Two investments and accumulation function

Question

Suppose that $a(t) = 0.1t^2 + 1$. At time 0, 1000 is invested. An additional investment of X is made at time 6. If the total accumulated value of these two investments at time t=8 is 18800, what is X?

My work: From original investment, at time t=6, it is valued at$A(6) = 1000a(6)= 4600$. Thus, we solve for $A(8) = 18800 = (4600+X) \frac{a(8)}{a(6)}?$

The correct answer is 6589.

The value of $X$ you have obtained in your method is not correct. I'm pointing this out because it's not immediately clear from the question that you mean this. — Toby Mak, Jul 02 '20 at 11:22

score 1 · Accepted Answer · answered Jul 02 '20 at 12:39

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Yes, you are right. But only if " the total accumulated value of these two investments at time $t=8$ is $\color{red}{18000}$". The equation is

$$(4600+X)\cdot (0.1\cdot 8^2+1)/(0.1\cdot 6^2+1)=18000$$

$$4600+X=18000\cdot (0.1\cdot 6^2+1)/(0.1\cdot 8^2+1)$$

$$X=18000\cdot (0.1\cdot 6^2+1)/(0.1\cdot 8^2+1)-4600=6589.\overline{189}\approx 6589$$

answered Jul 02 '20 at 12:39

callculus42

30,550

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oof thats a dumb mistake.. what happens when u study at 3am – MinYoung Kim Jul 02 '20 at 22:37
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The good news is that you have understood the subject in principle. (Posted at 1am) – callculus42 Jul 02 '20 at 23:06

Two investments and accumulation function

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