# Math Help - Joe Needs A House

1. ## Joe Needs A House

Joe wants to buy a house but does not want to get a loan. The average price of the dream house is $500,000 and its price is growing at 5 percent per year. How much should Joe invest in a project at the end of each year for the next 5 years in order to accumulate enough money to buy his dream house with cash at the end of the fifth year? Assume the project pays 12 percent rate of return. 2. Hello, magentarita! This is a messy one! Joe wants to buy a house but does not want to get a loan. The average price of the dream house is$500,000 and its price is growing at 5% per year.
How much should Joe invest in a project at the end of each year for the next 5 years in order
to accumulate enough money to buy his dream house with cash at the end of the 5th year?
Assume the project pays 12% rate of return.
The price of the house is now $500,000. Its price is increasing by 5% per year. In five years, its price will be: . $500,\!000(1.05)^5 \:\approx\;\638,\!140.78$ This is the amount he wants to have at the end of five years. Let's walk through his investment plan. At the end of year 1, he invests $x$ dollars. By the end of year 2, it has grown to $1.12x$ dollars . . and he deposits another $x$ dollars. He has $1.12x + x\:=\:(1.12+1)x$ dollars in his account. By the end of year 3, it has grown to $1.12(1.12 + 1)x \:=\:(1.12^2 + 1.12)x$ dollars . . and he deposits another $x$ dollars. He has $(1.12^2 + 1.12)x + x \:=\:(1.12^2 + 1.12 + 1)x$ dollars in his account. By the end of year 4, it has grown to $1.12(1.12^2+1.12 + 1)x$ dollars . . and he deposits another $x$ dollars. He has $(1.12^2 + 1.12 + 1)x + x \:=\:(1.12^3 + 1.12^2 + 1.12 + 1)x$ dollars in his account. By the end of year 5, it has grown to $1.12(1.12^3+1.12^2+1.12+1)x$ dollars . . and he deposits another $x$ dollars. He has $(1.12^4 + 1.12^3 + 1.12^2 + 1.12 + 1)x$ dollars in his account. And this should equal his desired goal:$638,140.78.

$\text{We have: }\;\underbrace{(1+1.12+1.12^2+1.12^3+1.12^4)}_{\te xt{geometric series}}x \;=\;638,\!140.78$ .[1]

The geometric series has first term $a = 1$, common ratio $r = 1.12$, and $n = 5$ terms.
. . Its sum is: . $\frac{1.12^5-1}{1.12 - 1} \:=\:\frac{1.12^5-1}{0.12}$

Then [1] becomes: . $\frac{1.12^5-1}{0.12}\,x \:=\:638,\!140.78 \quad\Rightarrow\quad x \:=\:\frac{0.12}{1.12^5-1}(638,\!140.78)$

Therefore: . $x \;=\;100,\!449.5691 \;\approx\;\boxed{\100,449.57}$

That is the amount he must invest yearly for 5 years to buy his dream house.

3. ## ok....

Originally Posted by Soroban
Hello, magentarita!

This is a messy one!

The price of the house is now $500,000. Its price is increasing by 5% per year. In five years, its price will be: . $500,\!000(1.05)^5 \:\approx\;\638,\!140.78$ This is the amount he wants to have at the end of five years. Let's walk through his investment plan. At the end of year 1, he invests $x$ dollars. By the end of year 2, it has grown to $1.12x$ dollars . . and he deposits another $x$ dollars. He has $1.12x + x\:=\1.12+1)x" alt="1.12x + x\:=\1.12+1)x" /> dollars in his account. By the end of year 3, it has grown to $1.12(1.12 + 1)x \:=\1.12^2 + 1.12)x" alt="1.12(1.12 + 1)x \:=\1.12^2 + 1.12)x" /> dollars . . and he deposits another $x$ dollars. He has $(1.12^2 + 1.12)x + x \:=\1.12^2 + 1.12 + 1)x" alt="(1.12^2 + 1.12)x + x \:=\1.12^2 + 1.12 + 1)x" /> dollars in his account. By the end of year 4, it has grown to $1.12(1.12^2+1.12 + 1)x$ dollars . . and he deposits another $x$ dollars. He has $(1.12^2 + 1.12 + 1)x + x \:=\1.12^3 + 1.12^2 + 1.12 + 1)x" alt="(1.12^2 + 1.12 + 1)x + x \:=\1.12^3 + 1.12^2 + 1.12 + 1)x" /> dollars in his account. By the end of year 5, it has grown to $1.12(1.12^3+1.12^2+1.12+1)x$ dollars . . and he deposits another $x$ dollars. He has $(1.12^4 + 1.12^3 + 1.12^2 + 1.12 + 1)x$ dollars in his account. And this should equal his desired goal:$638,140.78.

$\text{We have: }\;\underbrace{(1+1.12+1.12^2+1.12^3+1.12^4)}_{\te xt{geometric series}}x \;=\;638,\!140.78$ .[1]

The geometric series has first term $a = 1$, common ratio $r = 1.12$, and $n = 5$ terms.
. . Its sum is: . $\frac{1.12^5-1}{1.12 - 1} \:=\:\frac{1.12^5-1}{0.12}$

Then [1] becomes: . $\frac{1.12^5-1}{0.12}\,x \:=\:638,\!140.78 \quad\Rightarrow\quad x \:=\:\frac{0.12}{1.12^5-1}(638,\!140.78)$

Therefore: . $x \;=\;100,\!449.5691 \;\approx\;\boxed{\100,449.57}$

That is the amount he must invest yearly for 5 years to buy his dream house.
Fantastic work as always.