Results 1 to 3 of 3

Math Help - need help please!

  1. #1
    Newbie
    Joined
    May 2008
    Posts
    11

    Wink need help please!

    Machine A can be bought for $15,000, but machine B, the automated model, will cost $40,000 to buy and install. The automated machine will reduce annual cost by $5743.50 over machine A. Either machine will have a service life of seven years. The rate of return on the extra investment required by buy machine B is
    Follow Math Help Forum on Facebook and Google+

  2. #2
    A riddle wrapped in an enigma
    masters's Avatar
    Joined
    Jan 2008
    From
    Big Stone Gap, Virginia
    Posts
    2,551
    Thanks
    12
    Awards
    1
    Quote Originally Posted by JAMESANANE View Post
    Machine A can be bought for $15,000, but machine B, the automated model, will cost $40,000 to buy and install. The automated machine will reduce annual cost by $5743.50 over machine A. Either machine will have a service life of seven years. The rate of return on the extra investment required by buy machine B is
    Someone with more business saavy might approach this differently, but here goes:

    The $40,000 purchase will more than pay for itself in 7 years considering the annual reduction in operating costs (7 X $5,743.50 = $40,204.50).

    The return on investment (ROI) given just that information would be
    \frac{40204.50-40000}{40000}=.005 = .5\%

    If you figured from just the extra investment made ($40,000-$15,000=$25,000), then the ROI would be

    \frac{40204.50-25000}{40000}=.38=38\%
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor ebaines's Avatar
    Joined
    Jun 2008
    From
    Illinois
    Posts
    1,050
    Thanks
    288
    1. The rate of return on the extra investment is merely the extra profit you get from it each year divided by the additional initial investment. So that's 5743/25,000 = 23%

    2. The formula for compounding interest is:

    <br />
FV = PV*(1+i)^n<br />

    So here you have
    <br />
7500 = 5000*(1+0.05)^n<br />
    or
    <br />
1.5 = 1.05^n<br />

    Now solve for n. The way to do that is using logarithms. Take the log of both sides:

    <br />
log(1.5) = log(1.05^n)<br />

    Recall from highschool that  log(a^b) = b \cdot log(a) , so you have

    <br />
log(1.5) = n \cdot log(1.05) , and hence
    <br />
n = \frac {log(1.5)} {log(1.05)} = 8.3\ \text{years.}<br />
    Follow Math Help Forum on Facebook and Google+


/mathhelpforum @mathhelpforum