Question

A savings account is opened at time 0 with an initial deposit of $3,000. $2,000 is withdrawn at the end of year 1 and $1,000 is deposited at the end of year 2. The account grows by compound interest at a constant annual effective rate i. At the end of year 2, the account balance is $2,430. Find i.

Answer 1

I have made it clear in calculations sir, If you have any doubts please let me know in comments.

Solution Formula for Componend interest ix Azop (i+1)*t As final amount P= Initial principal amount interest rate g= per M= Number of times interest applied time period t= time periods elapsed (in yrs) Given constant annual effective rate i so nel. Given 3000, is deposited Now amount after first year > IXI Az + 3000 ( 1+4)* >\A= 3000 (it) We have withdrawn 2000 after first year, net amount 30001+3000-2000 = 2000't 1000 CS Scanned with CamScanner

Now, Net final amount after 2 yrs = 2,430 3000 + 1000 After first amount year and at end of second year, $ 1000 deposited. :: 2430 (20:00)[0+'] f 1000 I deposited amount (3000itiovo) (+) 2430-1000 = => 1430 3000 92 + 30001 +1000 +1000 E 300012+ 40001 +1000-1430 =0 > 300012 + 40001 430 = 0 001 and - 10433 .; negative Since interest rate cannot be interest rate (i) = 0.1 Scanned with CS CamScanner