<span>On the day of a child's birth, a sum of money is to be invested into a certificate of deposit (CD) that draws 6.2% annual interest compounded continuously. The plan is for the value of the CD to be at least $20,000 on the child's 18th birthday. </span>
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The substitution<span> method is used to eliminate one of the variables by replacement when solving a system of equations </span> <span> </span>
t = the number of years r = the interest rate A = the amount you have after 't' years
In this problem ...
't' is 18 years
'r' is 6.2% . . . that's 0.062
we want 'A' to be $20,000 by that time.
The question is: What is 'P' ? That's the amount we have to put in now.
20,000 = P e^(0.062 x 18)
20,000 = P e^(1.116)
Divide each side by e^(1.116): P = 20,000 / e^(1.116)
There's a key on your calculator for e^x . Punch in 1.116 and then hit that key, and you get 3.052 .
So P = 20,000 / 3.052 = $6,551.75
That's the real solution, but the question wants us to tweak it. It asks: "In units of $1,000, what's the minimum that must be invested ?"
$6,000 wouldn't be enough. So $7,000 must be invested now. _____________________________
I can do the second one, but I'm sure there' a much spiffier, cooler, better way to do it. I never learned this stuff in class, so here's my way, for what it's worth:
The money is all compounded continuously, so we'll use the same formula:
</span></span></span> <span>A = P (e^<span>rt)
t = the number of years r = the interest rate A = the amount you have after 't' years
He's going to put in $10,000 at the beginning of every month, for one year.
The original amount ( P ) is the same for each deposit . . . 10,000 . The interest rate is the same for each deposit . . . 7.6% = 0.076 .
What's different for each deposit ? Only the length of time it stays in the bank !
The first 10,000 stays in for 1 whole year. The second one stays in for only 11 months. t = 11/12 of 1 year. The third one stays in for 10 months. t = 10/12 of 1 year. . . The last deposit stays in for only 1 month. t = 1/12 of 1 year.
So it looks to me like we have to calculate the formula 12 times ...
once for t = 1/12, 2/12, 3/12, 4/12, 5/12, . . . . . 11/12, and 1 .
I'll do that over here on a piece of paper, but I'm only going to type the first one and the last one. If you need um, you can do um.
First deposit: A = 10,000 e^(0.076 x 1) = $10,789.62 . (it's in the bank . a whole year) 2nd deposit A = $10,721.50 3rd deposit A = $10,653.81 4th deposit A</span></span><span><span> = $10,586.55 </span>5th deposit </span><span><span>A = $10,519.72 </span>6th deposit </span><span> A = $10,453.30 7th deposit </span><span> A = $10,387.31 8th deposit </span><span> A = $10,321.73 9th deposit </span><span>A = $10,256.56 10th deposit </span><span> A = $10,191.81 11th deposit </span><span> A = $10,127.47 Last deposit : A = 10,000 e^(0.076 x 1/12) = $10,063.53 </span><span><span>(it's only in the bank for one month.)</span>
At the end of the year, each deposit is worth a little less than the one before it, because it's been in the bank a month less than the one before it.
So he deposited a total of (12 x $10,000) = $120,000 in the bank during the year. The amount he has at the end of the year (if he never goes to the bank at any time that year to take some out) is the sum of all the A's . I'm sure you can addum up without any problem. But they make me do everything around here and I'm, used to it, so I'll take a crack at it:
I get $125,072.91 at the end of the year.
I'm sloppy with my calculator so you ought to check this.</span>
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