Suppose that $780 is deposited at the end of every year into an account paying interest of 6% per year. At the end of fifteen (1
1 answer:
Answer:
The account will be worth approximately $1,869
Explanation:
First of all, note that 6% of an amount = 6/100 × the amount = 0.06×amount.
Next let us calculate the amounts gotten for the first 4 years, and establish a pattern that will will us for the remaining 11 years.
1st year total= deposit + (0.06×deposit) = 780 + (0.06 × 780)
= 780 + 46.8 = $826.8
2nd year total = 826.8 + (0.06 × 826.8) = $876.408
3rd year total = 876.408 + (0.06 × 876.408) = $928.992
4th year total = 928.992 + (0.06 × 928.992) = $984.732
Now, if we observe the total amounts as the year progresses, we notice that the next year increase by a certain constant factor which is 1.06; this is determined by dividing the amount in a year by the amount in the previous year. It is shown below;
Year 2 ÷ Year 1 = 876.408 ÷ 826.8 = 1.06
year 3 ÷ year 2 = 928.992 ÷ 876.408 = 1.06
year 4 ÷ year 3 = 984.732 ÷ 928.992 = 1.06
Now, to determine the amount in the next year, we will multiply the amount in the previous year by 1.06 (common increasing factor)
Year 5 = year 4 × 1.06 = 984.732 × 1.06 = $1,046.816
year 6 = 1046.816 × 1.06 = $1,106.445
year 7 = 1106.445 × 1.06 = $1,172.832
year 8 = 1172.832 × 1.06 = $1,243.202
year 9 = 1243.202 × 1.06 = $1,317.794
year 10 = 1317.794 × 1.06 = $1,396.862
year 11 = 1396.862 × 1.06 = $1,480.674
year 12 = 1480.674 × 1.06 = $1,569.514
year 13 = 1569.514 × 1.06 = $1,663.684
year 14 = 1663.685 × 1.06 = $1,763.506
Year 15 = 1763.506 × 1.06 = $1,869.316 which is approximately $1,869
Alternatively, you can count how many 1.06s are there from year 5 to year 15, and the answer is 11. then you can raise 1.06 to a power of 11 as shown
then multiply the amount in year 4 by 1.8983
= 984.732 × 1.8983 = $1,869.316. = approx. $1869
This second method is easier, but I wanted you to see what is going on that is why i did the details in the first method
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