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Can anyone Help me solve this?

In July 2010, Mike deposits $100 semiannually into an account compounding quarterly at 7% nominal annual interest for three years. In July 2013, he increases the deposit to $200 every 6 months, but the interest rates falls to 5% for that year. In July 2014, he deposits $2000 more (in addition to the previous deposits) in the bank for 5 years at a fixed rate of 6%. The interest rates from July 2014 to end of 2018 remain fixed at 6%. Find the value of his savings at the end of 2018.

  • 1
    What have you tried so far? Where are you stuck? – D Stanley Aug 29 at 16:09
  • 1
    Tip: Forget about finding a closed formula and just compute the interest earned each period. Add that to the prior balance and deposits to determine the new balance. – D Stanley Aug 29 at 16:10
  • Simple method: get Mike to check his bank statements. – Lawrence Aug 30 at 9:50
2

Thanks to @D Stanley for the hint

+------------------+-----------+--------------+--------------+---------+---------------------+
|                  |deposits   |principals**  |interest rate |interest*|accumulated interest |
+------------------+-----------+--------------+--------------+---------+---------------------+
|2nd half 2010     |100        |100           |0.07          |3.53     |3.53                 |
|1st half 2011     |200        |203.53        |0.07          |7.19     |10.72                |
|2nd half 2011     |300        |310.72        |0.07          |10.97    |21.69                |
|1st half 2012     |400        |421.69        |0.07          |14.89    |36.57                |
|2nd half 2012     |500        |536.57        |0.07          |18.94    |55.52                |
|1st half 2013     |600        |655.52        |0.07          |23.14    |78.66                |
|2nd half 2013     |800        |878.66        |0.05          |22.10    |100.77               |
|1st half 2014     |1000       |1100.77       |0.05          |27.69    |128.46               |
|2nd half 2014     |3200       |3328.46       |0.06          |100.60   |229.06               |
|1st half 2015     |3400       |3629.06       |0.06          |109.69   |338.75               |
|2nd half 2015     |3600       |3938.75       |0.06          |119.05   |457.80               |
|1st half 2016     |3800       |4257.80       |0.06          |128.69   |586.49               |
|2nd half 2016     |4000       |4586.49       |0.06          |138.63   |725.12               |
|1st half 2017     |4200       |4925.12       |0.06          |148.86   |873.98               |
|2nd half 2017     |4400       |5273.98       |0.06          |159.41   |1033.38              |
|1st half 2018     |4600       |5633.38       |0.06          |170.27   |1203.65              |
|2nd half 2018     |4800       |6003.65       |0.06          |181.46   |1385.11              |
+------------------+-----------+--------------+--------------+---------+---------------------+

*The formula used to calculate the interest column is

interest = (principal * (1 + (interest rate / 4) ) ^ (0.5 * 4) ) - principal
 0.5 is a half of a year
 4 is the number of compounding periods

** principals are the deposits + accumulated interest till the beginning of the period

According to my calculations he is supposed to has 6185.11 in his account by the end of 2018

  • 6185.11 = 4800 + 1385.11
  • 6185.11 = 6003.65 + 181.46
1

Using a formula for the final value of the compounding deposits

enter image description here

i = 0.07
d = 100
n = 6

r = (1 + i/4)^2 - 1 = 0.0353063
a1 = (d (1 + r) ((1 + r)^n - 1))/r = 678.663

i = 0.05
d = 200
n = 2

r = (1 + i/4)^2 - 1 = 0.0251562
a2 = (d (1 + r) ((1 + r)^n - 1))/r + a1 (1 + r)^n = 1128.46

i = 0.06
d = 200
x = 2000
n = 9

r = (1 + i/4)^2 - 1 = 0.030225
a3 = (d (1 + r) ((1 + r)^n - 1))/r + (a2 + x) (1 + r)^n = 6185.11

Value of savings at the end of 2018 is $6185.11

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