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lennon3

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Apr 11, 2006
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I am stuck on a couple math questions, from my practice test for the final so I'll post them on here and see if you guys can figure out how to set them up. I have the answers because it shows just the answer once you miss it but it doesn't show how to do it.

1. An individual retiremant account (ira) has $16,000 in it, the owner decides not to add any more money to the account other than interest earned at 7% compounded daily. How much will be in the account 31 years from now when the owner reaches retirement age?

The answer is 140,103.39

2. How many 8-letter code words are possible from the first 12 letters of the alphabet if no letter is repeated? If the letters can be repeated? If adjacent letters must be different?

A. if no letter is repeated- 19,958,400
B. If letters can be repeated- 429,981,696
C. If adjacent letters must be different- 233,846,052

Thanks to anyone who understands how to set these problems up!
 
2a) 12*11*10*9*8*7*6*5...you have 12 letters to pick from, then 11, then 10, etc.
b) 12^8...you can pick from all 12 in any spot
c) 12*11*11*11*11*11*11*11...you can pick from any of the numbers at first, but you can't select that number again, so you have 11 to pick from.
 
lennon3 said:
I am stuck on a couple math questions, from my practice test for the final so I'll post them on here and see if you guys can figure out how to set them up. I have the answers because it shows just the answer once you miss it but it doesn't show how to do it.

1. An individual retiremant account (ira) has $16,000 in it, the owner decides not to add any more money to the account other than interest earned at 7% compounded daily. How much will be in the account 31 years from now when the owner reaches retirement age?

The answer is 140,103.39

2. How many 8-letter code words are possible from the first 12 letters of the alphabet if no letter is repeated? If the letters can be repeated? If adjacent letters must be different?

A. if no letter is repeated- 19,958,400
B. If letters can be repeated- 429,981,696
C. If adjacent letters must be different- 233,846,052

Thanks to anyone who understands how to set these problems up!
Click to expand...

Okay, I can help you on number 2.

A. if no letters are repeated, for your first letter you have 12 options, for you next letter you have 11, then for the 3rd letter 10 options.... so you are looking at 12x11x10x9x8x7x6x5=19,958,400.

B. if letters can be repeated, then you have 12 options for each letter. 12x12x12x12x12x12x12x12=429,981,696.

C if adjacent letters must be different, then you have 12 choices for the first letter, then 11 for the letters after that.
12x11x11x11x11x11x11x11=233,846,052
 
Try this for number 1

M = P( 1 + i )^n ---- this is the quantity 1+i taken to the n'th power... not sure how to type the exponent.

M is the final amount including the principal.

P is the principal amount.

i is the rate of interest per time period.

n is the number of time periods invested.

You didn't mention it in the problem, but I would guess that 7% is an annual rate, so you would need to convert it to a daily rate if that is the case. And then n would be the number of days you are invested and P would be the initial 16k.
 
lennon3 said:
I am stuck on a couple math questions, from my practice test for the final so I'll post them on here and see if you guys can figure out how to set them up. I have the answers because it shows just the answer once you miss it but it doesn't show how to do it.

1. An individual retiremant account (ira) has $16,000 in it, the owner decides not to add any more money to the account other than interest earned at 7% compounded daily. How much will be in the account 31 years from now when the owner reaches retirement age?

The answer is 140,103.39

2. How many 8-letter code words are possible from the first 12 letters of the alphabet if no letter is repeated? If the letters can be repeated? If adjacent letters must be different?

A. if no letter is repeated- 19,958,400
B. If letters can be repeated- 429,981,696
C. If adjacent letters must be different- 233,846,052

Thanks to anyone who understands how to set these problems up!
Click to expand...

1. A = future amount
P = current principal = 16,000
r = annual compunding rate = 0.07
N = number of times compounded/year = 365
t = number of years = 31

A = P((1+ (r/N))^(Nt))

2. A) number of permutations of 8 letters from a set of 12. = 12!/(12 - 8)!
B) 12^8
C)12* (11^7) : 12 choices for first letter, 11 for each of the last 7.
 
