Question

Vavilen inflated a huge balloon in 19 minutes. Then he poked a hole in the balloon so that it slowly leaked air out until it was completely deflated.
The rate at which the air leaked out of the balloon (in liters of air per minute) was half of the rate at which Vavilen inflated the balloon. It took the balloon 38 minutes to completely deflate.
At what rate did Vavilen inflate the balloon, and at what rate did the balloon deflate?

Answer 1

Considering balloon to be an spherical object

Volume of sphere = $\frac{4}{3}\pi r^3$, where r is the radius of sphere.

After 19 minutes ,If radius of balloon changes from r to R, the Volume changes from V to V'.

V'= $\frac{4}{3}\pi R^3$,

As given,rate at which the air leaked out of the balloon (in liters of air per minute) was half of the rate at which Vavilen inflated the balloon.

Also, $\frac{dR}{dt}=2 \times \frac{dr}{dt}$

Now, again the radius changes from R to r.So, Volume changes from V' to Back to V.

Time taken by the balloon to deflate = 38 minutes

But there is no relation between time and volume of Balloon.

So, we can't predict at which rate balloon is deflating or inflating.Apart from the statement given in Question:   $\frac{dR}{dt}=2 \times \frac{dr}{dt}$

Answer 2

Answer: Hello there!

If V is the max volume of the balloon (in liters), we know that Vavilen inflated it in 19 minutes:

this means that he put V/19 liters of air per minute.

We also know that the ballon, with a volume V of air inside of it, deflated in 38 minutes, so the rate of deflation is the volume divided by the number of minutes it took to deflate:

the rate is -V/38 liters of air per minute. (here the minus sign appears because the air is leaving the ballon)

Where the missing part is the maximum volume of the ballon, which is not given in the problem.