Compound Interest Systems of Equations Problem

[ConstantineO]

Homework Statement

Romeo was given a gift of $10,000 when he turned 16. He invested it at 3% per annum. Three years later, Juliet was given $10,000, which she invested at 5% per annum. When will the two amounts be equal in value?

Homework Equations

Compound Interest Formula
Total = Capital(1+interest)^Years
t = c(1+i)^n

The Attempt at a Solution

Deciding how to create my formula is where things get fuzzy. The way I initially attempted this was to add a 3 year head start onto Romeo's interest formula. This is what results (Used wolfram to save time rewriting):

This eventually results in "4.61103." However, this is not the answer at the back of my textbook. They decided to opt for a different formula where the time is subtracted from Juliet's interest function. This is shown here:

This results in the correct answer of 7.61103. What I'm wondering, why do I have to subtract 3 years from Juliet's compound interest formula, and why does adding 3 years to Romeo's compound interest formula produce the wrong answer?

 

[BvU]

So clearly the book starts counting at the beginning of the story: when Romeo brings his money to the bank.
That means Julia's interest clock is at x-3 when Romeo's is at x.

 

[SammyS]

What is Romeo's age when then two accounts have equal value?

Both answers agree.

 

[ConstantineO]

So clearly the book starts counting at the beginning of the story: when Romeo brings his money to the bank.
That means Julia's interest clock is at x-3 when Romeo's is at x.

Is this always the convention when dealing with problems like this? From my experience with physics courses, I assumed that time shares a relation with the total amount of money generated by compound interest. The interest and initial capital is thus fixed, and you are only left with a relation between time and the total sum of money. If you were to treat time as a vector which you could progress backwards and forwards on, each snapshot through time depending on which way you were progressing would net either a larger or smaller amount of the total. I don't understand why the book is forcing me to start at the beginning of the story. The story is like pages of a book which can be turned towards or away from the end of the novel. The content of the novel does not change and its contents are already written.

What is Romeo's age when then two accounts have equal value?

Both answers agree.

I don't understand what Romeo's age has to do with this? Is this some kind of statement that alludes to time only progressing forward or something?

 

[BvU]

Nothing so fancy. The book is sloppy by not stating when it wants the clock (calendar in this case) to show t = 0. From the book answer it appears that it assumes t = 0 when Romeo turns 16. The exercise could equally well have assumed t = 0 when Julia receives the ten grand. If I were grading this, I'd have to allow both answers, certainly when the calculation steps are shown.

 

[ConstantineO]

Nothing so fancy. The book is sloppy by not stating when it wants the clock (calendar in this case) to show t = 0. From the book answer it appears that it assumes t = 0 when Romeo turns 16. The exercise could equally well have assumed t = 0 when Julia receives the ten grand. If I were grading this, I'd have to allow both answers, certainly when the calculation steps are shown.

I actually just figured it out while discussing this question with a friend just a few moments before you posted. I took the delta of the two times from both of the ways of solving this question and realized that this is nothing more than a semantics issue of where you measure time from. This question is poorly setup, and I am going to voice my irritation to the person marking.

 

[Ray Vickson]

I actually just figured it out while discussing this question with a friend just a few moments before you posted. I took the delta of the two times from both of the ways of solving this question and realized that this is nothing more than a semantics issue of where you measure time from. This question is poorly setup, and I am going to voice my irritation to the person marking.

A more serious problem is that when compounding annually, the answer is "never". At Romeo's 23rd birthday, Romeo's account contains more money than Juliet's, while at his 24th birthday, Juliet's account contains more than Romeo's. Unless there is something like continuous compounding, the two amounts never match exactly; that is, unless you can withdraw your money part-way through a year and earn a part-year's interest, you could never find a time where the two withdrawals are equal. (However, I would not raise this issue with your teacher if I were you; just "go with the flow".)

Also, instead of expressing irritation, it would be wiser for you to say that you will give two answers (depending on where to start measuring time) and point out (nicely) that the question is a bit ambiguous.

 

[SammyS]

A more serious problem is that when compounding annually, the answer is "never". At Romeo's 23rd birthday, Romeo's account contains more money than Juliet's, while at his 24th birthday, Juliet's account contains more than Romeo's. Unless there is something like continuous compounding, the two amounts never match exactly; that is, unless you can withdraw your money part-way through a year and earn a part-year's interest, you could never find a time where the two withdrawals are equal. (However, I would not raise this issue with your teacher if I were you; just "go with the flow".)

Also, instead of expressing irritation, it would be wiser for you to say that you will give two answers (depending on where to start measuring time) and point out (nicely) that the question is a bit ambiguous.

RGV makes an excellent point. This is a short-coming of many problems of this type.

It's often the case that for any fraction of a single compounding period that's "left over", simple interest is computed for that extra amount of time. With this method, find the value of the account at x = 4, then see if the values can be equal at any time in the 4th year using simple interest.

(Edited slightly.)

 

[ConstantineO]

RGV makes an excellent point. This is a short-coming of many problems of this type.

It's often the case that for any portion of a compounding period simple interest is computed for that extra amount of time. With this method, find the value of the account at x = 4, then see if the values can be equal at any time in the 4th year using simple interest.

I don't wish to sound like an imbecile, but I have no clue what that means. How would I set up a systems of equation if I subbed in 4 for both x's? I believe I would be left with only one unknown for either equation, or am I misunderstanding you? Please clarify what you mean by using x=4.

 

[SammyS]

I don't wish to sound like an imbecile, but I have no clue what that means. How would I set up a systems of equation if I subbed in 4 for both x's? I believe I would be left with only one unknown for either equation, or am I misunderstanding you? Please clarify what you mean by using x=4.

When x = 4, (7years for Romeo, 4 years for Juliet), what is the value in each in each account?

Start from that point for each & using simple interest, and see how long it takes for the values to be equal.

 

[ConstantineO]

When x = 4, (7years for Romeo, 4 years for Juliet), what is the value in each in each account?

