MATH SOLVE

10 months ago

Q:
# 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?

Accepted Solution

A:

Considering balloon to be an spherical objectVolume of sphere = [tex]\frac{4}{3}\pi r^3[/tex], 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'= [tex]\frac{4}{3}\pi R^3[/tex],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, [tex]\frac{dR}{dt}=2 \times \frac{dr}{dt}[/tex] 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 minutesBut 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: [tex]\frac{dR}{dt}=2 \times \frac{dr}{dt}[/tex]