Last edited:
dornstar44 said:
1. A = future amount
P = current principal = 16,000
r = annual compunding rate = 0.07
N = number of times compounded/year = 365
t = number of years = 31

A = P((1+ (r/N))^(Nt))
Click to expand...

This. 16,000 * (1+.07/365)^(31*365) = 140,103.39
 
IcSyU said:
This. 16,000 * (1+.07/365)^(31*365) = 140,103.39
Click to expand...

Alright thanks a lot guys, you helped a lot... I understand the second question completely and should be able to get the first one after I read over what you guys said again lol.
 
lennon3 said:
Alright thanks a lot guys, you helped a lot... I understand the second question completely and should be able to get the first one after I read over what you guys said again lol.
Click to expand...


For the first one, just make sure everything is using the same time period. Since the interest compounds daily, you need to convert the 31 years to days (365*31) and divide the APR by 365. Once everything's in the same units, you can just use F=P(1+i)^n. I think of it as future value=present value (1+i)^n, if you want to find the present value of a future return, use the same equation, but to the (-n) power
 
cs6804 said:
for #1 don't forget about leap years
Click to expand...

For leap years, the daily rate changes to offset the extra day, because the 7% annual rate is constant whether it's a leap year or not. Making this assumption may not be exact, but it's going to be a pretty doggone close approximation (well beyond any significant figures the professor could want).
 
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dornstar44 said:
For leap years, the daily rate changes to offset the extra day, because the 7% annual rate is constant whether it's a leap year or not. Making this assumption may not be exact, but it's going to be a pretty doggone close approximation (well beyond any significant figures the professor could want).
Click to expand...

In all my works with economics (engineering economics), 1 year= 365 days (even on the FE). I would use this on your exam, but if you feel your prof might question you on it, just write down this assumption. However, you should be fine.
 
lennon3 said:
I am stuck on a couple math questions, from my practice test for the final so I'll post them on here and see if you guys can figure out how to set them up. I have the answers because it shows just the answer once you miss it but it doesn't show how to do it.

1. An individual retiremant account (ira) has $16,000 in it, the owner decides not to add any more money to the account other than interest earned at 7% compounded daily. How much will be in the account 31 years from now when the owner reaches retirement age?

The answer is 140,103.39

2. How many 8-letter code words are possible from the first 12 letters of the alphabet if no letter is repeated? If the letters can be repeated? If adjacent letters must be different?

A. if no letter is repeated- 19,958,400
B. If letters can be repeated- 429,981,696
C. If adjacent letters must be different- 233,846,052

Thanks to anyone who understands how to set these problems up!
Click to expand...
I have never solved examples myself.
 
full
 
lennon3 said:
I am stuck on a couple math questions, from my practice test for the final so I'll post them on here and see if you guys can figure out how to set them up. I have the answers because it shows just the answer once you miss it but it doesn't show how to do it.

1. An individual retiremant account (ira) has $16,000 in it, the owner decides not to add any more money to the account other than interest earned at 7% compounded daily. How much will be in the account 31 years from now when the owner reaches retirement age?

The answer is 140,103.39

2. How many 8-letter code words are possible from the first 12 letters of the alphabet if no letter is repeated? If the letters can be repeated? If adjacent letters must be different?

A. if no letter is repeated- 19,958,400
B. If letters can be repeated- 429,981,696
C. If adjacent letters must be different- 233,846,052

Thanks to anyone who understands how to set these problems up!
Click to expand...

If the $16,000 is compounded DAILY at 7% you're going to have one hell of a lot more than $140,103.39. In fact, if it's compounded DAILY you'll pass $140k on day 32. You'd have nearly $850 TRILLION after just one year.
 
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A: doesn't have enough information to have an accurate answer.
B: has been solved.

3: Closing thread.
 
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