Suppose you take out a home mortgage for $160000$ at a monthly interest rate of $0.5$%. If you make payments...
$begingroup$
Suppose you take out a home mortgage for $160000$ at a monthly interest rate of $0.5$%. If you make payments of $1200$ per month, after how many months will the loan balance be zero. Estimate the answer by graphing the sequence of loan balances and then obtain an exact answer.
I have thought to do the following:
As you have to pay $0.5$% for interest then in the first month you have to pay $0.5$% of $160000=800$ for interest and as you paid for $1200$, then for the house you gave $1200-800=400$, so $160000-400=159600$ was already owed and to know how much is left of the second month one does the same, takes out $0,5$% of $159600=798$ and thus paid interest $798$ and was due $159198$. This reasoning is fine? How can I generalize this and do what they ask of me in the problem? Thank you.
calculus algebra-precalculus
asked Dec 1 '18 at 23:20
user482152user482152
987
$endgroup$
add a comment |
$begingroup$
Suppose you take out a home mortgage for $160000$ at a monthly interest rate of $0.5$%. If you make payments of $1200$ per month, after how many months will the loan balance be zero. Estimate the answer by graphing the sequence of loan balances and then obtain an exact answer.
I have thought to do the following:
As you have to pay $0.5$% for interest then in the first month you have to pay $0.5$% of $160000=800$ for interest and as you paid for $1200$, then for the house you gave $1200-800=400$, so $160000-400=159600$ was already owed and to know how much is left of the second month one does the same, takes out $0,5$% of $159600=798$ and thus paid interest $798$ and was due $159198$. This reasoning is fine? How can I generalize this and do what they ask of me in the problem? Thank you.
calculus algebra-precalculus
asked Dec 1 '18 at 23:20
user482152user482152
987
$endgroup$
add a comment |
$begingroup$
Suppose you take out a home mortgage for $160000$ at a monthly interest rate of $0.5$%. If you make payments of $1200$ per month, after how many months will the loan balance be zero. Estimate the answer by graphing the sequence of loan balances and then obtain an exact answer.
I have thought to do the following:
As you have to pay $0.5$% for interest then in the first month you have to pay $0.5$% of $160000=800$ for interest and as you paid for $1200$, then for the house you gave $1200-800=400$, so $160000-400=159600$ was already owed and to know how much is left of the second month one does the same, takes out $0,5$% of $159600=798$ and thus paid interest $798$ and was due $159198$. This reasoning is fine? How can I generalize this and do what they ask of me in the problem? Thank you.
calculus algebra-precalculus
asked Dec 1 '18 at 23:20
user482152user482152
987
$endgroup$
Suppose you take out a home mortgage for $160000$ at a monthly interest rate of $0.5$%. If you make payments of $1200$ per month, after how many months will the loan balance be zero. Estimate the answer by graphing the sequence of loan balances and then obtain an exact answer.
I have thought to do the following:
As you have to pay $0.5$% for interest then in the first month you have to pay $0.5$% of $160000=800$ for interest and as you paid for $1200$, then for the house you gave $1200-800=400$, so $160000-400=159600$ was already owed and to know how much is left of the second month one does the same, takes out $0,5$% of $159600=798$ and thus paid interest $798$ and was due $159198$. This reasoning is fine? How can I generalize this and do what they ask of me in the problem? Thank you.
calculus algebra-precalculus
calculus algebra-precalculus
asked Dec 1 '18 at 23:20
user482152user482152
987
asked Dec 1 '18 at 23:20
user482152user482152
987
asked Dec 1 '18 at 23:20
user482152user482152
987
asked Dec 1 '18 at 23:20
user482152user482152
987
asked Dec 1 '18 at 23:20
user482152user482152
987
987
add a comment |
add a comment |
2 Answers
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oldest
votes
$begingroup$
Let B(n) be the balance at the end of n months.
B(0) = a = 160,000.
B(n + 1) = 1.05.B(n) - 1200.
B(1) is easy to calculate. Calculate B(1) and B(2) to get a sense for B(n) and how complicated it will be. Don't use 160,000, use a to make this clearer.
All this mortgage stuff is so well known,
it's on the web, formulas and tables both.
answered Dec 1 '18 at 23:54
William ElliotWilliam Elliot
7,4062720
$endgroup$
add a comment |
$begingroup$
You're doing fine so far. The next step is to try to write it in symbols instead of numbers, so that you can see the pattern easier.
