# A balloon has more air added to it such that itsvolume is increased by 28%. what was the percent decreasein the thickness of the rubber?The answer is 15.2% but i dont know how to derive it. Thanks

mlehuzzah | Student, Graduate | (Level 1) Associate Educator

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Say the balloon has a radius of `r` before it has the extra air put in. Then assuming the balloon is perfectly spherical, the volume of the balloon before the extra air is put in, it `V_r = (4)/(3) pi r^3`

The volume of the balloon after the air is put in, `V_R` is `V_R = V_r * 1.28`

But we also have: `V_R = (4)/(3) pi R^3` where `R` is the radius after the extra air was added. Thus:

`(4)/(3) pi R^3 = (4)/(3) pi r^3 * 1.28`

`R^3 = r^3 * 1.28`

`R= r * root(3)(1.28) = r* 1.085767 `

The old (before the extra air) surface area of the balloon is:

`S_r = 4 pi r^2 `

The new surface area is:

`S_R = 4 pi R^2 = 4 pi (r * root(3)(1.28) )^2 = 4 pi r^2 (1.28)^(2/3)`

The volume of the rubber is approximately the surface area times the thickness. The volume of the rubber must be the same before and after the extra air is added. So, if `T_r` is the thickness of the rubber before the air is added, and `T_R` is the thickness after the air is added, we have:

`T_r S_r = T_R S_R`

`T_r 4 pi r^2 = T_R 4 pi r^2 (1.28)^(2/3)`

`T_r = T_R (1.28)^(2/3)`

`T_r (1.28)^(-2/3) = T_R`

`T_r * .8482555 = T_R`

Thus the new thickness is 84.8% of the old thickness, or a 15.2% decrease.