Start from that point for each & using simple interest, and see how long it takes for the values to be equal.

I am not familiar with the simple interest formula, so I googled it. I am operating under the assumption that it is different than the compound interest formula.

I am to assume this is correct?
Simple interest Formula
I = "the grown money"
p = the initial capital
r = interest rate per year
t = t
(I) = (p)(r)(t)

I think I know what you're getting at, so let me make an attempt.

Romeo Compound Interest Calculation Total
Ar = 10,000(1.03)^4+3
Ar = 10,000(1.03)^7
Ar = 12,298.73865

Julie Compound Interest Calculation Total
Aj = 10,000(1.05)^4
Aj = 12,155.0625

Romeo Simple Interest Calculation
(Ir) = (12,298.73865)(0.03)(t)
(Ir) = (368.94)(t)

Juliet Simple Interest Calculation
(Ir) = (12,155.0625)(0.05)(t)
(Ir) = 607.753125(t)

Equate the two to equal each other.

(368.94)(t) = 607.753125(t)
(368.94)(t) -607.753125(t) = 0
t(368.94-607.753125)=0

This doesn't work though...

 

[Ray Vickson]

I am not familiar with the simple interest formula, so I googled it. I am operating under the assumption that it is different than the compound interest formula.

I am to assume this is correct?
Simple interest Formula
I = "the grown money"
p = the initial capital
r = interest rate per year
t = t
(I) = (p)(r)(t)

I think I know what you're getting at, so let me make an attempt.

Romeo Compound Interest Calculation Total
Ar = 10,000(1.03)^4+3
Ar = 10,000(1.03)^7
Ar = 12,298.73865

Julie Compound Interest Calculation Total
Aj = 10,000(1.05)^4
Aj = 12,155.0625

Romeo Simple Interest Calculation
(Ir) = (12,298.73865)(0.03)(t)
(Ir) = (368.94)(t)

Juliet Simple Interest Calculation
(Ir) = (12,155.0625)(0.05)(t)
(Ir) = 607.753125(t)

Equate the two to equal each other.

(368.94)(t) = 607.753125(t)
(368.94)(t) -607.753125(t) = 0
t(368.94-607.753125)=0

This doesn't work though...

It's not that complicated. Let $R_t, J_t$ be the amounts in Romeo's and Juliet's accounts at time $t$ (in years, measured from Romeo's 16th birthday). We have:
$$\begin{array}{ccc} & t=7 & t=8 \\ R_t & 12298.74 & 12667.70 \\ J_t & 12155.06 & 12762.82 \end{array}$$
You can draw two straight-line plots, one going from x = 0 at (7,R7) to x = 1 at (8,R8) and another from x = 0 at (7,J7) to x = 1 at (8,J8). Where the two lines cross is the point where simple interest calculations (for the partial year) would give you equality of monetary values. This will be close to (but not exactly equal to) the value x = 0.611---that is, for time 7.611 ---that your previous calculation would have produced. (The difference is due to the slight curvature in the money vs.time graph over the year, due to continuous compounding over the year, as compared with the straight line obtained from simple interest accumulation over the year.)

 

[ConstantineO]

It's not that complicated. Let $R_t, J_t$ be the amounts in Romeo's and Juliet's accounts at time $t$ (in years, measured from Romeo's 16th birthday). We have:
$$\begin{array}{ccc} & t=7 & t=8 \\ R_t & 12298.74 & 12667.70 \\ J_t & 12155.06 & 12762.82 \end{array}$$
You can draw two straight-line plots, one going from x = 0 at (7,R7) to x = 1 at (8,R8) and another from x = 0 at (7,J7) to x = 1 at (8,J8). Where the two lines cross is the point where simple interest calculations (for the partial year) would give you equality of monetary values. This will be close to (but not exactly equal to) the value x = 0.611---that is, for time 7.611 ---that your previous calculation would have produced. (The difference is due to the slight curvature in the money vs.time graph over the year, due to continuous compounding over the year, as compared with the straight line obtained from simple interest accumulation over the year.)

Could you show me an example of this because I am simply not understanding what you're saying, or can you show me how to equate the two together using algebra?

 

[ConstantineO]

This is getting out of hand, so I am going to try to be as clear and as concise as possible. I know how to calculate the correct answer for this question, and I realize where I first made my mistake. I am now wondering what everyone is talking about when they mention using simple interest. I have literally 0 experience doing any simple interest calculations, and I have no idea what sammyS is talking about here.

When x = 4, (7years for Romeo, 4 years for Juliet), what is the value in each in each account?

Start from that point for each & using simple interest, and see how long it takes for the values to be equal.

If anyone is holding information back because they believe an attempt has not been made and it would violate forum rules, please stop. I understand how to do the book's question using the books method, and I know and where the discrepancy between the 4.61103 and 7.61103 came from. I have made a wholehearted attempt so please for sake of my sanity just give me an example or step by step breakdown. I learn through examples, I look at the process involved, and I create a model in my head how everything interacts and relates with each other.

What I am seeking is an explanation and a simple example of what this simple interest method is for calculating the 0.61103 year after the exact 7 years has passed.

What I am understanding so far.
- First use the compound interest formula to create outputs for both Romeo's and Juliet's accounts after being influenced by exactly 7 years of interest.
- This is done by assigning "4" to the variable x
- Get said products, in this case being:
Romeo's Total After Exactly 7 Years = 12,298.73865
Juliet's Total After Exactly 7 Years = 12,155.0625
- Since you know that the two amounts are still not equal, you must calculate the remaining time between the two using the Simple interest formula
- This is what I am having problems with and what I would appreciate having an example for. Preferably shown algebraically, so I am not plotting nearly straight lines in Desmos.

Not to sound contentious, but when you try to find the point of intersect of two nearly straight lines with a slope that is in hundredths of a unit, it is just plainly byzantine. I am probably doing this wrong, but graphically representing each simple interest formula and trying to find where they meet is insane.

