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A hot-air balloon was decommissioned after a successful flight by letting the 3000 m^3 of air in it leak out. The loss in the volume of air in the balloon was exponential, decreasing at x percent per minute. If there was 1626 m^3 of air in the balloon 15 minutes after it began to be decommissioned, at what percent per minute to the nearest percent did the balloon lose air?

A1%
B2%
C3%
D4%

Verified Answer :

dragonbro1015

for the first minute, the new volume of air is 3000 x \(\frac{X}{100}\)because X% volume of air is lost every minute.What would've happened is a total of 3000 x \((\frac{X}{100})^{15}\) would've been lost.

thus, what we have so far is:
3000 x \((\frac{100 - X}{100})^{15}\) = 1626
\((\frac{100 - X}{100})^{15}\) = \(\frac{1626}{3000}\)
\(\frac{100 - X}{100} = \sqrt[15]{\frac{1626}{3000} }\)


which is same as:
\(\frac{100 - X}{100} = 0.95999\) (rounded)
\(100 - X = 0.95999\) x 100
100 - X = 95.999
-X = 95.999 - 100
X ≈ 4

Therefore, 4% of air was lost from the balloon every minute.

BTW 
The reason I did \((\frac{X}{100})^{15}\) was because in multiplications, the order doesn't matter, so instead of doing
\(\frac{X}{100}(\frac{X}{100}(\frac{X}{100}3000))....\)
15 times, I just merged them into that equation

UPDATED sorry forgot some steps...

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