Answer: C) You choose account #2
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Explanation:
Let's compute the final balance for account 1.
We'll use the compound interest formula
A = P*(1+r/n)^(n*t)
where,
- P = principal or amount deposited
- r = decimal form of the annual interest rate
- n = number of times we compound the interest per year
- t = number of years
where in this case
- P = 10,000
- r = 0.035
- n = 4
- t = 10
So with all that in mind, we get
A = P*(1+r/n)^(n*t)
A = 10,000*(1+0.035/4)^(4*10)
A = 14,169.0883793113
A = 14,169.09
Account #1 gets us $14,169.09 which is over our goal of $14,000
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Let's repeat this idea for account 3. I'll get back to account 2 in the next section.
For account 3, we have these inputs
- P = 10,000
- r = 0.032
- n = 1
- t = 10
leading to...
A = P*(1+r/n)^(n*t)
A = 10,000*(1+0.032/1)^(1*10)
A = 13,702.4104633564
A = 13,702.41
This isn't larger than 14,000. So we cannot use account 3. The interest rate is too small and needs to be bumped up.
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Now let's compute account 2. We'll use the continuously compounded interest formula which is
A = P*e^(rt)
The P, r and t are the same as before. The 'e' is a constant and it's roughly e = 2.718... similar to how pi = 3.14... approximately.
I recommend using the calculator's stored value of 'e' to get as much accuracy as possible.
So,
A = P*e^(rt)
A = 10,000*e^(0.036*10)
A = 14,333.2941456052
A = 14,333.29
This exceeds 14,000 so we can use this account.
Comparing this to account 1's value (14,169.09), we can see that account 2 gets us the most money over the 10 year time period. Not only does account 2 have the higher interest rate, but it also has an infinitely higher compounding frequency as well. These two factors help contribute as to why account 2 beats out account 1.
This is why we should go for account 2.
