At Mr.Vid's Rentals you can join his VID Club for a monthly membership fee of $30.Club members pay only $.85 to rent each movie. If you are not a member of the VID Club, the movies rent for $2.25 each.How many videos would you have to rent each month to make a membership worth the investment?
Math - Henry, Wednesday, September 29, 2010 at 9:26pm
$30 / $.85 = 35.3 or 36 videos per m0.
Math (Correction) - PsyDAG, Thursday, September 30, 2010 at 1:50pm
30 + .85x < 2.25x
Solve for x.
Please only post your questions once. Repeating posts will not get a quicker response. In addition, it wastes our time looking over reposts that have already been answered in another post. Thank you.
Math - Jonathan, Thursday, January 24, 2013 at 8:12pm
I did this problem for homework so here is my homework lol but remember DON'T COPY THAT IS AGAINST THE LAW!
So I know the problem or question it is asking is “How many videos would you have to rent each month to make a membership worth investment?” and that's the gist now I just need to figure it out.
What I know:the monthly fee is $30, club members pay $0.85 to rent each, non members pay $2.25 to rent each.
What I need to know:# of videos to rent each month to make it worth the investment.
The strategy i'm using for this problem is setting it up in a simple mathematical solution.
30 + .85n <= 2.25n and solving it from there.
Solving The Problem:
so I started with 30 + .85n <= 2.25n and n=# of movies rented. Then I reduced it to 30<1.4n by doing 2.25-.85 and then plugging it back in to the equation. Then I reduced that to 21.4<n by doing 30÷1.4 and then plugging the answer back in to get a final answer of 21.4<n. Thus proving my answer is correct.
So the answer to the original problem “How many videos would you have to rent each month to make a membership worth investment?” is if u rent at least 22 videos pur month then u should join the club. This is a reasonable answer to the question because based on all the facts that I have shown on solving the problem proves that the answer is right making my answer a reasonable answer.