Cedagmat wrote:205. Mike invested a total of $ 5,000 at 5 % simple annual interest rate for n years. Is n > 4?
S1. At the end of n years, Mike's investment plus interest was more than $ 5, 500.
S2. At the end of n years, Mike's investment plus interest was less than $ 6,500.
[spoiler]Answer: E[/spoiler]

Set up the equation A=P(1+r/n)^(nt)
S1: 5,500=5000(1+0.05)^n
1.1=1.05^n : Solve for N

S1: 6,500=5000(1+0.05)^n
1.3=1.05^n : Solve for N

I thought that in both cases since we could solve for N that the answer would be D. Another follow up question is that I do not know how to solve for n in this case. Anyone can help?

First, this problem is about "simple interest", and not "compound interest". You're using the compound interest formula.

When interest is 'simple', then the interest is only applied to the initial investment, and not to any accumulated interest. So in this question, each year Mike earns 5% on his $5000 investment, and therefore earns $250 each year. So after n years, he will have 5000 + 250n dollars.

Second, the statements do not give you equations; they give you inequalities. Statement 1 tells you he has *more than* $5500 after n years. Since his investment will be worth 5000 + 250n after n years, from this statement we learn that 5000 + 250n > 5500. From that you can determine that n > 2. Similarly, the second statement tells you that 5000 + 250n < 6500, or that n < 6. So combined we know that 2 < n < 6, but we don't know whether n > 4, so the answer is E.

You could also just list the values of his investment each year:

year 0: $5000
year 1: $5250
year 2: $5500
year 3: $5750
year 4: $6000
year 5: $6250
year 6: $6500

from which you can see that the two statements are not sufficient; n could be 3, 4 or 5.

And to answer your followup question, you won't be asked to solve for n in an equation like 1.3=1.05^n, because you can't do that without a calculator.