Friday, July 12, 2013

crowdsourcing a math problem

I was stumped while trying to help a student with the following math problem. I suspect the solution is easier and more obvious than my conscious mind will allow. First, I'll describe the problem. Next, I'll talk about about how I fumbled around with it. Finally, I'll open the floor to any savvy commenters who can show me a better approach and give me the proper answer. I know what the answer to the problem is; I've seen the answer key. But I won't publish the answer here: this way, any commenter who hits upon the right answer will have done so legitimately, without the post hoc reverse-engineering of his reasoning.

The problem:

Each player on a basketball team attempts 30 foul shots in one practice. The average number of baskets made per player is 20. If the players with the highest and lowest number of baskets are removed from the group, the average is 19. If the team has 6 players, what is the lowest possible number of baskets made by a player?

My fumbling reasoning:

Let's label the basketball players, from worst to best, as A, B, C, D, E, and F.

What we know:

1. There are six players.
2. All six players take 30 shots, for a total of 180 shots.
3. If "baskets" means "successful shots," then the following equation expresses the average listed in the problem:

(A + B + C + D + E + F)/6 = 20

—which means that

A + B + C + D + E + F = 120.

If the worst and best players are A and F respectively, then the average (19 baskets) for the remaining four players can be expressed as

(B + C + D + E)/4 = 19

—which means that

B + C + D + E = 76

—which further means that, if we subtract 76 from 120, then

A + F = 120 - 76 = 44.

This is where I come unglued. How do I figure out the minimum number of shots made by a player? Despite having all the above data, I have no idea how to proceed. Let's see what I can come up with.

If Player A is the worst player, then A is the one making the lowest possible number of baskets. If we assume the other players all make an equal number of shots (an assumption that may not be warranted, especially since we know there's a best player), then perhaps we can try calculating A by seeing what value of A satisfies the following equation:

(A + 5X)/6 = 20

—which leads us to

A + 5X = 120

—at which point we start plugging in numbers.

If X = 24 (i.e., the other five players all get 24 baskets each), then A = 0. It would seem that "0 baskets" would be the answer to the question, but according to the answer key, it's not. I therefore assume my reasoning is faulty at one or more links in my logical chain.

Insights welcome.

UPDATE: And then a flash of satori came to me, and I figured out the answer. No need for comments, folks; I've got it. Obviously, A can't be 0, because one of the conditions to be satisfied is that A + F = 44. If A were 0, then F would be 44. But since each player took 30 shots, no player can have a score higher than 30. This means that if F = 30, the maximum possible number of successful baskets for any one player, then A must equal 14 (30 + 14 =44), and that's the answer in the book's answer key.

God, I love thinking out loud on a blog.


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