Gym grades for second term seniors

Algebra Level 2

There's a second term senior. Let's call him Bill.

As soon as second term starts, the first day of second term (Monday, Feb. 8) he receives 3/3 (100%) for his gym participation grade. So far so good.
However, the second day of second term (Tuesday, Feb. 9), it snowed really hard, so he wore boots instead to gym, so he got a 1/3 for participation (33.33...%).
As a result, his total gym grade dropped to a 66.66%.

If the senior shapes up after seeing that grade and gets perfect participation (3/3) every day, and the total gym grade is the average of all participation grades, how many days MINIMUM will it take for the senior to recuperate to at least a total grade of 90% in gym?

Assumptions:

You're not counting days 1 and 2 in your answer.
No school on weekends, but there are 7 days in a week.

5 days 7 days 8 days 6 days

1 solution

Arjen Vreugdenhil
Feb 11, 2016

Let $a_d$ be Bill's average grade after $d$ days. (We start with $d = 1$ on Wednesday.) Then $a_d = \frac{1 + \tfrac13 + \overbrace{1 + 1 + \cdots}^{d\ \text{times}}}{d+2}.$ Multiplying by the denominator, $(d+2)a_d = 1\tfrac13 + d.$ We want $a_d$ to become $\tfrac9{10}$ , so we solve $\tfrac9{10}(d+2) = 1\tfrac13 + d \\ \tfrac9{10}d + 1\tfrac45 = 1\tfrac13 + d \\ 1\tfrac45 - 1\tfrac13 = \tfrac1{10}d \\ 18 - 13\tfrac13 = d \\ d = 4\tfrac23.$ For higher values of $d$ , the average will be greater, of course. For a realistic answer we therefore round off to $d = 5$ .

Now for the tricky part: The fifth day after Tuesday is Sunday; but on Saturday and Sunday there is no gym. Therefore we must add two more days; bill will get his average above 90% on the $\boxed{{7}^\text{th}}$ day of the term.

Gym grades for second term seniors

1 solution

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