$a(t) = xt^{2} + yt + z $

$100a(2) = 152$ and $200a(4) = 240$

Using that, I found that $z = 1, $ $x = -0.105 $, and $y = 0.47$

The question asks for the accumulated value at time 8 of 1600 invested at time 6. So we need to find what was invested at time 0 for it to equal 1600 at time 6:

$1600= P(-0.105*6^{2} + 0.47*6 + 1) $

=> $P = 40,000$

This is already wrong... But I don't know why. I can't find A(8) without that P.


1 Answer 1


Actually everything you've done so far is right (however there's technically no need to calculate P, since the answer would simply be $1600\times\dfrac{A(8)}{A(6)}$).

With that in mind, if you analyze $A(t)$ with the coefficients you've already found, i.e. $A(t)=-0.105t^2+0.47t+1$, you'll notice there is a turning point at $t=2.238$, that $A(t)$ is decreasing for $t>2.238$, and that there is a zero at $t=6.05$. If you check, you'll get $A(6) = 0.04$, which matches up with what you found above (i.e. it gives $P=40,000$). Your conceptual error seems to be that you're assuming $A(t)$ must always be greater than 1 (which would usually be the case, but definitely isn't in this question).

So given all of the above, we can quite easily find that $A(6)=0.04$, and $A(8)=-1.96$, and therefore the answer is $1600\times\dfrac{A(8)}{A(6)}=1600\times\dfrac{-1.96}{0.04}=-\$78,400$

  • $\begingroup$ The answer I was given is $2057.14. I guess I will assume they made a typo. Also, why is the answer 1600*A(8)/A(6). How did you rearrange the formula to get that? $\endgroup$
    – Bob
    Feb 4, 2015 at 17:27
  • $\begingroup$ Hopefully I'm not jumping the gun on your learning here, but we can represent A(t) as A(0,t) as well, which indicates that it is the accumulation factor from time 0 to time t. Now if you think of A(2), you can also write it as A(0,1)*A(1,2) - that's how the accumulation works. So in general, A(0,t)=A(0,s)*A(s,t). We can rearrange that to get A(s,t)=A(0,t)/A(0,s), or in words, the accumulation factor from time 6 to time 8 is A(0,8)/A(0,6)=A(8)/A(6). So the 1600*A(8)/A(6) actually comes from 1600*A(6,8), which should seem much more logical. Hope that helps. $\endgroup$
    – mardat
    Feb 4, 2015 at 22:23

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