you use $1000 to open a saving account at your local bank on the first of Februa
ID: 3029556 • Letter: Y
Question
you use $1000 to open a saving account at your local bank on the first of February. the savings account has an interest rate of 1.5% per year and compounds monthly on the last day of month. You set up an auto-deposit of $100 from your paycheck to occur on the first of each month, starting with the second month(March). During April you withdraw $250 to purchase a new gameboy.
a)who's the lender and who's the borrower.
c) construct a polynomial in x to determine the account value on July 2.
d) what's the account value on July 2?
Explanation / Answer
3. The formula for compound interest is F = p( 1+r)n where F is the future value, P is the present value, r is the rate of interest per period in decimals and n is the number of periods. Now, x = 1+r/n so that r/n = x -1 and r = n(x-1). Thus, the value of $ 1000 after 2 months will be 1000[1 +2(x -1))2 = 1000(2x -1)2. The interest is 1000(2x -1)2 - 1000 = 1000[( 2x -1)2 -1] = 1000( 4x2 - 4x) = 4000( x2 - x)= 4000x2 -4000x. Similarly, the value of $750 after 3 months will become 750[1+3(x-1)3 = 750(3x-2)3 = 750(27x3 - 54x2 + 36x - 8) = 20250x3 - 40500x2 + 27000x - 6000.Thus, the value of the initial deposit of $ 1000, after a withdrawal of $ 250 in April, is 4000x2 -4000x + 20250x3 - 40500x2 + 27000x - 6000 = 20250x3 - 36500x2 + 23000x - 6000. Further, the formula for the future value of annuity is F = P [(1 +r)n -1]/r where P is the monthly payment, r is the rate of interest per period in decimals and n is the number of periods. Thus F = 100 + 100[{1+3(x -1)}3 -1] / 3(x-1)) = 100+100[(3x-2)3 -1]/3(x-1) = 100 +100(27x3 - 54x2 + 36x – 8-1) = 100 + 100(27x3 - 54x2 + 36x – 9)= 100 + 270x3 -5400x2 +3600x – 900 = 270x3 -5400x2 +3600x – 800. Finally, the value of the account on 2nd July is 20250x3 - 36500x2 + 23000x – 6000 + 270x3 -5400x2 +3600x – 800 = 20520x3 - 41900x2 +26600x -6800.
4.The value of $ 1000 after 2 months will be 1000(1 + 1.5/1200)2 = 1000(1.00125)2 = 1000*1.002501563 = $ 1002.50 ( on rounding off to the nearest cent). The interest is $ 1002.50 - $ 1000 = $ 2.50. Similarly, the value of $750 after 3 months will become 750( 1+ 1.5/1200)3 = 750(1.00125)3 = 750 * 1.003754689 = $ 752.82. Thus, the value of the initial deposit of $ 1000, after a withdrawal of $ 250 in April, is 2.50 + 752.82 = $755.32. Further, the formula for the future value of annuity is F = P [(1 +r)n -1]/r where P is the monthly payment, r is the rate of interest per period in decimals and n is the number of periods. Thus F = 100 + 100[ ( 1+ 1.5/100)3 -1] / (1.5/1200) = 100+ 100 [( 1.00125)3 -1]/0.00125 = 100+ 100*0.003754689/0.00125 = 100 + 300.38 = $ 400.38 ( on rounding off to the nearest cent) . Finally, the value of the account on 2nd July is $ 755.32+ $ 400.38 = $ 1155.70.
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