1. ## PreCalculus Challenge puzzler

i received this problem and worked through it, but my solution is flawed! it seems to be a pretty backwards solution.

The Problem
While watching a basketball game one day on television, I saw the following situation:

A player was fouled as he drove for the basket. At the bottom of the screen, it was stated that "Player X is making 78% of his free throws". The player missed the first free throw and made the second.

Later in the game, the player was again fouled. This time, at the bottom of the screen, it said "Player X is making 76% of his free throws". The statistic had been updated after his first two attempts.

Recognizing that the percentages presented on the screen were rouded to the nearest 1%, how many free throws had player X attempted and how many had he made? Be sure to find all possible solutions (if any).

First, I named things.

# Attempted at 2nd stat=f+2

# Made @ 1st stat + # made in game = 76% of attempts @ 2nd stat

0.78f+1=0.76(f+2)
SOLVE FOR F (# attempted @ 1st stat)
f=26

SO...
26 attempted at first stat
28 attempted at second stat

Right?? well...not really...

The Issue

78% of 26=20.28
20/26 rounds to 0.77 (wrong percentage)
76% of 28=21.28
21/28 rounds to 0.75 (wrong again)

the solution works when rounded from percentages to shots, but not from shots to percentages.

I need to find a solution in which the set considered for shots is whole numbers and I think I need a solution with graphs and zeros so I can be sure I can solve for all possible solutions (if there is more than one).

I'm stumped so any help is appreciated!!

-Mike

2. by the way I posted this in urgent help but I think it was moved...anyway this is sort of urgent, I need a fully worked out solution soon

so anyone who can help would be appreciated!!

3. Let $\displaystyle x$ be the total number of shots he made out of $\displaystyle n$ shots before he made the two shots.

thus, $\displaystyle \frac xn = 0.78$ ..................(1)

also, after he made the two shots, he made $\displaystyle n + 2$ shots in total. since he missed one, the number of shots he made becomes $\displaystyle x + 1$. at the end, he made 76% of his shots, so that

$\displaystyle \frac {x + 1}{n + 2} = 0.76$ ...................(2)

thus you have two equations and two unknowns. you can solve for $\displaystyle x$ and $\displaystyle n$

4. wow I made that so complicated - thanks a lot for the help.

-Mike

5. when solved, I got

n=38
x=29.64

same rounding problem occurs

30/38=0.789
29/38=0.763

31/40=0.775
30/40=0.75

???

solving this backwards creates problems, I need a new method! any help is appreciated as I am very desperate!

6. Originally Posted by mikedwd
when solved, I got

n=38
x=29.64

same rounding problem occurs

30/38=0.789
29/38=0.763

31/40=0.775
30/40=0.75

???

solving this backwards creates problems, I need a new method! any help is appreciated as I am very desperate!
oh yes, sorry, it should be n + 2 in the denominator of my last fraction, not n + 1

now those equations give you the right number

7. Originally Posted by Jhevon
oh no. you made a mistake. the solution will be the same as the one you got. you got the right answer. the slight discrepancy comes from the "rounding up" they did, we should be off by about 1% remember?
Yes but how is this rounding problem solved? the problem states that it will be rounded to the nearest 1%, which is a terrible coincidence for the numbers given (actually, I think the teacher did this on purpose to force us to use a different method to solve the problem)

Originally Posted by Jhevon
oh yes, sorry, it should be n + 2 in the denominator of my last fraction, not n + 1

now those equations give you the right number
oh yeah I thought that but I solved through anyway to see if it would be right, let me try this again

8. Originally Posted by mikedwd
Yes but how is this rounding problem solved? the problem states that it will be rounded to the nearest 1%, which is a terrible coincidence for the numbers given (actually, I think the teacher did this on purpose to force us to use a different method to solve the problem)
see my previous post. i made a correction

9. Yes, and I evaluated it.

x=59.28
n=76

59/76=0.776 (good)
60/78=0.769

The methodology is still backwards! i thought i had arrived at a correct answer

10. Originally Posted by mikedwd
Yes, and I evaluated it.

x=59.28
n=76

59/76=0.776 (good)
60/78=0.769

The methodology is still backwards! i thought i had arrived at a correct answer

how did you get n = 76?

11. Originally Posted by mikedwd
i received this problem and worked through it, but my solution is flawed! it seems to be a pretty backwards solution.

The Problem
While watching a basketball game one day on television, I saw the following situation:

A player was fouled as he drove for the basket. At the bottom of the screen, it was stated that "Player X is making 78% of his free throws". The player missed the first free throw and made the second.

Later in the game, the player was again fouled. This time, at the bottom of the screen, it said "Player X is making 76% of his free throws". The statistic had been updated after his first two attempts.

Recognizing that the percentages presented on the screen were rouded to the nearest 1%, how many free throws had player X attempted and how many had he made? Be sure to find all possible solutions (if any).
You have a pair of inequalities defined by this problem:

$\displaystyle 0.775 \le \frac{m}{a} < 0.785$

and:

$\displaystyle 0.755 \le \frac{m+1}{a+2} < 0.765$

Each of these defines a region between a pair of lines in the first quadrant, and all integer grid points in the region where these regions overlap are feasible solutions. This is a closed region in the form of a parallelogram.

RonL

12. This makes good sense, and solves my rounding problem

I'll try out some math on it (and not make any silly mistakes )

oh and should those "less than" and "less than or equal to" signs be switched around?

13. I need a little help solving this however.

I put all the inequalities in terms of m (so 4 inequalities)

so this is what I had:

m<0.785a
m(>or=)0.775a

m<0.765a+0.53
m(>or=)0.755a+0.51

(i switched around the < and the (<or=) symbols because i thought you made a typo)

i used the inequality application on my TI-84 Plus and graphed where the shades intersect, but I'm not sure how to find integer points within the defined region

14. Originally Posted by mikedwd
This makes good sense, and solves my rounding problem

I'll try out some math on it (and not make any silly mistakes )

oh and should those "less than" and "less than or equal to" signs be switched around?
Yes, that is the usual convention.

RonL

15. Originally Posted by mikedwd
I need a little help solving this however.

I put all the inequalities in terms of m (so 4 inequalities)

so this is what I had:

m<0.785a
m(>or=)0.775a

m<0.765a+0.53
m(>or=)0.755a+0.51

(i switched around the < and the (<or=) symbols because i thought you made a typo)

i used the inequality application on my TI-84 Plus and graphed where the shades intersect, but I'm not sure how to find integer points within the defined region
I would do it by search, the largest value that $\displaystyle a$ can take is less than $\displaystyle 54$, and the smallest more than $\displaystyle 16$ (find the points of intersection of the lines defining the feasible region to show this).

So for each $\displaystyle a$ between $\displaystyle 17$ and $\displaystyle 53$ I would check which values of $\displaystyle m$ give points in the feasible region (they will all be close to $\displaystyle 0.77a$).

RonL

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