Competition Mode Question

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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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.

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

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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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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