How to calculate Compound interest?

Thread Starter

zulfi100

Joined Jun 7, 2012
656
Hi,
I trying to solve the following question. I want to know have i calculated Amy's part correctly or not?

The question is:

Amy deposited $1000 into an account that earns 8% annual interest compounded every 6 months. Bod deposited $1000 into an account that earns 8% annual interest compounded quarterly. If neither Amy nor Bob makes any additional deposits or withdrawals in 6 months, how much more money will Bob have in his account than Amy?

For Amy:

P=1000, t=1, n=2, r=4%

V=1000(1+0.04)^2

Somebody please guide me.

Zulfi.
 

WBahn

Joined Mar 31, 2012
26,398
You calculated what Amy will have in her account over one year, not six months.

Remember, r is the rate per compounding period and n is the number of compounding periods.

There isn't going to be much difference after six months. But look at the difference after ten years or so.
 

WBahn

Joined Mar 31, 2012
26,398
If interest is compounded monthly and you leave it alone for three years, how many compounding periods are there?

If interest is compounded quarterly and you leave it alone for three quarters, how many compounding periods are there?

If interest is compounded every six months and you leave it alone for six months, how many compounding periods are there?

If interest is compound quarterly and you leave it alone for six months, how many compounding periods are there?
 

wayneh

Joined Sep 9, 2010
16,825
Hi,

Because it says:


Is it right or not?

In the same way n for Bob would be 4, because it says

Some body please guide me.

Zulfi.
The number you use for "n" in the equation is not the number of times per year that the interest is calculated. "n" is the number of compounding periods that have actually elapsed during the time period of interest.
 

Thread Starter

zulfi100

Joined Jun 7, 2012
656
Hi,
Wayneh I have provided answer to your query in my post#4. Kindly tell me if its right or not. If wrong what is the correct answer?

Zulfi.
 
Hi,
I trying to solve the following question. I want to know have i calculated Amy's part correctly or not?

The question is:

Amy deposited $1000 into an account that earns 8% annual interest compounded every 6 months. Bod deposited $1000 into an account that earns 8% annual interest compounded quarterly. If neither Amy nor Bob makes any additional deposits or withdrawals in 6 months, how much more money will Bob have in his account than Amy?

For Amy:

P=1000, t=1, n=2, r=4%

V=1000(1+0.04)^2

Somebody please guide me.

Zulfi.
Hi @zulfi100, good to see you grinding a way.

The answer to your question is, No, you have NOT calculated Amy's portion correctly.

I am sure that you have two formulas for interest in your book.

The simple interest formula is I = P x R x T.

The compound interest formula: A = P(1 + r/n) ^ nt.

I think that the problem is set up for you to learn to use both formulas or maybe they are just busting your chops :)

For Amy, in this word problem, just use the simple interest formula and add the interest to the principal. Use the compound interest formula for Bob.

You are trying to force the compound interest formula on Amy when there is no compounding.

For Amy P=1000, R=.04, T=1. Because the Annual (12 months) interest rate percentage is given, and the period is only 6 months, divide by 2 (.08/2=.04). The value for T is 1 because there is one period, that happens to be 6 months. For Amy, there is only one period for interest. There is no compounding, and so, Amy has 1000+I after 6 months or $1040.

If the problem is changed to how much money each will have in 10 years, then you would use the compound interest formula for both.
 

Thread Starter

zulfi100

Joined Jun 7, 2012
656
Hi,
The number you use for "n" in the equation is not the number of times per year that the interest is calculated. "n" is the number of compounding periods that have actually elapsed during the time period of interest.
and the question says:
Amy deposited $1000 into an account that earns 8% annual interest compounded every 6 months.
<earns 8% annual interest> annual interest means that interest is calculated per year and this is not 'n' (Right?) now <compounded every 6 months> this means 2 compounding periods have elapsed so n is 2.
For Amy, in this word problem, just use the simple interest formula
why??
It is using the term <compounded every 6 months>
Some body please guide me Why we should not be using the compound interest formula with Amy?

Zulfi.
 
For Amy, you can use either the compound interest formula or the simple interest formula. I thought it would be easier to see if you used the simple interest formula because there is only one "compound" - meaning interest is only added one time to her money, because the question asks how much she will have after 6 months and her interest is compounded every 6 months.

Maybe I am confusing you even more - here is how it works out for Amy using both formulas.

Compound interest formula: A = P (1 + r/n)^ (nt)

Where:

A = the amount of money that Amy has after 6 months, because the problem asks how much she has after 6 months.
P = the original amount of money that Amy invested (principal) - $1000
r = the annual interest rate = .08 (that is a 12 month rate)
n = the number of times that interest is compounded per year - that is 2 - because there are 2, 6-month periods in a year.
t = the number of years the money is invested - that is .5 because the problem asks how much she has after 6 months.

