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On January, 1, 2002, Michael invested a certain amount of money at r p

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On January, 1, 2002, Michael invested a certain amount of money at r p  [#permalink]

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New post 06 Sep 2019, 01:04
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A
B
C
D
E

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Question Stats:

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On January, 1, 2002, Michael invested a certain amount of money at r percent interest rate, compounded annually. What is the value of r?

(1) At the end of 2004, the investment plus the total interest was $ 5,955.
(2) At the end of 2006, the investment plus the total interest was $ 6,690.

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Re: On January, 1, 2002, Michael invested a certain amount of money at r p  [#permalink]

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New post 06 Sep 2019, 01:44
1
IMO answer is option C

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Re: On January, 1, 2002, Michael invested a certain amount of money at r p  [#permalink]

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New post 06 Sep 2019, 01:58
1
We have two unknown variables, that is r and amount invested, A. We know r is annual compound interest rate of the investment.

Statement 1 is not sufficient because we are able to generate the following equation from it.
5,955=A(1+r)^2......(1)
Since we don't know A the investment amount, we cannot determine r.

2. Similar to statement 1 above, we can derive the following equation
6,690=A(1+r)^4...... (2)
Since we don't know A, the investment amount, we are unable to determine r.

1+2 is however sufficient. We have two equations and two variables so we can substitute for A and find r or divide (2) by (1) and solve for r.

The answer is therefore C.
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Re: On January, 1, 2002, Michael invested a certain amount of money at r p  [#permalink]

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New post 06 Sep 2019, 02:33
Find r (interest rate)

1. At the end of 2004, total interest = 5955
Start investing from beginning of 2002 - n=3

(r+1)^3 = 5955
Can solve for r - sufficient

2. At the end of 2006, total interest =6690
N=5

(r+1)^5 = 6690
Can solve for r - sufficient

Therefore, D is the correct answer.

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Re: On January, 1, 2002, Michael invested a certain amount of money at r p  [#permalink]

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New post 06 Sep 2019, 02:43
On January, 1, 2002, Michael invested a certain amount of money at r percent interest rate, compounded annually. What is the value of r?

(1) At the end of 2004, the investment plus the total interest was $ 5,955.
(2) At the end of 2006, the investment plus the total interest was $ 6,690.

let principal amount =x
so interest = p*r*t/100
#1At the end of 2004, the investment plus the total interest was $ 5,955.
5955= interest at 2004 end "a" principal ; 5955=x+a ; a = 5955-x
5955-x=x*r*2/100
value of x & r is unkown insufficient
#2
At the end of 2006, the investment plus the total interest was $ 6,690.
6690-x=x*r*4/100
again in sufficient
from 1 &2 we can solve linearly value of r
IMO C
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Re: On January, 1, 2002, Michael invested a certain amount of money at r p  [#permalink]

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New post 06 Sep 2019, 08:09
Since the amount is compounded annually; final amount will be calculated using the formula =>

P(1 + r/100)^n

where P is invested Principle amount, r is rate of interest and n is no of years after which amount is being calculated

Statement 1:
Final Amount = 5955 = P(1+r/100)^2 => only 1 equation and 2 unknown variables (P and R)

Statement 2:
Final Amount = 6690 = P(1+r/100)^4 => only 1 equation and 2 unknown variables (P and R)

Combining both statements: Eq in Statement 2/ Eq in Statement 1 => 6690/5955 = (1 + r/100)^2 : thus r can be calculated.

So Ans: C
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Re: On January, 1, 2002, Michael invested a certain amount of money at r p  [#permalink]

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New post 06 Sep 2019, 11:01
Let Principle amount on January, 1, 2002 = P. Number of years = n
Interest rate = r = ?

(1) At the end of 2004, the investment plus the total interest was $ 5,955.

At the end of 2004, there are total 3 years i.e. n = 3. Now,

\(A2004 = P(1+\frac{r}{100})^3\)
 \(5955 = P(1+\frac{r}{100})^3\)

Two variable which can take any value.
INSUFFICIENT.

(2) At the end of 2006, the investment plus the total interest was $ 6,690.

At the end of 2006, there are total 5 years i.e. n = 5. Now,

\(A2006 = P(1+\frac{r}{100})^5\)
 \(6690 = P(1+\frac{r}{100})^5\)

Two variable which can take any value.
INSUFFICIENT.

Together 1) and 2)

\(A2006 - A2004 = P(1+\frac{r}{100})^5 - P(1+\frac{r}{100})^3\)

\(6690 – 5955 = P(1+\frac{r}{100})^5 - P(1+\frac{r}{100})^3\)

\(735 = P(1+\frac{r}{100})^5 - P(1+\frac{r}{100})^3\)

Again two variable which can take any value.

INSUFFICIENT.

Answer (E).
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Re: On January, 1, 2002, Michael invested a certain amount of money at r p  [#permalink]

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New post 06 Sep 2019, 21:08
IMO it's C.
Because from each of the statements we cannot get the value of Principal and Rate as two variables are there but on combining we can get the value of R and then we can get P.

On January, 1, 2002, Michael invested a certain amount of money at r percent interest rate, compounded annually. What is the value of r?

(1) At the end of 2004, the investment plus the total interest was $ 5,955.
(2) At the end of 2006, the investment plus the total interest was $ 6,690.
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Re: On January, 1, 2002, Michael invested a certain amount of money at r p  [#permalink]

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New post 08 Sep 2019, 04:39
Given; On January, 1, 2002, Michael invested a certain amount of money at r percent interest rate, compounded annually.

Asked: What is the value of r?

Let \(A = P (1+r)^n\)

(1) At the end of 2004, the investment plus the total interest was $ 5,955.
\(5955 = P (1+r)^2\)
Since P & r are unknown
NOT SUFFICIENT

(2) At the end of 2006, the investment plus the total interest was $ 6,690.
\(6690 = P(1+r)^4\)
Since P & r are unknown
NOT SUFFICIENT

Combining (1) & (2)
(1) At the end of 2004, the investment plus the total interest was $ 5,955.
\(5955 = P (1+r)^2\)
(2) At the end of 2006, the investment plus the total interest was $ 6,690.
\(6690 = P(1+r)^4\)
Since there 2 equations and 2 unknowns
We can get values of P & r
Divide equation 2 by equation 1, we get
6690/5955 = (1+r)^2
r = 6% approximately


IMO C
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Re: On January, 1, 2002, Michael invested a certain amount of money at r p   [#permalink] 08 Sep 2019, 04:39
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