The radius of a sphere is 9.4 cm. If it is increased by 10% by what percent will its volume increase

Question: The radius of a sphere is 9.4 cm. If it is increased by 10% by what percent will its volume increase?

• 33 %
• 35 %
• 40 %
• 67 %

Answer: 

The volume of a sphere is:

$$V = \frac{4}{3} \pi r^3$$

If the radius increases by 10%, the new radius is:

$$r' = 1.1r$$

The new volume:

$$V' = \frac{4}{3} \pi (1.1r)^3$$ $$V' = \frac{4}{3} \pi (1.331 r^3)$$ $$V' = 1.331 V$$

The percentage increase in volume is:

$$\frac{V' - V}{V} \times 100 = \frac{1.331V - V}{V} \times 100$$ $$= (1.331 - 1) \times 100$$ $$= 33.1\%$$

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Alternative Method: 

✅✅ Using the (AB Method)

We use the formula:

$$\% \text{ change in volume} \approx 3 \times \% \text{ change in radius} + \frac{3}{2} \times (\% \text{ change in radius})^2$$

Given that the radius increases by 10% (A = 10),

$$\% \text{ change in volume} \approx 3(10) + \frac{3}{2} (10)^2$$ $$= 30 + \frac{3}{2} \times 100$$ $$= 30 + 150$$ $$= 33\%$$

_'_ the volume increases by 33% (approximate)

which closely matches the exact value of 33.1%.

Additional Information :

✅✅Shortcut 1: 

📌 Direct Formula for Volume Change-

$$\text{Percentage Change in Volume} \approx 3 \times \text{Percentage Change in Radius} + \frac{3}{2} \times (\text{Percentage Change in Radius})^2$$

For a 10% increase in radius:

$$3(10) + \frac{3}{2} (10)^2 = 30 + 15 = 33\%$$

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✅✅Shortcut 2: 

📌Cube of Percentage Increase Factor-

Since volume depends on $r^3$, when the radius increases by 10%, the new radius is $1.1r$.
The volume ratio is:

$$\left(\frac{r'}{r}\right)^3 = (1.1)^3$$

Using the approximation:

$$(1.1)^3 \approx 1.331$$

Percentage increase:

$$(1.331 - 1) \times 100 = 33.1\%$$

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✅✅Shortcut 3:

📌Percentage Multiplier Rule-

For a small percentage $x$, the formula for volume increase is:

$$\text{Increase} \approx 3x + 0.03x^2$$

For $x = 10\%$:

$$3(10) + 0.03(100) = 30 + 3 = 33\%$$

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