**Algebra II**exam, but I've received enough requests that I've decided to post some of the open-ended questions.

Some questions were posted earlier here, here, here and here

### August 2016, Algebra II Part 4

**37.*** Seth’s parents gave him $5000 to invest for his 16th birthday. He is considering two investment options. Option A will pay him 4.5% interest compounded annually. Option B will pay him 4.6% compounded quarterly.
Write a function of option A and option B that calculates the value of each account after n years.
*

*Seth plans to use the money after he graduates from college in 6 years. Determine how much more money option B will earn than option A to the nearest cent.
*

*Algebraically determine, to the nearest tenth of a year, how long it would take for option B to double Seth’s initial investment.*

Not an overly difficult Part 4 question, but it is a little involved.

For the first part, be sure to write a function and not just an expression. The initial amount is $5000, the rates are 0.045 and 0.046, respectively. Because the second one is quarterly, you have to divide the rate by 4 and multiple the exponent, n, by 4. (You would not be incorrect if you did this in the first function, using 1.)

A = 5000(1 + 0.045)^{n}

B = 5000(1 + 0.046/4)^{4n}

Use the two functions you just wrote, with n = 6. Write down both amounts and then subtract.

A = 5000(1 + 0.045)^{6} = 6511.30062424 = 6511.30

B = 5000(1 + 0.046/4)^{(4)(6)} = 6578.86985121 = 6578.87

6578.87 - 6511.30 = **67.57**

Doubling the $5000 investment means the future value will be $10000. Put this in the second function, and solve for n.

10000 = 5000(1 + 0.046/4)^{4n} -- divide both sides by 5000

2 = (1 + 0.046/4)^{4n} -- change the rational expression into a decimal

2 = (1.0115)^{4n} -- solve using logs

4n = log_{1.0115}2 -- put this into your calculator

4n = 60.6196... -- divided by 4

n = 15.1549

In 15.2 years, Seth's investment will double.

*End of exam*

Any questions?

Thoughts, comments, concerns?

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