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02 Feb 2026, 14:12
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
60% (01:44) correct
40%
(01:27)
wrong
based on 122
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Two years ago, Melanie invested a total of $10,000. She invested part of the money at an annual interest rate of 8%, compounded annually, and the remainder at an annual interest rate of 6%, compounded semiannually. How much is Melanie’s investment worth now?
(1) One year ago, Melanie’s total investment was worth $10,647.20.
(2) Melanie invested $2,000 at the annual interest rate of 8%, compounded annually.
(1) One year ago, Melanie’s total investment was worth $10,647.20.
(2) Melanie invested $2,000 at the annual interest rate of 8%, compounded annually.
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02 Feb 2026, 20:31
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Solution:
Given: Invested total money is $10000
Invested a part of money at a annual rate of interest at 8% that is compunded annually.
Here we can apply the formula for computing compound interest as:
A = P(1+ R/n)^nt where
A = Future Value
P = Principle amount
R = Annual rate of interest
n = No. of compounding periods per year
T= Time in years
Suppose the amount invested for one year is x and for the second year would be 10000-x
For First part x(1 + 0.08)^2 => x(1.08)^2
For the Second Part 10000-x(1 + 0.06/2)^4 => 10000-x(1.03)^4
Statement 1. One year ago, Melanie’s total investment was worth $10,647.20. Sufficient
x(1.08)+ 10000-x(1.03)^2 = $10,647.20 => x = $2000.
We can put this value and the get our answer.
Statement 2 Melanie invested $2,000 at the annual interest rate of 8%, compounded annually. Sufficient
As value of x is already been given so, we can get our answer.
Hence, Option D
Given: Invested total money is $10000
Invested a part of money at a annual rate of interest at 8% that is compunded annually.
Here we can apply the formula for computing compound interest as:
A = P(1+ R/n)^nt where
A = Future Value
P = Principle amount
R = Annual rate of interest
n = No. of compounding periods per year
T= Time in years
Suppose the amount invested for one year is x and for the second year would be 10000-x
For First part x(1 + 0.08)^2 => x(1.08)^2
For the Second Part 10000-x(1 + 0.06/2)^4 => 10000-x(1.03)^4
Statement 1. One year ago, Melanie’s total investment was worth $10,647.20. Sufficient
x(1.08)+ 10000-x(1.03)^2 = $10,647.20 => x = $2000.
We can put this value and the get our answer.
Statement 2 Melanie invested $2,000 at the annual interest rate of 8%, compounded annually. Sufficient
As value of x is already been given so, we can get our answer.
Hence, Option D
kevincan
02 Feb 2026, 23:56
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The $10,000 is split between two vehicles or investments . Suppose the amount invested in the first vehicle is x , so the amount invested is 10000-x.
The value of the first investment two years later is x(1 + 0.08)^2 => x(1.08)^2
The value of the second investment two years later is (10000-x)(1 + 0.06/2)^4 => (10000-x)(1.03)^4
We can answer the question if and only if we can find the value of x.
Statement 1. One year ago, Melanie’s total investment was worth $10,647.20.
Thus, we can write
x(1.08)+ (10000-x)(1.03)^2 = $10,647.20
This is a linear equation with one variable , so there will be only one solution. We can declare this statement sufficient without even solving for x .
(2) gives us directly the value of x and thus is sufficient
The value of the first investment two years later is x(1 + 0.08)^2 => x(1.08)^2
The value of the second investment two years later is (10000-x)(1 + 0.06/2)^4 => (10000-x)(1.03)^4
We can answer the question if and only if we can find the value of x.
Statement 1. One year ago, Melanie’s total investment was worth $10,647.20.
Thus, we can write
x(1.08)+ (10000-x)(1.03)^2 = $10,647.20
This is a linear equation with one variable , so there will be only one solution. We can declare this statement sufficient without even solving for x .
(2) gives us directly the value of x and thus is sufficient