 

[Ray Vickson]

This is getting out of hand, so I am going to try to be as clear and as concise as possible. I know how to calculate the correct answer for this question, and I realize where I first made my mistake. I am now wondering what everyone is talking about when they mention using simple interest. I have literally 0 experience doing any simple interest calculations, and I have no idea what sammyS is talking about here.
If anyone is holding information back because they believe an attempt has not been made and it would violate forum rules, please stop. I understand how to do the book's question using the books method, and I know and where the discrepancy between the 4.61103 and 7.61103 came from. I have made a wholehearted attempt so please for sake of my sanity just give me an example or step by step breakdown. I learn through examples, I look at the process involved, and I create a model in my head how everything interacts and relates with each other.

What I am seeking is an explanation and a simple example of what this simple interest method is for calculating the 0.61103 year after the exact 7 years has passed.

What I am understanding so far.
- First use the compound interest formula to create outputs for both Romeo's and Juliet's accounts after being influenced by exactly 7 years of interest.
- This is done by assigning "4" to the variable x
- Get said products, in this case being:
Romeo's Total After Exactly 7 Years = 12,298.73865
Juliet's Total After Exactly 7 Years = 12,155.0625
- Since you know that the two amounts are still not equal, you must calculate the remaining time between the two using the Simple interest formula
- This is what I am having problems with and what I would appreciate having an example for. Preferably shown algebraically, so I am not plotting nearly straight lines in Desmos.

How could I have made my explanation any simpler? There was no hidden information or holding back of anything. I said: just draw two straight-line graphs for Romeo and Juliet, where the lines connect their (time,money) points at the start and end of the final year.

The true graphs of time vs. money will be curved, because of compound interest, but over a short period such as one year the "curvature" will be small, and the graph will look almost like a straight line; replacing the curve by a straight line ---only for that single year---would give you the "simple" interest schedule for that year. It will be almost the same as the compound interest schedule, because the curvature is small over short times such as a year.

If you don't believe me, just draw the graphs of money vs. time for Romeo and for Juliet. You will see that they curve up, but are almost straight over short periods. The only way to really understand it is to sit down and do it.

 

[ConstantineO]

How could I have made my explanation any simpler? There was no hidden information or holding back of anything. I said: just draw two straight-line graphs for Romeo and Juliet, where the lines connect their (time,money) points at the start and end of the final year.

The true graphs of time vs. money will be curved, because of compound interest, but over a short period such as one year the "curvature" will be small, and the graph will look almost like a straight line; replacing the curve by a straight line ---only for that single year---would give you the "simple" interest schedule for that year. It will be almost the same as the compound interest schedule, because the curvature is small over short times such as a year.

If you don't believe me, just draw the graphs of money vs. time for Romeo and for Juliet. You will see that they curve up, but are almost straight over short periods. The only way to really understand it is to sit down and do it.

I am going to assume the two lines would be plotted as
y=(12155.0625)(0.05)(x)
and
y=(12298.73865)(0.03)(x)

If that's the case, I don't understand how any human being is supposed to find the point of intersect graphically like this.

I did sit down and do it. I have been sitting down and ripping my hair out doing this for sometime.

Edit - In the event that anyone else would like to question how committed I am to "sitting down and doing it." I've had enough of that nonsense from teachers with superiority issues and over bloated opinions that like to question how hard I am making an attempt.

 

[Ray Vickson]

I am going to assume the two lines would be plotted as
y=(12155.0625)(0.05)(x)
and
y=(12298.73865)(0.03)(x)

If that's the case, I don't understand how any human being is supposed to find the point of intersect graphically like this.

I did sit down and do it. I have been sitting down and ripping my hair out doing this for sometime.

As long as you persist in trying to plot the wrong thing you will need to keep pulling out your hair. Go back and read what I wrote in #12. Do the plot exactly as I indicated there.

Better still plot the two curves y = 10000 * (1.03)^x and y = 10000 * (1.05)^(x-3) for x = from 3 to 8. They are curved, aren't they? Don't they look almost straight over the shorter interval 7 \leq x \leq 8? in each case if you were to plot a straight line from (7,y(7)) to (8,y(8)) it would look very close to the actual curve (x,y(x)) for x between 7 and 8. The "simple interest" effect between 7 and 8 would be the straight line, while the "compound interest" effect would be the curve. They both pass through the same points at 7 and 8, but they differ in between.

 

[ConstantineO]

As long as you persist in trying to plot the wrong thing you will need to keep pulling out your hair. Go back and read what I wrote in #12. Do the plot exactly as I indicated there.

Better still plot the two curves y = 10000 * (1.03)^x and y = 10000 * (1.05)^(x-3) for x = from 3 to 8. They are curved, aren't they? Don't they look almost straight over the shorter interval 7 \leq x \leq 8? in each case if you were to plot a straight line from (7,y(7)) to (8,y(8)) it would look very close to the actual curve (x,y(x)) for x between 7 and 8. The "simple interest" effect between 7 and 8 would be the straight line, while the "compound interest" effect would be the curve. They both pass through the same points at 7 and 8, but they differ in between.

As I have stated before.

I don't wish to sound like an imbecile, but I have no clue what that means. How would I set up a systems of equation if I subbed in 4 for both x's? I believe I would be left with only one unknown for either equation, or am I misunderstanding you? Please clarify what you mean by using x=4.

I was completely uncertain of what to plot. I am being told to use the simple interest formula, but the two formulas you have provided are compound interest formulas:
y = 10000 * (1.03)^x
y = 10000 * (1.05)^(x-3)
I plotted these long before I even got to this whole simple interest business when trying to figure out the original discrepancy between 7.61103 and 4.61103. I am fully aware that the compound interest lines are nearly straight. I don't see how the average rate of change between x =7 and x =8 is going to help me discover the remaining time of 0.61103 years. I am still unclear of what to plot? Do I plot the compound interest based functions for Romeo and Juliet, or do I plot the Simple interest functions? I have done both and I am not seeing anything that is giving me any kind of better understanding. Is this operation you describe too difficult represent algebraically?