Let $P_n$ be the amount of the loan outstanding after $n$ months. $P_0$is the amount of the loan after $0$ months, that is, at the beginning, so $P_0=160000.$ Now, if we know $P_n,$ how do we calculate $P_{n+1}?$ You already shown how to do it; take $0.5%$ of subtract that from $1200$, and the subtract the difference from $P_n$ $$
P_{n+1}=P_n-(1200-.05P_n)=1.05P_n-1200tag{1}$$
Now write out he same examples you did, but using symbols this time, by repeated applying equation $(1).$
$$begin{align}
P_1&=1.005P_0-1200\
P_2&=1.005P_1-1200=1.005(1.005P_0-1200)-1200 = 1.005^2P_0-1200(1+1.005)\
P_3&=1.005P_2-1200=1.005^3P_0-1200(1+1.005+1.005^2)\
&vdots
end{align}$$
As to what they want you to do, I'm guessing, because I don't know what tools you have available. I would think they want you to use equation $(1)$ to calculate the various values of $P_n$ perhaps using some computer spreadsheet program, and graph the values of $P_n$ over time, to see when the value first becomes negative, so that the loan has been paid off.
Then they want you to figure out the general formula for $P_n,$ a try to compute the value of $n$ for which $P_n$ = $0$. This will almost surely not be a whole number.
answered Dec 2 '18 at 0:13
saulspatzsaulspatz
14.1k21329
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
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$begingroup$
Let B(n) be the balance at the end of n months.
B(0) = a = 160,000.
B(n + 1) = 1.05.B(n) - 1200.
B(1) is easy to calculate. Calculate B(1) and B(2) to get a sense for B(n) and how complicated it will be. Don't use 160,000, use a to make this clearer.
All this mortgage stuff is so well known,
it's on the web, formulas and tables both.
answered Dec 1 '18 at 23:54
William ElliotWilliam Elliot
7,4062720
$endgroup$
add a comment |
$begingroup$
Let B(n) be the balance at the end of n months.
B(0) = a = 160,000.
B(n + 1) = 1.05.B(n) - 1200.
B(1) is easy to calculate. Calculate B(1) and B(2) to get a sense for B(n) and how complicated it will be. Don't use 160,000, use a to make this clearer.
All this mortgage stuff is so well known,
it's on the web, formulas and tables both.
answered Dec 1 '18 at 23:54
William ElliotWilliam Elliot
7,4062720
$endgroup$
add a comment |
$begingroup$
Let B(n) be the balance at the end of n months.
B(0) = a = 160,000.
B(n + 1) = 1.05.B(n) - 1200.
B(1) is easy to calculate. Calculate B(1) and B(2) to get a sense for B(n) and how complicated it will be. Don't use 160,000, use a to make this clearer.
All this mortgage stuff is so well known,
it's on the web, formulas and tables both.
answered Dec 1 '18 at 23:54
William ElliotWilliam Elliot
7,4062720
$endgroup$
Let B(n) be the balance at the end of n months.
B(0) = a = 160,000.
B(n + 1) = 1.05.B(n) - 1200.
B(1) is easy to calculate. Calculate B(1) and B(2) to get a sense for B(n) and how complicated it will be. Don't use 160,000, use a to make this clearer.
All this mortgage stuff is so well known,
it's on the web, formulas and tables both.
answered Dec 1 '18 at 23:54
William ElliotWilliam Elliot
7,4062720
answered Dec 1 '18 at 23:54
William ElliotWilliam Elliot
7,4062720
answered Dec 1 '18 at 23:54
William ElliotWilliam Elliot
7,4062720
answered Dec 1 '18 at 23:54
William ElliotWilliam Elliot
7,4062720
7,4062720
add a comment |
add a comment |
$begingroup$
You're doing fine so far. The next step is to try to write it in symbols instead of numbers, so that you can see the pattern easier.
Let $P_n$ be the amount of the loan outstanding after $n$ months. $P_0$is the amount of the loan after $0$ months, that is, at the beginning, so $P_0=160000.$ Now, if we know $P_n,$ how do we calculate $P_{n+1}?$ You already shown how to do it; take $0.5%$ of subtract that from $1200$, and the subtract the difference from $P_n$ $$
P_{n+1}=P_n-(1200-.05P_n)=1.05P_n-1200tag{1}$$
Now write out he same examples you did, but using symbols this time, by repeated applying equation $(1).$
$$begin{align}
P_1&=1.005P_0-1200\
P_2&=1.005P_1-1200=1.005(1.005P_0-1200)-1200 = 1.005^2P_0-1200(1+1.005)\
P_3&=1.005P_2-1200=1.005^3P_0-1200(1+1.005+1.005^2)\
&vdots
end{align}$$
As to what they want you to do, I'm guessing, because I don't know what tools you have available. I would think they want you to use equation $(1)$ to calculate the various values of $P_n$ perhaps using some computer spreadsheet program, and graph the values of $P_n$ over time, to see when the value first becomes negative, so that the loan has been paid off.
Then they want you to figure out the general formula for $P_n,$ a try to compute the value of $n$ for which $P_n$ = $0$. This will almost surely not be a whole number.
answered Dec 2 '18 at 0:13
saulspatzsaulspatz
14.1k21329
$endgroup$
add a comment |
$begingroup$
You're doing fine so far. The next step is to try to write it in symbols instead of numbers, so that you can see the pattern easier.
