Question

If radius of a sphere is increased by 20 percent, by what percent does it's volume increase? (Ans is 72.8, please show correct method)

Answer 1

Let the radius be = r

⇒The volume = 4πr³/3

When the radius is increased by 20%

 ⇒The new radius is r+20*r/100 =1.2r

⇒The new volume is = 4π(1.2r)³/3 = 1.728*(4πr³/3)

⇒The %increase in volume = (1.728(4πr³/3)-4πr³/3)*100/(4πr³/3)=72.8%

⇒The volume will increase by 72.8%

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Answer 2

The percentage increase in volume of sphere is "72.8 percent (%)".

Step-by-step explanation:

Let the radius of sphere = r

To find, the percent increase in volume of sphere = ?

We know that,

The volume of sphere = $\dfrac{4}{3}\pi r^{3}$

The new radius of sphere = $\dfrac{120}{100} r=\dfrac{6}{5}r$

∴ The new volume of sphere = $\dfrac{4}{3}\pi (\dfrac{6}{5}r)^{3}$

= $(\dfrac{4}{3}\pi r^{3})\dfrac{216}{125}$

∴ Increase in volume = The new volume of sphere - The volume of sphere

= $(\dfrac{4}{3}\pi r^{3})\dfrac{216}{125}$ - $\dfrac{4}{3}\pi r^{3}$

= $\dfrac{4}{3}\pi r^{3}$ ($\dfrac{216-125}{125}$)

= $\dfrac{4}{3}\pi r^{3}$$\dfrac{91}{125}$

∴ The percent increase in volume of sphere = ($\dfrac{4}{3}\pi r^{3}$$\dfrac{91}{125}$)/($\dfrac{4}{3}\pi r^{3}$) × 100

= $\dfrac{91}{125}\times 100$

= 0.728 × 100

= 72.8 %

Thus, the percentage increase in volume of sphere is "72.8 percent (%)".