I am getting far too irritated by this and far too frustrated and am quickly losing the desire to continue.

 

[ConstantineO]

I am not familiar with the simple interest formula, so I googled it. I am operating under the assumption that it is different than the compound interest formula.

Romeo Compound Interest Calculation Total
Ar = 10,000(1.03)^4+3
Ar = 10,000(1.03)^7
Ar = 12,298.73865

Julie Compound Interest Calculation Total
Aj = 10,000(1.05)^4
Aj = 12,155.0625

Romeo Simple Interest Calculation
(Ir) = (12,298.73865)(0.03)(t)
(Ir) = (368.94)(t)

Juliet Simple Interest Calculation
(Ir) = (12,155.0625)(0.05)(t)
(Ir) = 607.753125(t)

Equate the two to equal each other.

(368.94)(t) = 607.753125(t)
(368.94)(t) -607.753125(t) = 0
t(368.94-607.753125)=0

This doesn't work though...

I was on the right track here. There is no need for this graphing nonsense. I forgot a single number which was my "+ 3" in my simple interest formula. I was a fool not to recognize that the lack of a constant would render my simple interest system of equation unsolvable. The lack of anyone speaking up about this is worrisome. In the future, when referencing the slope of a secant line between the range of two x values of a curve, please write in the notation of 7 < x < 8. I had absolutely no clue what you were talking about.

So let's try that last bit again with a fixed formula.

Romeo Simple Interest Calculation
(Ir) = (12,298.73865)(0.03)(t+3)
(Ir) = (368.94)(t+3)

Juliet Simple Interest Calculation
(Ir) = (12,155.0625)(0.05)(t)
(Ir) = 607.753125(t)

Now the Systems of Equation
(607.753125)(t) = (368.94)(t+3)
(607.753125)(t) = (368.94)(t)+ 1106.82
t(607.753125-368.94) = 1106.82
t(238.813125) = (1106.82)
t = (1106.82)/(238.813125)
t = 4.634669891
Not too far away from 4.61103. I wonder why...

Let's try the simplified interest formula that measures time from Julias deposit date.

Romeo Simple Interest Calculation
(Ir) = (12,298.73865)(0.03)(t)
(Ir) = (368.94)(t)

Juliet Simple Interest Calculation
(Ir) = (12,155.0625)(0.05)(t-3)
(Ir) = 607.753125(t-3)

607.753125(t-3)= (368.94)(t)
(607.753125)(t) - 1823.259375 = (368.94)(t)
(607.753125)(t) - (368.94)(t) =1823.259375
t(607.753125 - 368.94) = 1823.259375
t(238.813125) = 1823.259375
t = 7.634.634669891
Not too far away from 7.61103. Surprise surprise...

Look Ma! No graphs.
7.634.634669891 - 4.634669891 = 3 just like 7.61103 - 4.61103 =3. I wonder if they're related?
I don't think I am wrong by stating that the time after the 7 year mark would actually be 0.634669891 of a year instead of 0.61103 if fractional compound interest is indeed calculated with simplified interest.

We can take things a step further and look at the secant line's slope of the curve that models interest from Romeo's deposit date during 7 < x < 8. Let's look at the average rate of change for both Romeo's and Juliet's account.

Romeo Compound Interest AROC: 7 < x < 8

While x = 4
Ar = 10,000(1.03)^4+3
Ar = 10,000(1.03)^7
Ar = 12,298.73865
x = 4
y =12,298.73865

While x =5
Ar = 10,000(1.03)^5+3
Ar = 10,000(1.03)^8
Ar = 12,667.70081
x = 5
y = 12,667.70081

Romeo's AROC
= (12,667.70081 - 12,298.73865) / (5 -4)
= 368.96216

Juliet's Compound Interest AROC 7 < x < 8

While x = 4
Aj = 10,000(1.05)^4
Aj = 12,155.0625
x = 4
y =12,155.0625

While x = 5
Aj = 10,000(1.05)^5
Aj = 12,762.81563
x = 5
y = 12,762.81563

Juliet's AROC
= (12,762.81563 - 12,155.0625) / (5-4)
=607.7531

After that huge bunch of calculations we now have slopes that we can create linear relations with.

Romeo's Linear Relation
y = (368.96216)(x + 3)

Juliet's Linear Relation
y = 607.7531(x)

Let's see what the Solution is for these in equations when put into a system.

(368.96216)(x + 3) = 607.7531(x)
(368.96216)(x) + 1106.88648 = 607.7531(x)
607.7531(x) -(368.96216)(x) = 1106.88648
x(607.7531-368.96216) = 1106.88648
x(238.783884) = 1106.88648
x = 4.635515854

Notice a pattern?

Time Measured from Romeo's Deposit Compound Calculation Product
x = 4.61103
Time Measured from Romeo's Deposit Simplified Interest Product
x = 4.634669891
Time Measured from Romeo's Deposit AROC: 7 < x< 8 Product
x = 4.635515854

I think its safe to say that they matched sometime in the range of 0.635515854 - 0.634669891- 0.61103 of the start of the 8th year. I think I made my point. I don't need any graphs to know how to rock, and I certainly showed that I sat down and did the work.

 

[Ray Vickson]

I was on the right track here. There is no need for this graphing nonsense. I forgot a single number which was my "+ 3" in my simple interest formula. I was a fool not to recognize that the lack of a constant would render my simple interest system of equation unsolvable. The lack of anyone speaking up about this is worrisome. In the future, when referencing the slope of a secant line between the range of two x values of a curve, please write in the notation of 7 < x < 8. I had absolutely no clue what you were talking about.

So let's try that last bit again with a fixed formula.

