manalq8 wrote:
The first flight out of Phoenix airport had a late departure. If the next three flights departed on-time, minimum how many subsequent flights need to depart from Phoenix on-time, for the airport's on-time departure rate to be higher than 90%?
A. 6
B. 7
C. 9
D. 10
E. 11
I will see what is the quickest way to solve it then I will provide the explanation
Official Solution:On a certain day, the first flight out of Phoenix airport had a delayed departure, while the next three flights departed on-time. What is the minimum number of subsequent flights that must depart from Phoenix on-time, to achieve an on-time departure rate of over 90% that day? A. 6
B. 7
C. 9
D. 10
E. 11
We want an on-time departure rate of over 90%, which means a delayed departure rate of less than 10%. Since we need the minimum number of subsequent flights that will make (delayed)/(total) less than 10%, then the one that was already delayed should be the only one.
Let \(x\) be the minimum number of subsequent flights that must depart on-time to achieve this. Then, the delayed departure rate is \(\frac{1}{x + 4} \), since there are already 4 flights, including the first delayed one. We want this rate to be less than 10%, so we have the inequality \(\frac{1}{x + 4} < \frac{1}{10}\). Solving for \(x\), we get x > 6. Therefore, the minimum number of subsequent flights that must depart on-time is 7.
Answer: B