Question

1 year(s) ago, Mack invested 6,490 dollars. In 1 year(s) from today, he expects to have 8,680 dollars. If Mack expects to earn the same annual return after 1 year(s) from today as the annual rate implied from the past and expected values given in the problem, then how much does Mack expect to have in 6 years from today?

3 year(s) ago, Carl had 248,400 dollars in his account. In 5 year(s), he expects to have 456,200 dollars. If he has earned and expects to earn the same return each year from 3 year(s) ago to 5 year(s) from today, then how much does he expect to have in 2 year(s) from today?

1 year(s) ago, Fatima invested 6,150 dollars. In 2 year(s) from today, she expects to have 7,900 dollars. If Fatima expects to earn the same annual return after 2 year(s) from today as the annual rate implied from the past and expected values given in the problem, then in how many years from today does she expect to have exactly 11,210 dollars? Round your answer to 2 decimal places (for example, 2.89, 14.70, or 6.00).

Answer 1

The amount of $6490 will grow to $8680 in 2 years (invested one year ago and expected value after one year)

So, rate of return (r) is given by

6490*(1+r)^2 =8680

=> r= 0.156478 or 15.65%

Amount expected 6 years from today = amount after one year * (1+r)^5

=8680*1.156478^5

=$17955.90

Carl's $248400 grew to $456200 in 8 years

growth rate = (456200/248400)^(1/8)-1 = 0.078947

Amount after 2 years from today = 248400* 1.078947^5 = $363206.74

Fatima's $6150 are expected to become $7900 in three years

growth rate = (7900/6150)^(1/3)-1 = 0.087052849

Let Fatima have $11210 after n years, then

6150* 1.087052849^(n+1) = 11210

=> 1.087052849^(n+1) = 1.822764

Taking natural log on both sides

=> (n+1) = 7.1924

=> n = 6.1924

So, Fatima can expect to have $11210 after 6.19 years from today