Romeo Simple Interest Calculation
(Ir) = (12,298.73865)(0.03)(t+3)
(Ir) = (368.94)(t+3)

Juliet Simple Interest Calculation
(Ir) = (12,155.0625)(0.05)(t)
(Ir) = 607.753125(t)

Now the Systems of Equation
(607.753125)(t) = (368.94)(t+3)
(607.753125)(t) = (368.94)(t)+ 1106.82
t(607.753125-368.94) = 1106.82
t(238.813125) = (1106.82)
t = (1106.82)/(238.813125)
t = 4.634669891
Not too far away from 4.61103. I wonder why...

Let's try the simplified interest formula that measures time from Julias deposit date.

Romeo Simple Interest Calculation
(Ir) = (12,298.73865)(0.03)(t)
(Ir) = (368.94)(t)

Juliet Simple Interest Calculation
(Ir) = (12,155.0625)(0.05)(t-3)
(Ir) = 607.753125(t-3)

607.753125(t-3)= (368.94)(t)
(607.753125)(t) - 1823.259375 = (368.94)(t)
(607.753125)(t) - (368.94)(t) =1823.259375
t(607.753125 - 368.94) = 1823.259375
t(238.813125) = 1823.259375
t = 7.634.634669891
Not too far away from 7.61103. Surprise surprise...

Look Ma! No graphs.
7.634.634669891 - 4.634669891 = 3 just like 7.61103 - 4.61103 =3. I wonder if they're related?
I don't think I am wrong by stating that the time after the 7 year mark would actually be 0.634669891 of a year instead of 0.61103 if fractional compound interest is indeed calculated with simplified interest.

We can take things a step further and look at the secant line's slope of the curve that models interest from Romeo's deposit date during 7 < x < 8. Let's look at the average rate of change for both Romeo's and Juliet's account.

Romeo Compound Interest AROC: 7 < x < 8

While x = 4
Ar = 10,000(1.03)^4+3
Ar = 10,000(1.03)^7
Ar = 12,298.73865
x = 4
y =12,298.73865

While x =5
Ar = 10,000(1.03)^5+3
Ar = 10,000(1.03)^8
Ar = 12,667.70081
x = 5
y = 12,667.70081

Romeo's AROC
= (12,667.70081 - 12,298.73865) / (5 -4)
= 368.96216

Juliet's Compound Interest AROC 7 < x < 8

While x = 4
Aj = 10,000(1.05)^4
Aj = 12,155.0625
x = 4
y =12,155.0625

While x = 5
Aj = 10,000(1.05)^5
Aj = 12,762.81563
x = 5
y = 12,762.81563

Juliet's AROC
= (12,762.81563 - 12,155.0625) / (5-4)
=607.7531

After that huge bunch of calculations we now have slopes that we can create linear relations with.

Romeo's Linear Relation
y = (368.96216)(x + 3)

Juliet's Linear Relation
y = 607.7531(x)

Let's see what the Solution is for these in equations when put into a system.

(368.96216)(x + 3) = 607.7531(x)
(368.96216)(x) + 1106.88648 = 607.7531(x)
607.7531(x) -(368.96216)(x) = 1106.88648
x(607.7531-368.96216) = 1106.88648
x(238.783884) = 1106.88648
x = 4.635515854

Notice a pattern?

Time Measured from Romeo's Deposit Compound Calculation Product
x = 4.61103
Time Measured from Romeo's Deposit Simplified Interest Product
x = 4.634669891
Time Measured from Romeo's Deposit AROC: 7 < x< 8 Product
x = 4.635515854

I think its safe to say that they matched sometime in the range of 0.635515854 - 0.634669891- 0.61103 of the start of the 8th year. I think I made my point. I don't need any graphs to know how to rock, and I certainly showed that I sat down and did the work.

Of course it can be done without graphs. Graphs can be of help in setting up the algebraic equations that must, eventually, be solved without graphs. However, you seemed to not understand how simple interest works (your claim, not mine) and to see that a graphical representation can sometimes be helpful---not always, just sometimes. If it was not helpful to you, fine. And, of course I suggested plotting the compound-interest graphs, but I guess you missed out the part where I said that during a short period, such as 1 year, the graph looks almost straight, and that a straight line replacement for the graph (but ONLY in that single year) gives you the simple-interest effect within the year.

Your writeup is almost incomprehensible to me, but your final answer seems OK. You say you still do not know why it is different from the original answer, and that is precisely what I was speaking of when I mentioned graphs: you could see right away why there is a difference.

 

[ConstantineO]

Of course it can be done without graphs. However, you seemed to not understand how simple interest works (your claim, not mine) and to see that a graphical representation can sometimes be helpful---not always, just sometimes. If it was not helpful to you, fine.

Your writeup is almost incomprehensible to me, but your final answer seems OK. You say you still do not know why it is different from the original answer, and that is precisely what I was speaking of when I mentioned graphs: you could see right away why there is a difference.

I showed the difference when I computed the average rate of change of the non-linear exponential function during the interval 7<x<8. Although the slope is negligible it has a significant enough effect to alter the x value by hundredths of a unit. Of course the linear function based on simplified interest is going to differ due to it being well "linear." I had no clue what any of this meant:

It's not that complicated. Let $R_t, J_t$ be the amounts in Romeo's and Juliet's accounts at time $t$ (in years, measured from Romeo's 16th birthday). We have:
$$\begin{array}{ccc} & t=7 & t=8 \\ R_t & 12298.74 & 12667.70 \\ J_t & 12155.06 & 12762.82 \end{array}$$
You can draw two straight-line plots, one going from x = 0 at (7,R7) to x = 1 at (8,R8) and another from x = 0 at (7,J7) to x = 1 at (8,J8). Where the two lines cross is the point where simple interest calculations (for the partial year) would give you equality of monetary values. This will be close to (but not exactly equal to) the value x = 0.611---that is, for time 7.611 ---that your previous calculation would have produced. (The difference is due to the slight curvature in the money vs.time graph over the year, due to continuous compounding over the year, as compared with the straight line obtained from simple interest accumulation over the year.)

I had no idea what to graph or what any of that statement meant so, I calculated three different ways of finding x to a reasonable certainty in my post above while borrowing some information from my initial post.

