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12 May 2020, 01:17
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On January 1, 2003, Jack invested $5,000 at r percent interest rate, compounded annually. What is the value of r?
(1) The investment plus the total interest at the end of 2004 was $550 more than the amount at the end of 2003.
(2) The investment plus the total interest at the end of 2004 increased by 21% from the original investment.
12 May 2020, 02:30
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On January 1, 2003, Jack invested $5,000 at r percent interest rate, compounded annually.
Q. What is the value of r?
(1) The investment plus the total interest at the end of 2004 was $550 more than the amount at the end of 2003.
At end of 2003,
X = 5000 * (1+r)
At end of 2004,
X + 550 = X * (1+r)
550 = X * r = 5000 * (1+r) * r
--> r = 10%
SUFFICIENT
(2) The investment plus the total interest at the end of 2004 increased by 21% from the original investment.
1.21 = (1+r)^2
--> r = 10%
SUFFICIENT
FINAL ANSWER IS (D)
Posted from my mobile device
Q. What is the value of r?
(1) The investment plus the total interest at the end of 2004 was $550 more than the amount at the end of 2003.
At end of 2003,
X = 5000 * (1+r)
At end of 2004,
X + 550 = X * (1+r)
550 = X * r = 5000 * (1+r) * r
--> r = 10%
SUFFICIENT
(2) The investment plus the total interest at the end of 2004 increased by 21% from the original investment.
1.21 = (1+r)^2
--> r = 10%
SUFFICIENT
FINAL ANSWER IS (D)
Posted from my mobile device
On January 1, 2003, Jack invested $5,000 at r percent interest rate, compounded annually. What is the value of r?
P = 5000, Rate, r = ?
Amount at the end of 2003, \(A_{2003} = P*(1+\frac{r}{100}) = 5000(1+\frac{r}{100})\)
(1) The investment plus the total interest at the end of 2004 was $550 more than the amount at the end of 2003.
\(A_{2004} - A_{2003} = 5000*(1+\frac{r}{100})^2 - 5000(1+\frac{r}{100}) = 550\)
\(5000*x^2 - 5000x = 550\)
'x' and r an be solved.
SUFFICIENT.
(2) The investment plus the total interest at the end of 2004 increased by 21% from the original investment.
Here 21% can be treated as simple interest, compounded annually only once or once in two years
\(A_{2004} = 5000*(1+\frac{21}{100})\)
\(5000*(1+\frac{21}{100}) = 5000*(1+\frac{r}{100})^2\)
'r' can be calculated.
SUFFICIENT.
Answer D.
P = 5000, Rate, r = ?
Amount at the end of 2003, \(A_{2003} = P*(1+\frac{r}{100}) = 5000(1+\frac{r}{100})\)
(1) The investment plus the total interest at the end of 2004 was $550 more than the amount at the end of 2003.
\(A_{2004} - A_{2003} = 5000*(1+\frac{r}{100})^2 - 5000(1+\frac{r}{100}) = 550\)
\(5000*x^2 - 5000x = 550\)
'x' and r an be solved.
SUFFICIENT.
(2) The investment plus the total interest at the end of 2004 increased by 21% from the original investment.
Here 21% can be treated as simple interest, compounded annually only once or once in two years
\(A_{2004} = 5000*(1+\frac{21}{100})\)
\(5000*(1+\frac{21}{100}) = 5000*(1+\frac{r}{100})^2\)
'r' can be calculated.
SUFFICIENT.
Answer D.
