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Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion


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September 3rd, 2016, 11:25 PM   #1
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financial math

My brother asked me to help him solve this problem. But I know nothing about financial math. I am going to write it down below.

(ex) How many monthly investments of $345 would have to be deposited into a savings account that pays 3% annual interest, compounded monthly, to obtain an amount of 10000 dollars?

The answer is 28.

I have no idea of how to approach it. Please help my brother and me. Thank you very much.
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September 4th, 2016, 05:01 AM   #2
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there is a formula for this sort of problem. It's complicated and won't really help you grasp what's going on.

If you are interested you can read about it here

http://www.thecalculatorsite.com/art...ula.php?page=2

what you can do is put this into excel as follows

the first cell, a1, is the initial payment \$345

the next cell is (1+r')a1+\$345 where r'=r/12, in this case 0.03/12 = 0.0025

and similarly cell a(n+1) = (1+r')a(n) + 345

you end up with a table like this, where the number on the left is the month number

0 \$345.00
1 \$690.86
2 \$1,037.59
3 \$1,385.18
4 \$1,733.65
5 \$2,082.98
6 \$2,433.19
7 \$2,784.27
8 \$3,136.23
9 \$3,489.07
10 \$3,842.80
11 \$4,197.40
12 \$4,552.90
13 \$4,909.28
14 \$5,266.55
15 \$5,624.72
16 \$5,983.78
17 \$6,343.74
18 \$6,704.60
19 \$7,066.36
20 \$7,429.03
21 \$7,792.60
22 \$8,157.08
23 \$8,522.47
24 \$8,888.78
25 \$9,256.00
26 \$9,624.14
27 \$9,993.20
28 \$10,363.18
29 \$10,734.09
30 \$11,105.93

and you can see that you cross the $10k line at month 28
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September 4th, 2016, 11:49 PM   #3
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Thank you very much, romsek. I really appreciate your help.
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September 5th, 2016, 02:09 AM   #4
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It's unsatisfactory to assert "you can see" without explaining how that is seen. Strictly speaking, the initial investment of \$345 suffices if you then wait long enough for the interest to accumulate. The problem doesn't state that further investments must be made.
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September 5th, 2016, 04:24 AM   #5
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I think you have to solve the equation

345.(1+0.03)^x=10000.

we find x=114 years !
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September 5th, 2016, 05:01 AM   #6
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Quote:
Originally Posted by abdallahhammam View Post
I think you have to solve the equation

345.(1+0.03)^x=10000.

we find x=114 years !
This calculates the amount of time a single payment of \$345 would take to mature to \$10000 if interest were paid once per year, not compounded monthly.

Read the original question again ...

Quote:
(ex) How many monthly investments of $345 would have to be deposited into a savings account that pays 3% annual interest, compounded monthly, to obtain an amount of 10000 dollars?
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September 5th, 2016, 05:21 AM   #7
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That's not correct. The 345 is deposited MONTHLY.

i = .03/12
p = 345
f = 10000

k = fi/p + 1

n = log(k) / log(1+i)

n = 28.01884....

That's assuming 1st 345 made at END of 1st month.
Romsek's assumes 1st 345 at beginning of 1st month.
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September 11th, 2016, 06:42 AM   #8
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Thanks everyone.
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