It was calculated by using the original compound interest formula,by modifying the simple interest formula and creating a systems of equations, and by interpolating the value of x using the secant line that is created by calculating the average rate of change between 7<x<8 and using the two gained slopes to create two new linear equations for use in a final system of equations.

Time Measured from Romeo's Deposit Compound Calculation Product (Initial Calculation from first post)
x = 4.61103
Time Measured from Romeo's Deposit Simplified Interest Product ( A modified and fixed version of my initial simple interest calculation and likely the most accurate)
x = 4.634669891
Time Measured from Romeo's Deposit AROC: 7 < x< 8 Product (Third method using intervals and average rate of change during (7<x<8) which was more or less done to numerically and logically support my modified simple interest calculation "A really messy way of checking my answer")
x = 4.635515854

 

[Ray Vickson]

I showed the difference when I computed the average rate of change of the non-linear exponential function during the interval 7<x<8. Although the slope is negligible it has a significant enough effect to alter the x value by hundredths of a unit. Of course the linear function based on simplified interest is going to differ due to it being well "linear." I had no clue what any of this meant:
I had no idea what to graph or what any of that statement meant so, I calculated three different ways of finding x to a reasonable certainty in my post above while borrowing some information from my initial post.

It was calculated by using the original compound interest formula,by modifying the simple interest formula and creating a systems of equations, and by interpolating the value of x using the secant line that is created by calculating the average rate of change between 7<x<8 and using the two gained slopes to create two new linear equations for use in a final system of equations.

Time Measured from Romeo's Deposit Compound Calculation Product (Initial Calculation from first post)
x = 4.61103
Time Measured from Romeo's Deposit Simplified Interest Product ( A modified and fixed version of my initial simple interest calculation and likely the most accurate)
x = 4.634669891
Time Measured from Romeo's Deposit AROC: 7 < x< 8 Product (Third method using intervals and average rate of change during (7<x<8) which was more or less done to numerically and logically support my modified simple interest calculation "A really messy way of checking my answer")
x = 4.635515854

If you draw the straight line from the point (7,12298.74) to (8,12667.70) you will get the simple-interest balance throughout year 7--8 for Romeo. Similarly for Juliet. That is what I wrote, but used symbols instead of numbers.

Of course, this amounts to using the linear function $R_t = 12298.74\, (1 + .03\,(t-7))$ for $7 \leq t \leq 8$. So, the simple-interest equality time for Romeo and Juliet is the solution of the linear equation
$$12298.70\,(1 + .03 \, (t-7)) = 12155.086\, (1 + 0.5 \, (t-7)),$$
while the compound-interest equality time is given by the solution of the nonlinear equation
$$12298.70 \, (1.03)^{t-7} = 12155.06 \, (1.05)^{t-7}.$$
The two solutions will be only slightly different because for $7 \leq t \leq 8$ the nonlinear function is almost linear---has small curvature over short intervals. Basicallly, we would be replacing geometric growth by arithmetic growth.

 

[ConstantineO]

Of course, this amounts to using the linear function $R_t = 12298.74\, (1 + .03\,(t-7))$ for $7 \leq t \leq 8$. So, the simple-interest equality time for Romeo and Juliet is the solution of the linear equation

Why have you arranged these formulas like this? A google search for me nets that the simple interest formula is

Interest = (Principal) (Rate) ( Time)

I am going to assume that.
Principal = 12298.74
Rate = 0.03
Time would = t-7

Why is there 1 + the Rate? Mathematically speaking it won't do much since you will only be subtracting either the Interest or in your case the Interest and the Principal combined from each other? This statement is unsolvable without one of the lines being shifted on the x-axis. As it stands, this is utterly useless due to both lines sharing the exact same solution of 7 to make them equal to each other. You've ultimately recited and made the exact same mistake as I did initially when I stated:

I
Romeo Compound Interest Calculation Total
Ar = 10,000(1.03)^4+3
Ar = 10,000(1.03)^7
Ar = 12,298.73865

Julie Compound Interest Calculation Total
Aj = 10,000(1.05)^4
Aj = 12,155.0625

Romeo Simple Interest Calculation
(Ir) = (12,298.73865)(0.03)(t)
(Ir) = (368.94)(t)

Juliet Simple Interest Calculation
(Ir) = (12,155.0625)(0.05)(t)
(Ir) = 607.753125(t)

Equate the two to equal each other.

(368.94)(t) = 607.753125(t)
(368.94)(t) -607.753125(t) = 0
t(368.94-607.753125)=0

This doesn't work though...

You cannot just equate Romeo's and Juliet's linear simplified interest lines like this. The logic even by graphing standards does not make sense. This statement gets killed off as soon as I sub a 7 in for either since it would instantly devolve into multiplying by 0. This algebraically speaking is nonsense, and they would both share a point at origin.

12298.70(1+.03(t−7))=12155.086(1+0.5(t−7))

The same here.

$$12298.70 \, (1.03)^{t-7} = 12155.06 \, (1.05)^{t-7}.$$

Now if you graph each function and see where the points intersect, yes you will find the answer but the above here has no basis in algebraic meaning. The new compound interest formula here is completely unnecessary since we can take the original two functions as described in the huge load of wolfram images and create a systems of equation from them that produces the same answer and doesn't require using intervals.

 

[BvU]

Why is there 1 + the Rate?

There is 1 + Rate * time, because when, as you write,
Interest = (Principal) (Rate) ( Time)
then amount is
Principal + Interest = (Principal) + (Principal) (Rate) ( Time) = (Principal) [ 1 + (Rate) ( Time) ]

 

[ConstantineO]

There is 1 + Rate * time, because when, as you write,
Interest = (Principal) (Rate) ( Time)
then amount is
Principal + Interest = (Principal) + (Principal) (Rate) ( Time) = (Principal) [ 1 + (Rate) ( Time) ]

Oh I realize that. I am just wondering why he decided to write it like that when you are just creating a ratio in the step that follows. I was just wondering if there was any certain reason that they opted to do this.