Let $P_n$ be the amount of the loan outstanding after $n$ months. $P_0$is the amount of the loan after $0$ months, that is, at the beginning, so $P_0=160000.$ Now, if we know $P_n,$ how do we calculate $P_{n+1}?$ You already shown how to do it; take $0.5%$ of subtract that from $1200$, and the subtract the difference from $P_n$ $$
P_{n+1}=P_n-(1200-.05P_n)=1.05P_n-1200tag{1}$$
Now write out he same examples you did, but using symbols this time, by repeated applying equation $(1).$
$$begin{align}
P_1&=1.005P_0-1200\
P_2&=1.005P_1-1200=1.005(1.005P_0-1200)-1200 = 1.005^2P_0-1200(1+1.005)\
P_3&=1.005P_2-1200=1.005^3P_0-1200(1+1.005+1.005^2)\
&vdots
end{align}$$
As to what they want you to do, I'm guessing, because I don't know what tools you have available. I would think they want you to use equation $(1)$ to calculate the various values of $P_n$ perhaps using some computer spreadsheet program, and graph the values of $P_n$ over time, to see when the value first becomes negative, so that the loan has been paid off.
Then they want you to figure out the general formula for $P_n,$ a try to compute the value of $n$ for which $P_n$ = $0$. This will almost surely not be a whole number.
answered Dec 2 '18 at 0:13
saulspatzsaulspatz
14.1k21329
$endgroup$
add a comment |
$begingroup$
You're doing fine so far. The next step is to try to write it in symbols instead of numbers, so that you can see the pattern easier.
Let $P_n$ be the amount of the loan outstanding after $n$ months. $P_0$is the amount of the loan after $0$ months, that is, at the beginning, so $P_0=160000.$ Now, if we know $P_n,$ how do we calculate $P_{n+1}?$ You already shown how to do it; take $0.5%$ of subtract that from $1200$, and the subtract the difference from $P_n$ $$
P_{n+1}=P_n-(1200-.05P_n)=1.05P_n-1200tag{1}$$
Now write out he same examples you did, but using symbols this time, by repeated applying equation $(1).$
$$begin{align}
P_1&=1.005P_0-1200\
P_2&=1.005P_1-1200=1.005(1.005P_0-1200)-1200 = 1.005^2P_0-1200(1+1.005)\
P_3&=1.005P_2-1200=1.005^3P_0-1200(1+1.005+1.005^2)\
&vdots
end{align}$$
As to what they want you to do, I'm guessing, because I don't know what tools you have available. I would think they want you to use equation $(1)$ to calculate the various values of $P_n$ perhaps using some computer spreadsheet program, and graph the values of $P_n$ over time, to see when the value first becomes negative, so that the loan has been paid off.
Then they want you to figure out the general formula for $P_n,$ a try to compute the value of $n$ for which $P_n$ = $0$. This will almost surely not be a whole number.
answered Dec 2 '18 at 0:13
saulspatzsaulspatz
14.1k21329
$endgroup$
You're doing fine so far. The next step is to try to write it in symbols instead of numbers, so that you can see the pattern easier.
Let $P_n$ be the amount of the loan outstanding after $n$ months. $P_0$is the amount of the loan after $0$ months, that is, at the beginning, so $P_0=160000.$ Now, if we know $P_n,$ how do we calculate $P_{n+1}?$ You already shown how to do it; take $0.5%$ of subtract that from $1200$, and the subtract the difference from $P_n$ $$
P_{n+1}=P_n-(1200-.05P_n)=1.05P_n-1200tag{1}$$
Now write out he same examples you did, but using symbols this time, by repeated applying equation $(1).$
$$begin{align}
P_1&=1.005P_0-1200\
P_2&=1.005P_1-1200=1.005(1.005P_0-1200)-1200 = 1.005^2P_0-1200(1+1.005)\
P_3&=1.005P_2-1200=1.005^3P_0-1200(1+1.005+1.005^2)\
&vdots
end{align}$$
As to what they want you to do, I'm guessing, because I don't know what tools you have available. I would think they want you to use equation $(1)$ to calculate the various values of $P_n$ perhaps using some computer spreadsheet program, and graph the values of $P_n$ over time, to see when the value first becomes negative, so that the loan has been paid off.
Then they want you to figure out the general formula for $P_n,$ a try to compute the value of $n$ for which $P_n$ = $0$. This will almost surely not be a whole number.
answered Dec 2 '18 at 0:13
saulspatzsaulspatz
14.1k21329
answered Dec 2 '18 at 0:13
saulspatzsaulspatz
14.1k21329
answered Dec 2 '18 at 0:13
saulspatzsaulspatz
14.1k21329
answered Dec 2 '18 at 0:13
saulspatzsaulspatz
14.1k21329
14.1k21329
add a comment |
add a comment |
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