathematical form.

Execution of Strategic Approach

1. Translate the problem requirements

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

2. Set up the volume 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:
\((r_{\text{new}})^3 = \left(1 - \frac{a}{100}\right) \times x^3\)

3. Solve for the radius relationship

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

4. Verify against answer choices

Looking at our result: \(r_{\text{new}} = x\sqrt[3]{1 - \frac{a}{100}}\)

Comparing with the answer choices:

1. \(x(100 - a)^3\) - This has the wrong power and wrong form
2. \(x\left(1 - \frac{a}{100}\right)^3\) - This cubes the factor instead of taking its cube root
3. \(x\sqrt[3]{100 - a}\) - This doesn't convert the percentage properly
4. \(x\sqrt[3]{1 - \frac{a}{100}}\) - This matches our result exactly!
5. \(x\left(1 - \sqrt[3]{\frac{a}{100}}\right)\) - This has the cube root in the wrong place

Final Answer

The radius of the partially deflated balloon is \(x\sqrt[3]{1 - \frac{a}{100}}\) centimeters.

Answer choice D is correct.

Common Faltering Points

Errors while devising the approach

1. Misinterpreting percentage decrease language

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.

2. Forgetting the relationship between volume and radius

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.

Errors while executing the approach

1. Incorrectly applying cube root properties

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.

2. Confusing cube and cube root operations

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.

Errors while selecting the answer

1. Choosing the wrong percentage representation

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 \(\left(1 - \frac{a}{100}\right)\).

Alternate Solutions

Smart Numbers Approach

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.