 

[Ray Vickson]

Why have you arranged these formulas like this? A google search for me nets that the simple interest formula is

Interest = (Principal) (Rate) ( Time)

I am going to assume that.
Principal = 12298.74
Rate = 0.03
Time would = t-7

Why is there 1 + the Rate? Mathematically speaking it won't do much since you will only be subtracting either the Interest or in your case the Interest and the Principal combined from each other? This statement is unsolvable without one of the lines being shifted on the x-axis. As it stands, this is utterly useless due to both lines sharing the exact same solution of 7 to make them equal to each other. You've ultimately recited and made the exact same mistake as I did initially when I stated:
You cannot just equate Romeo's and Juliet's linear simplified interest lines like this. The logic even by graphing standards does not make sense. This statement gets killed off as soon as I sub a 7 in for either since it would instantly devolve into multiplying by 0. This algebraically speaking is nonsense, and they would both share a point at origin.

12298.70(1+.03(t−7))=12155.086(1+0.5(t−7))

The same here.

***********************************************************

The left-hand-side L(t)= 12298.70 *(1 + .03*(t-7)) is Romeo's balance during the interval 7 <= t <= 8, while the right-hand-side R(t) = 12155.06 *(1 + .05*(t-7)) [there was a typo: 12155.086 should have been 12155.06] is Juliet's balance during the interval 7 <= t <= 8. Here, t = years after Romeo's 16th birthday, and is the same for both Romeo and Juliet---they are the same point on the calendar. Note that R(7) = 12298.70 [= 10000(1.03)^7] is Romeo's actual balance at time t = 7, while R(8) = 12298.70*1.03 = 12667.70 [= 10000 (1.03)^8] is his actual balance at t = 8. The formula interpolates linearly during fractional years between 7 and 8. You seem to object to this, so I ask: what do you think is wrong with it? Are you sure you have not just misunderstood?

If the bank allows withdrawal during a year but pays only a fraction of that year's interest, the linear interpolation method MAY be the correct way of getting it---it all depends on how the bank chooses to treat partial years. Different banks may do it differently. Bank A may reckon interest exactly in that linear fashion---basically, using "simple" interest within a year, but compound interest at year's end, year-after-year. Bank B may do it differently: it may do essentially continuous compounding throughout a year as well as between different years, so for Bank B the solution 1.03^t = 1.05^(t-3) would give the 100% correct solution. Bank C may do, for example, monthly compounding, using a monthly rate r given by (1+r_m)^(12) = 1.03 for Romeo's bank or (1+r_j)^(12) = 1.05 for Juliet's bank. Then, if x = time in months from Romeo's 16th birthday, the solution would be obtained from the equation (1+r_m)^x = (1 + r_j)^(x-36).

So, the correct procedure depends on some financing details that are written into the banking contract. We have been talking just about the two most probable scenarios.

***************************************************************

Now if you graph each function and see where the points intersect, yes you will find the answer but the above here has no basis in algebraic meaning. The new compound interest formula here is completely unnecessary since we can take the original two functions as described in the huge load of wolfram images and create a systems of equation from them that produces the same answer and doesn't require using intervals.

 

[ConstantineO]

Perfect. You cleared up what you were doing, and I understand now what you were trying to achieve.

 

[SteamKing]

What kind of school or course teaches compound interest problems without first going over simple interest?

It's like jumping straight into relativity without first understanding Galilean and Newtonian mechanics.

 

[ConstantineO]

What kind of school or course teaches compound interest problems without first going over simple interest?

It's like jumping straight into relativity without first understanding Galilean and Newtonian mechanics.

Your Newtonian and Galilean mechanics won't help you when your math moves from the Euclidean to non-Euclidean realm. All the vector and triangle rules in the world won't help you when your piece of paper or reality can be stretched or compressed like a piece of rubber or can inflate and deflate like a balloon. The reason why special relativity and quantum mechanics came to be is because the math and the down right logic of things of the macroscopic world do not apply to the very fast or very small. Now if you're in the deterministic, probabilistic or in the fringe emergent camp that's up to you. Now I ask, "What kind of school gave you the idea that Galilean and Newtonian mechanics will help you at all when understanding special relativity?"

I am an interesting example for a student. I'm a Canadian high school drop out that has recently returned to school to resume their studies and make it into McGill, UBC or SFU, hopefully. To state that my education is spotty would be putting it lightly, and I have to deal with woefully rushed adult education programs that are more or less designed to give a crash courses to get the bare minimum requirements to get into university or college. In retrospect, I have calculated common interest before when dealing with trying to understand how much to tip or how much something is going to cost, but I was more or less confused not by the formula but by how it was being used here.

I tried to approach the "simplified interest fiasco" by attempting to create a systems of equations using the 7 year account balance that Romeo and Juliet had accumulated through compound interest, and I tried creating a new set of linear systems using a slope that I derived from sampling the average rate of change during 7<x<8 . I will be frank, I don't like graphing, and I find the thought of using it to solve questions like this as rather messy and is not always practical. I don't like using interpolation, mostly due to my lack of fully understanding it and the error that it can introduce. Regardless, I tried using graphing as one the first ways to get an idea of what caused the initial differences that popped up in wolfram alpha, and I was fully aware of the line curvature characteristics of both Romeo's and Juliet's account.

I set about my problem using two methods.

The first method was by simply taking the totals after the 7 year period of compounding interest of Romeo's and Juliet's account and creating a linear "simplified" interest function. Since I was trying to do this entirely algebraically, I initially made the error of simply equating the two function to each other as shown here.

I am not familiar with the simple interest formula, so I googled it. I am operating under the assumption that it is different than the compound interest formula.

