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If $10,000 is invested at 8 percent annual interest, compounded semian

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If $10,000 is invested at 8 percent annual interest, compounded semian  [#permalink]

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New post 03 Jul 2017, 18:01
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If $10,000 is invested at 8 percent annual interest, compounded semiannually, what is the balance after 5 year, in terms of dollars?

A. \(10,000(1+ \frac{0.08}{2})^{10}\)
B. \(10,000(1+0.08)^5\)
C. \(10,000(1+0.08)\)
D. \(10,000(1-0.08)^2\)
E. \(10,000*1.08\)

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Re: If $10,000 is invested at 8 percent annual interest, compounded semian  [#permalink]

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New post 03 Jul 2017, 19:56
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MathRevolution wrote:
If $10,000 is invested at 8 percent annual interest, compounded semiannually, what is the balance after 5 year, in terms of dollars?

A. \(10,000(1+ \frac{0.08}{2})^{10}\)
B. \(10,000(1+0.08)^5\)
C. \(10,000(1+0.08)\)
D. \(10,000(1-0.08)^2\)
E. \(10,000*1.08\)



Whenever an amount is compounded semi-annually, the interest rate becomes half and TIME becomes two times..
So
1) Rate becomes 0.08/2
2) Time becomes 5*2=10

A is correct
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Re: If $10,000 is invested at 8 percent annual interest, compounded semian  [#permalink]

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New post 04 Jul 2017, 00:51
Compound interest = \(Amount * (1 + \frac{ROI}{100})^n\)
where
ROI is the rate of interest
n is the amount of years the amount is invested

When an amount is invested and interest is calculated semi-annually,
the rate of interest is halved and the number of years doubles.

The formula for compound interest is :
Compound interest = \(Amount * (1 + \frac{ROI}{200})^{2n}\)

Coming to the problem in hand,
Amount : 10000$
Interest(annual)-ROI : 8% = \(\frac{8}{100} = 0.08\)
Number of years-n : 5

Substituting these values, Interest = \(10,000(1+ \frac{0.08}{2})^{10}\) (Option A)

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If $10,000 is invested at 8 percent annual interest, compounded semian  [#permalink]

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New post 04 Jul 2017, 04:15
MathRevolution wrote:
If $10,000 is invested at 8 percent annual interest, compounded semiannually, what is the balance after 5 year, in terms of dollars?

A. \(10,000(1+ \frac{0.08}{2})^{10}\)
B. \(10,000(1+0.08)^5\)
C. \(10,000(1+0.08)\)
D. \(10,000(1-0.08)^2\)
E. \(10,000*1.08\)

The formula for compound interest can be written in a few ways. IMO, the version that matches the answer choices most closely is

A = P \((1+ \frac{r}{n})^{nt}\)

A= amount of money accumulated over t years, including interest (what the question asks for)
P = initial amount invested
r = annual rate of interest in decimal form
n = the number of times the interest is compounded annually
t = number of years the amount is deposited (or borrowed) for
nt = # of times per year interest is compounded times # of years

P = $10,000
r = .08
n = 2 (semi-annually = twice per year)
t = 5 (years)
\(\frac{r}{n}\) = \(\frac{.08}{2}\)
nt = 2*5 = 10

Using figures above, the equation becomes
A = \(10,000(1+ \frac{0.08}{2})^{10}\)

Answer A
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Re: If $10,000 is invested at 8 percent annual interest, compounded semian  [#permalink]

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New post 05 Jul 2017, 01:03
==> if it is invested at 8 percent annual interest and is compounded semiannually, From
10,000(1+8%\((\frac{1}{2}))^{2*5}= 10,000(1+)^{10}\) the answer is A

Answer: A
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Re: If $10,000 is invested at 8 percent annual interest, compounded semian  [#permalink]

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New post 05 Jul 2017, 01:30
MathRevolution wrote:
==> if it is invested at 8 percent annual interest and is compounded semiannually, From
10,000(1+8%\((\frac{1}{2}))^{2*5}= 10,000(1+)^{10}\) the answer is A

Answer: A


MathRevolution, your final step is incomplete. Please correct the solution!
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Re: If $10,000 is invested at 8 percent annual interest, compounded semian  [#permalink]

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New post 10 Sep 2018, 07:37
1
generis wrote:
MathRevolution wrote:
If $10,000 is invested at 8 percent annual interest, compounded semiannually, what is the balance after 5 year, in terms of dollars?

A. \(10,000(1+ \frac{0.08}{2})^{10}\)
B. \(10,000(1+0.08)^5\)
C. \(10,000(1+0.08)\)
D. \(10,000(1-0.08)^2\)
E. \(10,000*1.08\)

The formula for compound interest can be written in a few ways. IMO, the version that matches the answer choices most closely is

A = P \((1+ \frac{r}{n})^{nt}\)

A= amount of money accumulated over t years, including interest (what the question asks for)
P = initial amount invested
r = annual rate of interest in decimal form
n = the number of times the interest is compounded annually
t = number of years the amount is deposited (or borrowed) for
nt = # of times per year interest is compounded times # of years

P = $10,000
r = .08
n = 2 (semi-annually = twice per year)
t = 5 (years)
\(\frac{r}{n}\) = \(\frac{.08}{2}\)
nt = 2*5 = 10

Using figures above, the equation becomes
A = \(10,000(1+ \frac{0.08}{2})^{10}\)

Answer A





A slightly modified formula. Just for ease of use.

A = P \((1+ \frac{r}{100n})^{nt}\)
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Re: If $10,000 is invested at 8 percent annual interest, compounded semian   [#permalink] 10 Sep 2018, 07:37
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