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September 3rd, 2016, 11:25 PM  #1 
Senior Member Joined: Apr 2008 Posts: 176 Thanks: 3  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. 
September 4th, 2016, 05:01 AM  #2 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,488 Thanks: 749 
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 
September 4th, 2016, 11:49 PM  #3 
Senior Member Joined: Apr 2008 Posts: 176 Thanks: 3 
Thank you very much, romsek. I really appreciate your help.

September 5th, 2016, 02:09 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 18,037 Thanks: 1394 
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.

September 5th, 2016, 04:24 AM  #5 
Senior Member Joined: Aug 2016 From: morocco Posts: 273 Thanks: 32 
I think you have to solve the equation 345.(1+0.03)^x=10000. we find x=114 years ! 
September 5th, 2016, 05:01 AM  #6  
Math Team Joined: Jul 2011 From: Texas Posts: 2,639 Thanks: 1319  Quote:
Read the original question again ... Quote:
 
September 5th, 2016, 05:21 AM  #7 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 10,668 Thanks: 697 
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. 
September 11th, 2016, 06:42 AM  #8 
Senior Member Joined: Apr 2008 Posts: 176 Thanks: 3 
Thanks everyone.


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