Romeo Compound Interest Calculation Total
Ar = 10,000(1.03)^4+3
Ar = 10,000(1.03)^7
Ar = 12,298.73865

Julie Compound Interest Calculation Total
Aj = 10,000(1.05)^4
Aj = 12,155.0625

Romeo Simple Interest Calculation
(Ir) = (12,298.73865)(0.03)(t)
(Ir) = (368.94)(t)

Juliet Simple Interest Calculation
(Ir) = (12,155.0625)(0.05)(t)
(Ir) = 607.753125(t)

Equate the two to equal each other.

(368.94)(t) = 607.753125(t)
(368.94)(t) -607.753125(t) = 0
t(368.94-607.753125)=0

This doesn't work though...

I was rather foolish not to instantly recognize that the two would simply run through the origin and would be unsolvable. I thought Ray had made the same error here, and further adding to the confusion, his formula adds the interest to the principal which I was confused as to why that was necessary. My google search of "simplified interest" did not return any formulas that included such a step.

$$12298.70\,(1 + .03 \, (t-7)) = 12155.086\, (1 + 0.5 \, (t-7)),$$

Now you could use interpolation (which I am not entirely sure how) or sit there guessing all day what x is going to be, but I tried to instantly nail down what x was. My line of reasoning was that I simply forgot to horizontally translate one of the linear functions on the x-axis which would result in the two having a useful point of intersect. My attempt at fixing this was to simply add 3 to the years of Romeo's simplified interest function. This seemed to produce a result that was both numerically and logically sound as shown here. I did, however, make a typo, likely due to lack of sleep, and mislabeled this as being from Romeo's deposit date instead of Juliet's.

Romeo Simple Interest Calculation
(Ir) = (12,298.73865)(0.03)(t+3)
(Ir) = (368.94)(t+3)

Juliet Simple Interest Calculation
(Ir) = (12,155.0625)(0.05)(t)
(Ir) = 607.753125(t)

Now the Systems of Equation
(607.753125)(t) = (368.94)(t+3)
(607.753125)(t) = (368.94)(t)+ 1106.82
t(607.753125-368.94) = 1106.82
t(238.813125) = (1106.82)
t = (1106.82)/(238.813125)
t = 4.634669891
Not too far away from 4.61103. I wonder why...

If my method was applied to Ray"s model of simplified interest as shown here.

$$12298.70\,(1 + .03 \, (t-7)) = 12155.086\, (1 + 0.5 \, (t-7)),$$

It would likely look something like this. I hope this kind of gets across what I was trying to do and eliminate any confusion.
$$12298.70\,(0.03 \, (t+3)) = 12155.086\, (0.5 \, (t)),$$

My second method involved calculating the average rate of change of both Juliet's and Romeos account during 7<x<8 and using the "discovered" slopes to create two new simple linear functions that I could equate to each other. What I did was calculate the slope of a line drawn between x = 7 and x = 8 on each of their compounding curves. I think Ray mentioned something like this here, but I did not fully understand what he was talking about due to notation differences.

It's not that complicated. Let $R_t, J_t$ be the amounts in Romeo's and Juliet's accounts at time $t$ (in years, measured from Romeo's 16th birthday). We have:
$$\begin{array}{ccc} & t=7 & t=8 \\ R_t & 12298.74 & 12667.70 \\ J_t & 12155.06 & 12762.82 \end{array}$$
You can draw two straight-line plots, one going from x = 0 at (7,R7) to x = 1 at (8,R8) and another from x = 0 at (7,J7) to x = 1 at (8,J8). Where the two lines cross is the point where simple interest calculations (for the partial year) would give you equality of monetary values. This will be close to (but not exactly equal to) the value x = 0.611---that is, for time 7.611 ---that your previous calculation would have produced. (The difference is due to the slight curvature in the money vs.time graph over the year, due to continuous compounding over the year, as compared with the straight line obtained from simple interest accumulation over the year.)

This is what my version of this looked like. Yet again here I made the typo of messing up from which deposit date this was from. It should be clarified that this was from Juliet's deposit date, ergo the "+3" on Romeo's function.

We can take things a step further and look at the secant line's slope of the curve that models interest from Romeo's deposit date during 7 < x < 8. Let's look at the average rate of change for both Romeo's and Juliet's account.

Romeo Compound Interest AROC: 7 < x < 8

While x = 4
Ar = 10,000(1.03)^4+3
Ar = 10,000(1.03)^7
Ar = 12,298.73865
x = 4
y =12,298.73865

While x =5
Ar = 10,000(1.03)^5+3
Ar = 10,000(1.03)^8
Ar = 12,667.70081
x = 5
y = 12,667.70081

Romeo's AROC
= (12,667.70081 - 12,298.73865) / (5 -4)
= 368.96216

Juliet's Compound Interest AROC 7 < x < 8

While x = 4
Aj = 10,000(1.05)^4
Aj = 12,155.0625
x = 4
y =12,155.0625

While x = 5
Aj = 10,000(1.05)^5
Aj = 12,762.81563
x = 5
y = 12,762.81563

Juliet's AROC
= (12,762.81563 - 12,155.0625) / (5-4)
=607.7531

After that huge bunch of calculations we now have slopes that we can create linear relations with.

Romeo's Linear Relation
y = (368.96216)(x + 3)

Juliet's Linear Relation
y = 607.7531(x)

Let's see what the Solution is for these in equations when put into a system.

(368.96216)(x + 3) = 607.7531(x)
(368.96216)(x) + 1106.88648 = 607.7531(x)
607.7531(x) -(368.96216)(x) = 1106.88648
x(607.7531-368.96216) = 1106.88648
x(238.783884) = 1106.88648
x = 4.635515854

Notice a pattern?

Time Measured from Romeo's Deposit Compound Calculation Product
x = 4.61103
Time Measured from Romeo's Deposit Simplified Interest Product
x = 4.634669891
Time Measured from Romeo's Deposit AROC: 7 < x< 8 Product
x = 4.635515854

I hope this eliminates any more confusion. I'm willing to talk on skype or collaborative edit if there are still some grey areas.

 

[BvU]

Thank you for sharing this with us.