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I am really stuck with example 2-5 and 2-6. I don't really understand example 2-6 and example 2-5 I just can't figure out...I was able to do example 2-4 which was easy...

For example 2-4 I did F=P(1+ interest)^n


I solved for P and got the deposit value...Example 2-5 or 2-6 are different and don't work the same way so I am not sure what to do here..

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2 Answers 2

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For example 2-5: you want

$$ 6500 + Q = P (1.03)^4$$


$$3000 = Q (1.03)$$

so that

$$ = 6500 + \frac{3000}{1.03} = P (1.03)^4$$

Solve for $P$.

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thank you so much for explaining this to me....This is the first time I am having to take finance course and it is a pain in the rear to deal with these sort of questions....Normally I am doing calculus or algebra questions.. –  Raynos Jan 10 '13 at 16:52
Your welcome. Just think these through; interest questions like these I find easy because they are calculations I may do in real life anyway. –  Ron Gordon Jan 10 '13 at 17:05

2-5: $\frac{6500}{1.03^4} + \frac{3000}{1.03^5}$

2-6: $900,000 + 200,000[\frac{1}{1.04} + \frac{1}{1.04^2} + \frac{1}{1.04^3} + \frac{1}{1.04^4} + \frac{1}{1.04^5}]$ assuming costs due at end of each year and interest paid annually

2-7: $500,000 + 300,000[\frac{1}{1.04} + \frac{1}{1.04^2} + \frac{1}{1.04^3} + \frac{1}{1.04^4} + \frac{1}{1.04^5}]$

These arent good questions

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for 2-6 I am a bit confused...I have 900,000+ (200,000+Q)*(5/1.04) The reason I have it like this is because I am thinking after end of each year you get a return of 200,000 which you must pay but you also need to get back your amount you submitted at begining of year to make you the 200,000 for the next year...Because of this the below equation to me is the one for each year's output Year 1- 200,000+Q=Q(1.04) Year 2- 200,000+Q=Q(1.04) year 3- 200,000+Q=Q(1.04).........all the way to year 5..End of each year you must get 200k but extra to invest again for next year. –  Raynos Jan 11 '13 at 16:26
from the equation you have with the 1/0.04 it means that you are adding each year's interest and return without any subtraction of the 200k you must pay each year...The question says that basically end of each year you must have enough to pay 200k...So I am confused about that now.... –  Raynos Jan 11 '13 at 16:31
The $200,000 (\frac{1}{1.04})$ term will grow to 200,000 after one year and then get fully paid out. The $200,000 (\frac{1}{1.04^2})$ term will grow to 200,000 after two years before vanishing, and so on. I am just discounting each future payment obligation back to today. –  TheMathemagician Jan 11 '13 at 16:43
you are right in what you say, after you one year you have 200k to pay off, next year you have 200k to pay off....@TheMathemagician what I am saying is that you SHOULD not only have 200k at the end of year to pay but MORE because you have to REINVEST that remaining amount you have for it to grow till next year...Each year your return should be 200k + some additional number and this additional number is the remain after you pay 200k and you reinvest that remain to make 200k + remain for next year and so on... –  Raynos Jan 12 '13 at 22:15
Yes and if you work through the formula you'll see that's exactly what happens. After one year the first term will have grown to 200,000 and be paid. The other terms are all increased by 1.04. Now after another year the second term will now have grown to 200,000 and it gets paid off etc. By the fifth year only the final term will be left. –  TheMathemagician Jan 14 '13 at 10:10

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