Now we have:

(EDITED: to correct the error in how I initially wrote it - should be:)
1000 x (1+0.08/2)^(2x0.5)=1040

______________________________

Simple interest formula: A = P(1 + rt)

Where:
A = the amount of money that AMY has after 6 months, because the problem asks how much she has after 6 months.
P = the original amount of money that Amy invested (principal) - $1000
r = rate of Interest per year =.08
t = time period = .5 (1/2 of a year, 6 months)

A = 1000(1 + (0.08 × 0.5)) = 1040

OR

Where:
A = the amount of money that AMY has after 6 months, because the problem asks how much she has after 6 months.
P = the original amount of money that Amy invested (principal) - $1000
r = rate of Interest for 6 months =.04 (.08/2) - the problem gives the 12 month rate - the 6 month rate is 1/2 of that because it is compounded every 6 months - we only have a single compound because we want to know how much money she has after 6 months.
t = time period = 1 = one six month period

A = 1000(1 + (0.04 × 1)) = 1040


Any clearer?
 
Last edited:

wayneh

Joined Sep 9, 2010
16,825
<earns 8% annual interest> annual interest means that interest is calculated per year and this is not 'n' (Right?) now <compounded every 6 months> this means 2 compounding periods have elapsed so n is 2.
But the question is asking how much she has after only six months (n=1), not an entire year (n=2).
 

Thread Starter

zulfi100

Joined Jun 7, 2012
656
Hi,
Mr. Raymond, Thanks good work.
Compound interest formula: A = P (1 + r/n)^ (nt)

Where:

A = the amount of money that Amy has after 6 months, because the problem asks how much she has after 6 months.
P = the original amount of money that Amy invested (principal) - $1000
r = the annual interest rate = .08 (that is a 12 month rate)
n = the number of times that interest is compounded per year - that is 2 - because there are 2, 6-month periods in a year.
t = the number of years the money is invested - that is .5 because the problem asks how much she has after 6 months.

Now we have:
1000 ( 1 + 0.08 x 2 )^( 2 x 0.5 ) = 1040.00
After putting values should not we get:
1000(1+0.08 /2) ^(2 * .5) because the formula is r/n??

Zulfi.
 

Thread Starter

zulfi100

Joined Jun 7, 2012
656
Hi,
For Bob:

P=1000, n=4(compounded quarterly , r=0.08, t=0.6

Now A = P (1 + r/n)^ (nt)

A= 1000(1 + 0.08/4)^(0.4*.6)

= 1000(51/50)^(0.24)

=1004
Not correct?

Zulfi.
 

WBahn

Joined Mar 31, 2012
26,398
Zulfi, just walk it through transaction by transaction.

Bob deposits $1000 on Jan 1st. At the end of the 1st quarter (i.e., 1/4 of a year) how much interest has he earned on his $1000 at 8% annual interest?

Whatever that amount is, add it to his original $1000. That is his balance at the beginning of the second quarter.

How much interest on that new balance has he earned after another 1/4 year?

Whatever that amount is, add it to his balance at the start of the second quarter. That is his balance at the end of six months, which is the number you are looking for.
 
Hi,
Mr. Raymond, Thanks good work.

After putting values should not we get:
1000(1+0.08 /2) ^(2 * .5) because the formula is r/n??

Zulfi.
Your welcome. Yes, I made an error in how I wrote that line and I am glad that you followed it and were able to see it.

Answer the same questions for Bob:
Compound interest formula: A = P (1 + r/n)^ (nt)

Where:

A = the amount of money that Bob has after 6 months, because the problem asks how much he has after 6 months.
P = the original amount of money that Bob invested (principal) -??
r = the annual interest rate = ?? (that is a 12 month rate)
n = the number of times that interest is compounded per year - ??
t = the number of years the money is invested - ??


So, bottom line - what is your answer to the homework question - Amy and Bob?
 
Last edited:

WBahn

Joined Mar 31, 2012
26,398
Hi,
For Bob:

P=1000, n=4(compounded quarterly , r=0.08, t=0.6

Now A = P (1 + r/n)^ (nt)

A= 1000(1 + 0.08/4)^(0.4*.6)

= 1000(51/50)^(0.24)

=1004
Not correct?

Zulfi.
Does $1004 make sense?

At 8%/year, how much would he earn after a full year?

Wouldn't it make sense that he would make at least half of that amount after six months?

Again, answer this very simple question: If Bob's interest compounds every three months, how many times does it compound in six months? How many three-month periods are there in six months? THAT's how many compounding periods there are. Stop making it so much harder than it is!
 
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